3.205 \(\int \frac{(a+i a \tan (c+d x))^3}{\sqrt{e \sec (c+d x)}} \, dx\)
Optimal. Leaf size=124 \[ -\frac{26 i a^3}{3 d \sqrt{e \sec (c+d x)}}-\frac{2 i a^3 \tan ^2(c+d x)}{3 d \sqrt{e \sec (c+d x)}}-\frac{6 a^3 \tan (c+d x)}{d \sqrt{e \sec (c+d x)}}+\frac{14 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]
[Out]
(((-26*I)/3)*a^3)/(d*Sqrt[e*Sec[c + d*x]]) + (14*a^3*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*S
ec[c + d*x]]) - (6*a^3*Tan[c + d*x])/(d*Sqrt[e*Sec[c + d*x]]) - (((2*I)/3)*a^3*Tan[c + d*x]^2)/(d*Sqrt[e*Sec[c
+ d*x]])
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Rubi [A] time = 0.133643, antiderivative size = 146, normalized size of antiderivative =
1.18, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules
used = {3498, 3496, 3768, 3771, 2639} \[ -\frac{14 a^3 \sin (c+d x) \sqrt{e \sec (c+d x)}}{d e}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{e \sec (c+d x)}}+\frac{14 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt{e \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[c + d*x])^3/Sqrt[e*Sec[c + d*x]],x]
[Out]
(14*a^3*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (14*a^3*Sqrt[e*Sec[c + d*x]]*
Sin[c + d*x])/(d*e) + (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^2)/(d*Sqrt[e*Sec[c + d*x]]) - (((28*I)/3)*(a^3 + I*a
^3*Tan[c + d*x]))/(d*Sqrt[e*Sec[c + d*x]])
Rule 3498
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Rule 3496
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Rule 3768
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]
Rule 3771
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
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]
Rubi steps
\begin{align*} \int \frac{(a+i a \tan (c+d x))^3}{\sqrt{e \sec (c+d x)}} \, dx &=\frac{2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt{e \sec (c+d x)}}+\frac{1}{3} (7 a) \int \frac{(a+i a \tan (c+d x))^2}{\sqrt{e \sec (c+d x)}} \, dx\\ &=\frac{2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt{e \sec (c+d x)}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{e \sec (c+d x)}}-\frac{\left (7 a^3\right ) \int (e \sec (c+d x))^{3/2} \, dx}{e^2}\\ &=-\frac{14 a^3 \sqrt{e \sec (c+d x)} \sin (c+d x)}{d e}+\frac{2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt{e \sec (c+d x)}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{e \sec (c+d x)}}+\left (7 a^3\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx\\ &=-\frac{14 a^3 \sqrt{e \sec (c+d x)} \sin (c+d x)}{d e}+\frac{2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt{e \sec (c+d x)}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{e \sec (c+d x)}}+\frac{\left (7 a^3\right ) \int \sqrt{\cos (c+d x)} \, dx}{\sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{14 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{14 a^3 \sqrt{e \sec (c+d x)} \sin (c+d x)}{d e}+\frac{2 i a (a+i a \tan (c+d x))^2}{3 d \sqrt{e \sec (c+d x)}}-\frac{28 i \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{e \sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.56091, size = 101, normalized size = 0.81 \[ \frac{2 a^3 (\cos (c)+i \sin (c)) (\sin (d x)-i \cos (d x)) \sqrt{e \sec (c+d x)} \left (7 \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-i \tan (c+d x)-8\right )}{3 d e} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[c + d*x])^3/Sqrt[e*Sec[c + d*x]],x]
[Out]
(2*a^3*Sqrt[e*Sec[c + d*x]]*(Cos[c] + I*Sin[c])*((-I)*Cos[d*x] + Sin[d*x])*(-8 + 7*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] - I*Tan[c + d*x]))/(3*d*e)
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Maple [B] time = 0.317, size = 2552, normalized size = 20.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(1/2),x)
[Out]
1/3*a^3/d*(cos(d*x+c)-1)*(-18*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)+24*(-cos(d*x+c)/(cos(d*x+c)+1)^2
)^(3/2)*cos(d*x+c)^7+6*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^6+147*I*cos(d*x+c)^2*sin(d*x+c)*(-cos(d
*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+
c)-1)/sin(d*x+c),I)-2*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*sin(d*x+c)+2*I*cos(d*x+c)^4*sin(d*x+c)*(-cos(d*x+
c)/(cos(d*x+c)+1)^2)^(3/2)-44*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3*sin(d*x+c)+6*I*cos(d*x+c)^4*
sin(d*x+c)*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+
1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)-6*I*cos(d*x+c)^4*sin(d*x+c)*ln(-(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1
/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)-24*I*cos(d*
x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)-4*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)*sin
(d*x+c)-6*I*cos(d*x+c)^3*sin(d*x+c)*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*
(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)+6*I*cos(d*x+c)^3*sin(d*x+c)*ln(-(2*(-cos(d*
x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/
sin(d*x+c)^2)+24*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^6*sin(d*x+c)-66*(-cos(d*x+c)/(cos(d*x+c)+1)
^2)^(3/2)*cos(d*x+c)^5-12*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^4+60*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^
(3/2)*cos(d*x+c)^3+6*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^2+48*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/
2)*cos(d*x+c)^5*sin(d*x+c)+21*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-42*I*cos(d*x+c)*sin(d*x+c)*
(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(c
os(d*x+c)-1)/sin(d*x+c),I)+42*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^
(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-21*I*cos(d*x+c)^5*sin(d*x+c)*
(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(c
os(d*x+c)-1)/sin(d*x+c),I)+21*I*cos(d*x+c)^5*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1)
)^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-21*I*(-cos(d*x+c)/(cos(d*x+
c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*EllipticE(I*
(cos(d*x+c)-1)/sin(d*x+c),I)+21*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*cos(d*x+c)^6*sin(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-21*I*(-cos(d*x+c)/(cos(d*
x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^6*sin(d*x+c)*EllipticF(
I*(cos(d*x+c)-1)/sin(d*x+c),I)+42*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^5*sin(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-42*I*cos(d*x+c)^4*sin(d
*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elliptic
E(I*(cos(d*x+c)-1)/sin(d*x+c),I)-42*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^5*sin(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-42*I*(-cos(d*x+c)/(co
s(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^4*sin(d*x+c)*Ellipt
icE(I*(cos(d*x+c)-1)/sin(d*x+c),I)+42*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^4*sin(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-168*I*(-cos(d*x+c)/
(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*Ell
ipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)+168*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+42*I*cos(d*x+c)
^4*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-147*I*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1
/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I))*(cos(d*x+c)
+1)^5*(e/cos(d*x+c))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)/cos(d*x+c)^4/e/sin(d*x+c)^5
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt{e \sec \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((I*a*tan(d*x + c) + a)^3/sqrt(e*sec(d*x + c)), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-24 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 70 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 42 i \, a^{3}\right )} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )} + 3 \,{\left (d e e^{\left (3 i \, d x + 3 i \, c\right )} - d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e e^{\left (i \, d x + i \, c\right )} - d e\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-7 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - 7 i \, a^{3}\right )} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}}{d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e e^{\left (i \, d x + i \, c\right )}}, x\right )}{3 \,{\left (d e e^{\left (3 i \, d x + 3 i \, c\right )} - d e e^{\left (2 i \, d x + 2 i \, c\right )} + d e e^{\left (i \, d x + i \, c\right )} - d e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/3*(sqrt(2)*(-24*I*a^3*e^(4*I*d*x + 4*I*c) - 18*I*a^3*e^(3*I*d*x + 3*I*c) - 70*I*a^3*e^(2*I*d*x + 2*I*c) - 14
*I*a^3*e^(I*d*x + I*c) - 42*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 3*(d*e*e^(3*I*d
*x + 3*I*c) - d*e*e^(2*I*d*x + 2*I*c) + d*e*e^(I*d*x + I*c) - d*e)*integral(sqrt(2)*(-7*I*a^3*e^(2*I*d*x + 2*I
*c) - 14*I*a^3*e^(I*d*x + I*c) - 7*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e*e^(3*
I*d*x + 3*I*c) - 2*d*e*e^(2*I*d*x + 2*I*c) + d*e*e^(I*d*x + I*c)), x))/(d*e*e^(3*I*d*x + 3*I*c) - d*e*e^(2*I*d
*x + 2*I*c) + d*e*e^(I*d*x + I*c) - d*e)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \frac{1}{\sqrt{e \sec{\left (c + d x \right )}}}\, dx + \int - \frac{3 \tan ^{2}{\left (c + d x \right )}}{\sqrt{e \sec{\left (c + d x \right )}}}\, dx + \int \frac{3 i \tan{\left (c + d x \right )}}{\sqrt{e \sec{\left (c + d x \right )}}}\, dx + \int - \frac{i \tan ^{3}{\left (c + d x \right )}}{\sqrt{e \sec{\left (c + d x \right )}}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(1/2),x)
[Out]
a**3*(Integral(1/sqrt(e*sec(c + d*x)), x) + Integral(-3*tan(c + d*x)**2/sqrt(e*sec(c + d*x)), x) + Integral(3*
I*tan(c + d*x)/sqrt(e*sec(c + d*x)), x) + Integral(-I*tan(c + d*x)**3/sqrt(e*sec(c + d*x)), x))
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt{e \sec \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^3/sqrt(e*sec(d*x + c)), x)