3.677 \(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^5 \, dx\)

Optimal. Leaf size=99 \[ -\frac{a^2 c^5 (3 B+i A) (1-i \tan (e+f x))^6}{6 f}+\frac{2 a^2 c^5 (B+i A) (1-i \tan (e+f x))^5}{5 f}+\frac{a^2 B c^5 (1-i \tan (e+f x))^7}{7 f} \]

[Out]

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

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Rubi [A]  time = 0.167718, antiderivative size = 99, 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} \[ -\frac{a^2 c^5 (3 B+i A) (1-i \tan (e+f x))^6}{6 f}+\frac{2 a^2 c^5 (B+i A) (1-i \tan (e+f x))^5}{5 f}+\frac{a^2 B c^5 (1-i \tan (e+f x))^7}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^5,x]

[Out]

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

Mathematica [B]  time = 9.04328, size = 254, normalized size = 2.57 \[ \frac{a^2 c^5 \sec (e) \sec ^7(e+f x) (35 (3 B-7 i A) \cos (2 e+f x)+35 (3 B-7 i A) \cos (f x)-245 A \sin (2 e+f x)+189 A \sin (2 e+3 f x)-105 A \sin (4 e+3 f x)+98 A \sin (4 e+5 f x)+14 A \sin (6 e+7 f x)-105 i A \cos (2 e+3 f x)-105 i A \cos (4 e+3 f x)+245 A \sin (f x)-105 i B \sin (2 e+f x)+21 i B \sin (2 e+3 f x)-105 i B \sin (4 e+3 f x)+42 i B \sin (4 e+5 f x)+6 i B \sin (6 e+7 f x)+105 B \cos (2 e+3 f x)+105 B \cos (4 e+3 f x)+105 i B \sin (f x))}{840 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^2*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^5,x]

[Out]

(a^2*c^5*Sec[e]*Sec[e + f*x]^7*(35*((-7*I)*A + 3*B)*Cos[f*x] + 35*((-7*I)*A + 3*B)*Cos[2*e + f*x] - (105*I)*A*
Cos[2*e + 3*f*x] + 105*B*Cos[2*e + 3*f*x] - (105*I)*A*Cos[4*e + 3*f*x] + 105*B*Cos[4*e + 3*f*x] + 245*A*Sin[f*
x] + (105*I)*B*Sin[f*x] - 245*A*Sin[2*e + f*x] - (105*I)*B*Sin[2*e + f*x] + 189*A*Sin[2*e + 3*f*x] + (21*I)*B*
Sin[2*e + 3*f*x] - 105*A*Sin[4*e + 3*f*x] - (105*I)*B*Sin[4*e + 3*f*x] + 98*A*Sin[4*e + 5*f*x] + (42*I)*B*Sin[
4*e + 5*f*x] + 14*A*Sin[6*e + 7*f*x] + (6*I)*B*Sin[6*e + 7*f*x]))/(840*f)

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Maple [A]  time = 0.011, size = 147, normalized size = 1.5 \begin{align*}{\frac{{c}^{5}{a}^{2}}{f} \left ({\frac{i}{7}}B \left ( \tan \left ( fx+e \right ) \right ) ^{7}+{\frac{i}{6}}A \left ( \tan \left ( fx+e \right ) \right ) ^{6}-{\frac{2\,i}{5}}B \left ( \tan \left ( fx+e \right ) \right ) ^{5}-{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{6}}{2}}-{\frac{i}{2}}A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-{\frac{3\,A \left ( \tan \left ( fx+e \right ) \right ) ^{5}}{5}}-iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}-{\frac{B \left ( \tan \left ( fx+e \right ) \right ) ^{4}}{2}}-{\frac{3\,i}{2}}A \left ( \tan \left ( fx+e \right ) \right ) ^{2}-{\frac{2\,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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^5,x)

[Out]

1/f*c^5*a^2*(1/7*I*B*tan(f*x+e)^7+1/6*I*A*tan(f*x+e)^6-2/5*I*B*tan(f*x+e)^5-1/2*B*tan(f*x+e)^6-1/2*I*A*tan(f*x
+e)^4-3/5*A*tan(f*x+e)^5-I*B*tan(f*x+e)^3-1/2*B*tan(f*x+e)^4-3/2*I*A*tan(f*x+e)^2-2/3*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 = 1.7265, size = 201, normalized size = 2.03 \begin{align*} -\frac{-60 i \, B a^{2} c^{5} \tan \left (f x + e\right )^{7} - 70 \,{\left (i \, A - 3 \, B\right )} a^{2} c^{5} \tan \left (f x + e\right )^{6} +{\left (252 \, A + 168 i \, B\right )} a^{2} c^{5} \tan \left (f x + e\right )^{5} - 210 \,{\left (-i \, A - B\right )} a^{2} c^{5} \tan \left (f x + e\right )^{4} +{\left (280 \, A + 420 i \, B\right )} a^{2} c^{5} \tan \left (f x + e\right )^{3} - 210 \,{\left (-3 i \, A + B\right )} a^{2} c^{5} \tan \left (f x + e\right )^{2} - 420 \, A a^{2} c^{5} \tan \left (f x + e\right )}{420 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^5,x, algorithm="maxima")

[Out]

-1/420*(-60*I*B*a^2*c^5*tan(f*x + e)^7 - 70*(I*A - 3*B)*a^2*c^5*tan(f*x + e)^6 + (252*A + 168*I*B)*a^2*c^5*tan
(f*x + e)^5 - 210*(-I*A - B)*a^2*c^5*tan(f*x + e)^4 + (280*A + 420*I*B)*a^2*c^5*tan(f*x + e)^3 - 210*(-3*I*A +
B)*a^2*c^5*tan(f*x + e)^2 - 420*A*a^2*c^5*tan(f*x + e))/f

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Fricas [A]  time = 1.2768, size = 441, normalized size = 4.45 \begin{align*} \frac{{\left (1344 i \, A + 1344 \, B\right )} a^{2} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (1568 i \, A - 672 \, B\right )} a^{2} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (224 i \, A - 96 \, B\right )} a^{2} c^{5}}{105 \,{\left (f e^{\left (14 i \, f x + 14 i \, e\right )} + 7 \, f e^{\left (12 i \, f x + 12 i \, e\right )} + 21 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 35 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 35 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 21 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 7 \, 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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^5,x, algorithm="fricas")

[Out]

1/105*((1344*I*A + 1344*B)*a^2*c^5*e^(4*I*f*x + 4*I*e) + (1568*I*A - 672*B)*a^2*c^5*e^(2*I*f*x + 2*I*e) + (224
*I*A - 96*B)*a^2*c^5)/(f*e^(14*I*f*x + 14*I*e) + 7*f*e^(12*I*f*x + 12*I*e) + 21*f*e^(10*I*f*x + 10*I*e) + 35*f
*e^(8*I*f*x + 8*I*e) + 35*f*e^(6*I*f*x + 6*I*e) + 21*f*e^(4*I*f*x + 4*I*e) + 7*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [B]  time = 71.3342, size = 231, normalized size = 2.33 \begin{align*} \frac{\frac{\left (64 i A a^{2} c^{5} + 64 B a^{2} c^{5}\right ) e^{- 10 i e} e^{4 i f x}}{5 f} + \frac{\left (224 i A a^{2} c^{5} - 96 B a^{2} c^{5}\right ) e^{- 12 i e} e^{2 i f x}}{15 f} + \frac{\left (224 i A a^{2} c^{5} - 96 B a^{2} c^{5}\right ) e^{- 14 i e}}{105 f}}{e^{14 i f x} + 7 e^{- 2 i e} e^{12 i f x} + 21 e^{- 4 i e} e^{10 i f x} + 35 e^{- 6 i e} e^{8 i f x} + 35 e^{- 8 i e} e^{6 i f x} + 21 e^{- 10 i e} e^{4 i f x} + 7 e^{- 12 i e} e^{2 i f x} + e^{- 14 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**5,x)

[Out]

((64*I*A*a**2*c**5 + 64*B*a**2*c**5)*exp(-10*I*e)*exp(4*I*f*x)/(5*f) + (224*I*A*a**2*c**5 - 96*B*a**2*c**5)*ex
p(-12*I*e)*exp(2*I*f*x)/(15*f) + (224*I*A*a**2*c**5 - 96*B*a**2*c**5)*exp(-14*I*e)/(105*f))/(exp(14*I*f*x) + 7
*exp(-2*I*e)*exp(12*I*f*x) + 21*exp(-4*I*e)*exp(10*I*f*x) + 35*exp(-6*I*e)*exp(8*I*f*x) + 35*exp(-8*I*e)*exp(6
*I*f*x) + 21*exp(-10*I*e)*exp(4*I*f*x) + 7*exp(-12*I*e)*exp(2*I*f*x) + exp(-14*I*e))

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Giac [B]  time = 2.17806, size = 258, normalized size = 2.61 \begin{align*} \frac{1344 i \, A a^{2} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 1344 \, B a^{2} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 1568 i \, A a^{2} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 672 \, B a^{2} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 224 i \, A a^{2} c^{5} - 96 \, B a^{2} c^{5}}{105 \,{\left (f e^{\left (14 i \, f x + 14 i \, e\right )} + 7 \, f e^{\left (12 i \, f x + 12 i \, e\right )} + 21 \, f e^{\left (10 i \, f x + 10 i \, e\right )} + 35 \, f e^{\left (8 i \, f x + 8 i \, e\right )} + 35 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 21 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 7 \, 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))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^5,x, algorithm="giac")

[Out]

1/105*(1344*I*A*a^2*c^5*e^(4*I*f*x + 4*I*e) + 1344*B*a^2*c^5*e^(4*I*f*x + 4*I*e) + 1568*I*A*a^2*c^5*e^(2*I*f*x
+ 2*I*e) - 672*B*a^2*c^5*e^(2*I*f*x + 2*I*e) + 224*I*A*a^2*c^5 - 96*B*a^2*c^5)/(f*e^(14*I*f*x + 14*I*e) + 7*f
*e^(12*I*f*x + 12*I*e) + 21*f*e^(10*I*f*x + 10*I*e) + 35*f*e^(8*I*f*x + 8*I*e) + 35*f*e^(6*I*f*x + 6*I*e) + 21
*f*e^(4*I*f*x + 4*I*e) + 7*f*e^(2*I*f*x + 2*I*e) + f)