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

Optimal. Leaf size=86 \[ \frac {1}{3} \sqrt {\pi } e^{-a} b^{3/2} \text {erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {1}{3} \sqrt {\pi } e^a b^{3/2} \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+\frac {2}{3} b x \cosh \left (a+\frac {b}{x^2}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right ) \]

[Out]

2/3*b*x*cosh(a+b/x^2)+1/3*x^3*sinh(a+b/x^2)+1/3*b^(3/2)*erf(b^(1/2)/x)*Pi^(1/2)/exp(a)-1/3*b^(3/2)*exp(a)*erfi
(b^(1/2)/x)*Pi^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5346, 5326, 5327, 5298, 2204, 2205} \[ \frac {1}{3} \sqrt {\pi } e^{-a} b^{3/2} \text {Erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {1}{3} \sqrt {\pi } e^a b^{3/2} \text {Erfi}\left (\frac {\sqrt {b}}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )+\frac {2}{3} b x \cosh \left (a+\frac {b}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b/x^2],x]

[Out]

(2*b*x*Cosh[a + b/x^2])/3 + (b^(3/2)*Sqrt[Pi]*Erf[Sqrt[b]/x])/(3*E^a) - (b^(3/2)*E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])
/3 + (x^3*Sinh[a + b/x^2])/3

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 5298

Int[Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^n), x], x] - Dist[1/2, Int[E^(-c - d*
x^n), x], x] /; FreeQ[{c, d}, x] && IGtQ[n, 1]

Rule 5326

Int[((e_.)*(x_))^(m_)*Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[((e*x)^(m + 1)*Sinh[c + d*x^n])/(e*(m +
1)), x] - Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Cosh[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[
n, 0] && LtQ[m, -1]

Rule 5327

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[((e*x)^(m + 1)*Cosh[c + d*x^n])/(e*(m +
1)), x] - Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Sinh[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[
n, 0] && LtQ[m, -1]

Rule 5346

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sinh[c + d/
x^n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IntegerQ[p] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx &=-\operatorname {Subst}\left (\int \frac {\sinh \left (a+b x^2\right )}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{3} (2 b) \operatorname {Subst}\left (\int \frac {\cosh \left (a+b x^2\right )}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{3} b x \cosh \left (a+\frac {b}{x^2}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{3} \left (4 b^2\right ) \operatorname {Subst}\left (\int \sinh \left (a+b x^2\right ) \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{3} b x \cosh \left (a+\frac {b}{x^2}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )+\frac {1}{3} \left (2 b^2\right ) \operatorname {Subst}\left (\int e^{-a-b x^2} \, dx,x,\frac {1}{x}\right )-\frac {1}{3} \left (2 b^2\right ) \operatorname {Subst}\left (\int e^{a+b x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2}{3} b x \cosh \left (a+\frac {b}{x^2}\right )+\frac {1}{3} b^{3/2} e^{-a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {1}{3} b^{3/2} e^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 84, normalized size = 0.98 \[ \frac {1}{3} \left (\sqrt {\pi } b^{3/2} (\cosh (a)-\sinh (a)) \text {erf}\left (\frac {\sqrt {b}}{x}\right )-\sqrt {\pi } b^{3/2} (\sinh (a)+\cosh (a)) \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+2 b x \cosh \left (a+\frac {b}{x^2}\right )+x^3 \sinh \left (a+\frac {b}{x^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b/x^2],x]

[Out]

(2*b*x*Cosh[a + b/x^2] + b^(3/2)*Sqrt[Pi]*Erf[Sqrt[b]/x]*(Cosh[a] - Sinh[a]) - b^(3/2)*Sqrt[Pi]*Erfi[Sqrt[b]/x
]*(Cosh[a] + Sinh[a]) + x^3*Sinh[a + b/x^2])/3

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fricas [B]  time = 0.43, size = 267, normalized size = 3.10 \[ -\frac {x^{3} - {\left (x^{3} + 2 \, b x\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 2 \, \sqrt {\pi } {\left (b \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (b \cosh \relax (a) + b \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {-b} \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right ) - 2 \, \sqrt {\pi } {\left (b \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (b \cosh \relax (a) - b \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {b} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) - 2 \, {\left (x^{3} + 2 \, b x\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) - {\left (x^{3} + 2 \, b x\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 2 \, b x}{6 \, {\left (\cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(x^3 - (x^3 + 2*b*x)*cosh((a*x^2 + b)/x^2)^2 - 2*sqrt(pi)*(b*cosh(a)*cosh((a*x^2 + b)/x^2) + b*cosh((a*x^
2 + b)/x^2)*sinh(a) + (b*cosh(a) + b*sinh(a))*sinh((a*x^2 + b)/x^2))*sqrt(-b)*erf(sqrt(-b)/x) - 2*sqrt(pi)*(b*
cosh(a)*cosh((a*x^2 + b)/x^2) - b*cosh((a*x^2 + b)/x^2)*sinh(a) + (b*cosh(a) - b*sinh(a))*sinh((a*x^2 + b)/x^2
))*sqrt(b)*erf(sqrt(b)/x) - 2*(x^3 + 2*b*x)*cosh((a*x^2 + b)/x^2)*sinh((a*x^2 + b)/x^2) - (x^3 + 2*b*x)*sinh((
a*x^2 + b)/x^2)^2 - 2*b*x)/(cosh((a*x^2 + b)/x^2) + sinh((a*x^2 + b)/x^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*sinh(a + b/x^2), x)

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maple [A]  time = 0.05, size = 103, normalized size = 1.20 \[ -\frac {{\mathrm e}^{-a} x^{3} {\mathrm e}^{-\frac {b}{x^{2}}}}{6}+\frac {{\mathrm e}^{-a} \erf \left (\frac {\sqrt {b}}{x}\right ) b^{\frac {3}{2}} \sqrt {\pi }}{3}+\frac {{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}} b x}{3}+\frac {{\mathrm e}^{a} x^{3} {\mathrm e}^{\frac {b}{x^{2}}}}{6}+\frac {{\mathrm e}^{a} b \,{\mathrm e}^{\frac {b}{x^{2}}} x}{3}-\frac {{\mathrm e}^{a} b^{2} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b}}{x}\right )}{3 \sqrt {-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*exp(-a)*x^3*exp(-b/x^2)+1/3*exp(-a)*erf(b^(1/2)/x)*b^(3/2)*Pi^(1/2)+1/3*exp(-a)*exp(-b/x^2)*b*x+1/6*exp(a
)*x^3*exp(b/x^2)+1/3*exp(a)*b*exp(b/x^2)*x-1/3*exp(a)*b^2*Pi^(1/2)/(-b)^(1/2)*erf((-b)^(1/2)/x)

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maxima [A]  time = 0.39, size = 58, normalized size = 0.67 \[ \frac {1}{3} \, x^{3} \sinh \left (a + \frac {b}{x^{2}}\right ) + \frac {1}{6} \, {\left (x \sqrt {\frac {b}{x^{2}}} e^{\left (-a\right )} \Gamma \left (-\frac {1}{2}, \frac {b}{x^{2}}\right ) + x \sqrt {-\frac {b}{x^{2}}} e^{a} \Gamma \left (-\frac {1}{2}, -\frac {b}{x^{2}}\right )\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {sinh}\left (a+\frac {b}{x^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sinh(a + b/x^2),x)

[Out]

int(x^2*sinh(a + b/x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sinh(a + b/x**2), x)

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