3.504 \(\int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{1-n} \, dx\)
Optimal. Leaf size=40 \[ \frac{i a (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n}}{f n} \]
[Out]
(I*a*(d*Sec[e + f*x])^(2*n))/(f*n*(a + I*a*Tan[e + f*x])^n)
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Rubi [A] time = 0.0574708, antiderivative size = 40, normalized size of antiderivative = 1.,
number of steps used = 1, number of rules used = 1, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.031, Rules used =
{3493} \[ \frac{i a (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n}}{f n} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(1 - n),x]
[Out]
(I*a*(d*Sec[e + f*x])^(2*n))/(f*n*(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]
Rubi steps
\begin{align*} \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{1-n} \, dx &=\frac{i a (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n}}{f n}\\ \end{align*}
Mathematica [A] time = 0.411444, size = 40, normalized size = 1. \[ \frac{i a (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n}}{f n} \]
Antiderivative was successfully verified.
[In]
Integrate[(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(1 - n),x]
[Out]
(I*a*(d*Sec[e + f*x])^(2*n))/(f*n*(a + I*a*Tan[e + f*x])^n)
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Maple [C] time = 0.596, size = 1294, normalized size = 32.4 \begin{align*} \text{result too large to display} \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))^(1-n),x)
[Out]
I/f*a^(-n)*2^n*a*d^(2*n)*exp(1/2*I*Pi*(-n*csgn(I*a)*csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I*(f*x+e)))^2-n*csgn(I
/(exp(2*I*(f*x+e))+1))*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^2-csgn(I/(exp(2*I*(f*x+e))+1))*csgn(I*exp
(2*I*(f*x+e)))*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))-2*n*csgn(I*exp(I*(f*x+e)))*csgn(I*exp(2*I*(f*x+e)
))^2-n*csgn(I*exp(2*I*(f*x+e)))*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^2+n*csgn(I*exp(I*(f*x+e)))^2*csg
n(I*exp(2*I*(f*x+e)))-n*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))*csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I*(f
*x+e)))^2-csgn(I*a)*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))*csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I*(f*x+e
)))+csgn(I*exp(2*I*(f*x+e)))*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^2+csgn(I*exp(2*I*(f*x+e))/(exp(2*I*
(f*x+e))+1))*csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I*(f*x+e)))^2+csgn(I*a)*csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I
*(f*x+e)))^2+csgn(I/(exp(2*I*(f*x+e))+1))*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^2-csgn(I*exp(I*(f*x+e)
))^2*csgn(I*exp(2*I*(f*x+e)))+n*csgn(I*exp(2*I*(f*x+e)))^3+n*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^3+n
*csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I*(f*x+e)))^3-2*n*csgn(I*d/(exp(2*I*(f*x+e))+1)*exp(I*(f*x+e)))^3-2*n*csg
n(I*exp(I*(f*x+e))/(exp(2*I*(f*x+e))+1))^3-csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^3-csgn(I*exp(2*I*(f*x
+e)))^3-csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I*(f*x+e)))^3+2*csgn(I*exp(I*(f*x+e)))*csgn(I*exp(2*I*(f*x+e)))^2+
n*csgn(I*a)*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))*csgn(I*a/(exp(2*I*(f*x+e))+1)*exp(2*I*(f*x+e)))-2*n*
csgn(I/(exp(2*I*(f*x+e))+1))*csgn(I*exp(I*(f*x+e)))*csgn(I*exp(I*(f*x+e))/(exp(2*I*(f*x+e))+1))+n*csgn(I/(exp(
2*I*(f*x+e))+1))*csgn(I*exp(2*I*(f*x+e)))*csgn(I*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))-2*n*csgn(I*exp(I*(f*x+
e))/(exp(2*I*(f*x+e))+1))*csgn(I*d)*csgn(I*d/(exp(2*I*(f*x+e))+1)*exp(I*(f*x+e)))+2*n*csgn(I*d)*csgn(I*d/(exp(
2*I*(f*x+e))+1)*exp(I*(f*x+e)))^2+2*n*csgn(I*exp(I*(f*x+e)))*csgn(I*exp(I*(f*x+e))/(exp(2*I*(f*x+e))+1))^2+2*n
*csgn(I*exp(I*(f*x+e))/(exp(2*I*(f*x+e))+1))*csgn(I*d/(exp(2*I*(f*x+e))+1)*exp(I*(f*x+e)))^2+2*n*csgn(I/(exp(2
*I*(f*x+e))+1))*csgn(I*exp(I*(f*x+e))/(exp(2*I*(f*x+e))+1))^2))/n*(exp(2*I*(f*x+e))+1)^(-n)
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Maxima [B] time = 2.47156, size = 185, normalized size = 4.62 \begin{align*} \frac{i \, a^{-n + 1} d^{2 \, n} e^{\left (-n \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - n \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right ) - n \log \left (-\frac{2 i \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right ) + 2 \, n \log \left (-\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )\right )}}{f n} \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))^(1-n),x, algorithm="maxima")
[Out]
I*a^(-n + 1)*d^(2*n)*e^(-n*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - n*log(sin(f*x + e)/(cos(f*x + e) + 1) -
1) - n*log(-2*I*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1) + 2*n*log(-sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2 - 1))/(f*n)
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Fricas [B] time = 1.90458, size = 236, normalized size = 5.9 \begin{align*} \frac{\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 + 1} \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}{\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, f n} \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))^(1-n),x, algorithm="fricas")
[Out]
1/2*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^(-n + 1)*(2*d*e^(I*f*x + I*e)/(e^(2*I*f*x + 2*I*e) + 1
))^(2*n)*(I*e^(2*I*f*x + 2*I*e) + I)*e^(-2*I*f*x - 2*I*e)/(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))**(1-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 + 1}\,{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))^(1-n),x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^(2*n)*(I*a*tan(f*x + e) + a)^(-n + 1), x)