3.1.30 \(\int x \sinh (a+\frac {b}{x}) \, dx\) [30]

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

[Out]

1/2*b*x*cosh(a+b/x)-1/2*b^2*cosh(a)*Shi(b/x)-1/2*b^2*Chi(b/x)*sinh(a)+1/2*x^2*sinh(a+b/x)

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Rubi [A]
time = 0.08, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5428, 3378, 3384, 3379, 3382} \begin {gather*} -\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 ) \end {gather*}

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 3378

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 3379

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 3382

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]

Rule 3384

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 5428

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]))

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 54, normalized size = 0.90 \begin {gather*} \frac {1}{2} \left (-b^2 \text {Chi}\left (\frac {b}{x}\right ) \sinh (a)+x \left (b \cosh \left (a+\frac {b}{x}\right )+x \sinh \left (a+\frac {b}{x}\right )\right )-b^2 \cosh (a) \text {Shi}\left (\frac {b}{x}\right )\right ) \end {gather*}

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

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Maple [A]
time = 0.84, size = 93, normalized size = 1.55

method	result	size
risch	\(\frac {b \,{\mathrm e}^{-\frac {a x +b}{x}} x}{4}-\frac {{\mathrm e}^{-\frac {a x +b}{x}} x^{2}}{4}-\frac {b^{2} {\mathrm e}^{-a} \expIntegral \left (1, \frac {b}{x}\right )}{4}+\frac {{\mathrm e}^{\frac {a x +b}{x}} x^{2}}{4}+\frac {b \,{\mathrm e}^{\frac {a x +b}{x}} x}{4}+\frac {b^{2} {\mathrm e}^{a} \expIntegral \left (1, -\frac {b}{x}\right )}{4}\)	\(93\)
meijerg	\(-\frac {i b^{2} \sqrt {\pi }\, \cosh \left (a \right ) \left (\frac {4 i x \cosh \left (\frac {b}{x}\right )}{b \sqrt {\pi }}+\frac {4 i x^{2} \sinh \left (\frac {b}{x}\right )}{b^{2} \sqrt {\pi }}-\frac {4 i \hyperbolicSineIntegral \left (\frac {b}{x}\right )}{\sqrt {\pi }}\right )}{8}+\frac {b^{2} \sqrt {\pi }\, \sinh \left (a \right ) \left (-\frac {4 x^{2} \left (\frac {9 b^{2}}{2 x^{2}}+3\right )}{3 \sqrt {\pi }\, b^{2}}+\frac {4 x^{2} \cosh \left (\frac {b}{x}\right )}{\sqrt {\pi }\, b^{2}}+\frac {4 x \sinh \left (\frac {b}{x}\right )}{\sqrt {\pi }\, b}-\frac {4 \left (\hyperbolicCosineIntegral \left (\frac {b}{x}\right )-\ln \left (\frac {b}{x}\right )-\gamma \right )}{\sqrt {\pi }}-\frac {2 \left (2 \gamma -3-2 \ln \left (x \right )+2 \ln \left (i b \right )\right )}{\sqrt {\pi }}+\frac {4 x^{2}}{\sqrt {\pi }\, b^{2}}\right )}{8}\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(a+b/x),x,method=_RETURNVERBOSE)

[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)

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Maxima [A]
time = 0.29, size = 44, normalized size = 0.73 \begin {gather*} \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 {gather*}

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

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Fricas [A]
time = 0.37, size = 83, normalized size = 1.38 \begin {gather*} \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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sinh {\left (a + \frac {b}{x} \right )}\, dx \end {gather*}

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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (52) = 104\).
time = 0.42, size = 313, normalized size = 5.22 \begin {gather*} \frac {a^{2} b^{3} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )} - a^{2} b^{3} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a} - \frac {2 \, {\left (a x + b\right )} a b^{3} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )}}{x} + \frac {2 \, {\left (a x + b\right )} a b^{3} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a}}{x} + \frac {{\left (a x + b\right )}^{2} b^{3} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )}}{x^{2}} - \frac {{\left (a x + b\right )}^{2} b^{3} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a}}{x^{2}} - a b^{3} e^{\left (\frac {a x + b}{x}\right )} - a b^{3} e^{\left (-\frac {a x + b}{x}\right )} + b^{3} e^{\left (\frac {a x + b}{x}\right )} + \frac {{\left (a x + b\right )} b^{3} e^{\left (\frac {a x + b}{x}\right )}}{x} - b^{3} e^{\left (-\frac {a x + b}{x}\right )} + \frac {{\left (a x + b\right )} b^{3} e^{\left (-\frac {a x + b}{x}\right )}}{x}}{4 \, {\left (a^{2} - \frac {2 \, {\left (a x + b\right )} a}{x} + \frac {{\left (a x + b\right )}^{2}}{x^{2}}\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2*b^3*Ei(a - (a*x + b)/x)*e^(-a) - a^2*b^3*Ei(-a + (a*x + b)/x)*e^a - 2*(a*x + b)*a*b^3*Ei(a - (a*x + b
)/x)*e^(-a)/x + 2*(a*x + b)*a*b^3*Ei(-a + (a*x + b)/x)*e^a/x + (a*x + b)^2*b^3*Ei(a - (a*x + b)/x)*e^(-a)/x^2
- (a*x + b)^2*b^3*Ei(-a + (a*x + b)/x)*e^a/x^2 - a*b^3*e^((a*x + b)/x) - a*b^3*e^(-(a*x + b)/x) + b^3*e^((a*x
+ b)/x) + (a*x + b)*b^3*e^((a*x + b)/x)/x - b^3*e^(-(a*x + b)/x) + (a*x + b)*b^3*e^(-(a*x + b)/x)/x)/((a^2 - 2
*(a*x + b)*a/x + (a*x + b)^2/x^2)*b)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,\mathrm {sinh}\left (a+\frac {b}{x}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(a + b/x),x)

[Out]

int(x*sinh(a + b/x), x)

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