3.1.29 \(\int x^2 \sinh (a+\frac {b}{x}) \, dx\) [29]

Optimal. Leaf size=78 \[ \frac {1}{6} b x^2 \cosh \left (a+\frac {b}{x}\right )-\frac {1}{6} b^3 \cosh (a) \text {Chi}\left (\frac {b}{x}\right )+\frac {1}{6} b^2 x \sinh \left (a+\frac {b}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x}\right )-\frac {1}{6} b^3 \sinh (a) \text {Shi}\left (\frac {b}{x}\right ) \]

[Out]

-1/6*b^3*Chi(b/x)*cosh(a)+1/6*b*x^2*cosh(a+b/x)-1/6*b^3*Shi(b/x)*sinh(a)+1/6*b^2*x*sinh(a+b/x)+1/3*x^3*sinh(a+
b/x)

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Rubi [A]
time = 0.10, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5428, 3378, 3384, 3379, 3382} \begin {gather*} -\frac {1}{6} b^3 \cosh (a) \text {Chi}\left (\frac {b}{x}\right )-\frac {1}{6} b^3 \sinh (a) \text {Shi}\left (\frac {b}{x}\right )+\frac {1}{6} b^2 x \sinh \left (a+\frac {b}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x}\right )+\frac {1}{6} b x^2 \cosh \left (a+\frac {b}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b/x],x]

[Out]

(b*x^2*Cosh[a + b/x])/6 - (b^3*Cosh[a]*CoshIntegral[b/x])/6 + (b^2*x*Sinh[a + b/x])/6 + (x^3*Sinh[a + b/x])/3
- (b^3*Sinh[a]*SinhIntegral[b/x])/6

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^2 \sinh \left (a+\frac {b}{x}\right ) \, dx &=-\text {Subst}\left (\int \frac {\sinh (a+b x)}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x}\right )-\frac {1}{3} b \text {Subst}\left (\int \frac {\cosh (a+b x)}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{6} b x^2 \cosh \left (a+\frac {b}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x}\right )-\frac {1}{6} b^2 \text {Subst}\left (\int \frac {\sinh (a+b x)}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{6} b x^2 \cosh \left (a+\frac {b}{x}\right )+\frac {1}{6} b^2 x \sinh \left (a+\frac {b}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x}\right )-\frac {1}{6} b^3 \text {Subst}\left (\int \frac {\cosh (a+b x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{6} b x^2 \cosh \left (a+\frac {b}{x}\right )+\frac {1}{6} b^2 x \sinh \left (a+\frac {b}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x}\right )-\frac {1}{6} \left (b^3 \cosh (a)\right ) \text {Subst}\left (\int \frac {\cosh (b x)}{x} \, dx,x,\frac {1}{x}\right )-\frac {1}{6} \left (b^3 \sinh (a)\right ) \text {Subst}\left (\int \frac {\sinh (b x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{6} b x^2 \cosh \left (a+\frac {b}{x}\right )-\frac {1}{6} b^3 \cosh (a) \text {Chi}\left (\frac {b}{x}\right )+\frac {1}{6} b^2 x \sinh \left (a+\frac {b}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x}\right )-\frac {1}{6} b^3 \sinh (a) \text {Shi}\left (\frac {b}{x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 70, normalized size = 0.90 \begin {gather*} \frac {1}{6} \left (-b^3 \cosh (a) \text {Chi}\left (\frac {b}{x}\right )+x \left (b x \cosh \left (a+\frac {b}{x}\right )+b^2 \sinh \left (a+\frac {b}{x}\right )+2 x^2 \sinh \left (a+\frac {b}{x}\right )\right )-b^3 \sinh (a) \text {Shi}\left (\frac {b}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b/x],x]

[Out]

(-(b^3*Cosh[a]*CoshIntegral[b/x]) + x*(b*x*Cosh[a + b/x] + b^2*Sinh[a + b/x] + 2*x^2*Sinh[a + b/x]) - b^3*Sinh
[a]*SinhIntegral[b/x])/6

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Maple [A]
time = 0.88, size = 130, normalized size = 1.67

method	result	size
risch	\(-\frac {b^{2} {\mathrm e}^{-\frac {a x +b}{x}} x}{12}+\frac {b \,{\mathrm e}^{-\frac {a x +b}{x}} x^{2}}{12}-\frac {{\mathrm e}^{-\frac {a x +b}{x}} x^{3}}{6}+\frac {b^{3} {\mathrm e}^{-a} \expIntegral \left (1, \frac {b}{x}\right )}{12}+\frac {{\mathrm e}^{\frac {a x +b}{x}} x^{3}}{6}+\frac {b \,{\mathrm e}^{\frac {a x +b}{x}} x^{2}}{12}+\frac {b^{2} {\mathrm e}^{\frac {a x +b}{x}} x}{12}+\frac {b^{3} {\mathrm e}^{a} \expIntegral \left (1, -\frac {b}{x}\right )}{12}\)	\(130\)
meijerg	\(\frac {b^{3} \sqrt {\pi }\, \cosh \left (a \right ) \left (-\frac {8 x^{2} \left (\frac {55 b^{2}}{2 x^{2}}+45\right )}{45 \sqrt {\pi }\, b^{2}}+\frac {8 x^{2} \cosh \left (\frac {b}{x}\right )}{3 \sqrt {\pi }\, b^{2}}+\frac {16 x^{3} \left (\frac {5 b^{2}}{2 x^{2}}+5\right ) \sinh \left (\frac {b}{x}\right )}{15 \sqrt {\pi }\, b^{3}}-\frac {8 \left (\hyperbolicCosineIntegral \left (\frac {b}{x}\right )-\ln \left (\frac {b}{x}\right )-\gamma \right )}{3 \sqrt {\pi }}-\frac {4 \left (2 \gamma -\frac {11}{3}-2 \ln \left (x \right )+2 \ln \left (i b \right )\right )}{3 \sqrt {\pi }}+\frac {8 x^{2}}{\sqrt {\pi }\, b^{2}}\right )}{16}+\frac {i b^{3} \sqrt {\pi }\, \sinh \left (a \right ) \left (-\frac {8 i \left (\frac {b^{2}}{x^{2}}+2\right ) x^{3} \cosh \left (\frac {b}{x}\right )}{3 b^{3} \sqrt {\pi }}-\frac {8 i x^{2} \sinh \left (\frac {b}{x}\right )}{3 b^{2} \sqrt {\pi }}+\frac {8 i \hyperbolicSineIntegral \left (\frac {b}{x}\right )}{3 \sqrt {\pi }}\right )}{16}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*b^2*exp(-(a*x+b)/x)*x+1/12*b*exp(-(a*x+b)/x)*x^2-1/6*exp(-(a*x+b)/x)*x^3+1/12*b^3*exp(-a)*Ei(1,b/x)+1/6*
exp((a*x+b)/x)*x^3+1/12*b*exp((a*x+b)/x)*x^2+1/12*b^2*exp((a*x+b)/x)*x+1/12*b^3*exp(a)*Ei(1,-b/x)

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Maxima [A]
time = 0.31, size = 47, normalized size = 0.60 \begin {gather*} \frac {1}{3} \, x^{3} \sinh \left (a + \frac {b}{x}\right ) + \frac {1}{6} \, {\left (b^{2} e^{\left (-a\right )} \Gamma \left (-2, \frac {b}{x}\right ) + b^{2} e^{a} \Gamma \left (-2, -\frac {b}{x}\right )\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*sinh(a + b/x) + 1/6*(b^2*e^(-a)*gamma(-2, b/x) + b^2*e^a*gamma(-2, -b/x))*b

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Fricas [A]
time = 0.35, size = 93, normalized size = 1.19 \begin {gather*} \frac {1}{6} \, b x^{2} \cosh \left (\frac {a x + b}{x}\right ) - \frac {1}{12} \, {\left (b^{3} {\rm Ei}\left (\frac {b}{x}\right ) + b^{3} {\rm Ei}\left (-\frac {b}{x}\right )\right )} \cosh \left (a\right ) - \frac {1}{12} \, {\left (b^{3} {\rm Ei}\left (\frac {b}{x}\right ) - b^{3} {\rm Ei}\left (-\frac {b}{x}\right )\right )} \sinh \left (a\right ) + \frac {1}{6} \, {\left (b^{2} x + 2 \, x^{3}\right )} \sinh \left (\frac {a x + b}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*x^2*cosh((a*x + b)/x) - 1/12*(b^3*Ei(b/x) + b^3*Ei(-b/x))*cosh(a) - 1/12*(b^3*Ei(b/x) - b^3*Ei(-b/x))*si
nh(a) + 1/6*(b^2*x + 2*x^3)*sinh((a*x + b)/x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sinh {\left (a + \frac {b}{x} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sinh(a + b/x), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (68) = 136\).
time = 0.42, size = 534, normalized size = 6.85 \begin {gather*} -\frac {a^{3} b^{4} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )} + a^{3} b^{4} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a} - \frac {3 \, {\left (a x + b\right )} a^{2} b^{4} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )}}{x} - \frac {3 \, {\left (a x + b\right )} a^{2} b^{4} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a}}{x} + \frac {3 \, {\left (a x + b\right )}^{2} a b^{4} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )}}{x^{2}} + \frac {3 \, {\left (a x + b\right )}^{2} a b^{4} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a}}{x^{2}} + a^{2} b^{4} e^{\left (\frac {a x + b}{x}\right )} - a^{2} b^{4} e^{\left (-\frac {a x + b}{x}\right )} - \frac {{\left (a x + b\right )}^{3} b^{4} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )}}{x^{3}} - \frac {{\left (a x + b\right )}^{3} b^{4} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a}}{x^{3}} - a b^{4} e^{\left (\frac {a x + b}{x}\right )} - \frac {2 \, {\left (a x + b\right )} a b^{4} e^{\left (\frac {a x + b}{x}\right )}}{x} - a b^{4} e^{\left (-\frac {a x + b}{x}\right )} + \frac {2 \, {\left (a x + b\right )} a b^{4} e^{\left (-\frac {a x + b}{x}\right )}}{x} + 2 \, b^{4} e^{\left (\frac {a x + b}{x}\right )} + \frac {{\left (a x + b\right )}^{2} b^{4} e^{\left (\frac {a x + b}{x}\right )}}{x^{2}} + \frac {{\left (a x + b\right )} b^{4} e^{\left (\frac {a x + b}{x}\right )}}{x} - 2 \, b^{4} e^{\left (-\frac {a x + b}{x}\right )} - \frac {{\left (a x + b\right )}^{2} b^{4} e^{\left (-\frac {a x + b}{x}\right )}}{x^{2}} + \frac {{\left (a x + b\right )} b^{4} e^{\left (-\frac {a x + b}{x}\right )}}{x}}{12 \, {\left (a^{3} - \frac {3 \, {\left (a x + b\right )} a^{2}}{x} + \frac {3 \, {\left (a x + b\right )}^{2} a}{x^{2}} - \frac {{\left (a x + b\right )}^{3}}{x^{3}}\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(a^3*b^4*Ei(a - (a*x + b)/x)*e^(-a) + a^3*b^4*Ei(-a + (a*x + b)/x)*e^a - 3*(a*x + b)*a^2*b^4*Ei(a - (a*x
+ b)/x)*e^(-a)/x - 3*(a*x + b)*a^2*b^4*Ei(-a + (a*x + b)/x)*e^a/x + 3*(a*x + b)^2*a*b^4*Ei(a - (a*x + b)/x)*e
^(-a)/x^2 + 3*(a*x + b)^2*a*b^4*Ei(-a + (a*x + b)/x)*e^a/x^2 + a^2*b^4*e^((a*x + b)/x) - a^2*b^4*e^(-(a*x + b)
/x) - (a*x + b)^3*b^4*Ei(a - (a*x + b)/x)*e^(-a)/x^3 - (a*x + b)^3*b^4*Ei(-a + (a*x + b)/x)*e^a/x^3 - a*b^4*e^
((a*x + b)/x) - 2*(a*x + b)*a*b^4*e^((a*x + b)/x)/x - a*b^4*e^(-(a*x + b)/x) + 2*(a*x + b)*a*b^4*e^(-(a*x + b)
/x)/x + 2*b^4*e^((a*x + b)/x) + (a*x + b)^2*b^4*e^((a*x + b)/x)/x^2 + (a*x + b)*b^4*e^((a*x + b)/x)/x - 2*b^4*
e^(-(a*x + b)/x) - (a*x + b)^2*b^4*e^(-(a*x + b)/x)/x^2 + (a*x + b)*b^4*e^(-(a*x + b)/x)/x)/((a^3 - 3*(a*x + b
)*a^2/x + 3*(a*x + b)^2*a/x^2 - (a*x + b)^3/x^3)*b)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\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^2*sinh(a + b/x),x)

[Out]

int(x^2*sinh(a + b/x), x)

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