3.41 \(\int x^4 \sinh (a+\frac{b}{x^2}) \, dx\)

Optimal. Leaf size=104 \[ -\frac{2}{15} \sqrt{\pi } e^{-a} b^{5/2} \text{Erf}\left (\frac{\sqrt{b}}{x}\right )-\frac{2}{15} \sqrt{\pi } e^a b^{5/2} \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )+\frac{4}{15} b^2 x \sinh \left (a+\frac{b}{x^2}\right )+\frac{1}{5} x^5 \sinh \left (a+\frac{b}{x^2}\right )+\frac{2}{15} b x^3 \cosh \left (a+\frac{b}{x^2}\right ) \]

[Out]

(2*b*x^3*Cosh[a + b/x^2])/15 - (2*b^(5/2)*Sqrt[Pi]*Erf[Sqrt[b]/x])/(15*E^a) - (2*b^(5/2)*E^a*Sqrt[Pi]*Erfi[Sqr
t[b]/x])/15 + (4*b^2*x*Sinh[a + b/x^2])/15 + (x^5*Sinh[a + b/x^2])/5

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Rubi [A]  time = 0.0903547, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5346, 5326, 5327, 5299, 2204, 2205} \[ -\frac{2}{15} \sqrt{\pi } e^{-a} b^{5/2} \text{Erf}\left (\frac{\sqrt{b}}{x}\right )-\frac{2}{15} \sqrt{\pi } e^a b^{5/2} \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )+\frac{4}{15} b^2 x \sinh \left (a+\frac{b}{x^2}\right )+\frac{1}{5} x^5 \sinh \left (a+\frac{b}{x^2}\right )+\frac{2}{15} b x^3 \cosh \left (a+\frac{b}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sinh[a + b/x^2],x]

[Out]

(2*b*x^3*Cosh[a + b/x^2])/15 - (2*b^(5/2)*Sqrt[Pi]*Erf[Sqrt[b]/x])/(15*E^a) - (2*b^(5/2)*E^a*Sqrt[Pi]*Erfi[Sqr
t[b]/x])/15 + (4*b^2*x*Sinh[a + b/x^2])/15 + (x^5*Sinh[a + b/x^2])/5

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]

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 5299

Int[Cosh[(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 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]

Rubi steps

\begin{align*} \int x^4 \sinh \left (a+\frac{b}{x^2}\right ) \, dx &=-\operatorname{Subst}\left (\int \frac{\sinh \left (a+b x^2\right )}{x^6} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{5} x^5 \sinh \left (a+\frac{b}{x^2}\right )-\frac{1}{5} (2 b) \operatorname{Subst}\left (\int \frac{\cosh \left (a+b x^2\right )}{x^4} \, dx,x,\frac{1}{x}\right )\\ &=\frac{2}{15} b x^3 \cosh \left (a+\frac{b}{x^2}\right )+\frac{1}{5} x^5 \sinh \left (a+\frac{b}{x^2}\right )-\frac{1}{15} \left (4 b^2\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (a+b x^2\right )}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{2}{15} b x^3 \cosh \left (a+\frac{b}{x^2}\right )+\frac{4}{15} b^2 x \sinh \left (a+\frac{b}{x^2}\right )+\frac{1}{5} x^5 \sinh \left (a+\frac{b}{x^2}\right )-\frac{1}{15} \left (8 b^3\right ) \operatorname{Subst}\left (\int \cosh \left (a+b x^2\right ) \, dx,x,\frac{1}{x}\right )\\ &=\frac{2}{15} b x^3 \cosh \left (a+\frac{b}{x^2}\right )+\frac{4}{15} b^2 x \sinh \left (a+\frac{b}{x^2}\right )+\frac{1}{5} x^5 \sinh \left (a+\frac{b}{x^2}\right )-\frac{1}{15} \left (4 b^3\right ) \operatorname{Subst}\left (\int e^{-a-b x^2} \, dx,x,\frac{1}{x}\right )-\frac{1}{15} \left (4 b^3\right ) \operatorname{Subst}\left (\int e^{a+b x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{2}{15} b x^3 \cosh \left (a+\frac{b}{x^2}\right )-\frac{2}{15} b^{5/2} e^{-a} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{b}}{x}\right )-\frac{2}{15} b^{5/2} e^a \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b}}{x}\right )+\frac{4}{15} b^2 x \sinh \left (a+\frac{b}{x^2}\right )+\frac{1}{5} x^5 \sinh \left (a+\frac{b}{x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.110222, size = 102, normalized size = 0.98 \[ \frac{1}{15} \left (2 \sqrt{\pi } b^{5/2} (\sinh (a)-\cosh (a)) \text{Erf}\left (\frac{\sqrt{b}}{x}\right )-2 \sqrt{\pi } b^{5/2} (\sinh (a)+\cosh (a)) \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )+4 b^2 x \sinh \left (a+\frac{b}{x^2}\right )+3 x^5 \sinh \left (a+\frac{b}{x^2}\right )+2 b x^3 \cosh \left (a+\frac{b}{x^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sinh[a + b/x^2],x]

[Out]

(2*b*x^3*Cosh[a + b/x^2] + 2*b^(5/2)*Sqrt[Pi]*Erf[Sqrt[b]/x]*(-Cosh[a] + Sinh[a]) - 2*b^(5/2)*Sqrt[Pi]*Erfi[Sq
rt[b]/x]*(Cosh[a] + Sinh[a]) + 4*b^2*x*Sinh[a + b/x^2] + 3*x^5*Sinh[a + b/x^2])/15

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Maple [A]  time = 0.059, size = 138, normalized size = 1.3 \begin{align*} -{\frac{{{\rm e}^{-a}}{x}^{5}}{10}{{\rm e}^{-{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{-a}}b{x}^{3}}{15}{{\rm e}^{-{\frac{b}{{x}^{2}}}}}}-{\frac{2\,{{\rm e}^{-a}}\sqrt{\pi }}{15}{b}^{{\frac{5}{2}}}{\it Erf} \left ({\frac{1}{x}\sqrt{b}} \right ) }-{\frac{2\,{{\rm e}^{-a}}{b}^{2}x}{15}{{\rm e}^{-{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{a}}{x}^{5}}{10}{{\rm e}^{{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{a}}b{x}^{3}}{15}{{\rm e}^{{\frac{b}{{x}^{2}}}}}}+{\frac{2\,{{\rm e}^{a}}{b}^{2}x}{15}{{\rm e}^{{\frac{b}{{x}^{2}}}}}}-{\frac{2\,{{\rm e}^{a}}{b}^{3}\sqrt{\pi }}{15}{\it Erf} \left ({\frac{1}{x}\sqrt{-b}} \right ){\frac{1}{\sqrt{-b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*sinh(a+b/x^2),x)

[Out]

-1/10*exp(-a)*x^5*exp(-b/x^2)+1/15*exp(-a)*b*x^3*exp(-b/x^2)-2/15*exp(-a)*b^(5/2)*erf(b^(1/2)/x)*Pi^(1/2)-2/15
*exp(-a)*exp(-b/x^2)*b^2*x+1/10*exp(a)*x^5*exp(b/x^2)+1/15*exp(a)*b*exp(b/x^2)*x^3+2/15*exp(a)*b^2*exp(b/x^2)*
x-2/15*exp(a)*b^3*Pi^(1/2)/(-b)^(1/2)*erf((-b)^(1/2)/x)

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Maxima [A]  time = 1.24816, size = 84, normalized size = 0.81 \begin{align*} \frac{1}{5} \, x^{5} \sinh \left (a + \frac{b}{x^{2}}\right ) + \frac{1}{10} \,{\left (x^{3} \left (\frac{b}{x^{2}}\right )^{\frac{3}{2}} e^{\left (-a\right )} \Gamma \left (-\frac{3}{2}, \frac{b}{x^{2}}\right ) + x^{3} \left (-\frac{b}{x^{2}}\right )^{\frac{3}{2}} e^{a} \Gamma \left (-\frac{3}{2}, -\frac{b}{x^{2}}\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*sinh(a+b/x^2),x, algorithm="maxima")

[Out]

1/5*x^5*sinh(a + b/x^2) + 1/10*(x^3*(b/x^2)^(3/2)*e^(-a)*gamma(-3/2, b/x^2) + x^3*(-b/x^2)^(3/2)*e^a*gamma(-3/
2, -b/x^2))*b

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Fricas [B]  time = 1.80541, size = 795, normalized size = 7.64 \begin{align*} -\frac{3 \, x^{5} - 2 \, b x^{3} + 4 \, b^{2} x -{\left (3 \, x^{5} + 2 \, b x^{3} + 4 \, b^{2} x\right )} \cosh \left (\frac{a x^{2} + b}{x^{2}}\right )^{2} - 4 \, \sqrt{\pi }{\left (b^{2} \cosh \left (a\right ) \cosh \left (\frac{a x^{2} + b}{x^{2}}\right ) + b^{2} \cosh \left (\frac{a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) +{\left (b^{2} \cosh \left (a\right ) + b^{2} \sinh \left (a\right )\right )} \sinh \left (\frac{a x^{2} + b}{x^{2}}\right )\right )} \sqrt{-b} \operatorname{erf}\left (\frac{\sqrt{-b}}{x}\right ) + 4 \, \sqrt{\pi }{\left (b^{2} \cosh \left (a\right ) \cosh \left (\frac{a x^{2} + b}{x^{2}}\right ) - b^{2} \cosh \left (\frac{a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) +{\left (b^{2} \cosh \left (a\right ) - b^{2} \sinh \left (a\right )\right )} \sinh \left (\frac{a x^{2} + b}{x^{2}}\right )\right )} \sqrt{b} \operatorname{erf}\left (\frac{\sqrt{b}}{x}\right ) - 2 \,{\left (3 \, x^{5} + 2 \, b x^{3} + 4 \, b^{2} x\right )} \cosh \left (\frac{a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac{a x^{2} + b}{x^{2}}\right ) -{\left (3 \, x^{5} + 2 \, b x^{3} + 4 \, b^{2} x\right )} \sinh \left (\frac{a x^{2} + b}{x^{2}}\right )^{2}}{30 \,{\left (\cosh \left (\frac{a x^{2} + b}{x^{2}}\right ) + \sinh \left (\frac{a x^{2} + b}{x^{2}}\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*sinh(a+b/x^2),x, algorithm="fricas")

[Out]

-1/30*(3*x^5 - 2*b*x^3 + 4*b^2*x - (3*x^5 + 2*b*x^3 + 4*b^2*x)*cosh((a*x^2 + b)/x^2)^2 - 4*sqrt(pi)*(b^2*cosh(
a)*cosh((a*x^2 + b)/x^2) + b^2*cosh((a*x^2 + b)/x^2)*sinh(a) + (b^2*cosh(a) + b^2*sinh(a))*sinh((a*x^2 + b)/x^
2))*sqrt(-b)*erf(sqrt(-b)/x) + 4*sqrt(pi)*(b^2*cosh(a)*cosh((a*x^2 + b)/x^2) - b^2*cosh((a*x^2 + b)/x^2)*sinh(
a) + (b^2*cosh(a) - b^2*sinh(a))*sinh((a*x^2 + b)/x^2))*sqrt(b)*erf(sqrt(b)/x) - 2*(3*x^5 + 2*b*x^3 + 4*b^2*x)
*cosh((a*x^2 + b)/x^2)*sinh((a*x^2 + b)/x^2) - (3*x^5 + 2*b*x^3 + 4*b^2*x)*sinh((a*x^2 + b)/x^2)^2)/(cosh((a*x
^2 + b)/x^2) + sinh((a*x^2 + b)/x^2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \sinh{\left (a + \frac{b}{x^{2}} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*sinh(a+b/x**2),x)

[Out]

Integral(x**4*sinh(a + b/x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \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^4*sinh(a+b/x^2),x, algorithm="giac")

[Out]

integrate(x^4*sinh(a + b/x^2), x)