\(\int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx\) [281]

   Optimal result
   Rubi [N/A]
   Mathematica [F(-1)]
   Maple [N/A] (verified)
   Fricas [N/A]
   Sympy [N/A]
   Maxima [N/A]
   Giac [N/A]
   Mupad [N/A]

Optimal result

Integrand size = 31, antiderivative size = 31 \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=\text {Int}\left (\frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))},x\right ) \]

[Out]

Unintegrable(sech(d*x+c)^2/(f*x+e)/(a+I*a*sinh(d*x+c)),x)

Rubi [N/A]

Not integrable

Time = 0.05 (sec) , antiderivative size = 31, 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 = {} \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=\int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx \]

[In]

Int[Sech[c + d*x]^2/((e + f*x)*(a + I*a*Sinh[c + d*x])),x]

[Out]

Defer[Int][Sech[c + d*x]^2/((e + f*x)*(a + I*a*Sinh[c + d*x])), x]

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

Mathematica [F(-1)]

Timed out. \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=\text {\$Aborted} \]

[In]

Integrate[Sech[c + d*x]^2/((e + f*x)*(a + I*a*Sinh[c + d*x])),x]

[Out]

$Aborted

Maple [N/A] (verified)

Not integrable

Time = 1.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94

\[\int \frac {\operatorname {sech}\left (d x +c \right )^{2}}{\left (f x +e \right ) \left (a +i a \sinh \left (d x +c \right )\right )}d x\]

[In]

int(sech(d*x+c)^2/(f*x+e)/(a+I*a*sinh(d*x+c)),x)

[Out]

int(sech(d*x+c)^2/(f*x+e)/(a+I*a*sinh(d*x+c)),x)

Fricas [N/A]

Not integrable

Time = 0.26 (sec) , antiderivative size = 768, normalized size of antiderivative = 24.77 \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{2}}{{\left (f x + e\right )} {\left (i \, a \sinh \left (d x + c\right ) + a\right )}} \,d x } \]

[In]

integrate(sech(d*x+c)^2/(f*x+e)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-1/3*(4*d^2*f^2*x^2 + 8*d^2*e*f*x + 4*d^2*e^2 - 2*f^2*e^(2*d*x + 2*c) - 2*f^2 + (I*d*f^2*x + I*d*e*f - 2*I*f^2
)*e^(3*d*x + 3*c) + (8*I*d^2*f^2*x^2 + 8*I*d^2*e^2 + I*d*e*f - 2*I*f^2 + (16*I*d^2*e*f + I*d*f^2)*x)*e^(d*x +
c) - 3*(a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + a*d^3*e^3 - (a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 +
 3*a*d^3*e^2*f*x + a*d^3*e^3)*e^(4*d*x + 4*c) + 2*(I*a*d^3*f^3*x^3 + 3*I*a*d^3*e*f^2*x^2 + 3*I*a*d^3*e^2*f*x +
 I*a*d^3*e^3)*e^(3*d*x + 3*c) + 2*(I*a*d^3*f^3*x^3 + 3*I*a*d^3*e*f^2*x^2 + 3*I*a*d^3*e^2*f*x + I*a*d^3*e^3)*e^
(d*x + c))*integral(-1/3*(4*d^2*f^3*x^2 + 8*d^2*e*f^2*x + 4*d^2*e^2*f - 6*f^3 - (I*d^2*f^3*x^2 + 2*I*d^2*e*f^2
*x + I*d^2*e^2*f - 6*I*f^3)*e^(d*x + c))/(a*d^3*f^4*x^4 + 4*a*d^3*e*f^3*x^3 + 6*a*d^3*e^2*f^2*x^2 + 4*a*d^3*e^
3*f*x + a*d^3*e^4 + (a*d^3*f^4*x^4 + 4*a*d^3*e*f^3*x^3 + 6*a*d^3*e^2*f^2*x^2 + 4*a*d^3*e^3*f*x + a*d^3*e^4)*e^
(2*d*x + 2*c)), x))/(a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + a*d^3*e^3 - (a*d^3*f^3*x^3 + 3*a*d^
3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + a*d^3*e^3)*e^(4*d*x + 4*c) + 2*(I*a*d^3*f^3*x^3 + 3*I*a*d^3*e*f^2*x^2 + 3*I*a*
d^3*e^2*f*x + I*a*d^3*e^3)*e^(3*d*x + 3*c) + 2*(I*a*d^3*f^3*x^3 + 3*I*a*d^3*e*f^2*x^2 + 3*I*a*d^3*e^2*f*x + I*
a*d^3*e^3)*e^(d*x + c))

Sympy [N/A]

Not integrable

Time = 4.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=- \frac {i \int \frac {\operatorname {sech}^{2}{\left (c + d x \right )}}{e \sinh {\left (c + d x \right )} - i e + f x \sinh {\left (c + d x \right )} - i f x}\, dx}{a} \]

[In]

integrate(sech(d*x+c)**2/(f*x+e)/(a+I*a*sinh(d*x+c)),x)

[Out]

-I*Integral(sech(c + d*x)**2/(e*sinh(c + d*x) - I*e + f*x*sinh(c + d*x) - I*f*x), x)/a

Maxima [N/A]

Not integrable

Time = 0.66 (sec) , antiderivative size = 632, normalized size of antiderivative = 20.39 \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{2}}{{\left (f x + e\right )} {\left (i \, a \sinh \left (d x + c\right ) + a\right )}} \,d x } \]

[In]

integrate(sech(d*x+c)^2/(f*x+e)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-4*I*f*integrate(1/(8*I*a*d*f^2*x^2 + 16*I*a*d*e*f*x + 8*I*a*d*e^2 + 8*(a*d*f^2*x^2*e^c + 2*a*d*e*f*x*e^c + a*
d*e^2*e^c)*e^(d*x)), x) - 1/3*(4*d^2*f^2*x^2 + 8*d^2*e*f*x + 4*d^2*e^2 - 2*f^2*e^(2*d*x + 2*c) - 2*f^2 + (I*d*
f^2*x*e^(3*c) + (I*d*e*f - 2*I*f^2)*e^(3*c))*e^(3*d*x) + (8*I*d^2*f^2*x^2*e^c + (16*I*d^2*e*f + I*d*f^2)*x*e^c
 + (8*I*d^2*e^2 + I*d*e*f - 2*I*f^2)*e^c)*e^(d*x))/(a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + a*d^
3*e^3 - (a*d^3*f^3*x^3*e^(4*c) + 3*a*d^3*e*f^2*x^2*e^(4*c) + 3*a*d^3*e^2*f*x*e^(4*c) + a*d^3*e^3*e^(4*c))*e^(4
*d*x) + 2*(I*a*d^3*f^3*x^3*e^(3*c) + 3*I*a*d^3*e*f^2*x^2*e^(3*c) + 3*I*a*d^3*e^2*f*x*e^(3*c) + I*a*d^3*e^3*e^(
3*c))*e^(3*d*x) + 2*(I*a*d^3*f^3*x^3*e^c + 3*I*a*d^3*e*f^2*x^2*e^c + 3*I*a*d^3*e^2*f*x*e^c + I*a*d^3*e^3*e^c)*
e^(d*x)) - 4*integrate(1/24*(5*d^2*f^3*x^2 + 10*d^2*e*f^2*x + 5*d^2*e^2*f - 12*f^3)/(a*d^3*f^4*x^4 + 4*a*d^3*e
*f^3*x^3 + 6*a*d^3*e^2*f^2*x^2 + 4*a*d^3*e^3*f*x + a*d^3*e^4 - (-I*a*d^3*f^4*x^4*e^c - 4*I*a*d^3*e*f^3*x^3*e^c
 - 6*I*a*d^3*e^2*f^2*x^2*e^c - 4*I*a*d^3*e^3*f*x*e^c - I*a*d^3*e^4*e^c)*e^(d*x)), x)

Giac [N/A]

Not integrable

Time = 53.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{2}}{{\left (f x + e\right )} {\left (i \, a \sinh \left (d x + c\right ) + a\right )}} \,d x } \]

[In]

integrate(sech(d*x+c)^2/(f*x+e)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate(sech(d*x + c)^2/((f*x + e)*(I*a*sinh(d*x + c) + a)), x)

Mupad [N/A]

Not integrable

Time = 1.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {\text {sech}^2(c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^2\,\left (e+f\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

[In]

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

[Out]

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