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

Optimal. Leaf size=109 \[ \frac{2 a^2 (B+i A) (c-i c \tan (e+f x))^n}{f n}-\frac{a^2 (3 B+i A) (c-i c \tan (e+f x))^{n+1}}{c f (n+1)}+\frac{a^2 B (c-i c \tan (e+f x))^{n+2}}{c^2 f (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.163657, antiderivative size = 109, 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} \[ \frac{2 a^2 (B+i A) (c-i c \tan (e+f x))^n}{f n}-\frac{a^2 (3 B+i A) (c-i c \tan (e+f x))^{n+1}}{c f (n+1)}+\frac{a^2 B (c-i c \tan (e+f x))^{n+2}}{c^2 f (n+2)} \]

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])^n,x]

[Out]

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

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))^n \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x) (A+B x) (c-i c x)^{-1+n} \, 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)^{-1+n}-\frac{a (A-3 i B) (c-i c x)^n}{c}-\frac{i a B (c-i c x)^{1+n}}{c^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 a^2 (i A+B) (c-i c \tan (e+f x))^n}{f n}-\frac{a^2 (i A+3 B) (c-i c \tan (e+f x))^{1+n}}{c f (1+n)}+\frac{a^2 B (c-i c \tan (e+f x))^{2+n}}{c^2 f (2+n)}\\ \end{align*}

Mathematica [A]  time = 6.62668, size = 146, normalized size = 1.34 \[ \frac{a^2 \sec ^2(e+f x) (c \sec (e+f x))^n \left (\left (B \left (n^2+2 n+4\right )+i A (n+2)^2\right ) \cos (2 (e+f x))-n (A (n+2)-i B (n+4)) \sin (2 (e+f x))+(n+2) (-B (n-2)+i A (n+2))\right ) \exp (n (-\log (c \sec (e+f x))+\log (c-i c \tan (e+f x))))}{2 f n (n+1) (n+2)} \]

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])^n,x]

[Out]

(a^2*E^(n*(-Log[c*Sec[e + f*x]] + Log[c - I*c*Tan[e + f*x]]))*Sec[e + f*x]^2*(c*Sec[e + f*x])^n*((2 + n)*(-(B*
(-2 + n)) + I*A*(2 + n)) + (I*A*(2 + n)^2 + B*(4 + 2*n + n^2))*Cos[2*(e + f*x)] - n*(A*(2 + n) - I*B*(4 + n))*
Sin[2*(e + f*x)]))/(2*f*n*(1 + n)*(2 + n))

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Maple [B]  time = 0.49, size = 280, normalized size = 2.6 \begin{align*}{\frac{in{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}A{a}^{2}}{f \left ( 1+n \right ) \left ( 2+n \right ) }}+{\frac{4\,i{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}A{a}^{2}}{f \left ( 1+n \right ) \left ( 2+n \right ) }}+{\frac{4\,i{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}A{a}^{2}}{fn \left ( 1+n \right ) \left ( 2+n \right ) }}+{\frac{{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}{a}^{2}B}{f \left ( 1+n \right ) \left ( 2+n \right ) }}+4\,{\frac{{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}{a}^{2}B}{fn \left ( 1+n \right ) \left ( 2+n \right ) }}-{\frac{{a}^{2}B \left ( \tan \left ( fx+e \right ) \right ) ^{2}{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}}{f \left ( 2+n \right ) }}-{\frac{{a}^{2} \left ( -iBn+An-4\,iB+2\,A \right ) \tan \left ( fx+e \right ){{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}}{f \left ( 1+n \right ) \left ( 2+n \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))^n,x)

[Out]

I/f*n/(1+n)/(2+n)*exp(n*ln(c-I*c*tan(f*x+e)))*A*a^2+4*I/f/(1+n)/(2+n)*exp(n*ln(c-I*c*tan(f*x+e)))*A*a^2+4*I/f/
n/(1+n)/(2+n)*exp(n*ln(c-I*c*tan(f*x+e)))*A*a^2+1/f/(1+n)/(2+n)*exp(n*ln(c-I*c*tan(f*x+e)))*a^2*B+4/f/n/(1+n)/
(2+n)*exp(n*ln(c-I*c*tan(f*x+e)))*a^2*B-a^2*B/f/(2+n)*tan(f*x+e)^2*exp(n*ln(c-I*c*tan(f*x+e)))-a^2*(-I*B*n+A*n
-4*I*B+2*A)/f/(1+n)/(2+n)*tan(f*x+e)*exp(n*ln(c-I*c*tan(f*x+e)))

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Maxima [B]  time = 2.42606, size = 902, normalized size = 8.28 \begin{align*} \text{result too large to display} \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))^n,x, algorithm="maxima")

[Out]

(((2*A + 2*I*B)*a^2*c^n*n^2 + 8*A*a^2*c^n*n + (8*A - 8*I*B)*a^2*c^n)*2^n*cos(-2*f*x + n*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e) + 1) - 2*e) + ((2*A - 2*I*B)*a^2*c^n*n^2 + (6*A - 6*I*B)*a^2*c^n*n + (4*A - 4*I*B)*a^2*c^
n)*2^n*cos(-4*f*x + n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*e) + ((2*A + 2*I*B)*a^2*c^n*n + (4*A
- 4*I*B)*a^2*c^n)*2^n*cos(n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2*((I*A - B)*a^2*c^n*n^2 + 4*I
*A*a^2*c^n*n + 4*(I*A + B)*a^2*c^n)*2^n*sin(-2*f*x + n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 2*e)
- 2*((I*A + B)*a^2*c^n*n^2 + 3*(I*A + B)*a^2*c^n*n + 2*(I*A + B)*a^2*c^n)*2^n*sin(-4*f*x + n*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e) + 1) - 4*e) - 2*((I*A - B)*a^2*c^n*n + 2*(I*A + B)*a^2*c^n)*2^n*sin(n*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e) + 1)))/(((-I*n^3 - 3*I*n^2 - 2*I*n)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*
cos(2*f*x + 2*e) + 1)^(1/2*n)*cos(4*f*x + 4*e) + (n^3 + 3*n^2 + 2*n)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2
+ 2*cos(2*f*x + 2*e) + 1)^(1/2*n)*sin(4*f*x + 4*e) + (-I*n^3 - 3*I*n^2 + (-2*I*n^3 - 6*I*n^2 - 4*I*n)*cos(2*f*
x + 2*e) + 2*(n^3 + 3*n^2 + 2*n)*sin(2*f*x + 2*e) - 2*I*n)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*
f*x + 2*e) + 1)^(1/2*n))*f)

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Fricas [B]  time = 1.43118, size = 498, normalized size = 4.57 \begin{align*} \frac{{\left ({\left (2 i \, A - 2 \, B\right )} a^{2} n +{\left (4 i \, A + 4 \, B\right )} a^{2} +{\left ({\left (2 i \, A + 2 \, B\right )} a^{2} n^{2} +{\left (6 i \, A + 6 \, B\right )} a^{2} n +{\left (4 i \, A + 4 \, B\right )} a^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left ({\left (2 i \, A - 2 \, B\right )} a^{2} n^{2} + 8 i \, A a^{2} n +{\left (8 i \, A + 8 \, B\right )} a^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \left (\frac{2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}}{f n^{3} + 3 \, f n^{2} + 2 \, f n +{\left (f n^{3} + 3 \, f n^{2} + 2 \, f n\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \,{\left (f n^{3} + 3 \, f n^{2} + 2 \, f n\right )} e^{\left (2 i \, f x + 2 i \, e\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))^n,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))**n,x)

[Out]

Exception raised: AttributeError

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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 )}^{2}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \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))^n,x, algorithm="giac")

[Out]

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