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]

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

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Rubi [A]  time = 0.19, antiderivative size = 135, normalized size of antiderivative = 1.00, 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 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]))))

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]

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*}

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Mathematica [A]  time = 10.89, 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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fricas [A]  time = 0.82, size = 185, normalized size = 1.37 \[ \frac {{\left (2688 i \, A + 2688 \, B\right )} a^{3} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (3584 i \, A - 896 \, B\right )} a^{3} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (1024 i \, A - 256 \, B\right )} a^{3} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (128 i \, A - 32 \, B\right )} a^{3} c^{5}}{105 \, {\left (f e^{\left (16 i \, f x + 16 i \, e\right )} + 8 \, f e^{\left (14 i \, f x + 14 i \, e\right )} + 28 \, f e^{\left (12 i \, f x + 12 i \, e\right )} + 56 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 70 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 56 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 28 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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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giac [B]  time = 4.89, size = 242, normalized size = 1.79 \[ \frac {2688 i \, A a^{3} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + 2688 \, B a^{3} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + 3584 i \, A a^{3} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} - 896 \, B a^{3} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 1024 i \, A a^{3} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 256 \, B a^{3} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 128 i \, A a^{3} c^{5} - 32 \, B a^{3} c^{5}}{105 \, {\left (f e^{\left (16 i \, f x + 16 i \, e\right )} + 8 \, f e^{\left (14 i \, f x + 14 i \, e\right )} + 28 \, f e^{\left (12 i \, f x + 12 i \, e\right )} + 56 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 70 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 56 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 28 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

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)

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maple [A]  time = 0.03, size = 169, normalized size = 1.25 \[ \frac {c^{5} a^{3} \left (-\frac {2 i B \left (\tan ^{7}\left (f x +e \right )\right )}{7}-\frac {B \left (\tan ^{8}\left (f x +e \right )\right )}{8}-\frac {i A \left (\tan ^{6}\left (f x +e \right )\right )}{3}-\frac {A \left (\tan ^{7}\left (f x +e \right )\right )}{7}-\frac {4 i B \left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {B \left (\tan ^{6}\left (f x +e \right )\right )}{6}-i A \left (\tan ^{4}\left (f x +e \right )\right )-\frac {A \left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {2 i B \left (\tan ^{3}\left (f x +e \right )\right )}{3}+\frac {B \left (\tan ^{4}\left (f x +e \right )\right )}{4}-i A \left (\tan ^{2}\left (f x +e \right )\right )+\frac {A \left (\tan ^{3}\left (f x +e \right )\right )}{3}+\frac {B \left (\tan ^{2}\left (f x +e \right )\right )}{2}+A \tan \left (f x +e \right )\right )}{f} \]

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*c^5*a^3*(-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 = 0.90, size = 167, normalized size = 1.24 \[ -\frac {105 \, B a^{3} c^{5} \tan \left (f x + e\right )^{8} + 120 \, {\left (A + 2 i \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{7} + {\left (280 i \, A + 140 \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{6} + 168 \, {\left (A + 4 i \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{5} + {\left (840 i \, A - 210 \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{4} - 280 \, {\left (A - 2 i \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{3} + {\left (840 i \, A - 420 \, B\right )} a^{3} c^{5} \tan \left (f x + e\right )^{2} - 840 \, A a^{3} c^{5} \tan \left (f x + e\right )}{840 \, f} \]

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 + 2*I*B)*a^3*c^5*tan(f*x + e)^7 + (280*I*A + 140*B)*a^3*c^5*tan(
f*x + e)^6 + 168*(A + 4*I*B)*a^3*c^5*tan(f*x + e)^5 + (840*I*A - 210*B)*a^3*c^5*tan(f*x + e)^4 - 280*(A - 2*I*
B)*a^3*c^5*tan(f*x + e)^3 + (840*I*A - 420*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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mupad [B]  time = 8.65, size = 174, normalized size = 1.29 \[ -\frac {\frac {a^3\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-B+A\,2{}\mathrm {i}\right )}{2}+\frac {a^3\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (-B+A\,4{}\mathrm {i}\right )}{4}-A\,a^3\,c^5\,\mathrm {tan}\left (e+f\,x\right )-\frac {a^3\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (A-B\,2{}\mathrm {i}\right )}{3}+\frac {a^3\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (B+A\,2{}\mathrm {i}\right )}{6}+\frac {a^3\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (A+B\,4{}\mathrm {i}\right )}{5}+\frac {a^3\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (A+B\,2{}\mathrm {i}\right )}{7}+\frac {B\,a^3\,c^5\,{\mathrm {tan}\left (e+f\,x\right )}^8}{8}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^5,x)

[Out]

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

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sympy [B]  time = 1.83, size = 323, normalized size = 2.39 \[ \frac {128 A a^{3} c^{5} + 32 i B a^{3} c^{5} + \left (1024 A a^{3} c^{5} e^{2 i e} + 256 i B a^{3} c^{5} e^{2 i e}\right ) e^{2 i f x} + \left (3584 A a^{3} c^{5} e^{4 i e} + 896 i B a^{3} c^{5} e^{4 i e}\right ) e^{4 i f x} + \left (2688 A a^{3} c^{5} e^{6 i e} - 2688 i B a^{3} c^{5} e^{6 i e}\right ) e^{6 i f x}}{- 105 i f e^{16 i e} e^{16 i f x} - 840 i f e^{14 i e} e^{14 i f x} - 2940 i f e^{12 i e} e^{12 i f x} - 5880 i f e^{10 i e} e^{10 i f x} - 7350 i f e^{8 i e} e^{8 i f x} - 5880 i f e^{6 i e} e^{6 i f x} - 2940 i f e^{4 i e} e^{4 i f x} - 840 i f e^{2 i e} e^{2 i f x} - 105 i f} \]

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]

(128*A*a**3*c**5 + 32*I*B*a**3*c**5 + (1024*A*a**3*c**5*exp(2*I*e) + 256*I*B*a**3*c**5*exp(2*I*e))*exp(2*I*f*x
) + (3584*A*a**3*c**5*exp(4*I*e) + 896*I*B*a**3*c**5*exp(4*I*e))*exp(4*I*f*x) + (2688*A*a**3*c**5*exp(6*I*e) -
2688*I*B*a**3*c**5*exp(6*I*e))*exp(6*I*f*x))/(-105*I*f*exp(16*I*e)*exp(16*I*f*x) - 840*I*f*exp(14*I*e)*exp(14
*I*f*x) - 2940*I*f*exp(12*I*e)*exp(12*I*f*x) - 5880*I*f*exp(10*I*e)*exp(10*I*f*x) - 7350*I*f*exp(8*I*e)*exp(8*
I*f*x) - 5880*I*f*exp(6*I*e)*exp(6*I*f*x) - 2940*I*f*exp(4*I*e)*exp(4*I*f*x) - 840*I*f*exp(2*I*e)*exp(2*I*f*x)
- 105*I*f)

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