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

Optimal. Leaf size=80 \[ \frac{b c \sinh \left (\frac{2 b}{d}\right ) \text{Chi}\left (\frac{2 b c}{d (c+d x)}\right )}{d^2}-\frac{b c \cosh \left (\frac{2 b}{d}\right ) \text{Shi}\left (\frac{2 b c}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh ^2\left (\frac{b x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.145793, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {5607, 3313, 12, 3303, 3298, 3301} \[ \frac{b c \sinh \left (\frac{2 b}{d}\right ) \text{Chi}\left (\frac{2 b c}{d (c+d x)}\right )}{d^2}-\frac{b c \cosh \left (\frac{2 b}{d}\right ) \text{Shi}\left (\frac{2 b c}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh ^2\left (\frac{b x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[(b*x)/(c + d*x)]^2,x]

[Out]

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

Rule 5607

Int[Sinh[((e_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_))]^(n_.), x_Symbol] :> -Dist[d^(-1), Subst[Int[Sinh[(
b*e)/d - (e*(b*c - a*d)*x)/d]^n/x^2, x], x, 1/(c + d*x)], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && NeQ[b*
c - a*d, 0]

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

Mathematica [A]  time = 0.358976, size = 85, normalized size = 1.06 \[ \frac{2 b c \sinh \left (\frac{2 b}{d}\right ) \text{Chi}\left (\frac{2 b c}{d (c+d x)}\right )-2 b c \cosh \left (\frac{2 b}{d}\right ) \text{Shi}\left (\frac{2 b c}{d (c+d x)}\right )+d \left ((c+d x) \cosh \left (\frac{2 b x}{c+d x}\right )-d x\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[(b*x)/(c + d*x)]^2,x]

[Out]

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

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Maple [A]  time = 0.087, size = 120, normalized size = 1.5 \begin{align*} -{\frac{x}{2}}+{\frac{dx+c}{4\,d}{{\rm e}^{-2\,{\frac{bx}{dx+c}}}}}+{\frac{cb}{2\,{d}^{2}}{{\rm e}^{-2\,{\frac{b}{d}}}}{\it Ei} \left ( 1,-2\,{\frac{cb}{d \left ( dx+c \right ) }} \right ) }+{\frac{x}{4}{{\rm e}^{2\,{\frac{bx}{dx+c}}}}}+{\frac{c}{4\,d}{{\rm e}^{2\,{\frac{bx}{dx+c}}}}}-{\frac{cb}{2\,{d}^{2}}{{\rm e}^{2\,{\frac{b}{d}}}}{\it Ei} \left ( 1,2\,{\frac{cb}{d \left ( dx+c \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, b c \int \frac{x e^{\left (\frac{2 \, b c}{d^{2} x + c d}\right )}}{d^{2} x^{2} e^{\left (\frac{2 \, b}{d}\right )} + 2 \, c d x e^{\left (\frac{2 \, b}{d}\right )} + c^{2} e^{\left (\frac{2 \, b}{d}\right )}}\,{d x} - \frac{1}{2} \, b c \int \frac{x e^{\left (-\frac{2 \, b c}{d^{2} x + c d} + \frac{2 \, b}{d}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} + \frac{1}{4} \,{\left (x e^{\left (\frac{2 \, b c}{d^{2} x + c d}\right )} + x e^{\left (-\frac{2 \, b c}{d^{2} x + c d} + \frac{4 \, b}{d}\right )}\right )} e^{\left (-\frac{2 \, b}{d}\right )} - \frac{1}{2} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.11719, size = 589, normalized size = 7.36 \begin{align*} -\frac{d^{2} x -{\left (d^{2} x + c d\right )} \cosh \left (\frac{b x}{d x + c}\right )^{2} +{\left (b c{\rm Ei}\left (-\frac{2 \, b c}{d^{2} x + c d}\right ) \cosh \left (\frac{2 \, b}{d}\right ) - d^{2} x - c d\right )} \sinh \left (\frac{b x}{d x + c}\right )^{2} -{\left (b c{\rm Ei}\left (-\frac{2 \, b c}{d^{2} x + c d}\right ) \cosh \left (\frac{b x}{d x + c}\right )^{2} - b c{\rm Ei}\left (\frac{2 \, b c}{d^{2} x + c d}\right )\right )} \cosh \left (\frac{2 \, b}{d}\right ) -{\left (b c{\rm Ei}\left (-\frac{2 \, b c}{d^{2} x + c d}\right ) \cosh \left (\frac{b x}{d x + c}\right )^{2} - b c{\rm Ei}\left (-\frac{2 \, b c}{d^{2} x + c d}\right ) \sinh \left (\frac{b x}{d x + c}\right )^{2} + b c{\rm Ei}\left (\frac{2 \, b c}{d^{2} x + c d}\right )\right )} \sinh \left (\frac{2 \, b}{d}\right )}{2 \,{\left (d^{2} \cosh \left (\frac{b x}{d x + c}\right )^{2} - d^{2} \sinh \left (\frac{b x}{d x + c}\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x/(d*x + c))^2, x)