3.194 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=240 \[ -\frac{(317 A+67 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{151 A+41 i B}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]
[Out]
((1/8 + I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) +
(A + I*B)/(5*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + (17*A + (7*I)*B)/(30*a*d*Sqrt[Tan[c + d*x]]
*(a + I*a*Tan[c + d*x])^(3/2)) + (151*A + (41*I)*B)/(60*a^2*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) -
((317*A + (67*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Sqrt[Tan[c + d*x]])
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Rubi [A] time = 0.802976, antiderivative size = 240, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used =
{3596, 3598, 12, 3544, 205} \[ -\frac{(317 A+67 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{151 A+41 i B}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]
[Out]
((1/8 + I/8)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) +
(A + I*B)/(5*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + (17*A + (7*I)*B)/(30*a*d*Sqrt[Tan[c + d*x]]
*(a + I*a*Tan[c + d*x])^(3/2)) + (151*A + (41*I)*B)/(60*a^2*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) -
((317*A + (67*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Sqrt[Tan[c + d*x]])
Rule 3596
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n,
0]
Rule 3598
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3544
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (11 A+i B)-3 a (i A-B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 (83 A+13 i B)-a^2 (17 i A-7 B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151 A+41 i B}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{8} a^3 (317 A+67 i B)-\frac{1}{4} a^3 (151 i A-41 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151 A+41 i B}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{2 \int \frac{15 a^4 (i A+B) \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{15 a^7}\\ &=\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151 A+41 i B}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{(i A+B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151 A+41 i B}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{A+i B}{5 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac{17 A+7 i B}{30 a d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{151 A+41 i B}{60 a^2 d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \sqrt{a+i a \tan (c+d x)}}{60 a^3 d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 9.88676, size = 288, normalized size = 1.2 \[ \frac{\sec ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x)) \left (\frac{\sqrt{2} (B+i A) e^{3 i (c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )}{\sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}+\frac{2 ((19 B-149 i A) \tan (c+d x)+\cos (2 (c+d x)) ((86 B-466 i A) \tan (c+d x)-20 (23 A+4 i B))+340 A+80 i B)}{15 \sqrt{\tan (c+d x)} \sqrt{\sec (c+d x)}}\right )}{8 d (a+i a \tan (c+d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]
[Out]
(Sec[c + d*x]^(3/2)*(A + B*Tan[c + d*x])*((Sqrt[2]*(I*A + B)*E^((3*I)*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c
+ d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(Sqrt[-1 + E^((2
*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]) + (2*(340*A + (80*I)*B + ((-149*I)*A + 19*B)*
Tan[c + d*x] + Cos[2*(c + d*x)]*(-20*(23*A + (4*I)*B) + ((-466*I)*A + 86*B)*Tan[c + d*x])))/(15*Sqrt[Sec[c + d
*x]]*Sqrt[Tan[c + d*x]])))/(8*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2))
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Maple [B] time = 0.059, size = 1158, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)
[Out]
-1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(-1060*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+
c)^2+1268*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4+268*I*B*(a*tan(d*x+c)*(1+I*tan(d*x
+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4-15*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c))
)^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a+908*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^
(1/2)*tan(d*x+c)^3+60*I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*t
an(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a+15*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d
*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a+60*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a-5660*A*(a*tan(d*x+c)*(1+
I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+15*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*
tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a+2940*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(
1/2)*(-I*a)^(1/2)*tan(d*x+c)+90*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I
*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a-60*I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1
+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-4468*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3-60*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-90*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a-15*A*2^(1/2)*ln(-(-2*2^(1/2)*(
-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a-420*B*(a*ta
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+480*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1
/2))/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/(-I*a)^(1/2)
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [B] time = 2.189, size = 1534, normalized size = 6.39 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
1/120*(sqrt(2)*((-463*I*A + 83*B)*e^(8*I*d*x + 8*I*c) + (-269*I*A + 19*B)*e^(6*I*d*x + 6*I*c) + (220*I*A - 80*
B)*e^(4*I*d*x + 4*I*c) + (29*I*A - 19*B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*
sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 15*sqrt(1/2)*(a^3*d*e^(8*I*d*x
+ 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*log((2*sqrt(1/2)*a^3*d*sqrt((I*A
^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/
(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I
*d*x - I*c)/(4*I*A + 4*B)) - 15*sqrt(1/2)*(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((I*A^2
+ 2*A*B - I*B^2)/(a^5*d^2))*log(-(2*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c
) - sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x +
2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)))/(a^3*d*e^(8*I*d*x + 8
*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [A] time = 1.52889, size = 236, normalized size = 0.98 \begin{align*} \frac{-\left (i + 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} +{\left (\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - \left (2 i - 2\right ) \, a^{3}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a + 8 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} - 10 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 4 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4}\right )} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
[Out]
(-(I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^3 + ((2*I - 2)*(I*a*tan(d*x + c)
+ a)*a^2 - (2*I - 2)*a^3)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/((-2*I*(I*a*
tan(d*x + c) + a)^6*a + 8*I*(I*a*tan(d*x + c) + a)^5*a^2 - 10*I*(I*a*tan(d*x + c) + a)^4*a^3 + 4*I*(I*a*tan(d*
x + c) + a)^3*a^4)*d)