3.432 \(\int \frac{(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=38 \[ \frac{2 i (e \sec (c+d x))^{5/2}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((2*I)/5)*(e*Sec[c + d*x])^(5/2))/(d*(a + I*a*Tan[c + d*x])^(5/2))

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Rubi [A]  time = 0.0783137, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.033, Rules used = {3488} \[ \frac{2 i (e \sec (c+d x))^{5/2}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((2*I)/5)*(e*Sec[c + d*x])^(5/2))/(d*(a + I*a*Tan[c + d*x])^(5/2))

Rule 3488

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)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps

\begin{align*} \int \frac{(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{2 i (e \sec (c+d x))^{5/2}}{5 d (a+i a \tan (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.220448, size = 38, normalized size = 1. \[ \frac{2 i (e \sec (c+d x))^{5/2}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((2*I)/5)*(e*Sec[c + d*x])^(5/2))/(d*(a + I*a*Tan[c + d*x])^(5/2))

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Maple [B]  time = 0.281, size = 105, normalized size = 2.8 \begin{align*}{\frac{{\frac{2\,i}{5}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d{a}^{3} \left ( 4\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-i\sin \left ( dx+c \right ) -3\,\cos \left ( dx+c \right ) \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2/5*I/d/a^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^3*(e/cos(d*x+c))^(5/2)/(4*I*sin(d*x+c)*c
os(d*x+c)^2+4*cos(d*x+c)^3-I*sin(d*x+c)-3*cos(d*x+c))

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Maxima [B]  time = 1.61859, size = 103, normalized size = 2.71 \begin{align*} \frac{2 i \, e^{\frac{5}{2}}{\left (-\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac{5}{2}}}{5 \, a^{\frac{5}{2}} d{\left (-\frac{2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2/5*I*e^(5/2)*(-sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(5/2)/(a^(5/2)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1
) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(5/2))

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Fricas [B]  time = 2.08998, size = 192, normalized size = 5.05 \begin{align*} \frac{2 \,{\left (i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac{5}{2} i \, d x - \frac{5}{2} i \, c\right )}}{5 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/5*(I*e^2*e^(2*I*d*x + 2*I*c) + I*e^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^
(-5/2*I*d*x - 5/2*I*c)/(a^3*d)

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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((e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(5/2)/(I*a*tan(d*x + c) + a)^(5/2), x)