3.31 \(\int \sinh (a+\frac{b}{x}) \, dx\)

Optimal. Leaf size=33 \[ -b \cosh (a) \text{Chi}\left (\frac{b}{x}\right )-b \sinh (a) \text{Shi}\left (\frac{b}{x}\right )+x \sinh \left (a+\frac{b}{x}\right ) \]

[Out]

-(b*Cosh[a]*CoshIntegral[b/x]) + x*Sinh[a + b/x] - b*Sinh[a]*SinhIntegral[b/x]

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Rubi [A]  time = 0.0757435, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5302, 3297, 3303, 3298, 3301} \[ -b \cosh (a) \text{Chi}\left (\frac{b}{x}\right )-b \sinh (a) \text{Shi}\left (\frac{b}{x}\right )+x \sinh \left (a+\frac{b}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x],x]

[Out]

-(b*Cosh[a]*CoshIntegral[b/x]) + x*Sinh[a + b/x] - b*Sinh[a]*SinhIntegral[b/x]

Rule 5302

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

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

Mathematica [A]  time = 0.0196536, size = 33, normalized size = 1. \[ -b \cosh (a) \text{Chi}\left (\frac{b}{x}\right )-b \sinh (a) \text{Shi}\left (\frac{b}{x}\right )+x \sinh \left (a+\frac{b}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x],x]

[Out]

-(b*Cosh[a]*CoshIntegral[b/x]) + x*Sinh[a + b/x] - b*Sinh[a]*SinhIntegral[b/x]

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Maple [A]  time = 0.029, size = 56, normalized size = 1.7 \begin{align*}{\frac{b{{\rm e}^{-a}}}{2}{\it Ei} \left ( 1,{\frac{b}{x}} \right ) }-{\frac{x}{2}{{\rm e}^{-{\frac{ax+b}{x}}}}}+{\frac{{{\rm e}^{a}}b}{2}{\it Ei} \left ( 1,-{\frac{b}{x}} \right ) }+{\frac{x}{2}{{\rm e}^{{\frac{ax+b}{x}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b/x),x)

[Out]

1/2*b*exp(-a)*Ei(1,b/x)-1/2*exp(-(a*x+b)/x)*x+1/2*b*exp(a)*Ei(1,-b/x)+1/2*exp((a*x+b)/x)*x

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Maxima [A]  time = 1.15812, size = 49, normalized size = 1.48 \begin{align*} -\frac{1}{2} \,{\left ({\rm Ei}\left (-\frac{b}{x}\right ) e^{\left (-a\right )} +{\rm Ei}\left (\frac{b}{x}\right ) e^{a}\right )} b + x \sinh \left (a + \frac{b}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x),x, algorithm="maxima")

[Out]

-1/2*(Ei(-b/x)*e^(-a) + Ei(b/x)*e^a)*b + x*sinh(a + b/x)

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Fricas [A]  time = 1.6999, size = 135, normalized size = 4.09 \begin{align*} -\frac{1}{2} \,{\left (b{\rm Ei}\left (\frac{b}{x}\right ) + b{\rm Ei}\left (-\frac{b}{x}\right )\right )} \cosh \left (a\right ) - \frac{1}{2} \,{\left (b{\rm Ei}\left (\frac{b}{x}\right ) - b{\rm Ei}\left (-\frac{b}{x}\right )\right )} \sinh \left (a\right ) + x \sinh \left (\frac{a x + b}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x),x, algorithm="fricas")

[Out]

-1/2*(b*Ei(b/x) + b*Ei(-b/x))*cosh(a) - 1/2*(b*Ei(b/x) - b*Ei(-b/x))*sinh(a) + x*sinh((a*x + b)/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x),x)

[Out]

Integral(sinh(a + b/x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x),x, algorithm="giac")

[Out]

integrate(sinh(a + b/x), x)