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

Optimal. Leaf size=171 \[ \frac{a^{3/2} (11 A-12 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a (4 B+5 i A) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d} \]

[Out]

(a^(3/2)*(11*A - (12*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (2*Sqrt[2]*a^(3/2)*(A - I*B)*A
rcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (a*((5*I)*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c +
d*x]])/(4*d) - (a*A*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(2*d)

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Rubi [A]  time = 0.58305, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3593, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac{a^{3/2} (11 A-12 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a (4 B+5 i A) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/2)*(11*A - (12*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (2*Sqrt[2]*a^(3/2)*(A - I*B)*A
rcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (a*((5*I)*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c +
d*x]])/(4*d) - (a*A*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(2*d)

Rule 3593

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

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 3600

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

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

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

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 208

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

Rubi steps

\begin{align*} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{1}{2} \int \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (5 i A+4 B)-\frac{1}{2} a (3 A-4 i B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a (5 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{\int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (11 A-12 i B)-\frac{1}{4} a^2 (5 i A+4 B) \tan (c+d x)\right ) \, dx}{2 a}\\ &=-\frac{a (5 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{1}{8} (-11 A+12 i B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx-(2 a (i A+B)) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{a (5 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}-\frac{\left (4 a^2 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}-\frac{\left (a^2 (11 A-12 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a (5 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}+\frac{(a (11 i A+12 B)) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 d}\\ &=\frac{a^{3/2} (11 A-12 i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{2 \sqrt{2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a (5 i A+4 B) \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{a A \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 d}\\ \end{align*}

Mathematica [B]  time = 5.65381, size = 400, normalized size = 2.34 \[ \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac{-2 (11 A-12 i B) \left (\log \left (\left (-1+e^{i (c+d x)}\right )^2\right )-\log \left (\left (1+e^{i (c+d x)}\right )^2\right )+\log \left (-2 e^{i (c+d x)} \left (1+\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt{2} \sqrt{1+e^{2 i (c+d x)}}+3\right )-\log \left (2 e^{i (c+d x)} \left (1+\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt{2} \sqrt{1+e^{2 i (c+d x)}}+3\right )\right )-64 \sqrt{2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}+\frac{8 i (\tan (c+d x)+i) \csc (c+d x) (2 A \csc (c+d x)+(4 B+5 i A) \sec (c+d x))}{\sec ^{\frac{5}{2}}(c+d x)}\right )}{32 d \sec ^{\frac{5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*((-64*Sqrt[2]*(A - I*B)*ArcSinh[E^(I*(c + d*x))] - 2*(11*A
- (12*I)*B)*(Log[(-1 + E^(I*(c + d*x)))^2] - Log[(1 + E^(I*(c + d*x)))^2] + Log[3 + 3*E^((2*I)*(c + d*x)) + 2*
Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] - 2*E^(I*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - Log[3
+ 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] + 2*E^(I*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E
^((2*I)*(c + d*x))])]))/((E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)) +
((8*I)*Csc[c + d*x]*(2*A*Csc[c + d*x] + ((5*I)*A + 4*B)*Sec[c + d*x])*(I + Tan[c + d*x]))/Sec[c + d*x]^(5/2)))
/(32*d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.45, size = 1290, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d*a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(11*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-12*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(
1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+14*I*A*cos(d*x+c)^2*sin(d*x+c)-10*I*A*cos(d*x+c)*sin(d*x+c)
+12*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-
1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)-4*A*cos(d*x+c)^2-10*A*cos(d*x+c)+14*A*cos(d*x+c)^3-8*I*B*cos(d*x+c)^3-8
*B*cos(d*x+c)*sin(d*x+c)+8*B*cos(d*x+c)^2*sin(d*x+c)+8*I*B*cos(d*x+c)+16*A*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)-16*B*2^(1
/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+1
1*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/s
in(d*x+c))*sin(d*x+c)-16*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2))*2^(1/2)-16*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*2^(1/2)-16*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arc
tanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+1
6*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)
+1))^(1/2))*2^(1/2)-11*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d
*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+12*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-11*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(
1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-12*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(-(-(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+16*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+16*I*B*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c
))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2))/(cos(d*x+c)-1)/(I*sin(d*x+c)+cos(d*x+c)-1)/(cos(d*x+c)+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.98599, size = 2080, normalized size = 12.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*((7*A - 4*I*B)*a*e^(4*I*d*x + 4*I*c) + 4*A*a*e^(2*I*d*x + 2*I*c) - (3*A - 4*I*B)*a)*sqrt(a/(e^(
2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + sqrt((121*A^2 - 264*I*A*B - 144*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c)
- 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((11*I*A + 12*B)*a*e^(2*I*d*x + 2*I*c) + (11*I*A + 12*B)*a)*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 2*I*sqrt((121*A^2 - 264*I*A*B - 144*B^2)*a^3/d^2)*d*e^(2*I*d*x +
2*I*c))*e^(-2*I*d*x - 2*I*c)/((11*I*A + 12*B)*a)) - sqrt((121*A^2 - 264*I*A*B - 144*B^2)*a^3/d^2)*(d*e^(4*I*d
*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((11*I*A + 12*B)*a*e^(2*I*d*x + 2*I*c) + (11*I*A + 12*
B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 2*I*sqrt((121*A^2 - 264*I*A*B - 144*B^2)*a^3/d^2)*d*
e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((11*I*A + 12*B)*a)) - 4*sqrt((8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*(d
*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A
+ 2*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + I*sqrt((8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*d*e^
(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) + 4*sqrt((8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*(d*e^(
4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2
*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - I*sqrt((8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*d*e^(2*I
*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + 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(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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