3.313 \(\int \frac {(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=740 \[ -\frac {e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}+\frac {e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a b^2 d}-\frac {e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}+\frac {e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}-\frac {2 \sqrt {2} e^2 \sqrt {a^2-b^2} \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} e^2 \sqrt {a^2-b^2} \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {a e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}-\frac {a e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} b^2 d}+\frac {a e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}-\frac {a e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}+\frac {e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{b d \sqrt {e \tan (c+d x)}} \]

[Out]

1/2*a*e^(3/2)*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^(3/2)*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^(3/2)*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^(3/2)*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^(3
/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^(3/2)*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^(3/2)*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^(3/2)*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((-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)*(a^2-b^2)^(1/2)*sin(d*x+c)^(1/2)/a/b/d/(-cos(d*x+c))^(1/2)/(e*tan(d*x+c))^(1/2)+2*e^2*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)*(a^2-b^2)^(1/2)*sin(d*x+c)^(1/2)/a/b
/d/(-cos(d*x+c))^(1/2)/(e*tan(d*x+c))^(1/2)-e^2*(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)/b/d/(e*tan(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.01, antiderivative size = 740, normalized size of antiderivative = 1.00, number of steps used = 35, number of rules used = 19, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.760, Rules used = {3891, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641, 3892, 2733, 2729, 2907, 1213, 537} \[ -\frac {e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a b^2 d}+\frac {e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a b^2 d}-\frac {e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}+\frac {e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a b^2 d}-\frac {2 \sqrt {2} e^2 \sqrt {a^2-b^2} \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} e^2 \sqrt {a^2-b^2} \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {a e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} b^2 d}-\frac {a e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} b^2 d}+\frac {a e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}-\frac {a e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} b^2 d}+\frac {e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{b d \sqrt {e \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Tan[c + d*x])^(3/2)/(a + b*Sec[c + d*x]),x]

[Out]

(a*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*b^2*d) - ((a^2 - b^2)*e^(3/2)*ArcTan[1
- (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*b^2*d) - (a*e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c +
d*x]])/Sqrt[e]])/(Sqrt[2]*b^2*d) + ((a^2 - b^2)*e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(S
qrt[2]*a*b^2*d) + (a*e^(3/2)*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^(3/2)*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^(3/2)*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^(3/2)*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^2 - b^2]*e^2*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*b*d*Sqrt[-Cos[c + d*x]]*Sqrt[e*Tan[c + d*x]]) + (2*Sqrt[2]*Sqrt[a^2 - b^2]*e
^2*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*b*d*Sqrt[-Cos[c + d*x]]*Sqrt[e*Tan[c + d*x]]) + (e^2*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Si
n[2*c + 2*d*x]])/(b*d*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 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 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]

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]

Rubi steps

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

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Mathematica [C]  time = 6.91, size = 202, normalized size = 0.27 \[ \frac {2 e \sqrt {1-\tan \left (\frac {1}{2} (c+d x)\right )} \cot \left (\frac {1}{2} (c+d x)\right ) \sqrt {e \tan (c+d x)} \sqrt {\frac {-\sin (c+d x)+\cos (c+d x)-1}{\cos (c+d x)+1}} \left (-\Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\sqrt {-\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )-\Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\sqrt {-\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )+\Pi \left (-i;\left .\sin ^{-1}\left (\sqrt {-\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )+\Pi \left (i;\left .\sin ^{-1}\left (\sqrt {-\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Tan[c + d*x])^(3/2)/(a + b*Sec[c + d*x]),x]

[Out]

(2*e*Cot[(c + d*x)/2]*(EllipticPi[-I, ArcSin[Sqrt[-Tan[(c + d*x)/2]]], -1] + EllipticPi[I, ArcSin[Sqrt[-Tan[(c
+ d*x)/2]]], -1] - EllipticPi[-(Sqrt[a - b]/Sqrt[a + b]), ArcSin[Sqrt[-Tan[(c + d*x)/2]]], -1] - EllipticPi[S
qrt[a - b]/Sqrt[a + b], ArcSin[Sqrt[-Tan[(c + d*x)/2]]], -1])*Sqrt[(-1 + Cos[c + d*x] - Sin[c + d*x])/(1 + Cos
[c + d*x])]*Sqrt[1 - Tan[(c + d*x)/2]]*Sqrt[e*Tan[c + d*x]])/(a*d)

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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))^(3/2)/(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 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^(3/2)/(b*sec(d*x + c) + a), x)

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maple [B]  time = 1.64, size = 1801, normalized size = 2.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x)

[Out]

1/2/d*(1+cos(d*x+c))^2*(e*sin(d*x+c)/cos(d*x+c))^(3/2)*(-1+cos(d*x+c))*cos(d*x+c)*((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))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(-I*(a^2-b^2
)^(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*(a^2-b^2)^(1/2)*E
llipticPi(((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))*a^2+I*(a^2-b^2)^(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*(a^2-b^2)^(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))*a^2+2*(a^2-b^2)^(1/2)*EllipticPi(((1-cos(d*x+c)+si
n(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b+2*(a^2-b^2)^(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))*a*b+I*(a^2-b^2)^(3/2)*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))-I*(a^2-b^2)^(3/2)*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))-(a^2-b^2)^(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))*a^2-(a^2-b^2)^(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*
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*b-2*E
llipticPi(((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*b^2+2*El
lipticPi(((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*b^2-2*(
a^2-b^2)^(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*(a^2-b^2)^(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^2-2*(a^2-b^2)^(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*(a^2-b^2)^(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-(a^2-b^2)^(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))*a^2-(a^2-b^2)^(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*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*b+2*I*(a^2-b^2)^(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))*a*b-2*I*(a^2-b^2)^(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))*a*b+(a^2-b^2)^(3/2)*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
))+(a^2-b^2)^(3/2)*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))+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^3+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^3-2*EllipticPi(((1-c
os(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^3-2*EllipticPi(((1-co
s(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^3)/sin(d*x+c)^4*2^(1/2
)/((a^2-b^2)^(1/2)-a+b)/((a^2-b^2)^(1/2)+a-b)/(a^2-b^2)^(1/2)/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^(3/2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*tan(d*x + c))^(3/2)/(b*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{b+a\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^(3/2)/(a + b/cos(c + d*x)),x)

[Out]

int((cos(c + d*x)*(e*tan(c + d*x))^(3/2))/(b + a*cos(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{a + b \sec {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**(3/2)/(a+b*sec(d*x+c)),x)

[Out]

Integral((e*tan(c + d*x))**(3/2)/(a + b*sec(c + d*x)), x)

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