\(\int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx\) [149]

Optimal result

Integrand size = 21, antiderivative size = 21 \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\text {Int}\left (\frac {1}{x (a+i a \sinh (c+d x))^{5/2}},x\right ) \]

[Out]

Unintegrable(1/x/(a+I*a*sinh(d*x+c))^(5/2),x)

Rubi [N/A]

Not integrable

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx \]

[In]

Int[1/(x*(a + I*a*Sinh[c + d*x])^(5/2)),x]

[Out]

Defer[Int][1/(x*(a + I*a*Sinh[c + d*x])^(5/2)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 36.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx \]

[In]

Integrate[1/(x*(a + I*a*Sinh[c + d*x])^(5/2)),x]

[Out]

Integrate[1/(x*(a + I*a*Sinh[c + d*x])^(5/2)), x]

Maple [N/A] (verified)

Not integrable

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

\[\int \frac {1}{x \left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.25 (sec) , antiderivative size = 396, normalized size of antiderivative = 18.86 \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} x} \,d x } \]

[In]

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

[Out]

1/24*(24*(a^3*d^4*x^4*e^(4*d*x + 4*c) - 4*I*a^3*d^4*x^4*e^(3*d*x + 3*c) - 6*a^3*d^4*x^4*e^(2*d*x + 2*c) + 4*I*
a^3*d^4*x^4*e^(d*x + c) + a^3*d^4*x^4)*integral(1/48*(-9*I*d^4*x^4 + 80*I*d^2*x^2 - 384*I)*sqrt(1/2*I*a*e^(-d*
x - c))*e^(d*x + c)/(a^3*d^4*x^5*e^(d*x + c) - I*a^3*d^4*x^5), x) + ((-9*I*d^3*x^3 + 18*I*d^2*x^2 + 8*I*d*x -
48*I)*e^(4*d*x + 4*c) - (33*d^3*x^3 - 70*d^2*x^2 - 8*d*x + 144)*e^(3*d*x + 3*c) + (-33*I*d^3*x^3 - 70*I*d^2*x^
2 + 8*I*d*x + 144*I)*e^(2*d*x + 2*c) - (9*d^3*x^3 + 18*d^2*x^2 - 8*d*x - 48)*e^(d*x + c))*sqrt(1/2*I*a*e^(-d*x
- c)))/(a^3*d^4*x^4*e^(4*d*x + 4*c) - 4*I*a^3*d^4*x^4*e^(3*d*x + 3*c) - 6*a^3*d^4*x^4*e^(2*d*x + 2*c) + 4*I*a
^3*d^4*x^4*e^(d*x + c) + a^3*d^4*x^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/x/(a+I*a*sinh(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} x} \,d x } \]

[In]

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

[Out]

integrate(1/((I*a*sinh(d*x + c) + a)^(5/2)*x), x)

Giac [N/A]

Not integrable

Time = 0.49 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} x} \,d x } \]

[In]

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

[Out]

integrate(1/((I*a*sinh(d*x + c) + a)^(5/2)*x), x)

Mupad [N/A]

Not integrable

Time = 2.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x (a+i a \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{x\,{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[In]

int(1/(x*(a + a*sinh(c + d*x)*1i)^(5/2)),x)

[Out]

int(1/(x*(a + a*sinh(c + d*x)*1i)^(5/2)), x)