3.506 \(\int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{3-n} \, dx\)

Optimal. Leaf size=148 \[ \frac{4 i a^2 (a+i a \tan (e+f x))^{1-n} (d \sec (e+f x))^{2 n}}{f \left (n^2+3 n+2\right )}+\frac{8 i a^3 (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n}}{f n \left (n^2+3 n+2\right )}+\frac{i a (a+i a \tan (e+f x))^{2-n} (d \sec (e+f x))^{2 n}}{f (n+2)} \]

[Out]

((4*I)*a^2*(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(1 - n))/(f*(2 + 3*n + n^2)) + (I*a*(d*Sec[e + f*x])^
(2*n)*(a + I*a*Tan[e + f*x])^(2 - n))/(f*(2 + n)) + ((8*I)*a^3*(d*Sec[e + f*x])^(2*n))/(f*n*(2 + 3*n + n^2)*(a
+ I*a*Tan[e + f*x])^n)

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Rubi [A]  time = 0.211018, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3494, 3493} \[ \frac{4 i a^2 (a+i a \tan (e+f x))^{1-n} (d \sec (e+f x))^{2 n}}{f \left (n^2+3 n+2\right )}+\frac{8 i a^3 (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n}}{f n \left (n^2+3 n+2\right )}+\frac{i a (a+i a \tan (e+f x))^{2-n} (d \sec (e+f x))^{2 n}}{f (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(3 - n),x]

[Out]

((4*I)*a^2*(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(1 - n))/(f*(2 + 3*n + n^2)) + (I*a*(d*Sec[e + f*x])^
(2*n)*(a + I*a*Tan[e + f*x])^(2 - n))/(f*(2 + n)) + ((8*I)*a^3*(d*Sec[e + f*x])^(2*n))/(f*n*(2 + 3*n + n^2)*(a
+ I*a*Tan[e + f*x])^n)

Rule 3494

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

Rule 3493

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

Rubi steps

\begin{align*} \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{3-n} \, dx &=\frac{i a (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{2-n}}{f (2+n)}+\frac{(4 a) \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{2-n} \, dx}{2+n}\\ &=\frac{4 i a^2 (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{1-n}}{f \left (2+3 n+n^2\right )}+\frac{i a (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{2-n}}{f (2+n)}+\frac{\left (8 a^2\right ) \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{1-n} \, dx}{2+3 n+n^2}\\ &=\frac{4 i a^2 (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{1-n}}{f \left (2+3 n+n^2\right )}+\frac{i a (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{2-n}}{f (2+n)}+\frac{8 i a^3 (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n}}{f n \left (2+3 n+n^2\right )}\\ \end{align*}

Mathematica [A]  time = 1.84091, size = 129, normalized size = 0.87 \[ \frac{i a^3 \sec ^2(e+f x) (\cos (3 f x)+i \sin (3 f x)) (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \left (\left (n^2+3 n+4\right ) \cos (2 (e+f x))+i n (n+3) \sin (2 (e+f x))+2 (n+2)\right )}{f n (n+1) (n+2) (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(3 - n),x]

[Out]

(I*a^3*Sec[e + f*x]^2*(d*Sec[e + f*x])^(2*n)*(Cos[3*f*x] + I*Sin[3*f*x])*(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)*(Cos[f*x] + I*Sin[f*x])^3*(a + I*a*Tan[e + f*x]
)^n)

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Maple [F]  time = 0.862, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{2\,n} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{3-n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(3-n),x)

[Out]

int((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(3-n),x)

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Maxima [B]  time = 10.3702, size = 833, normalized size = 5.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(3-n),x, algorithm="maxima")

[Out]

(2^(n + 3)*a^3*d^(2*n)*cos(n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - I*2^(n + 3)*a^3*d^(2*n)*sin(n*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 8*(a^3*d^(2*n)*n + 2*a^3*d^(2*n))*2^n*cos(-2*f*x + n*arctan
2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 2*e) + 4*(a^3*d^(2*n)*n^2 + 3*a^3*d^(2*n)*n + 2*a^3*d^(2*n))*2^n*c
os(-4*f*x + n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*e) - (8*I*a^3*d^(2*n)*n + 16*I*a^3*d^(2*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*d^(2*n)*n^2 + 12*I*a^3*d^
(2*n)*n + 8*I*a^3*d^(2*n))*2^n*sin(-4*f*x + n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*e))/(((-I*a^
n*n^3 - 3*I*a^n*n^2 - 2*I*a^n*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)*co
s(4*f*x + 4*e) + (a^n*n^3 + 3*a^n*n^2 + 2*a^n*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*a^n*n^3 - 3*I*a^n*n^2 - 2*I*a^n*n + (-2*I*a^n*n^3 - 6*I*a^n*n^2 - 4*I*a^n
*n)*cos(2*f*x + 2*e) + 2*(a^n*n^3 + 3*a^n*n^2 + 2*a^n*n)*sin(2*f*x + 2*e))*(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 = 1.93509, size = 401, normalized size = 2.71 \begin{align*} \frac{{\left ({\left (i \, n^{2} + 3 i \, n + 2 i\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (i \, n^{2} + 5 i \, n + 6 i\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (2 i \, n + 6 i\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} \left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{-n + 3} \left (\frac{2 \, d e^{\left (i \, f x + i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{2 \, n} e^{\left (-6 i \, f x - 6 i \, e\right )}}{2 \,{\left (f n^{3} + 3 \, f n^{2} + 2 \, f n\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(3-n),x, algorithm="fricas")

[Out]

1/2*((I*n^2 + 3*I*n + 2*I)*e^(6*I*f*x + 6*I*e) + (I*n^2 + 5*I*n + 6*I)*e^(4*I*f*x + 4*I*e) + (2*I*n + 6*I)*e^(
2*I*f*x + 2*I*e) + 2*I)*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^(-n + 3)*(2*d*e^(I*f*x + I*e)/(e^(
2*I*f*x + 2*I*e) + 1))^(2*n)*e^(-6*I*f*x - 6*I*e)/(f*n^3 + 3*f*n^2 + 2*f*n)

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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((d*sec(f*x+e))**(2*n)*(a+I*a*tan(f*x+e))**(3-n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{2 \, n}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{-n + 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(3-n),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(2*n)*(I*a*tan(f*x + e) + a)^(-n + 3), x)