3.29 \(\int x^2 \sinh (a+\frac{b}{x}) \, dx\)

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

[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

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Rubi [A]  time = 0.143453, antiderivative size = 78, normalized size of antiderivative = 1., 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 = {5320, 3297, 3303, 3298, 3301} \[ -\frac{1}{6} b^3 \cosh (a) \text{Chi}\left (\frac{b}{x}\right )-\frac{1}{6} b^3 \sinh (a) \text{Shi}\left (\frac{b}{x}\right )+\frac{1}{6} b^2 x \sinh \left (a+\frac{b}{x}\right )+\frac{1}{3} x^3 \sinh \left (a+\frac{b}{x}\right )+\frac{1}{6} b x^2 \cosh \left (a+\frac{b}{x}\right ) \]

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 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^2 \sinh \left (a+\frac{b}{x}\right ) \, dx &=-\operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 ) \operatorname{Subst}\left (\int \frac{\cosh (b x)}{x} \, dx,x,\frac{1}{x}\right )-\frac{1}{6} \left (b^3 \sinh (a)\right ) \operatorname{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*}

Mathematica [A]  time = 0.0649958, size = 70, normalized size = 0.9 \[ \frac{1}{6} \left (b^3 (-\cosh (a)) \text{Chi}\left (\frac{b}{x}\right )-b^3 \sinh (a) \text{Shi}\left (\frac{b}{x}\right )+x \left (b^2 \sinh \left (a+\frac{b}{x}\right )+2 x^2 \sinh \left (a+\frac{b}{x}\right )+b x \cosh \left (a+\frac{b}{x}\right )\right )\right ) \]

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 = 0.036, size = 130, normalized size = 1.7 \begin{align*} -{\frac{{b}^{2}x}{12}{{\rm e}^{-{\frac{ax+b}{x}}}}}+{\frac{b{x}^{2}}{12}{{\rm e}^{-{\frac{ax+b}{x}}}}}-{\frac{{x}^{3}}{6}{{\rm e}^{-{\frac{ax+b}{x}}}}}+{\frac{{b}^{3}{{\rm e}^{-a}}}{12}{\it Ei} \left ( 1,{\frac{b}{x}} \right ) }+{\frac{{x}^{3}}{6}{{\rm e}^{{\frac{ax+b}{x}}}}}+{\frac{b{x}^{2}}{12}{{\rm e}^{{\frac{ax+b}{x}}}}}+{\frac{{b}^{2}x}{12}{{\rm e}^{{\frac{ax+b}{x}}}}}+{\frac{{b}^{3}{{\rm e}^{a}}}{12}{\it Ei} \left ( 1,-{\frac{b}{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 1.16563, size = 63, normalized size = 0.81 \begin{align*} \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{align*}

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 = 1.7202, size = 212, normalized size = 2.72 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sinh{\left (a + \frac{b}{x} \right )}\, dx \end{align*}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*sinh(a + b/x), x)