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

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

[Out]

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

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Rubi [A]  time = 0.200447, 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} \[ \frac{2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{7/2}}{7 c^2 f}-\frac{8 a^3 (2 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac{8 a^3 (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^3 B (c-i c \tan (e+f x))^{9/2}}{9 c^3 f} \]

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])^(3/2),x]

[Out]

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

Mathematica [A]  time = 8.49487, size = 130, normalized size = 0.9 \[ -\frac{2 a^3 c \sec ^3(e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (e-2 f x)-i \sin (e-2 f x)) ((81 A-62 i B) \tan (e+f x)+\cos (2 (e+f x)) ((81 A-97 i B) \tan (e+f x)-129 i A-113 B)+7 (B-12 i A))}{315 f (\cos (f x)+i \sin (f x))^3} \]

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])^(3/2),x]

[Out]

(-2*a^3*c*Sec[e + f*x]^3*(Cos[e - 2*f*x] - I*Sin[e - 2*f*x])*Sqrt[c - I*c*Tan[e + f*x]]*(7*((-12*I)*A + B) + (
81*A - (62*I)*B)*Tan[e + f*x] + Cos[2*(e + f*x)]*((-129*I)*A - 113*B + (81*A - (97*I)*B)*Tan[e + f*x])))/(315*
f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]  time = 0.074, size = 121, normalized size = 0.8 \begin{align*}{\frac{2\,i{a}^{3}}{f{c}^{3}} \left ({\frac{i}{9}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}+{\frac{-5\,iBc+Ac}{7} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{-4\, \left ( -iBc+Ac \right ) c+4\,iB{c}^{2}}{5} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{ \left ( -4\,iBc+4\,Ac \right ){c}^{2}}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{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))^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.19415, size = 146, normalized size = 1.01 \begin{align*} \frac{2 i \,{\left (35 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{9}{2}} B a^{3} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}}{\left (45 \, A - 225 i \, B\right )} a^{3} c -{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}{\left (252 \, A - 504 i \, B\right )} a^{3} c^{2} +{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (420 \, A - 420 i \, B\right )} a^{3} c^{3}\right )}}{315 \, 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))^(3/2),x, algorithm="maxima")

[Out]

2/315*I*(35*I*(-I*c*tan(f*x + e) + c)^(9/2)*B*a^3 + (-I*c*tan(f*x + e) + c)^(7/2)*(45*A - 225*I*B)*a^3*c - (-I
*c*tan(f*x + e) + c)^(5/2)*(252*A - 504*I*B)*a^3*c^2 + (-I*c*tan(f*x + e) + c)^(3/2)*(420*A - 420*I*B)*a^3*c^3
)/(c^3*f)

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Fricas [A]  time = 1.49056, size = 437, normalized size = 3.03 \begin{align*} \frac{\sqrt{2}{\left ({\left (1680 i \, A + 1680 \, B\right )} a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (3024 i \, A + 1008 \, B\right )} a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (1728 i \, A + 576 \, B\right )} a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (384 i \, A + 128 \, B\right )} a^{3} c\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{315 \,{\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, 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))^(3/2),x, algorithm="fricas")

[Out]

1/315*sqrt(2)*((1680*I*A + 1680*B)*a^3*c*e^(6*I*f*x + 6*I*e) + (3024*I*A + 1008*B)*a^3*c*e^(4*I*f*x + 4*I*e) +
(1728*I*A + 576*B)*a^3*c*e^(2*I*f*x + 2*I*e) + (384*I*A + 128*B)*a^3*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*
e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int A c \sqrt{- i c \tan{\left (e + f x \right )} + c}\, dx + \int - A c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\, dx + \int B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int - B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )}\, dx + \int 2 i A c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan{\left (e + f x \right )}\, dx + \int 2 i A c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\, dx + \int 2 i B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int 2 i B c \sqrt{- i c \tan{\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\, dx\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))**(3/2),x)

[Out]

a**3*(Integral(A*c*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4
, x) + Integral(B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(-B*c*sqrt(-I*c*tan(e + f*x) + c)*t
an(e + f*x)**5, x) + Integral(2*I*A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(2*I*A*c*sqrt(-I*
c*tan(e + f*x) + c)*tan(e + f*x)**3, x) + Integral(2*I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2, x) + I
ntegral(2*I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \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))^(3/2),x, algorithm="giac")

[Out]

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