3.756 \(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=144 \[ \frac{2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{11/2}}{11 c^2 f}-\frac{8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{9/2}}{9 c f}+\frac{8 a^3 (B+i A) (c-i c \tan (e+f x))^{7/2}}{7 f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{13/2}}{13 c^3 f} \]

[Out]

(8*a^3*(I*A + B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*f) - (8*a^3*(I*A + 2*B)*(c - I*c*Tan[e + f*x])^(9/2))/(9*c*f
) + (2*a^3*(I*A + 5*B)*(c - I*c*Tan[e + f*x])^(11/2))/(11*c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(13/2))/(13
*c^3*f)

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Rubi [A]  time = 0.210195, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.047, Rules used = {3588, 77} \[ \frac{2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{11/2}}{11 c^2 f}-\frac{8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{9/2}}{9 c f}+\frac{8 a^3 (B+i A) (c-i c \tan (e+f x))^{7/2}}{7 f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{13/2}}{13 c^3 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(8*a^3*(I*A + B)*(c - I*c*Tan[e + f*x])^(7/2))/(7*f) - (8*a^3*(I*A + 2*B)*(c - I*c*Tan[e + f*x])^(9/2))/(9*c*f
) + (2*a^3*(I*A + 5*B)*(c - I*c*Tan[e + f*x])^(11/2))/(11*c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(13/2))/(13
*c^3*f)

Rule 3588

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x)^2 (A+B x) (c-i c x)^{5/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (4 a^2 (A-i B) (c-i c x)^{5/2}-\frac{4 a^2 (A-2 i B) (c-i c x)^{7/2}}{c}+\frac{a^2 (A-5 i B) (c-i c x)^{9/2}}{c^2}+\frac{i a^2 B (c-i c x)^{11/2}}{c^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{8 a^3 (i A+B) (c-i c \tan (e+f x))^{7/2}}{7 f}-\frac{8 a^3 (i A+2 B) (c-i c \tan (e+f x))^{9/2}}{9 c f}+\frac{2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{11/2}}{11 c^2 f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{13/2}}{13 c^3 f}\\ \end{align*}

Mathematica [A]  time = 13.2048, size = 127, normalized size = 0.88 \[ -\frac{2 a^3 c^3 (\cos (3 e)-i \sin (3 e)) \sec ^5(e+f x) \sqrt{c-i c \tan (e+f x)} (7 (169 A-86 i B) \tan (e+f x)+\cos (2 (e+f x)) (7 (169 A-185 i B) \tan (e+f x)-1391 i A-1279 B)-572 i A+737 B)}{9009 f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(-2*a^3*c^3*Sec[e + f*x]^5*(Cos[3*e] - I*Sin[3*e])*Sqrt[c - I*c*Tan[e + f*x]]*((-572*I)*A + 737*B + 7*(169*A -
(86*I)*B)*Tan[e + f*x] + Cos[2*(e + f*x)]*((-1391*I)*A - 1279*B + 7*(169*A - (185*I)*B)*Tan[e + f*x])))/(9009
*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.075, size = 121, normalized size = 0.8 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{3}} \left ({\frac{i}{13}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{13}{2}}}+{\frac{-5\,iBc+Ac}{11} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{11}{2}}}}+{\frac{-4\, \left ( -iBc+Ac \right ) c+4\,iB{c}^{2}}{9} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}+{\frac{ \left ( -4\,iBc+4\,Ac \right ){c}^{2}}{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x)

[Out]

2*I/f*a^3/c^3*(1/13*I*B*(c-I*c*tan(f*x+e))^(13/2)+1/11*(-5*I*B*c+A*c)*(c-I*c*tan(f*x+e))^(11/2)+1/9*(-4*(-I*B*
c+A*c)*c+4*I*B*c^2)*(c-I*c*tan(f*x+e))^(9/2)+4/7*(-I*B*c+A*c)*c^2*(c-I*c*tan(f*x+e))^(7/2))

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Maxima [A]  time = 1.21198, size = 146, normalized size = 1.01 \begin{align*} \frac{2 i \,{\left (693 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{13}{2}} B a^{3} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{11}{2}}{\left (819 \, A - 4095 i \, B\right )} a^{3} c -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{9}{2}}{\left (4004 \, A - 8008 i \, B\right )} a^{3} c^{2} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}{\left (5148 \, A - 5148 i \, B\right )} a^{3} c^{3}\right )}}{9009 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

2/9009*I*(693*I*(-I*c*tan(f*x + e) + c)^(13/2)*B*a^3 + (-I*c*tan(f*x + e) + c)^(11/2)*(819*A - 4095*I*B)*a^3*c
- (-I*c*tan(f*x + e) + c)^(9/2)*(4004*A - 8008*I*B)*a^3*c^2 + (-I*c*tan(f*x + e) + c)^(7/2)*(5148*A - 5148*I*
B)*a^3*c^3)/(c^3*f)

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Fricas [A]  time = 3.02649, size = 539, normalized size = 3.74 \begin{align*} \frac{\sqrt{2}{\left ({\left (82368 i \, A + 82368 \, B\right )} a^{3} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (118976 i \, A - 9152 \, B\right )} a^{3} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (43264 i \, A - 3328 \, B\right )} a^{3} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (6656 i \, A - 512 \, B\right )} a^{3} c^{3}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{9009 \,{\left (f e^{\left (12 i \, f x + 12 i \, e\right )} + 6 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 6 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/9009*sqrt(2)*((82368*I*A + 82368*B)*a^3*c^3*e^(6*I*f*x + 6*I*e) + (118976*I*A - 9152*B)*a^3*c^3*e^(4*I*f*x +
4*I*e) + (43264*I*A - 3328*B)*a^3*c^3*e^(2*I*f*x + 2*I*e) + (6656*I*A - 512*B)*a^3*c^3)*sqrt(c/(e^(2*I*f*x +
2*I*e) + 1))/(f*e^(12*I*f*x + 12*I*e) + 6*f*e^(10*I*f*x + 10*I*e) + 15*f*e^(8*I*f*x + 8*I*e) + 20*f*e^(6*I*f*x
+ 6*I*e) + 15*f*e^(4*I*f*x + 4*I*e) + 6*f*e^(2*I*f*x + 2*I*e) + f)

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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+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [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+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

Timed out