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

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

[Out]

4*I*a^2*(d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(1-n)/f/(n^2+3*n+2)+I*a*(d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e)
)^(2-n)/f/(2+n)+8*I*a^3*(d*sec(f*x+e))^(2*n)/f/n/(n^2+3*n+2)/((a+I*a*tan(f*x+e))^n)

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Rubi [A]  time = 0.21, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 1.96, 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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fricas [A]  time = 0.61, size = 171, normalized size = 1.16 \[ \frac {{\left ({\left (i \, n^{2} + 3 i \, n + 2 i\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (i \, n^{2} + 5 i \, n + 6 i\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (2 i \, n + 6 i\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} \left (\frac {2 \, d e^{\left (i \, f x + i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{2 \, n} e^{\left (-i \, e n + {\left (-i \, f n + 3 i \, f\right )} x - 6 i \, f x - {\left (n - 3\right )} \log \left (\frac {2 \, d e^{\left (i \, f x + i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) - {\left (n - 3\right )} \log \left (\frac {a}{d}\right ) - 3 i \, e\right )}}{2 \, {\left (f n^{3} + 3 \, f n^{2} + 2 \, f n\right )}} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{2 \, n} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{-n + 3}\,{d x} \]

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)

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maple [F]  time = 3.78, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{2 n} \left (a +i a \tan \left (f x +e \right )\right )^{3-n}\, dx \]

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 = 1.59, size = 617, normalized size = 4.17 \[ \frac {2^{n + 3} a^{3} d^{2 \, n} \cos \left (n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) - i \cdot 2^{n + 3} a^{3} d^{2 \, n} \sin \left (n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + 8 \, {\left (a^{3} d^{2 \, n} n + 2 \, a^{3} d^{2 \, n}\right )} 2^{n} \cos \left (-2 \, f x + n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \, e\right ) + 4 \, {\left (a^{3} d^{2 \, n} n^{2} + 3 \, a^{3} d^{2 \, n} n + 2 \, a^{3} d^{2 \, n}\right )} 2^{n} \cos \left (-4 \, f x + n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 4 \, e\right ) - {\left (8 i \, a^{3} d^{2 \, n} n + 16 i \, a^{3} d^{2 \, n}\right )} 2^{n} \sin \left (-2 \, f x + n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \, e\right ) - {\left (4 i \, a^{3} d^{2 \, n} n^{2} + 12 i \, a^{3} d^{2 \, n} n + 8 i \, a^{3} d^{2 \, n}\right )} 2^{n} \sin \left (-4 \, f x + n \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 4 \, e\right )}{{\left ({\left (-i \, a^{n} n^{3} - 3 i \, a^{n} n^{2} - 2 i \, a^{n} n\right )} {\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {1}{2} \, n} \cos \left (4 \, f x + 4 \, e\right ) + {\left (a^{n} n^{3} + 3 \, a^{n} n^{2} + 2 \, a^{n} n\right )} {\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {1}{2} \, n} \sin \left (4 \, f x + 4 \, e\right ) + {\left (-i \, a^{n} n^{3} - 3 i \, a^{n} n^{2} - 2 i \, a^{n} n + {\left (-2 i \, a^{n} n^{3} - 6 i \, a^{n} n^{2} - 4 i \, a^{n} n\right )} \cos \left (2 \, f x + 2 \, e\right ) + 2 \, {\left (a^{n} n^{3} + 3 \, a^{n} n^{2} + 2 \, a^{n} n\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} {\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {1}{2} \, n}\right )} f} \]

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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mupad [B]  time = 12.47, size = 321, normalized size = 2.17 \[ -\left (\cos \left (6\,e+6\,f\,x\right )-\sin \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}\right )\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2\,n}\,\left (\frac {{\left (a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}\right )}^{3-n}}{f\,n\,\left (n^2\,1{}\mathrm {i}+n\,3{}\mathrm {i}+2{}\mathrm {i}\right )}+\frac {\left (\cos \left (4\,e+4\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )\,{\left (a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}\right )}^{3-n}\,\left (n^2+5\,n+6\right )}{2\,f\,n\,\left (n^2\,1{}\mathrm {i}+n\,3{}\mathrm {i}+2{}\mathrm {i}\right )}+\frac {\left (\cos \left (6\,e+6\,f\,x\right )+\sin \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}\right )\,{\left (a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}\right )}^{3-n}\,\left (n^2+3\,n+2\right )}{2\,f\,n\,\left (n^2\,1{}\mathrm {i}+n\,3{}\mathrm {i}+2{}\mathrm {i}\right )}+\frac {\left (2\,n+6\right )\,\left (\cos \left (2\,e+2\,f\,x\right )+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )\,{\left (a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}\right )}^{3-n}}{2\,f\,n\,\left (n^2\,1{}\mathrm {i}+n\,3{}\mathrm {i}+2{}\mathrm {i}\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(2*n)*(a + a*tan(e + f*x)*1i)^(3 - n),x)

[Out]

-(cos(6*e + 6*f*x) - sin(6*e + 6*f*x)*1i)*(d/cos(e + f*x))^(2*n)*((a + (a*sin(e + f*x)*1i)/cos(e + f*x))^(3 -
n)/(f*n*(n*3i + n^2*1i + 2i)) + ((cos(4*e + 4*f*x) + sin(4*e + 4*f*x)*1i)*(a + (a*sin(e + f*x)*1i)/cos(e + f*x
))^(3 - n)*(5*n + n^2 + 6))/(2*f*n*(n*3i + n^2*1i + 2i)) + ((cos(6*e + 6*f*x) + sin(6*e + 6*f*x)*1i)*(a + (a*s
in(e + f*x)*1i)/cos(e + f*x))^(3 - n)*(3*n + n^2 + 2))/(2*f*n*(n*3i + n^2*1i + 2i)) + ((2*n + 6)*(cos(2*e + 2*
f*x) + sin(2*e + 2*f*x)*1i)*(a + (a*sin(e + f*x)*1i)/cos(e + f*x))^(3 - n))/(2*f*n*(n*3i + n^2*1i + 2i)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{2 n} \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{3 - n}\, dx \]

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]

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

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