3.42 \(\int x^3 \sinh (a+\frac{b}{x^2}) \, dx\)

Optimal. Leaf size=62 \[ -\frac{1}{4} b^2 \sinh (a) \text{Chi}\left (\frac{b}{x^2}\right )-\frac{1}{4} b^2 \cosh (a) \text{Shi}\left (\frac{b}{x^2}\right )+\frac{1}{4} x^4 \sinh \left (a+\frac{b}{x^2}\right )+\frac{1}{4} b x^2 \cosh \left (a+\frac{b}{x^2}\right ) \]

[Out]

(b*x^2*Cosh[a + b/x^2])/4 - (b^2*CoshIntegral[b/x^2]*Sinh[a])/4 + (x^4*Sinh[a + b/x^2])/4 - (b^2*Cosh[a]*SinhI
ntegral[b/x^2])/4

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Rubi [A]  time = 0.113255, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{1}{4} b^2 \sinh (a) \text{Chi}\left (\frac{b}{x^2}\right )-\frac{1}{4} b^2 \cosh (a) \text{Shi}\left (\frac{b}{x^2}\right )+\frac{1}{4} x^4 \sinh \left (a+\frac{b}{x^2}\right )+\frac{1}{4} b x^2 \cosh \left (a+\frac{b}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sinh[a + b/x^2],x]

[Out]

(b*x^2*Cosh[a + b/x^2])/4 - (b^2*CoshIntegral[b/x^2]*Sinh[a])/4 + (x^4*Sinh[a + b/x^2])/4 - (b^2*Cosh[a]*SinhI
ntegral[b/x^2])/4

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

Mathematica [A]  time = 0.0415229, size = 56, normalized size = 0.9 \[ \frac{1}{4} \left (-b^2 \sinh (a) \text{Chi}\left (\frac{b}{x^2}\right )-b^2 \cosh (a) \text{Shi}\left (\frac{b}{x^2}\right )+x^4 \sinh \left (a+\frac{b}{x^2}\right )+b x^2 \cosh \left (a+\frac{b}{x^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sinh[a + b/x^2],x]

[Out]

(b*x^2*Cosh[a + b/x^2] - b^2*CoshIntegral[b/x^2]*Sinh[a] + x^4*Sinh[a + b/x^2] - b^2*Cosh[a]*SinhIntegral[b/x^
2])/4

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Maple [A]  time = 0.031, size = 93, normalized size = 1.5 \begin{align*} -{\frac{{{\rm e}^{-a}}{x}^{4}}{8}{{\rm e}^{-{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{-a}}b{x}^{2}}{8}{{\rm e}^{-{\frac{b}{{x}^{2}}}}}}-{\frac{{{\rm e}^{-a}}{b}^{2}}{8}{\it Ei} \left ( 1,{\frac{b}{{x}^{2}}} \right ) }+{\frac{{{\rm e}^{a}}{x}^{4}}{8}{{\rm e}^{{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{a}}b{x}^{2}}{8}{{\rm e}^{{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{a}}{b}^{2}}{8}{\it Ei} \left ( 1,-{\frac{b}{{x}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinh(a+b/x^2),x)

[Out]

-1/8*exp(-a)*x^4*exp(-b/x^2)+1/8*exp(-a)*b*x^2*exp(-b/x^2)-1/8*exp(-a)*b^2*Ei(1,b/x^2)+1/8*exp(a)*x^4*exp(b/x^
2)+1/8*exp(a)*b*exp(b/x^2)*x^2+1/8*exp(a)*b^2*Ei(1,-b/x^2)

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Maxima [A]  time = 1.24101, size = 59, normalized size = 0.95 \begin{align*} \frac{1}{4} \, x^{4} \sinh \left (a + \frac{b}{x^{2}}\right ) + \frac{1}{8} \,{\left (b e^{\left (-a\right )} \Gamma \left (-1, \frac{b}{x^{2}}\right ) - b e^{a} \Gamma \left (-1, -\frac{b}{x^{2}}\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(a+b/x^2),x, algorithm="maxima")

[Out]

1/4*x^4*sinh(a + b/x^2) + 1/8*(b*e^(-a)*gamma(-1, b/x^2) - b*e^a*gamma(-1, -b/x^2))*b

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Fricas [A]  time = 1.77274, size = 215, normalized size = 3.47 \begin{align*} \frac{1}{4} \, x^{4} \sinh \left (\frac{a x^{2} + b}{x^{2}}\right ) + \frac{1}{4} \, b x^{2} \cosh \left (\frac{a x^{2} + b}{x^{2}}\right ) - \frac{1}{8} \,{\left (b^{2}{\rm Ei}\left (\frac{b}{x^{2}}\right ) - b^{2}{\rm Ei}\left (-\frac{b}{x^{2}}\right )\right )} \cosh \left (a\right ) - \frac{1}{8} \,{\left (b^{2}{\rm Ei}\left (\frac{b}{x^{2}}\right ) + b^{2}{\rm Ei}\left (-\frac{b}{x^{2}}\right )\right )} \sinh \left (a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(a+b/x^2),x, algorithm="fricas")

[Out]

1/4*x^4*sinh((a*x^2 + b)/x^2) + 1/4*b*x^2*cosh((a*x^2 + b)/x^2) - 1/8*(b^2*Ei(b/x^2) - b^2*Ei(-b/x^2))*cosh(a)
- 1/8*(b^2*Ei(b/x^2) + b^2*Ei(-b/x^2))*sinh(a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sinh{\left (a + \frac{b}{x^{2}} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*sinh(a+b/x**2),x)

[Out]

Integral(x**3*sinh(a + b/x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sinh \left (a + \frac{b}{x^{2}}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(a+b/x^2),x, algorithm="giac")

[Out]

integrate(x^3*sinh(a + b/x^2), x)