3.185 \(\int \frac{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=203 \[ -\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{2 (-1)^{3/4} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{(-B+i A) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(2*(-1)^(3/4)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) - ((1/
4 - I/4)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + ((I
*A - B)*Tan[c + d*x]^(3/2))/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((A + (3*I)*B)*Sqrt[Tan[c + d*x]])/(2*a*d*Sqr
t[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.668543, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{\left (\frac{1}{4}-\frac{i}{4}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{2 (-1)^{3/4} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{(-B+i A) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^(3/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*(-1)^(3/4)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) - ((1/
4 - I/4)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + ((I
*A - B)*Tan[c + d*x]^(3/2))/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((A + (3*I)*B)*Sqrt[Tan[c + d*x]])/(2*a*d*Sqr
t[a + I*a*Tan[c + d*x]])

Rule 3595

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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rule 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

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]

Rule 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\sqrt{\tan (c+d x)} \left (\frac{3}{2} a (i A-B)+3 i a B \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{4} a^2 (A+3 i B)-3 a^2 B \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(A-i B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{4 a^2}-\frac{(i B) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{a^3}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(i B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{a d}+\frac{(i A+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 )}{2 d}\\ &=\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (i A+B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 i B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a d}\\ &=\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (i A+B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 i B) \operatorname{Subst}\left (\int \frac{1}{1-i a x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a d}\\ &=\frac{2 (-1)^{3/4} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (i A+B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{(A+3 i B) \sqrt{\tan (c+d x)}}{2 a d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 6.14975, size = 285, normalized size = 1.4 \[ \frac{e^{-i (c+d x)} \sqrt{\tan (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \left (\sqrt{-1+e^{2 i (c+d x)}} \left (-4 i A e^{2 i (c+d x)}+i A+10 B e^{2 i (c+d x)}-B\right )+3 (B+i A) e^{3 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-12 \sqrt{2} B e^{3 i (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{6 \sqrt{2} a d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^(3/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((Sqrt[-1 + E^((2*I)*(c + d*x))]*(I*A - B - (4*I)*A*E^((2*I)*(c + d*x)) + 10*B*E^((2*I)*(c + d*x))) + 3*(I*A +
B)*E^((3*I)*(c + d*x))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] - 12*Sqrt[2]*B*E^((3*I)*(c + d
*x))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]^(3/2)*Sqrt[Tan[c + d*x]])
/(6*Sqrt[2]*a*d*E^(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)))]
*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.068, size = 1223, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

1/24*I/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(-72*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+
I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a+20*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan
(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+9*I*A*(I*a)^(1/2)*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-3*A*(I*a)^(1/2)*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-36*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+3*I*B*(I*a)^(1/2)*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+44
*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+9*B*(I*a)^(1/2)*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+24*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1
/2)*tan(d*x+c)^3*a+24*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a
)^(1/2))*(-I*a)^(1/2)*a-3*I*A*(I*a)^(1/2)*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))*a+9*A*(I*a)^(1/2)*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-32*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-9*I*B*(I*a)^(1/2)*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-3*B*(I*a)^(1/2)*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))*a-72*B*ln(1/2*(
2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)*a
+80*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-12*A*(a*tan(d*x+c)*(1+I*tan(d*
x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^3/(I*a)^(1/2)/(-I
*a)^(1/2)

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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(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.86375, size = 2169, normalized size = 10.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log((2*I*sqrt(1/2)*a^2*d
*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*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)) - 3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(4*I*d*x
+ 4*I*c)*log((-2*I*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*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)) - 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(4
*I*d*x + 4*I*c)*log(1/605*(208*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) + 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) + (156*I*a^2*d*e^(2*I*d*x + 2*I*c) - 52*I*
a^2*d)*sqrt(4*I*B^2/(a^3*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) + 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(4*I*d*x + 4*
I*c)*log(1/605*(208*sqrt(2)*(B*e^(2*I*d*x + 2*I*c) + 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) + (-156*I*a^2*d*e^(2*I*d*x + 2*I*c) + 52*I*a^2*d)*sqr
t(4*I*B^2/(a^3*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) - sqrt(2)*(2*(2*A + 5*I*B)*e^(4*I*d*x + 4*I*c) + 3*(A + 3*I
*B)*e^(2*I*d*x + 2*I*c) - A - I*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^(-4*I*d*x - 4*I*c)/(a^2*d)

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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(tan(d*x+c)**(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError