3.292 \(\int \sinh (\frac{b x}{c+d x}) \, dx\)

Optimal. Leaf size=74 \[ \frac{b c \cosh \left (\frac{b}{d}\right ) \text{Chi}\left (\frac{b c}{d (c+d x)}\right )}{d^2}-\frac{b c \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh \left (\frac{b x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.129571, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {5607, 3297, 3303, 3298, 3301} \[ \frac{b c \cosh \left (\frac{b}{d}\right ) \text{Chi}\left (\frac{b c}{d (c+d x)}\right )}{d^2}-\frac{b c \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh \left (\frac{b x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[(b*x)/(c + d*x)],x]

[Out]

(b*c*Cosh[b/d]*CoshIntegral[(b*c)/(d*(c + d*x))])/d^2 + ((c + d*x)*Sinh[(b*x)/(c + d*x)])/d - (b*c*Sinh[b/d]*S
inhIntegral[(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 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 \sinh \left (\frac{b x}{c+d x}\right ) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\sinh \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 \left (\frac{b x}{c+d x}\right )}{d}+\frac{(b c) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{b}{d}-\frac{b c x}{d}\right )}{x} \, dx,x,\frac{1}{c+d x}\right )}{d^2}\\ &=\frac{(c+d x) \sinh \left (\frac{b x}{c+d x}\right )}{d}+\frac{\left (b c \cosh \left (\frac{b}{d}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{b c x}{d}\right )}{x} \, dx,x,\frac{1}{c+d x}\right )}{d^2}-\frac{\left (b c \sinh \left (\frac{b}{d}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{b c x}{d}\right )}{x} \, dx,x,\frac{1}{c+d x}\right )}{d^2}\\ &=\frac{b c \cosh \left (\frac{b}{d}\right ) \text{Chi}\left (\frac{b c}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh \left (\frac{b x}{c+d x}\right )}{d}-\frac{b c \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c}{d (c+d x)}\right )}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.282876, size = 70, normalized size = 0.95 \[ \frac{b c \cosh \left (\frac{b}{d}\right ) \text{Chi}\left (\frac{b c}{d (c+d x)}\right )-b c \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c}{d (c+d x)}\right )+d (c+d x) \sinh \left (\frac{b x}{c+d x}\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[(b*x)/(c + d*x)],x]

[Out]

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

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Maple [A]  time = 0.032, size = 113, normalized size = 1.5 \begin{align*} -{\frac{dx+c}{2\,d}{{\rm e}^{-{\frac{bx}{dx+c}}}}}-{\frac{cb}{2\,{d}^{2}}{{\rm e}^{-{\frac{b}{d}}}}{\it Ei} \left ( 1,-{\frac{cb}{d \left ( dx+c \right ) }} \right ) }+{\frac{x}{2}{{\rm e}^{{\frac{bx}{dx+c}}}}}+{\frac{c}{2\,d}{{\rm e}^{{\frac{bx}{dx+c}}}}}-{\frac{cb}{2\,{d}^{2}}{{\rm e}^{{\frac{b}{d}}}}{\it Ei} \left ( 1,{\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)),x)

[Out]

-1/2/d*exp(-b*x/(d*x+c))*(d*x+c)-1/2*c*b/d^2*exp(-b/d)*Ei(1,-b*c/d/(d*x+c))+1/2*exp(b*x/(d*x+c))*x+1/2*c/d*exp
(b*x/(d*x+c))-1/2*c*b/d^2*exp(b/d)*Ei(1,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{b c}{d^{2} x + c d}\right )}}{d^{2} x^{2} e^{\frac{b}{d}} + 2 \, c d x e^{\frac{b}{d}} + c^{2} e^{\frac{b}{d}}}\,{d x} - \frac{1}{2} \, b c \int \frac{x e^{\left (-\frac{b c}{d^{2} x + c d} + \frac{b}{d}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} - \frac{1}{2} \,{\left (x e^{\left (\frac{b c}{d^{2} x + c d}\right )} - x e^{\left (-\frac{b c}{d^{2} x + c d} + \frac{2 \, b}{d}\right )}\right )} e^{\left (-\frac{b}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.97991, size = 532, normalized size = 7.19 \begin{align*} -\frac{b c{\rm Ei}\left (-\frac{b c}{d^{2} x + c d}\right ) \cosh \left (\frac{b}{d}\right ) \sinh \left (\frac{b x}{d x + c}\right )^{2} -{\left (b c{\rm Ei}\left (-\frac{b c}{d^{2} x + c d}\right ) \cosh \left (\frac{b x}{d x + c}\right )^{2} + b c{\rm Ei}\left (\frac{b c}{d^{2} x + c d}\right )\right )} \cosh \left (\frac{b}{d}\right ) - 2 \,{\left (d^{2} x + c d\right )} \sinh \left (\frac{b x}{d x + c}\right ) -{\left (b c{\rm Ei}\left (-\frac{b c}{d^{2} x + c d}\right ) \cosh \left (\frac{b x}{d x + c}\right )^{2} - b c{\rm Ei}\left (-\frac{b c}{d^{2} x + c d}\right ) \sinh \left (\frac{b x}{d x + c}\right )^{2} - b c{\rm Ei}\left (\frac{b c}{d^{2} x + c d}\right )\right )} \sinh \left (\frac{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)),x, algorithm="fricas")

[Out]

-1/2*(b*c*Ei(-b*c/(d^2*x + c*d))*cosh(b/d)*sinh(b*x/(d*x + c))^2 - (b*c*Ei(-b*c/(d^2*x + c*d))*cosh(b*x/(d*x +
c))^2 + b*c*Ei(b*c/(d^2*x + c*d)))*cosh(b/d) - 2*(d^2*x + c*d)*sinh(b*x/(d*x + c)) - (b*c*Ei(-b*c/(d^2*x + c*
d))*cosh(b*x/(d*x + c))^2 - b*c*Ei(-b*c/(d^2*x + c*d))*sinh(b*x/(d*x + c))^2 - b*c*Ei(b*c/(d^2*x + c*d)))*sinh
(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (\frac{b x}{c + d x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(b*x/(c + d*x)), x)

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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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x/(d*x + c)), x)