3.1044 \(\int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx\)

Optimal. Leaf size=99 \[ \frac{i a^3 (c-i c \tan (e+f x))^{n+2}}{c^2 f (n+2)}+\frac{4 i a^3 (c-i c \tan (e+f x))^n}{f n}-\frac{4 i a^3 (c-i c \tan (e+f x))^{n+1}}{c f (n+1)} \]

[Out]

((4*I)*a^3*(c - I*c*Tan[e + f*x])^n)/(f*n) - ((4*I)*a^3*(c - I*c*Tan[e + f*x])^(1 + n))/(c*f*(1 + n)) + (I*a^3
*(c - I*c*Tan[e + f*x])^(2 + n))/(c^2*f*(2 + n))

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Rubi [A]  time = 0.14088, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ \frac{i a^3 (c-i c \tan (e+f x))^{n+2}}{c^2 f (n+2)}+\frac{4 i a^3 (c-i c \tan (e+f x))^n}{f n}-\frac{4 i a^3 (c-i c \tan (e+f x))^{n+1}}{c f (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^n,x]

[Out]

((4*I)*a^3*(c - I*c*Tan[e + f*x])^n)/(f*n) - ((4*I)*a^3*(c - I*c*Tan[e + f*x])^(1 + n))/(c*f*(1 + n)) + (I*a^3
*(c - I*c*Tan[e + f*x])^(2 + n))/(c^2*f*(2 + n))

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx &=\left (a^3 c^3\right ) \int \sec ^6(e+f x) (c-i c \tan (e+f x))^{-3+n} \, dx\\ &=\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int (c-x)^2 (c+x)^{-1+n} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \left (4 c^2 (c+x)^{-1+n}-4 c (c+x)^n+(c+x)^{1+n}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac{4 i a^3 (c-i c \tan (e+f x))^n}{f n}-\frac{4 i a^3 (c-i c \tan (e+f x))^{1+n}}{c f (1+n)}+\frac{i a^3 (c-i c \tan (e+f x))^{2+n}}{c^2 f (2+n)}\\ \end{align*}

Mathematica [A]  time = 4.05234, size = 110, normalized size = 1.11 \[ \frac{i a^3 \sec ^2(e+f x) (c \sec (e+f x))^n \left (\left (n^2+3 n+4\right ) \cos (2 (e+f x))+i n (n+3) \sin (2 (e+f x))+2 (n+2)\right ) \exp (n (-\log (c \sec (e+f x))+\log (c-i c \tan (e+f x))))}{f n (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^n,x]

[Out]

(I*a^3*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*(2 + n) +
(4 + 3*n + n^2)*Cos[2*(e + f*x)] + I*n*(3 + n)*Sin[2*(e + f*x)]))/(f*n*(1 + n)*(2 + n))

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Maple [B]  time = 0.385, size = 192, normalized size = 1.9 \begin{align*}{\frac{i{a}^{3}n{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}}{ \left ( 1+n \right ) f \left ( 2+n \right ) }}+{\frac{5\,i{a}^{3}{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}}{ \left ( 1+n \right ) f \left ( 2+n \right ) }}+{\frac{8\,i{a}^{3}{{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}}{ \left ( 1+n \right ) \left ( 2+n \right ) fn}}-{\frac{i{a}^{3} \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 ) }}-2\,{\frac{{a}^{3} \left ( 3+n \right ) \tan \left ( fx+e \right ){{\rm e}^{n\ln \left ( c-ic\tan \left ( fx+e \right ) \right ) }}}{ \left ( 1+n \right ) f \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))^3*(c-I*c*tan(f*x+e))^n,x)

[Out]

I*a^3/(1+n)/(2+n)/f*n*exp(n*ln(c-I*c*tan(f*x+e)))+5*I*a^3/(1+n)/(2+n)/f*exp(n*ln(c-I*c*tan(f*x+e)))+8*I*a^3/(1
+n)/(2+n)/f/n*exp(n*ln(c-I*c*tan(f*x+e)))-I/f/(2+n)*a^3*tan(f*x+e)^2*exp(n*ln(c-I*c*tan(f*x+e)))-2*a^3*(3+n)/(
1+n)/f/(2+n)*tan(f*x+e)*exp(n*ln(c-I*c*tan(f*x+e)))

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Maxima [B]  time = 1.87847, size = 737, normalized size = 7.44 \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))^3*(c-I*c*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

(2^(n + 3)*a^3*c^n*cos(n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - I*2^(n + 3)*a^3*c^n*sin(n*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 8*(a^3*c^n*n + 2*a^3*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) + 4*(a^3*c^n*n^2 + 3*a^3*c^n*n + 2*a^3*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) - (8*I*a^3*c^n*n + 16*I*a^3*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) - (4*I*a^3*c^n*n^2 + 12*I*a^3*c^n*n + 8*I*a^3*c^n)*2^n*sin(-4*f*x + n*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*e))/(((-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 + s
in(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 [A]  time = 1.65345, size = 369, normalized size = 3.73 \begin{align*} \frac{{\left (8 i \, a^{3} +{\left (4 i \, a^{3} n^{2} + 12 i \, a^{3} n + 8 i \, a^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (8 i \, a^{3} n + 16 i \, a^{3}\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))^3*(c-I*c*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

(8*I*a^3 + (4*I*a^3*n^2 + 12*I*a^3*n + 8*I*a^3)*e^(4*I*f*x + 4*I*e) + (8*I*a^3*n + 16*I*a^3)*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))**3*(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 (i \, a \tan \left (f x + e\right ) + a\right )}^{3}{\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))^3*(c-I*c*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*(-I*c*tan(f*x + e) + c)^n, x)