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

Optimal. Leaf size=248 \[ \frac{(-1)^{3/4} a^{3/2} (23 B+22 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}+\frac{(2+2 i) a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{a (7 B+6 i A) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d} \]

[Out]

((-1)^(3/4)*a^(3/2)*((22*I)*A + 23*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]
]])/(8*d) + ((2 + 2*I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/d + (a*(10*A - (9*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) + (a*((6*I)*A + 7*B)*Tan[c +
d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(12*d) + ((I/3)*a*B*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d

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Rubi [A]  time = 0.902085, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.237, Rules used = {3594, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{(-1)^{3/4} a^{3/2} (23 B+22 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}+\frac{(2+2 i) a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{a (7 B+6 i A) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(3/4)*a^(3/2)*((22*I)*A + 23*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]
]])/(8*d) + ((2 + 2*I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/d + (a*(10*A - (9*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) + (a*((6*I)*A + 7*B)*Tan[c +
d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(12*d) + ((I/3)*a*B*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && GtQ[m, 1] &&  !LtQ[n, -1]

Rule 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && 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 \tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{1}{3} \int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (6 A-5 i B)+\frac{1}{2} a (6 i A+7 B) \tan (c+d x)\right ) \, dx\\ &=\frac{a (6 i A+7 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{4} a^2 (6 i A+7 B)+\frac{3}{4} a^2 (10 A-9 i B) \tan (c+d x)\right ) \, dx}{6 a}\\ &=\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{a (6 i A+7 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{8} a^3 (10 A-9 i B)-\frac{3}{8} a^3 (22 i A+23 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{6 a^2}\\ &=\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{a (6 i A+7 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}-(2 a (A-i B)) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx+\frac{1}{16} (22 A-23 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\\ &=\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{a (6 i A+7 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\left (a^2 (22 A-23 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}+\frac{\left (4 a^3 (i A+B)\right ) \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 )}{d}\\ &=\frac{(2+2 i) a^{3/2} (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 )}{d}+\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{a (6 i A+7 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\left (a^2 (22 A-23 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{8 d}\\ &=\frac{(2+2 i) a^{3/2} (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 )}{d}+\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{a (6 i A+7 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{\left (a^2 (22 A-23 i B)\right ) \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 )}{8 d}\\ &=-\frac{\sqrt [4]{-1} a^{3/2} (22 A-23 i B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{8 d}+\frac{(2+2 i) a^{3/2} (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 )}{d}+\frac{a (10 A-9 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{a (6 i A+7 B) \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{i a B \tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}\\ \end{align*}

Mathematica [A]  time = 7.18527, size = 420, normalized size = 1.69 \[ \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac{\sqrt{2} e^{-i (c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt{2} (22 A-23 i B) \left (\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )-128 (A-i B) \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right )}{\sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}+\frac{4 (\cos (c)-i \sin (c)) \sqrt{\tan (c+d x)} \sec ^{\frac{5}{2}}(c+d x) (2 (7 B+6 i A) \sin (2 (c+d x))+5 (6 A-7 i B) \cos (2 (c+d x))+30 A-19 i B)}{3 \cos (d x)+3 i \sin (d x)}\right )}{64 d \sec ^{\frac{5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(-128*(A - I*B)*Log[E^(I*(c + d*x
)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*(22*A - (23*I)*B)*(Log[1 - 3*E^((2*I)*(c + d*x)) - 2*Sqrt[2]*E^
(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] - Log[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt
[-1 + E^((2*I)*(c + d*x))]])))/(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)))]) + (4*Sec[c + d*x]^(5/2)*(Cos[c] - I*Sin[c])*(30*A - (19*I)*B + 5*(6*A - (7*I)*B)*Cos[2*(c +
d*x)] + 2*((6*I)*A + 7*B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/(3*Cos[d*x] + (3*I)*Sin[d*x]))*(a + I*a*Tan[c
+ d*x])^(3/2)*(A + B*Tan[c + d*x]))/(64*d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.062, size = 652, normalized size = 2.6 \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+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

1/48/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(16*I*B*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2
)*(-I*a)^(1/2)*(I*a)^(1/2)+24*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)+27
*I*B*(-I*a)^(1/2)*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))
*a-54*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)+28*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)-24*I*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))*(I*a)^(1/2)*2^(1/2)*a-30*A*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+60*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-
I*a)^(1/2)*(I*a)^(1/2)-48*I*(-I*a)^(1/2)*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))*a+24*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*(I*a)^(1/2)-48*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))*a*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)/(I*a)^(1/2)

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

[Out]

Timed out

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Fricas [B]  time = 2.04483, size = 2564, normalized size = 10.34 \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+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(2*sqrt(2)*(7*(6*A - 7*I*B)*a*e^(4*I*d*x + 4*I*c) + 2*(30*A - 19*I*B)*a*e^(2*I*d*x + 2*I*c) + 3*(6*A - 7*
I*B)*a)*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) + 3*sqrt((-484*I*A^2 - 1012*A*B + 529*I*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)*((22*I*A + 23*B)*a*e^(2*I*d*x + 2*I*c) + (22*I*A + 23*B)*a)*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) + 2*I*sqrt((-484*I*A^2 - 1012
*A*B + 529*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((22*I*A + 23*B)*a)) - 3*sqrt((-484*I*A
^2 - 1012*A*B + 529*I*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)*((22*I*
A + 23*B)*a*e^(2*I*d*x + 2*I*c) + (22*I*A + 23*B)*a)*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) - 2*I*sqrt((-484*I*A^2 - 1012*A*B + 529*I*B^2)*a^3/d^2)*
d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((22*I*A + 23*B)*a)) - 24*sqrt((-8*I*A^2 - 16*A*B + 8*I*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))*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) + I*sqrt((-8*I*A^2 - 16*A*B + 8*I*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)) + 24*sqrt((-8*I*A^2 - 16*A*B + 8*I*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))*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) - I*sqrt((-8*I*A^2 - 16*A*B
+ 8*I*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(tan(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.3871, size = 412, normalized size = 1.66 \begin{align*} \frac{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} + 4 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a +{\left (2 \, a \tan \left (d x + c\right ) - 2 i \, a\right )} a^{2}\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 (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )} +{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a -{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2}\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 )}{\left (\frac{-i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + i \, a^{2}}{\sqrt{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + a^{4}}} + 1\right )}}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 2 \, a^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((-2*I*(I*a*tan(d*x + c) + a)^3 + 4*I*(I*a*tan(d*x + c) + a)^2*a + (2*a*tan(d*x + c) - 2*I*a)*a^2)*sqrt(-2
*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B*((-I*(I*a*tan(d*x + c) + a)*a + I*a^2)/sqrt((I
*a*tan(d*x + c) + a)^2*a^2 - 2*(I*a*tan(d*x + c) + a)*a^3 + a^4) + 1) + ((I*a*tan(d*x + c) + a)^2*a - (I*a*tan
(d*x + c) + a)*a^2)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*((-I*(I*a*tan(d*x + c) +
a)*a + I*a^2)/sqrt((I*a*tan(d*x + c) + a)^2*a^2 - 2*(I*a*tan(d*x + c) + a)*a^3 + a^4) + 1))/(((I*a*tan(d*x + c
) + a)*a^2 - 2*a^3)*d)