3.13 \(\int \frac{\sinh ^2(a+b x)}{(c+d x)^2} \, dx\)

Optimal. Leaf size=81 \[ \frac{b \sinh \left (2 a-\frac{2 b c}{d}\right ) \text{Chi}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}+\frac{b \cosh \left (2 a-\frac{2 b c}{d}\right ) \text{Shi}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{\sinh ^2(a+b x)}{d (c+d x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.154983, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3313, 12, 3303, 3298, 3301} \[ \frac{b \sinh \left (2 a-\frac{2 b c}{d}\right ) \text{Chi}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}+\frac{b \cosh \left (2 a-\frac{2 b c}{d}\right ) \text{Shi}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{\sinh ^2(a+b x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^2/(c + d*x)^2,x]

[Out]

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

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.44441, size = 75, normalized size = 0.93 \[ \frac{b \sinh \left (2 a-\frac{2 b c}{d}\right ) \text{Chi}\left (\frac{2 b (c+d x)}{d}\right )+b \cosh \left (2 a-\frac{2 b c}{d}\right ) \text{Shi}\left (\frac{2 b (c+d x)}{d}\right )-\frac{d \sinh ^2(a+b x)}{c+d x}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.072, size = 152, normalized size = 1.9 \begin{align*}{\frac{1}{2\,d \left ( dx+c \right ) }}-{\frac{b{{\rm e}^{-2\,bx-2\,a}}}{ \left ( 4\,bdx+4\,cb \right ) d}}+{\frac{b}{2\,{d}^{2}}{{\rm e}^{-2\,{\frac{da-cb}{d}}}}{\it Ei} \left ( 1,2\,bx+2\,a-2\,{\frac{da-cb}{d}} \right ) }-{\frac{b{{\rm e}^{2\,bx+2\,a}}}{4\,{d}^{2}} \left ({\frac{cb}{d}}+bx \right ) ^{-1}}-{\frac{b}{2\,{d}^{2}}{{\rm e}^{2\,{\frac{da-cb}{d}}}}{\it Ei} \left ( 1,-2\,bx-2\,a-2\,{\frac{-da+cb}{d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^2/(d*x+c)^2,x)

[Out]

1/2/d/(d*x+c)-1/4*b*exp(-2*b*x-2*a)/(b*d*x+b*c)/d+1/2*b/d^2*exp(-2*(a*d-b*c)/d)*Ei(1,2*b*x+2*a-2*(a*d-b*c)/d)-
1/4*b/d^2*exp(2*b*x+2*a)/(b*c/d+b*x)-1/2*b/d^2*exp(2*(a*d-b*c)/d)*Ei(1,-2*b*x-2*a-2*(-a*d+b*c)/d)

________________________________________________________________________________________

Maxima [A]  time = 1.38888, size = 119, normalized size = 1.47 \begin{align*} -\frac{e^{\left (-2 \, a + \frac{2 \, b c}{d}\right )} E_{2}\left (\frac{2 \,{\left (d x + c\right )} b}{d}\right )}{4 \,{\left (d x + c\right )} d} - \frac{e^{\left (2 \, a - \frac{2 \, b c}{d}\right )} E_{2}\left (-\frac{2 \,{\left (d x + c\right )} b}{d}\right )}{4 \,{\left (d x + c\right )} d} + \frac{1}{2 \,{\left (d^{2} x + c d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*e^(-2*a + 2*b*c/d)*exp_integral_e(2, 2*(d*x + c)*b/d)/((d*x + c)*d) - 1/4*e^(2*a - 2*b*c/d)*exp_integral_
e(2, -2*(d*x + c)*b/d)/((d*x + c)*d) + 1/2/(d^2*x + c*d)

________________________________________________________________________________________

Fricas [B]  time = 2.72063, size = 365, normalized size = 4.51 \begin{align*} -\frac{d \cosh \left (b x + a\right )^{2} + d \sinh \left (b x + a\right )^{2} -{\left ({\left (b d x + b c\right )}{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) -{\left (b d x + b c\right )}{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cosh \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) -{\left ({\left (b d x + b c\right )}{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\right )}{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sinh \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) - d}{2 \,{\left (d^{3} x + c d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*cosh(b*x + a)^2 + d*sinh(b*x + a)^2 - ((b*d*x + b*c)*Ei(2*(b*d*x + b*c)/d) - (b*d*x + b*c)*Ei(-2*(b*d*
x + b*c)/d))*cosh(-2*(b*c - a*d)/d) - ((b*d*x + b*c)*Ei(2*(b*d*x + b*c)/d) + (b*d*x + b*c)*Ei(-2*(b*d*x + b*c)
/d))*sinh(-2*(b*c - a*d)/d) - d)/(d^3*x + c*d^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)**2/(d*x+c)**2,x)

[Out]

Integral(sinh(a + b*x)**2/(c + d*x)**2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^2/(d*x + c)^2, x)