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

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

[Out]

1/4*b*x^2*cosh(a+b/x^2)-1/4*b^2*cosh(a)*Shi(b/x^2)-1/4*b^2*Chi(b/x^2)*sinh(a)+1/4*x^4*sinh(a+b/x^2)

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Rubi [A]  time = 0.11, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 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 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]

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 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]))

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*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.90 \[ \frac {1}{4} \left (-b^2 \sinh (a) \text {Chi}\left (\frac {b}{x^2}\right )-b^2 \cosh (a) \text {Shi}\left (\frac {b}{x^2}\right )+b x^2 \cosh \left (a+\frac {b}{x^2}\right )+x^4 \sinh \left (a+\frac {b}{x^2}\right )\right ) \]

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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fricas [A]  time = 0.43, size = 89, normalized size = 1.44 \[ \frac {1}{4} \, x^{4} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) + \frac {1}{4} \, b x^{2} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - \frac {1}{8} \, {\left (b^{2} {\rm Ei}\left (\frac {b}{x^{2}}\right ) - b^{2} {\rm Ei}\left (-\frac {b}{x^{2}}\right )\right )} \cosh \relax (a) - \frac {1}{8} \, {\left (b^{2} {\rm Ei}\left (\frac {b}{x^{2}}\right ) + b^{2} {\rm Ei}\left (-\frac {b}{x^{2}}\right )\right )} \sinh \relax (a) \]

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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giac [B]  time = 0.20, size = 353, normalized size = 5.69 \[ \frac {a^{2} b^{3} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )} - a^{2} b^{3} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a} - \frac {2 \, {\left (a x^{2} + b\right )} a b^{3} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )}}{x^{2}} + \frac {2 \, {\left (a x^{2} + b\right )} a b^{3} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a}}{x^{2}} - a b^{3} e^{\left (\frac {a x^{2} + b}{x^{2}}\right )} - a b^{3} e^{\left (-\frac {a x^{2} + b}{x^{2}}\right )} + b^{3} e^{\left (\frac {a x^{2} + b}{x^{2}}\right )} - b^{3} e^{\left (-\frac {a x^{2} + b}{x^{2}}\right )} + \frac {{\left (a x^{2} + b\right )}^{2} b^{3} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )}}{x^{4}} - \frac {{\left (a x^{2} + b\right )}^{2} b^{3} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a}}{x^{4}} + \frac {{\left (a x^{2} + b\right )} b^{3} e^{\left (\frac {a x^{2} + b}{x^{2}}\right )}}{x^{2}} + \frac {{\left (a x^{2} + b\right )} b^{3} e^{\left (-\frac {a x^{2} + b}{x^{2}}\right )}}{x^{2}}}{8 \, {\left (a^{2} - \frac {2 \, {\left (a x^{2} + b\right )} a}{x^{2}} + \frac {{\left (a x^{2} + b\right )}^{2}}{x^{4}}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2*b^3*Ei(a - (a*x^2 + b)/x^2)*e^(-a) - a^2*b^3*Ei(-a + (a*x^2 + b)/x^2)*e^a - 2*(a*x^2 + b)*a*b^3*Ei(a
- (a*x^2 + b)/x^2)*e^(-a)/x^2 + 2*(a*x^2 + b)*a*b^3*Ei(-a + (a*x^2 + b)/x^2)*e^a/x^2 - a*b^3*e^((a*x^2 + b)/x^
2) - a*b^3*e^(-(a*x^2 + b)/x^2) + b^3*e^((a*x^2 + b)/x^2) - b^3*e^(-(a*x^2 + b)/x^2) + (a*x^2 + b)^2*b^3*Ei(a
- (a*x^2 + b)/x^2)*e^(-a)/x^4 - (a*x^2 + b)^2*b^3*Ei(-a + (a*x^2 + b)/x^2)*e^a/x^4 + (a*x^2 + b)*b^3*e^((a*x^2
+ b)/x^2)/x^2 + (a*x^2 + b)*b^3*e^(-(a*x^2 + b)/x^2)/x^2)/((a^2 - 2*(a*x^2 + b)*a/x^2 + (a*x^2 + b)^2/x^4)*b)

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maple [A]  time = 0.04, size = 93, normalized size = 1.50 \[ -\frac {{\mathrm e}^{-a} x^{4} {\mathrm e}^{-\frac {b}{x^{2}}}}{8}+\frac {{\mathrm e}^{-a} b \,x^{2} {\mathrm e}^{-\frac {b}{x^{2}}}}{8}-\frac {{\mathrm e}^{-a} b^{2} \Ei \left (1, \frac {b}{x^{2}}\right )}{8}+\frac {{\mathrm e}^{a} x^{4} {\mathrm e}^{\frac {b}{x^{2}}}}{8}+\frac {{\mathrm e}^{a} b \,{\mathrm e}^{\frac {b}{x^{2}}} x^{2}}{8}+\frac {{\mathrm e}^{a} b^{2} \Ei \left (1, -\frac {b}{x^{2}}\right )}{8} \]

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 = 0.51, size = 44, normalized size = 0.71 \[ \frac {1}{4} \, x^{4} \sinh \left (a + \frac {b}{x^{2}}\right ) + \frac {1}{8} \, {\left (b e^{\left (-a\right )} \Gamma \left (-1, \frac {b}{x^{2}}\right ) - b e^{a} \Gamma \left (-1, -\frac {b}{x^{2}}\right )\right )} b \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^3\,\mathrm {sinh}\left (a+\frac {b}{x^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinh(a + b/x^2),x)

[Out]

int(x^3*sinh(a + b/x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sinh {\left (a + \frac {b}{x^{2}} \right )}\, dx \]

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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