3.262 \(\int \frac{1}{(a \cos (c+d x)+i a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=33 \[ \frac{2 i}{5 d (a \cos (c+d x)+i a \sin (c+d x))^{5/2}} \]

[Out]

((2*I)/5)/(d*(a*Cos[c + d*x] + I*a*Sin[c + d*x])^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0163566, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {3071} \[ \frac{2 i}{5 d (a \cos (c+d x)+i a \sin (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^(-5/2),x]

[Out]

((2*I)/5)/(d*(a*Cos[c + d*x] + I*a*Sin[c + d*x])^(5/2))

Rule 3071

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^n)/(b*d*n), x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a \cos (c+d x)+i a \sin (c+d x))^{5/2}} \, dx &=\frac{2 i}{5 d (a \cos (c+d x)+i a \sin (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.036128, size = 32, normalized size = 0.97 \[ \frac{2 i}{5 d (a (\cos (c+d x)+i \sin (c+d x)))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^(-5/2),x]

[Out]

((2*I)/5)/(d*(a*(Cos[c + d*x] + I*Sin[c + d*x]))^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 28, normalized size = 0.9 \begin{align*}{\frac{{\frac{2\,i}{5}}}{d} \left ( a\cos \left ( dx+c \right ) +ia\sin \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.58497, size = 69, normalized size = 2.09 \begin{align*} \frac{2 i \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i\right )}^{\frac{5}{2}}}{5 \, a^{\frac{5}{2}} d{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.92852, size = 59, normalized size = 1.79 \begin{align*} \frac{2 i \, e^{\left (-\frac{5}{2} i \, d x - \frac{5}{2} i \, c\right )}}{5 \, a^{\frac{5}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/5*I*e^(-5/2*I*d*x - 5/2*I*c)/(a^(5/2)*d)

________________________________________________________________________________________

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(1/(a*cos(d*x+c)+I*a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.79466, size = 50, normalized size = 1.52 \begin{align*} \frac{2 i}{5 \, d \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i}\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+I*a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

2/5*I/(d*(-(a*tan(1/2*d*x + 1/2*c) - I*a)/(tan(1/2*d*x + 1/2*c) + I))^(5/2))