3.312 \(\int \frac {(e \tan (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=761 \[ -\frac {e^{5/2} \left (a^2-b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}+\frac {e^{5/2} \left (a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a b^2 d}+\frac {e^{5/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}-\frac {e^{5/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}+\frac {a e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}-\frac {a e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} b^2 d}-\frac {a e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}+\frac {a e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}+\frac {2 \sqrt {2} e^2 \sqrt {a-b} \sqrt {a+b} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} e^2 \sqrt {a-b} \sqrt {a+b} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {\sin (c+d x)}}-\frac {2 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{b d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d} \]

[Out]

1/2*a*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/b^2/d*2^(1/2)-1/2*(a^2-b^2)*e^(5/2)*arctan(1-2^(1
/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/b^2/d*2^(1/2)-1/2*a*e^(5/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))
/b^2/d*2^(1/2)+1/2*(a^2-b^2)*e^(5/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/b^2/d*2^(1/2)-1/4*a*e^(5
/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/b^2/d*2^(1/2)+1/4*(a^2-b^2)*e^(5/2)*ln(e^(1/2)
-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/b^2/d*2^(1/2)+1/4*a*e^(5/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+
c))^(1/2)+e^(1/2)*tan(d*x+c))/b^2/d*2^(1/2)-1/4*(a^2-b^2)*e^(5/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1
/2)*tan(d*x+c))/a/b^2/d*2^(1/2)+2*e^2*EllipticPi(sin(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2),-(a-b)^(1/2)/(a+b)^(1/2
),I)*2^(1/2)*(a-b)^(1/2)*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(e*tan(d*x+c))^(1/2)/a/b/d/sin(d*x+c)^(1/2)-2*e^2*Ellipt
icPi(sin(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2),(a-b)^(1/2)/(a+b)^(1/2),I)*2^(1/2)*(a-b)^(1/2)*(a+b)^(1/2)*cos(d*x+
c)^(1/2)*(e*tan(d*x+c))^(1/2)/a/b/d/sin(d*x+c)^(1/2)+2*e^2*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi
+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))*(e*tan(d*x+c))^(1/2)/b/d/sin(2*d*x+2*c)^(1/2)+2*e*cos(d*x+c)*(e*tan
(d*x+c))^(3/2)/b/d

________________________________________________________________________________________

Rubi [A]  time = 1.17, antiderivative size = 761, normalized size of antiderivative = 1.00, number of steps used = 38, number of rules used = 22, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.880, Rules used = {3891, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639, 3890, 2733, 2730, 2906, 2905, 490, 1213, 537} \[ -\frac {e^{5/2} \left (a^2-b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}+\frac {e^{5/2} \left (a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a b^2 d}+\frac {e^{5/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}-\frac {e^{5/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}+\frac {a e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}-\frac {a e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} b^2 d}-\frac {a e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}+\frac {a e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}+\frac {2 \sqrt {2} e^2 \sqrt {a-b} \sqrt {a+b} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} e^2 \sqrt {a-b} \sqrt {a+b} \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {\sin (c+d x)}}-\frac {2 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{b d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Tan[c + d*x])^(5/2)/(a + b*Sec[c + d*x]),x]

[Out]

(a*e^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*b^2*d) - ((a^2 - b^2)*e^(5/2)*ArcTan[1
- (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*b^2*d) - (a*e^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c +
d*x]])/Sqrt[e]])/(Sqrt[2]*b^2*d) + ((a^2 - b^2)*e^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(S
qrt[2]*a*b^2*d) - (a*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*b^
2*d) + ((a^2 - b^2)*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*a*b
^2*d) + (a*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*b^2*d) - ((a
^2 - b^2)*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*a*b^2*d) + (2
*Sqrt[2]*Sqrt[a - b]*Sqrt[a + b]*e^2*Sqrt[Cos[c + d*x]]*EllipticPi[-(Sqrt[a - b]/Sqrt[a + b]), ArcSin[Sqrt[Sin
[c + d*x]]/Sqrt[1 + Cos[c + d*x]]], -1]*Sqrt[e*Tan[c + d*x]])/(a*b*d*Sqrt[Sin[c + d*x]]) - (2*Sqrt[2]*Sqrt[a -
b]*Sqrt[a + b]*e^2*Sqrt[Cos[c + d*x]]*EllipticPi[Sqrt[a - b]/Sqrt[a + b], ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 +
Cos[c + d*x]]], -1]*Sqrt[e*Tan[c + d*x]])/(a*b*d*Sqrt[Sin[c + d*x]]) - (2*e^2*Cos[c + d*x]*EllipticE[c - Pi/4
+ d*x, 2]*Sqrt[e*Tan[c + d*x]])/(b*d*Sqrt[Sin[2*c + 2*d*x]]) + (2*e*Cos[c + d*x]*(e*Tan[c + d*x])^(3/2))/(b*d)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2730

Int[1/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(g_)*tan[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[
e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[g*Tan[e + f*x]]), Int[Sqrt[Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]*(a + b*Sin[e +
f*x])), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2733

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[g^(2*
IntPart[p])*(g*Cot[e + f*x])^FracPart[p]*(g*Tan[e + f*x])^FracPart[p], Int[(a + b*Sin[e + f*x])^m/(g*Tan[e + f
*x])^p, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rule 2905

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> Dist[(-4*Sqrt[2]*g)/f, Subst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sq
rt[g*Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2906

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x
_)])), x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]], Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]
*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rule 3890

Int[Sqrt[cot[(c_.) + (d_.)*(x_)]*(e_.)]/(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/a, Int[Sqr
t[e*Cot[c + d*x]], x], x] - Dist[b/a, Int[Sqrt[e*Cot[c + d*x]]/(b + a*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c,
d, e}, x] && NeQ[a^2 - b^2, 0]

Rule 3891

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)/(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Dist[e^2/b^2, I
nt[(e*Cot[c + d*x])^(m - 2)*(a - b*Csc[c + d*x]), x], x] + Dist[(e^2*(a^2 - b^2))/b^2, Int[(e*Cot[c + d*x])^(m
- 2)/(a + b*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0]

Rubi steps

\begin {align*} \int \frac {(e \tan (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx &=-\frac {e^2 \int (a-b \sec (c+d x)) \sqrt {e \tan (c+d x)} \, dx}{b^2}+\frac {\left (\left (a^2-b^2\right ) e^2\right ) \int \frac {\sqrt {e \tan (c+d x)}}{a+b \sec (c+d x)} \, dx}{b^2}\\ &=-\frac {\left (a e^2\right ) \int \sqrt {e \tan (c+d x)} \, dx}{b^2}+\frac {e^2 \int \sec (c+d x) \sqrt {e \tan (c+d x)} \, dx}{b}+\frac {\left (\left (a^2-b^2\right ) e^2\right ) \int \sqrt {e \tan (c+d x)} \, dx}{a b^2}-\frac {\left (\left (a^2-b^2\right ) e^2\right ) \int \frac {\sqrt {e \tan (c+d x)}}{b+a \cos (c+d x)} \, dx}{a b}\\ &=\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d}-\frac {\left (2 e^2\right ) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx}{b}-\frac {\left (a e^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{b^2 d}+\frac {\left (\left (a^2-b^2\right ) e^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a b^2 d}-\frac {\left (\left (a^2-b^2\right ) e^2 \sqrt {e \cot (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {1}{(b+a \cos (c+d x)) \sqrt {e \cot (c+d x)}} \, dx}{a b}\\ &=\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d}-\frac {\left (2 a e^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{b^2 d}+\frac {\left (2 \left (a^2-b^2\right ) e^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a b^2 d}-\frac {\left (\left (a^2-b^2\right ) e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {-\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a b \sqrt {\sin (c+d x)}}-\frac {\left (2 e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)} \, dx}{b \sqrt {\sin (c+d x)}}\\ &=\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d}+\frac {\left (a e^3\right ) \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{b^2 d}-\frac {\left (a e^3\right ) \operatorname {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{b^2 d}-\frac {\left (\left (a^2-b^2\right ) e^3\right ) \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a b^2 d}+\frac {\left (\left (a^2-b^2\right ) e^3\right ) \operatorname {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a b^2 d}-\frac {\left (\left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a b \sqrt {\sin (c+d x)}}-\frac {\left (2 e^2 \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{b \sqrt {\sin (2 c+2 d x)}}\\ &=-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{b d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d}-\frac {\left (a e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}-\frac {\left (a e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}+\frac {\left (\left (a^2-b^2\right ) e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}+\frac {\left (\left (a^2-b^2\right ) e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}-\frac {\left (a e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 b^2 d}-\frac {\left (a e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 b^2 d}+\frac {\left (\left (a^2-b^2\right ) e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a b^2 d}+\frac {\left (\left (a^2-b^2\right ) e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a b^2 d}-\frac {\left (4 \sqrt {2} \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^4} \left (a+b+(-a+b) x^4\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a b d \sqrt {\sin (c+d x)}}\\ &=-\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}+\frac {\left (a^2-b^2\right ) e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}+\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}-\frac {\left (a^2-b^2\right ) e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{b d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d}-\frac {\left (a e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}+\frac {\left (a e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}+\frac {\left (\left (a^2-b^2\right ) e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {\left (\left (a^2-b^2\right ) e^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {\left (2 \sqrt {2} \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a+b}-\sqrt {a-b} x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} b d \sqrt {\sin (c+d x)}}+\frac {\left (2 \sqrt {2} \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a+b}+\sqrt {a-b} x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} b d \sqrt {\sin (c+d x)}}\\ &=\frac {a e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}-\frac {\left (a^2-b^2\right ) e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {a e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}+\frac {\left (a^2-b^2\right ) e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}+\frac {\left (a^2-b^2\right ) e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}+\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}-\frac {\left (a^2-b^2\right ) e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{b d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d}-\frac {\left (2 \sqrt {2} \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a+b}-\sqrt {a-b} x^2\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} b d \sqrt {\sin (c+d x)}}+\frac {\left (2 \sqrt {2} \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a+b}+\sqrt {a-b} x^2\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} b d \sqrt {\sin (c+d x)}}\\ &=\frac {a e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}-\frac {\left (a^2-b^2\right ) e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {a e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}+\frac {\left (a^2-b^2\right ) e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}-\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}+\frac {\left (a^2-b^2\right ) e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}+\frac {a e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} b^2 d}-\frac {\left (a^2-b^2\right ) e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a b^2 d}+\frac {2 \sqrt {2} \sqrt {a-b} \sqrt {a+b} e^2 \sqrt {\cos (c+d x)} \Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )\right |-1\right ) \sqrt {e \tan (c+d x)}}{a b d \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} \sqrt {a-b} \sqrt {a+b} e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )\right |-1\right ) \sqrt {e \tan (c+d x)}}{a b d \sqrt {\sin (c+d x)}}-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{b d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{b d}\\ \end {align*}

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Mathematica [C]  time = 26.44, size = 1846, normalized size = 2.43 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Tan[c + d*x])^(5/2)/(a + b*Sec[c + d*x]),x]

[Out]

(2*(b + a*Cos[c + d*x])*Cot[c + d*x]*(e*Tan[c + d*x])^(5/2))/(b*d*(a + b*Sec[c + d*x])) - ((b + a*Cos[c + d*x]
)*Sec[c + d*x]*(e*Tan[c + d*x])^(5/2)*((4*a*Sec[c + d*x]^2*((-2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Tan[c + d*x]]
)/(-a^2 + b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Tan[c + d*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2
+ b^2] - Sqrt[2]*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Tan[c + d*x]] + b*Tan[c + d*x]] - Log[Sqrt[-a^2 + b^2] + Sqrt
[2]*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Tan[c + d*x]] + b*Tan[c + d*x]])/(4*Sqrt[2]*Sqrt[b]*(-a^2 + b^2)^(1/4)) +
(a*AppellF1[3/4, 1/2, 1, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(3/2))/(3*a^2 -
3*b^2))*(a + b*Sqrt[1 + Tan[c + d*x]^2]))/((b + a*Cos[c + d*x])*(1 + Tan[c + d*x]^2)^(3/2)) - (b*Sec[c + d*x]*
(6*Sqrt[2]*(a^2 - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 6*Sqrt[2]*a^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c +
d*x]]] + 6*Sqrt[2]*b^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] - (6 + 6*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 -
((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] + (6 + 6*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 + ((1
+ I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] - 3*Sqrt[2]*a^2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c
+ d*x]] + 3*Sqrt[2]*b^2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 3*Sqrt[2]*a^2*Log[1 + Sqrt[2]*Sq
rt[Tan[c + d*x]] + Tan[c + d*x]] - 3*Sqrt[2]*b^2*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + (3 + 3*I
)*Sqrt[b]*(a^2 - b^2)^(3/4)*Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*T
an[c + d*x]] - (3 + 3*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqr
t[Tan[c + d*x]] + I*b*Tan[c + d*x]] + 8*a*b*AppellF1[3/4, 1/2, 1, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(
a^2 - b^2)]*Tan[c + d*x]^(3/2))*(a + b*Sqrt[1 + Tan[c + d*x]^2]))/(4*(a^3 - a*b^2)*(b + a*Cos[c + d*x])*(1 + T
an[c + d*x]^2)) + (Cos[2*(c + d*x)]*Sec[c + d*x]^2*(-84*Sqrt[2]*b*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] + 84*
Sqrt[2]*b*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + ((42 + 42*I)*(-a^2 + 2*b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqr
t[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(Sqrt[b]*(a^2 - b^2)^(1/4)) + ((42 + 42*I)*(a^2 - 2*b^2)*ArcTan[1 + ((1 +
I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(Sqrt[b]*(a^2 - b^2)^(1/4)) + 42*Sqrt[2]*b*Log[1 - Sqrt[2]
*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - 42*Sqrt[2]*b*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + ((21 +
21*I)*(a^2 - 2*b^2)*Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c +
d*x]])/(Sqrt[b]*(a^2 - b^2)^(1/4)) + ((21 + 21*I)*(-a^2 + 2*b^2)*Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[b]*(a^2 -
b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]])/(Sqrt[b]*(a^2 - b^2)^(1/4)) + (112*a^3*AppellF1[3/4, 1/2, 1
, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(3/2))/(a^2 - b^2) - (168*a*b^2*AppellF
1[3/4, 1/2, 1, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(3/2))/(a^2 - b^2) - (24*a
*b^2*AppellF1[7/4, 1/2, 1, 11/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(7/2))/(a^2 -
b^2) - (168*a*Tan[c + d*x]^(3/2))/Sqrt[1 + Tan[c + d*x]^2])*(a + b*Sqrt[1 + Tan[c + d*x]^2]))/(84*a*(b + a*Co
s[c + d*x])*(-1 + Tan[c + d*x]^2)*Sqrt[1 + Tan[c + d*x]^2])))/(b*d*(a + b*Sec[c + d*x])*Tan[c + d*x]^(5/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^(5/2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^(5/2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^(5/2)/(b*sec(d*x + c) + a), x)

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maple [B]  time = 1.76, size = 3747, normalized size = 4.92 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^(5/2)/(a+b*sec(d*x+c)),x)

[Out]

-1/d*(a-b)*(-1+cos(d*x+c))^2*(-I*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(
d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1
/2*I,1/2*2^(1/2))*b^2+2*cos(d*x+c)*2^(1/2)*a*b-(a^2-b^2)^(1/2)*cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c
))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b+2*cos(d*x+c)*((1-cos(d*x+
c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*a*b-4*cos(d*x+c)*((1-cos(d*x+c)+sin(d*
x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*Ellip
ticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*a*b-I*cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1
-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b^2-cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1
-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a^2+cos(d*x+c)*((1-cos(
d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b
^2-cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((a-b)*(a+b
))^(1/2)),1/2*2^(1/2))*a^2+cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))
/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),
-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b^2-cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-
1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*
x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b^2+I*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x
+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b^2-cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x
+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b^2+(a^2-b^2)^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c)
)/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a+(a^2-b^2)^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/
sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(
((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b-(a^2-b^2)^(1/2)*((
1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(
d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^
(1/2))*a-(a^2-b^2)^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^
(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+
((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b+2*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))
/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1
/2*2^(1/2))*a*b-4*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*a*b+I*co
s(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b^2+(a
^2-b^2)^(1/2)*cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^
(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((
a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a+(a^2-b^2)^(1/2)*cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1
+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x
+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b-(a^2-b^2)^(1/2)*cos(d*x+c)*((1-cos(d*x+c
)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a-2*
2^(1/2)*a*b-((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b^2
-((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/s
in(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2
^(1/2))*a^2+((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(
1/2)),1/2*2^(1/2))*b^2-((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1
/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((
a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a^2+((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(
a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b^2-((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+
sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x
+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b^2)*cos(d*x+c)^2*(1+cos(d*x+c))^2*(e*sin(d*x+c)/cos(d*x+c))^(5/2)/sin(d*x+c
)^7*2^(1/2)/b/((a^2-b^2)^(1/2)-a+b)/((a^2-b^2)^(1/2)+a-b)/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^(5/2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*tan(d*x + c))^(5/2)/(b*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{b+a\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^(5/2)/(a + b/cos(c + d*x)),x)

[Out]

int((cos(c + d*x)*(e*tan(c + d*x))^(5/2))/(b + a*cos(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{a + b \sec {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**(5/2)/(a+b*sec(d*x+c)),x)

[Out]

Integral((e*tan(c + d*x))**(5/2)/(a + b*sec(c + d*x)), x)

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