3.1.44 \(\int x \sinh (a+\frac {b}{x^2}) \, dx\) [44]

Optimal. Leaf size=42 \[ -\frac {1}{2} b \cosh (a) \text {Chi}\left (\frac {b}{x^2}\right )+\frac {1}{2} x^2 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{2} b \sinh (a) \text {Shi}\left (\frac {b}{x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, 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 \cosh (a) \text {Chi}\left (\frac {b}{x^2}\right )-\frac {1}{2} b \sinh (a) \text {Shi}\left (\frac {b}{x^2}\right )+\frac {1}{2} x^2 \sinh \left (a+\frac {b}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b/x^2],x]

[Out]

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

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Mathematica [A]
time = 0.02, size = 39, normalized size = 0.93 \begin {gather*} \frac {1}{2} \left (-b \cosh (a) \text {Chi}\left (\frac {b}{x^2}\right )+x^2 \sinh \left (a+\frac {b}{x^2}\right )-b \sinh (a) \text {Shi}\left (\frac {b}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b/x^2],x]

[Out]

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

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Maple [A]
time = 0.29, size = 58, normalized size = 1.38

method	result	size
risch	\(-\frac {{\mathrm e}^{-a} x^{2} {\mathrm e}^{-\frac {b}{x^{2}}}}{4}+\frac {{\mathrm e}^{-a} b \expIntegral \left (1, \frac {b}{x^{2}}\right )}{4}+\frac {{\mathrm e}^{a} {\mathrm e}^{\frac {b}{x^{2}}} x^{2}}{4}+\frac {{\mathrm e}^{a} b \expIntegral \left (1, -\frac {b}{x^{2}}\right )}{4}\)	\(58\)
meijerg	\(-\frac {b \sqrt {\pi }\, \cosh \left (a \right ) \left (\frac {4}{\sqrt {\pi }}-\frac {4 x^{2} \sinh \left (\frac {b}{x^{2}}\right )}{\sqrt {\pi }\, b}+\frac {4 \hyperbolicCosineIntegral \left (\frac {b}{x^{2}}\right )-4 \ln \left (\frac {b}{x^{2}}\right )-4 \gamma }{\sqrt {\pi }}+\frac {4 \gamma -4-8 \ln \left (x \right )+4 \ln \left (i b \right )}{\sqrt {\pi }}\right )}{8}-\frac {i b \sqrt {\pi }\, \sinh \left (a \right ) \left (\frac {4 i x^{2} \cosh \left (\frac {b}{x^{2}}\right )}{b \sqrt {\pi }}-\frac {4 i \hyperbolicSineIntegral \left (\frac {b}{x^{2}}\right )}{\sqrt {\pi }}\right )}{8}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(a+b/x^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 39, normalized size = 0.93 \begin {gather*} \frac {1}{2} \, x^{2} \sinh \left (a + \frac {b}{x^{2}}\right ) - \frac {1}{4} \, {\left ({\rm Ei}\left (-\frac {b}{x^{2}}\right ) e^{\left (-a\right )} + {\rm Ei}\left (\frac {b}{x^{2}}\right ) e^{a}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 63, normalized size = 1.50 \begin {gather*} \frac {1}{2} \, x^{2} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) - \frac {1}{4} \, {\left (b {\rm Ei}\left (\frac {b}{x^{2}}\right ) + b {\rm Ei}\left (-\frac {b}{x^{2}}\right )\right )} \cosh \left (a\right ) - \frac {1}{4} \, {\left (b {\rm Ei}\left (\frac {b}{x^{2}}\right ) - b {\rm Ei}\left (-\frac {b}{x^{2}}\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^2),x, algorithm="fricas")

[Out]

1/2*x^2*sinh((a*x^2 + b)/x^2) - 1/4*(b*Ei(b/x^2) + b*Ei(-b/x^2))*cosh(a) - 1/4*(b*Ei(b/x^2) - b*Ei(-b/x^2))*si
nh(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^{2}} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(a+b/x**2),x)

[Out]

Integral(x*sinh(a + b/x**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (36) = 72\).
time = 0.43, size = 193, normalized size = 4.60 \begin {gather*} -\frac {a b^{2} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )} - \frac {{\left (a x^{2} + b\right )} b^{2} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )}}{x^{2}} - b^{2} e^{\left (-\frac {a x^{2} + b}{x^{2}}\right )}}{4 \, {\left (a - \frac {a x^{2} + b}{x^{2}}\right )} b} - \frac {a b^{2} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a} - \frac {{\left (a x^{2} + b\right )} b^{2} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a}}{x^{2}} + b^{2} e^{\left (\frac {a x^{2} + b}{x^{2}}\right )}}{4 \, {\left (a - \frac {a x^{2} + b}{x^{2}}\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*b^2*Ei(a - (a*x^2 + b)/x^2)*e^(-a) - (a*x^2 + b)*b^2*Ei(a - (a*x^2 + b)/x^2)*e^(-a)/x^2 - b^2*e^(-(a*x
^2 + b)/x^2))/((a - (a*x^2 + b)/x^2)*b) - 1/4*(a*b^2*Ei(-a + (a*x^2 + b)/x^2)*e^a - (a*x^2 + b)*b^2*Ei(-a + (a
*x^2 + b)/x^2)*e^a/x^2 + b^2*e^((a*x^2 + b)/x^2))/((a - (a*x^2 + b)/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^2}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(a + b/x^2),x)

[Out]

int(x*sinh(a + b/x^2), x)

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