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

Optimal. Leaf size=60 \[ -\frac{1}{2} b^2 \sinh (a) \text{Chi}\left (\frac{b}{x}\right )-\frac{1}{2} b^2 \cosh (a) \text{Shi}\left (\frac{b}{x}\right )+\frac{1}{2} x^2 \sinh \left (a+\frac{b}{x}\right )+\frac{1}{2} b x \cosh \left (a+\frac{b}{x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.107096, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5320, 3297, 3303, 3298, 3301} \[ -\frac{1}{2} b^2 \sinh (a) \text{Chi}\left (\frac{b}{x}\right )-\frac{1}{2} b^2 \cosh (a) \text{Shi}\left (\frac{b}{x}\right )+\frac{1}{2} x^2 \sinh \left (a+\frac{b}{x}\right )+\frac{1}{2} b x \cosh \left (a+\frac{b}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b/x],x]

[Out]

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

Rule 5320

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sinh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 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 x \sinh \left (a+\frac{b}{x}\right ) \, dx &=-\operatorname{Subst}\left (\int \frac{\sinh (a+b x)}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} x^2 \sinh \left (a+\frac{b}{x}\right )-\frac{1}{2} b \operatorname{Subst}\left (\int \frac{\cosh (a+b x)}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} b x \cosh \left (a+\frac{b}{x}\right )+\frac{1}{2} x^2 \sinh \left (a+\frac{b}{x}\right )-\frac{1}{2} b^2 \operatorname{Subst}\left (\int \frac{\sinh (a+b x)}{x} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} b x \cosh \left (a+\frac{b}{x}\right )+\frac{1}{2} x^2 \sinh \left (a+\frac{b}{x}\right )-\frac{1}{2} \left (b^2 \cosh (a)\right ) \operatorname{Subst}\left (\int \frac{\sinh (b x)}{x} \, dx,x,\frac{1}{x}\right )-\frac{1}{2} \left (b^2 \sinh (a)\right ) \operatorname{Subst}\left (\int \frac{\cosh (b x)}{x} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} b x \cosh \left (a+\frac{b}{x}\right )-\frac{1}{2} b^2 \text{Chi}\left (\frac{b}{x}\right ) \sinh (a)+\frac{1}{2} x^2 \sinh \left (a+\frac{b}{x}\right )-\frac{1}{2} b^2 \cosh (a) \text{Shi}\left (\frac{b}{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0440879, size = 54, normalized size = 0.9 \[ \frac{1}{2} \left (b^2 \sinh (a) \left (-\text{Chi}\left (\frac{b}{x}\right )\right )-b^2 \cosh (a) \text{Shi}\left (\frac{b}{x}\right )+x \left (x \sinh \left (a+\frac{b}{x}\right )+b \cosh \left (a+\frac{b}{x}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b/x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.03, size = 93, normalized size = 1.6 \begin{align*}{\frac{bx}{4}{{\rm e}^{-{\frac{ax+b}{x}}}}}-{\frac{{x}^{2}}{4}{{\rm e}^{-{\frac{ax+b}{x}}}}}-{\frac{{b}^{2}{{\rm e}^{-a}}}{4}{\it Ei} \left ( 1,{\frac{b}{x}} \right ) }+{\frac{{x}^{2}}{4}{{\rm e}^{{\frac{ax+b}{x}}}}}+{\frac{bx}{4}{{\rm e}^{{\frac{ax+b}{x}}}}}+{\frac{{b}^{2}{{\rm e}^{a}}}{4}{\it Ei} \left ( 1,-{\frac{b}{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.15938, size = 59, normalized size = 0.98 \begin{align*} \frac{1}{2} \, x^{2} \sinh \left (a + \frac{b}{x}\right ) + \frac{1}{4} \,{\left (b e^{\left (-a\right )} \Gamma \left (-1, \frac{b}{x}\right ) - b e^{a} \Gamma \left (-1, -\frac{b}{x}\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*sinh(a + b/x) + 1/4*(b*e^(-a)*gamma(-1, b/x) - b*e^a*gamma(-1, -b/x))*b

________________________________________________________________________________________

Fricas [A]  time = 1.72197, size = 190, normalized size = 3.17 \begin{align*} \frac{1}{2} \, b x \cosh \left (\frac{a x + b}{x}\right ) + \frac{1}{2} \, x^{2} \sinh \left (\frac{a x + b}{x}\right ) - \frac{1}{4} \,{\left (b^{2}{\rm Ei}\left (\frac{b}{x}\right ) - b^{2}{\rm Ei}\left (-\frac{b}{x}\right )\right )} \cosh \left (a\right ) - \frac{1}{4} \,{\left (b^{2}{\rm Ei}\left (\frac{b}{x}\right ) + b^{2}{\rm Ei}\left (-\frac{b}{x}\right )\right )} \sinh \left (a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x*cosh((a*x + b)/x) + 1/2*x^2*sinh((a*x + b)/x) - 1/4*(b^2*Ei(b/x) - b^2*Ei(-b/x))*cosh(a) - 1/4*(b^2*Ei
(b/x) + b^2*Ei(-b/x))*sinh(a)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sinh(a + b/x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sinh(a + b/x), x)