3.317 \(\int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=836 \[ -\frac {2 \sqrt {2} \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right ) \sqrt {\sin (c+d x)} b^3}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right ) \sqrt {\sin (c+d x)} b^3}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right ) b^2}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right ) b^2}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}-\sqrt {2} \sqrt {e \tan (c+d x)}\right ) b^2}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right ) b^2}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {F\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)} b}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}}+\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}} \]

[Out]

1/2*a*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/(a^2-b^2)/d/e^(5/2)*2^(1/2)-1/2*b^2*arctan(1-2^(1/2)*(e*t
an(d*x+c))^(1/2)/e^(1/2))/a/(a^2-b^2)/d/e^(5/2)*2^(1/2)-1/2*a*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/(
a^2-b^2)/d/e^(5/2)*2^(1/2)+1/2*b^2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/(a^2-b^2)/d/e^(5/2)*2^(1/2
)+1/4*a*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/(a^2-b^2)/d/e^(5/2)*2^(1/2)-1/4*b^2*ln(e^(
1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/(a^2-b^2)/d/e^(5/2)*2^(1/2)-1/4*a*ln(e^(1/2)+2^(1/2)*(
e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/(a^2-b^2)/d/e^(5/2)*2^(1/2)+1/4*b^2*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^
(1/2)+e^(1/2)*tan(d*x+c))/a/(a^2-b^2)/d/e^(5/2)*2^(1/2)-2*b^3*EllipticPi((-cos(d*x+c))^(1/2)/(1+sin(d*x+c))^(1
/2),b/(a-(a^2-b^2)^(1/2)),I)*2^(1/2)*sin(d*x+c)^(1/2)/a/(a^2-b^2)^(3/2)/d/e^2/(-cos(d*x+c))^(1/2)/(e*tan(d*x+c
))^(1/2)+2*b^3*EllipticPi((-cos(d*x+c))^(1/2)/(1+sin(d*x+c))^(1/2),b/(a+(a^2-b^2)^(1/2)),I)*2^(1/2)*sin(d*x+c)
^(1/2)/a/(a^2-b^2)^(3/2)/d/e^2/(-cos(d*x+c))^(1/2)/(e*tan(d*x+c))^(1/2)-1/3*b*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(
c+1/4*Pi+d*x)*EllipticF(cos(c+1/4*Pi+d*x),2^(1/2))*sec(d*x+c)*sin(2*d*x+2*c)^(1/2)/(a^2-b^2)/d/e^2/(e*tan(d*x+
c))^(1/2)-2/3*(a-b*sec(d*x+c))/(a^2-b^2)/d/e/(e*tan(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 1.06, antiderivative size = 836, normalized size of antiderivative = 1.00, number of steps used = 36, number of rules used = 20, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3893, 3882, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641, 3892, 2733, 2729, 2907, 1213, 537} \[ -\frac {2 \sqrt {2} \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right ) \sqrt {\sin (c+d x)} b^3}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right ) \sqrt {\sin (c+d x)} b^3}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right ) b^2}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right ) b^2}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}-\sqrt {2} \sqrt {e \tan (c+d x)}\right ) b^2}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right ) b^2}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {F\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)} b}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}}+\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Sec[c + d*x])*(e*Tan[c + d*x])^(5/2)),x]

[Out]

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

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{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 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 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2641

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

Rule 2729

Int[Sqrt[(g_.)*tan[(e_.) + (f_.)*(x_)]]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[Cos[e
+ f*x]]*Sqrt[g*Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[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 2907

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Dist[(2*Sqrt[2]*d*(b + q))/(f*q), Subst[Int[1/((d*(b + q) + a*x^2
)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - Dist[(2*Sqrt[2]*d*(b - q))/(f*
q), Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]],
x]] /; FreeQ[{a, b, d, e, f}, 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 3882

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

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 3892

Int[1/(Sqrt[cot[(c_.) + (d_.)*(x_)]*(e_.)]*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[1/a, Int
[1/Sqrt[e*Cot[c + d*x]], x], x] - Dist[b/a, Int[1/(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 3893

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

Rubi steps

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

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Mathematica [C]  time = 23.90, size = 2169, normalized size = 2.59 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Sec[c + d*x])*(e*Tan[c + d*x])^(5/2)),x]

[Out]

((b + a*Cos[c + d*x])*((2*a)/(3*(a^2 - b^2)) - (2*(-a + b*Cos[c + d*x])*Csc[c + d*x]^2)/(3*(-a^2 + b^2)))*Sec[
c + d*x]*Tan[c + d*x]^3)/(d*(a + b*Sec[c + d*x])*(e*Tan[c + d*x])^(5/2)) - ((b + a*Cos[c + d*x])*Sec[c + d*x]*
Tan[c + d*x]^(5/2)*((2*(3*a^2 - 5*b^2)*Sec[c + d*x]^3*(a + b*Sqrt[1 + Tan[c + d*x]^2])*(((-1/8 + I/8)*a*(2*Arc
Tan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d
*x]])/(a^2 - b^2)^(1/4)] + Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Ta
n[c + d*x]] - 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)^(3/4)) + (5*b*(-a^2 + b^2)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x
]^2)/(a^2 - b^2)]*Sqrt[Tan[c + d*x]]*Sqrt[1 + Tan[c + d*x]^2])/((5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, -Ta
n[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, -Tan[c + d*x]^2, (b^2*T
an[c + d*x]^2)/(a^2 - b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^
2 - b^2)])*Tan[c + d*x]^2)*(a^2 - b^2*(1 + Tan[c + d*x]^2)))))/((b + a*Cos[c + d*x])*(1 + Tan[c + d*x]^2)^2) +
(8*a*b*Sec[c + d*x]^2*(a + b*Sqrt[1 + Tan[c + d*x]^2])*((Sqrt[b]*(-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]*(-a^2 + b^2)^(3/4)) + (
5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Tan[c + d*x
]])/(Sqrt[1 + Tan[c + d*x]^2]*(-5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)
/(a^2 - b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2
+ b^2)*AppellF1[5/4, 3/2, 1, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2)*(-a^2 +
b^2*(1 + Tan[c + d*x]^2)))))/((b + a*Cos[c + d*x])*(1 + Tan[c + d*x]^2)^(3/2)) + ((3*a^2 - 3*b^2)*Cos[2*(c +
d*x)]*Sec[c + d*x]^3*(a + b*Sqrt[1 + Tan[c + d*x]^2])*((-20*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/a
+ (20*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/a + ((10 - 10*I)*(a^2 - 2*b^2)*ArcTan[1 - ((1 + I)*Sqrt[
b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - ((10 - 10*I)*(a^2 - 2*b^2)*ArcTan[1
+ ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - (10*Sqrt[2]*Log[1
- Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/a + (10*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x
]])/a + ((5 - 5*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]])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - ((5 - 5*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]])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - (8*b*AppellF1[5/
4, 1/2, 1, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(5/2))/(-a^2 + b^2) - (200*b*(
-a^2 + b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Tan[c + d*x]])/
(Sqrt[1 + Tan[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2
- b^2)] + 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^
2)*AppellF1[5/4, 3/2, 1, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2)*(-a^2 + b^2*
(1 + Tan[c + d*x]^2)))))/(20*(b + a*Cos[c + d*x])*(1 - Tan[c + d*x]^2)*(1 + Tan[c + d*x]^2))))/(6*(a - b)*(a +
b)*d*(a + b*Sec[c + d*x])*(e*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(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c) + a)*(e*tan(d*x + c))^(5/2)), x)

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maple [B]  time = 1.93, size = 16178, normalized size = 19.35 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(d*x + c) + a)*(e*tan(d*x + c))^(5/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((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 \[ \int \frac {1}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (a + b \sec {\left (c + d x \right )}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))**(5/2),x)

[Out]

Integral(1/((e*tan(c + d*x))**(5/2)*(a + b*sec(c + d*x))), x)

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