3.691 \(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^5 \, dx\)

Optimal. Leaf size=135 \[ \frac{a^3 c^5 (5 B+i A) (1-i \tan (e+f x))^7}{7 f}-\frac{2 a^3 c^5 (2 B+i A) (1-i \tan (e+f x))^6}{3 f}+\frac{4 a^3 c^5 (B+i A) (1-i \tan (e+f x))^5}{5 f}-\frac{a^3 B c^5 (1-i \tan (e+f x))^8}{8 f} \]

[Out]

(4*a^3*(I*A + B)*c^5*(1 - I*Tan[e + f*x])^5)/(5*f) - (2*a^3*(I*A + 2*B)*c^5*(1 - I*Tan[e + f*x])^6)/(3*f) + (a
^3*(I*A + 5*B)*c^5*(1 - I*Tan[e + f*x])^7)/(7*f) - (a^3*B*c^5*(1 - I*Tan[e + f*x])^8)/(8*f)

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Rubi [A]  time = 0.190149, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 77} \[ \frac{a^3 c^5 (5 B+i A) (1-i \tan (e+f x))^7}{7 f}-\frac{2 a^3 c^5 (2 B+i A) (1-i \tan (e+f x))^6}{3 f}+\frac{4 a^3 c^5 (B+i A) (1-i \tan (e+f x))^5}{5 f}-\frac{a^3 B c^5 (1-i \tan (e+f x))^8}{8 f} \]

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])^5,x]

[Out]

(4*a^3*(I*A + B)*c^5*(1 - I*Tan[e + f*x])^5)/(5*f) - (2*a^3*(I*A + 2*B)*c^5*(1 - I*Tan[e + f*x])^6)/(3*f) + (a
^3*(I*A + 5*B)*c^5*(1 - I*Tan[e + f*x])^7)/(7*f) - (a^3*B*c^5*(1 - I*Tan[e + f*x])^8)/(8*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))^5 \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x)^2 (A+B x) (c-i c x)^4 \, 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)^4-\frac{4 a^2 (A-2 i B) (c-i c x)^5}{c}+\frac{a^2 (A-5 i B) (c-i c x)^6}{c^2}+\frac{i a^2 B (c-i c x)^7}{c^3}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{4 a^3 (i A+B) c^5 (1-i \tan (e+f x))^5}{5 f}-\frac{2 a^3 (i A+2 B) c^5 (1-i \tan (e+f x))^6}{3 f}+\frac{a^3 (i A+5 B) c^5 (1-i \tan (e+f x))^7}{7 f}-\frac{a^3 B c^5 (1-i \tan (e+f x))^8}{8 f}\\ \end{align*}

Mathematica [A]  time = 10.4736, size = 215, normalized size = 1.59 \[ \frac{a^3 c^5 \sec (e) \sec ^8(e+f x) (70 (B-i A) \cos (e+2 f x)+35 (B-4 i A) \cos (e)+154 A \sin (e+2 f x)-70 A \sin (3 e+2 f x)+112 A \sin (3 e+4 f x)+32 A \sin (5 e+6 f x)+4 A \sin (7 e+8 f x)-70 i A \cos (3 e+2 f x)-140 A \sin (e)-14 i B \sin (e+2 f x)-70 i B \sin (3 e+2 f x)+28 i B \sin (3 e+4 f x)+8 i B \sin (5 e+6 f x)+i B \sin (7 e+8 f x)+70 B \cos (3 e+2 f x)-35 i B \sin (e))}{840 f} \]

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])^5,x]

[Out]

(a^3*c^5*Sec[e]*Sec[e + f*x]^8*(35*((-4*I)*A + B)*Cos[e] + 70*((-I)*A + B)*Cos[e + 2*f*x] - (70*I)*A*Cos[3*e +
2*f*x] + 70*B*Cos[3*e + 2*f*x] - 140*A*Sin[e] - (35*I)*B*Sin[e] + 154*A*Sin[e + 2*f*x] - (14*I)*B*Sin[e + 2*f
*x] - 70*A*Sin[3*e + 2*f*x] - (70*I)*B*Sin[3*e + 2*f*x] + 112*A*Sin[3*e + 4*f*x] + (28*I)*B*Sin[3*e + 4*f*x] +
32*A*Sin[5*e + 6*f*x] + (8*I)*B*Sin[5*e + 6*f*x] + 4*A*Sin[7*e + 8*f*x] + I*B*Sin[7*e + 8*f*x]))/(840*f)

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Maple [A]  time = 0.012, size = 169, normalized size = 1.3 \begin{align*}{\frac{{a}^{3}{c}^{5}}{f} \left ( -{\frac{2\,i}{7}}B \left ( \tan \left ( fx+e \right ) \right ) ^{7}-{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{8}}{8}}-{\frac{i}{3}}A \left ( \tan \left ( fx+e \right ) \right ) ^{6}-{\frac{A \left ( \tan \left ( fx+e \right ) \right ) ^{7}}{7}}-{\frac{4\,i}{5}}B \left ( \tan \left ( fx+e \right ) \right ) ^{5}-{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{6}}{6}}-iA \left ( \tan \left ( fx+e \right ) \right ) ^{4}-{\frac{A \left ( \tan \left ( fx+e \right ) \right ) ^{5}}{5}}-{\frac{2\,i}{3}}B \left ( \tan \left ( fx+e \right ) \right ) ^{3}+{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{4}}{4}}-iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}+{\frac{A \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3}}+{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2}}+A\tan \left ( fx+e \right ) \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))^5,x)

[Out]

1/f*a^3*c^5*(-2/7*I*B*tan(f*x+e)^7-1/8*B*tan(f*x+e)^8-1/3*I*A*tan(f*x+e)^6-1/7*A*tan(f*x+e)^7-4/5*I*B*tan(f*x+
e)^5-1/6*B*tan(f*x+e)^6-I*A*tan(f*x+e)^4-1/5*A*tan(f*x+e)^5-2/3*I*B*tan(f*x+e)^3+1/4*B*tan(f*x+e)^4-I*A*tan(f*
x+e)^2+1/3*A*tan(f*x+e)^3+1/2*B*tan(f*x+e)^2+A*tan(f*x+e))

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Maxima [A]  time = 1.70261, size = 230, normalized size = 1.7 \begin{align*} -\frac{105 \, B a^{3} c^{5} \tan \left (f x + e\right )^{8} +{\left (120 \, A + 240 i \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{7} - 140 \,{\left (-2 i \, A - B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{6} +{\left (168 \, A + 672 i \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{5} - 210 \,{\left (-4 i \, A + B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{4} -{\left (280 \, A - 560 i \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{3} - 420 \,{\left (-2 i \, A + B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{2} - 840 \, A a^{3} c^{5} \tan \left (f x + e\right )}{840 \, 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))^5,x, algorithm="maxima")

[Out]

-1/840*(105*B*a^3*c^5*tan(f*x + e)^8 + (120*A + 240*I*B)*a^3*c^5*tan(f*x + e)^7 - 140*(-2*I*A - B)*a^3*c^5*tan
(f*x + e)^6 + (168*A + 672*I*B)*a^3*c^5*tan(f*x + e)^5 - 210*(-4*I*A + B)*a^3*c^5*tan(f*x + e)^4 - (280*A - 56
0*I*B)*a^3*c^5*tan(f*x + e)^3 - 420*(-2*I*A + B)*a^3*c^5*tan(f*x + e)^2 - 840*A*a^3*c^5*tan(f*x + e))/f

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Fricas [A]  time = 1.29634, size = 547, normalized size = 4.05 \begin{align*} \frac{{\left (2688 i \, A + 2688 \, B\right )} a^{3} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (3584 i \, A - 896 \, B\right )} a^{3} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (1024 i \, A - 256 \, B\right )} a^{3} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (128 i \, A - 32 \, B\right )} a^{3} c^{5}}{105 \,{\left (f e^{\left (16 i \, f x + 16 i \, e\right )} + 8 \, f e^{\left (14 i \, f x + 14 i \, e\right )} + 28 \, f e^{\left (12 i \, f x + 12 i \, e\right )} + 56 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 70 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 56 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 28 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, 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))^5,x, algorithm="fricas")

[Out]

1/105*((2688*I*A + 2688*B)*a^3*c^5*e^(6*I*f*x + 6*I*e) + (3584*I*A - 896*B)*a^3*c^5*e^(4*I*f*x + 4*I*e) + (102
4*I*A - 256*B)*a^3*c^5*e^(2*I*f*x + 2*I*e) + (128*I*A - 32*B)*a^3*c^5)/(f*e^(16*I*f*x + 16*I*e) + 8*f*e^(14*I*
f*x + 14*I*e) + 28*f*e^(12*I*f*x + 12*I*e) + 56*f*e^(10*I*f*x + 10*I*e) + 70*f*e^(8*I*f*x + 8*I*e) + 56*f*e^(6
*I*f*x + 6*I*e) + 28*f*e^(4*I*f*x + 4*I*e) + 8*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))**5,x)

[Out]

Timed out

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Giac [B]  time = 2.41563, size = 327, normalized size = 2.42 \begin{align*} \frac{2688 i \, A a^{3} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + 2688 \, B a^{3} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + 3584 i \, A a^{3} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} - 896 \, B a^{3} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 1024 i \, A a^{3} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 256 \, B a^{3} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 128 i \, A a^{3} c^{5} - 32 \, B a^{3} c^{5}}{105 \,{\left (f e^{\left (16 i \, f x + 16 i \, e\right )} + 8 \, f e^{\left (14 i \, f x + 14 i \, e\right )} + 28 \, f e^{\left (12 i \, f x + 12 i \, e\right )} + 56 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 70 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 56 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 28 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, 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))^5,x, algorithm="giac")

[Out]

1/105*(2688*I*A*a^3*c^5*e^(6*I*f*x + 6*I*e) + 2688*B*a^3*c^5*e^(6*I*f*x + 6*I*e) + 3584*I*A*a^3*c^5*e^(4*I*f*x
+ 4*I*e) - 896*B*a^3*c^5*e^(4*I*f*x + 4*I*e) + 1024*I*A*a^3*c^5*e^(2*I*f*x + 2*I*e) - 256*B*a^3*c^5*e^(2*I*f*
x + 2*I*e) + 128*I*A*a^3*c^5 - 32*B*a^3*c^5)/(f*e^(16*I*f*x + 16*I*e) + 8*f*e^(14*I*f*x + 14*I*e) + 28*f*e^(12
*I*f*x + 12*I*e) + 56*f*e^(10*I*f*x + 10*I*e) + 70*f*e^(8*I*f*x + 8*I*e) + 56*f*e^(6*I*f*x + 6*I*e) + 28*f*e^(
4*I*f*x + 4*I*e) + 8*f*e^(2*I*f*x + 2*I*e) + f)