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

Optimal. Leaf size=309 \[ \frac{(2 B+(1+i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{(2 B+(1+i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^3 d}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{(2 B-(1-i) A) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{(2 B-(1-i) A) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{(-B+i A) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2} \]

[Out]

(((1 + I)*A + 2*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - (((1 + I)*A + 2*B)*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - (((-1 + I)*A + 2*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan
[c + d*x]])/(32*Sqrt[2]*a^3*d) + (((-1 + I)*A + 2*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(32*S
qrt[2]*a^3*d) + ((I*A - B)*Tan[c + d*x]^(3/2))/(6*d*(a + I*a*Tan[c + d*x])^3) + ((I/4)*B*Sqrt[Tan[c + d*x]])/(
a*d*(a + I*a*Tan[c + d*x])^2) + ((A - (2*I)*B)*Sqrt[Tan[c + d*x]])/(8*d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.627202, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3595, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{(2 B+(1+i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{(2 B+(1+i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^3 d}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{(2 B-(1-i) A) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{(2 B-(1-i) A) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{(-B+i A) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2} \]

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,x]

[Out]

(((1 + I)*A + 2*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - (((1 + I)*A + 2*B)*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - (((-1 + I)*A + 2*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan
[c + d*x]])/(32*Sqrt[2]*a^3*d) + (((-1 + I)*A + 2*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(32*S
qrt[2]*a^3*d) + ((I*A - B)*Tan[c + d*x]^(3/2))/(6*d*(a + I*a*Tan[c + d*x])^3) + ((I/4)*B*Sqrt[Tan[c + d*x]])/(
a*d*(a + I*a*Tan[c + d*x])^2) + ((A - (2*I)*B)*Sqrt[Tan[c + d*x]])/(8*d*(a^3 + I*a^3*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 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 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 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} \, dx &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\sqrt{\tan (c+d x)} \left (\frac{3}{2} a (i A-B)-\frac{3}{2} a (A-3 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{-3 i a^2 B-3 a^2 (2 i A+3 B) \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{24 a^4}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\int \frac{-3 a^3 A-3 a^3 (i A+2 B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{-3 a^3 A-3 a^3 (i A+2 B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{((-1+i) A+2 B) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^3 d}-\frac{((1+i) A+2 B) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^3 d}\\ &=\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{((-1+i) A+2 B) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 \sqrt{2} a^3 d}-\frac{((-1+i) A+2 B) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 \sqrt{2} a^3 d}-\frac{((1+i) A+2 B) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^3 d}-\frac{((1+i) A+2 B) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^3 d}\\ &=-\frac{((-1+i) A+2 B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{((-1+i) A+2 B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{((1+i) A+2 B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}+\frac{((1+i) A+2 B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}\\ &=\frac{((1+i) A+2 B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{((1+i) A+2 B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{((-1+i) A+2 B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{((-1+i) A+2 B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(i A-B) \tan ^{\frac{3}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{i B \sqrt{\tan (c+d x)}}{4 a d (a+i a \tan (c+d x))^2}+\frac{(A-2 i B) \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 3.77691, size = 274, normalized size = 0.89 \[ \frac{e^{-4 i (c+d x)} \sqrt{\tan (c+d x)} \csc (c+d x) (\cos (3 (c+d x))-i \sin (3 (c+d x))) \left (\left (-2 e^{2 i (c+d x)}-e^{4 i (c+d x)}+2 e^{6 i (c+d x)}+1\right ) \left (-i A \left (1+2 e^{2 i (c+d x)}\right )+B \left (-e^{2 i (c+d x)}\right )+B\right )+6 (B+i A) e^{6 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )-3 B e^{6 i (c+d x)} \sqrt{-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt{-1+e^{4 i (c+d x)}}\right )\right )}{96 a^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((1 - 2*E^((2*I)*(c + d*x)) - E^((4*I)*(c + d*x)) + 2*E^((6*I)*(c + d*x)))*(B - B*E^((2*I)*(c + d*x)) - I*A*(
1 + 2*E^((2*I)*(c + d*x)))) - 3*B*E^((6*I)*(c + d*x))*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)
*(c + d*x))]] + 6*(I*A + B)*E^((6*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*A
rcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Csc[c + d*x]*(Cos[3*(c + d*x)] - I*Sin[3*(
c + d*x)])*Sqrt[Tan[c + d*x]])/(96*a^3*d*E^((4*I)*(c + d*x)))

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Maple [A]  time = 0.05, size = 278, normalized size = 0.9 \begin{align*} -{\frac{B}{4\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{{\frac{i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{5\,A}{12\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{i}{12}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{B}{4\,{a}^{3}d \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }-{\frac{{\frac{i}{4}}A}{{a}^{3}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }-{\frac{B}{4\,{a}^{3}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) } \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,x)

[Out]

-1/4/d/a^3/(tan(d*x+c)-I)^3*B*tan(d*x+c)^(5/2)-1/8*I/d/a^3/(tan(d*x+c)-I)^3*A*tan(d*x+c)^(5/2)-5/12/d/a^3/(tan
(d*x+c)-I)^3*tan(d*x+c)^(3/2)*A+1/12*I/d/a^3/(tan(d*x+c)-I)^3*tan(d*x+c)^(3/2)*B+1/8*I/d/a^3/(tan(d*x+c)-I)^3*
A*tan(d*x+c)^(1/2)-1/4/d/a^3*B/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))-1/4*I/d/a^3/
(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*A-1/4/d/a^3/(2^(1/2)+I*2^(1/2))*arctan(2*ta
n(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*B

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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,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.61822, size = 1713, normalized size = 5.54 \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,x, algorithm="fricas")

[Out]

-1/96*(3*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*((16*I*a^3*d*e^(2*I*d*x +
2*I*c) + 16*I*a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2
)/(a^6*d^2)) + 16*(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 3*a^3*d*sqrt((-I*A^2 - 2*A*
B + I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*((-16*I*a^3*d*e^(2*I*d*x + 2*I*c) - 16*I*a^3*d)*sqrt((-I*e^(
2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^6*d^2)) + 16*(A - I*B)*e^(2*
I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 24*a^3*d*sqrt(-1/64*I*B^2/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log
(1/8*(8*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(
-1/64*I*B^2/(a^6*d^2)) + B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 24*a^3*d*sqrt(-1/64*I*B^2/(a^6*d^2))*e^(6*I*d*x +
6*I*c)*log(-1/8*(8*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c)
+ 1))*sqrt(-1/64*I*B^2/(a^6*d^2)) - B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*(2*(2*A - I*B)*e^(6*I*d*x + 6*I*c) +
(4*A + I*B)*e^(4*I*d*x + 4*I*c) - (A - 2*I*B)*e^(2*I*d*x + 2*I*c) - A - I*B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)
/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-6*I*d*x - 6*I*c)/(a^3*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,x)

[Out]

Timed out

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Giac [A]  time = 1.26051, size = 177, normalized size = 0.57 \begin{align*} -\frac{\left (i + 1\right ) \, \sqrt{2} B \arctan \left (\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac{\left (i - 1\right ) \, \sqrt{2}{\left (i \, A + B\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac{3 i \, A \tan \left (d x + c\right )^{\frac{5}{2}} + 6 \, B \tan \left (d x + c\right )^{\frac{5}{2}} + 10 \, A \tan \left (d x + c\right )^{\frac{3}{2}} - 2 i \, B \tan \left (d x + c\right )^{\frac{3}{2}} - 3 i \, A \sqrt{\tan \left (d x + c\right )}}{24 \, a^{3} d{\left (\tan \left (d x + c\right ) - i\right )}^{3}} \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,x, algorithm="giac")

[Out]

-(1/16*I + 1/16)*sqrt(2)*B*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) + (1/16*I - 1/16)*sqrt(2)*
(I*A + B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) - 1/24*(3*I*A*tan(d*x + c)^(5/2) + 6*B*tan
(d*x + c)^(5/2) + 10*A*tan(d*x + c)^(3/2) - 2*I*B*tan(d*x + c)^(3/2) - 3*I*A*sqrt(tan(d*x + c)))/(a^3*d*(tan(d
*x + c) - I)^3)