3.264 \(\int \frac{\cosh ^2(c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx\)

Optimal. Leaf size=103 \[ -\frac{i d \cosh \left (c-\frac{d e}{f}\right ) \text{Chi}\left (\frac{d e}{f}+d x\right )}{a f^2}-\frac{i d \sinh \left (c-\frac{d e}{f}\right ) \text{Shi}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{i \sinh (c+d x)}{a f (e+f x)}-\frac{1}{a f (e+f x)} \]

[Out]

-(1/(a*f*(e + f*x))) - (I*d*Cosh[c - (d*e)/f]*CoshIntegral[(d*e)/f + d*x])/(a*f^2) + (I*Sinh[c + d*x])/(a*f*(e
+ f*x)) - (I*d*Sinh[c - (d*e)/f]*SinhIntegral[(d*e)/f + d*x])/(a*f^2)

________________________________________________________________________________________

Rubi [A]  time = 0.220979, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {5563, 32, 3297, 3303, 3298, 3301} \[ -\frac{i d \cosh \left (c-\frac{d e}{f}\right ) \text{Chi}\left (\frac{d e}{f}+d x\right )}{a f^2}-\frac{i d \sinh \left (c-\frac{d e}{f}\right ) \text{Shi}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{i \sinh (c+d x)}{a f (e+f x)}-\frac{1}{a f (e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a*f*(e + f*x))) - (I*d*Cosh[c - (d*e)/f]*CoshIntegral[(d*e)/f + d*x])/(a*f^2) + (I*Sinh[c + d*x])/(a*f*(e
+ f*x)) - (I*d*Sinh[c - (d*e)/f]*SinhIntegral[(d*e)/f + d*x])/(a*f^2)

Rule 5563

Int[(Cosh[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symb
ol] :> Dist[1/a, Int[(e + f*x)^m*Cosh[c + d*x]^(n - 2), x], x] + Dist[1/b, Int[(e + f*x)^m*Cosh[c + d*x]^(n -
2)*Sinh[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 1] && EqQ[a^2 + b^2, 0]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{\cosh ^2(c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx &=-\frac{i \int \frac{\sinh (c+d x)}{(e+f x)^2} \, dx}{a}+\frac{\int \frac{1}{(e+f x)^2} \, dx}{a}\\ &=-\frac{1}{a f (e+f x)}+\frac{i \sinh (c+d x)}{a f (e+f x)}-\frac{(i d) \int \frac{\cosh (c+d x)}{e+f x} \, dx}{a f}\\ &=-\frac{1}{a f (e+f x)}+\frac{i \sinh (c+d x)}{a f (e+f x)}-\frac{\left (i d \cosh \left (c-\frac{d e}{f}\right )\right ) \int \frac{\cosh \left (\frac{d e}{f}+d x\right )}{e+f x} \, dx}{a f}-\frac{\left (i d \sinh \left (c-\frac{d e}{f}\right )\right ) \int \frac{\sinh \left (\frac{d e}{f}+d x\right )}{e+f x} \, dx}{a f}\\ &=-\frac{1}{a f (e+f x)}-\frac{i d \cosh \left (c-\frac{d e}{f}\right ) \text{Chi}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{i \sinh (c+d x)}{a f (e+f x)}-\frac{i d \sinh \left (c-\frac{d e}{f}\right ) \text{Shi}\left (\frac{d e}{f}+d x\right )}{a f^2}\\ \end{align*}

Mathematica [A]  time = 0.528068, size = 85, normalized size = 0.83 \[ -\frac{i \left (d (e+f x) \cosh \left (c-\frac{d e}{f}\right ) \text{Chi}\left (d \left (\frac{e}{f}+x\right )\right )+d (e+f x) \sinh \left (c-\frac{d e}{f}\right ) \text{Shi}\left (d \left (\frac{e}{f}+x\right )\right )-f (\sinh (c+d x)+i)\right )}{a f^2 (e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(d*(e + f*x)*Cosh[c - (d*e)/f]*CoshIntegral[d*(e/f + x)] - f*(I + Sinh[c + d*x]) + d*(e + f*x)*Sinh[c -
(d*e)/f]*SinhIntegral[d*(e/f + x)]))/(a*f^2*(e + f*x))

________________________________________________________________________________________

Maple [A]  time = 0.095, size = 164, normalized size = 1.6 \begin{align*} -{\frac{1}{af \left ( fx+e \right ) }}+{\frac{{\frac{i}{2}}d{{\rm e}^{dx+c}}}{a{f}^{2}} \left ({\frac{de}{f}}+dx \right ) ^{-1}}+{\frac{{\frac{i}{2}}d}{a{f}^{2}}{{\rm e}^{{\frac{cf-de}{f}}}}{\it Ei} \left ( 1,-dx-c-{\frac{-cf+de}{f}} \right ) }-{\frac{{\frac{i}{2}}d{{\rm e}^{-dx-c}}}{af \left ( dfx+de \right ) }}+{\frac{{\frac{i}{2}}d}{a{f}^{2}}{{\rm e}^{-{\frac{cf-de}{f}}}}{\it Ei} \left ( 1,dx+c-{\frac{cf-de}{f}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/f/(f*x+e)+1/2*I*d/a/f^2*exp(d*x+c)/(d*e/f+d*x)+1/2*I*d/a/f^2*exp((c*f-d*e)/f)*Ei(1,-d*x-c-(-c*f+d*e)/f)-1
/2*I/a*d*exp(-d*x-c)/f/(d*f*x+d*e)+1/2*I/a*d/f^2*exp(-(c*f-d*e)/f)*Ei(1,d*x+c-(c*f-d*e)/f)

________________________________________________________________________________________

Maxima [A]  time = 1.65693, size = 124, normalized size = 1.2 \begin{align*} -\frac{1}{a f^{2} x + a e f} - \frac{i \, e^{\left (-c + \frac{d e}{f}\right )} E_{2}\left (\frac{{\left (f x + e\right )} d}{f}\right )}{2 \,{\left (f x + e\right )} a f} + \frac{i \, e^{\left (c - \frac{d e}{f}\right )} E_{2}\left (-\frac{{\left (f x + e\right )} d}{f}\right )}{2 \,{\left (f x + e\right )} a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*f^2*x + a*e*f) - 1/2*I*e^(-c + d*e/f)*exp_integral_e(2, (f*x + e)*d/f)/((f*x + e)*a*f) + 1/2*I*e^(c - d*
e/f)*exp_integral_e(2, -(f*x + e)*d/f)/((f*x + e)*a*f)

________________________________________________________________________________________

Fricas [A]  time = 2.16562, size = 278, normalized size = 2.7 \begin{align*} \frac{{\left (i \, f e^{\left (2 \, d x + 2 \, c\right )} +{\left ({\left (-i \, d f x - i \, d e\right )}{\rm Ei}\left (-\frac{d f x + d e}{f}\right ) e^{\left (\frac{d e - c f}{f}\right )} +{\left (-i \, d f x - i \, d e\right )}{\rm Ei}\left (\frac{d f x + d e}{f}\right ) e^{\left (-\frac{d e - c f}{f}\right )} - 2 \, f\right )} e^{\left (d x + c\right )} - i \, f\right )} e^{\left (-d x - c\right )}}{2 \,{\left (a f^{3} x + a e f^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError