3.1.41 \(\int x^4 \sinh (a+\frac {b}{x^2}) \, dx\) [41]

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

[Out]

2/15*b*x^3*cosh(a+b/x^2)+4/15*b^2*x*sinh(a+b/x^2)+1/5*x^5*sinh(a+b/x^2)-2/15*b^(5/2)*erf(b^(1/2)/x)*Pi^(1/2)/e
xp(a)-2/15*b^(5/2)*exp(a)*erfi(b^(1/2)/x)*Pi^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5454, 5434, 5435, 5407, 2235, 2236} \begin {gather*} -\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 ) \end {gather*}

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 2235

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 2236

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 5407

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 5434

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 5435

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 5454

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^4 \sinh \left (a+\frac {b}{x^2}\right ) \, dx &=-\text {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) \text {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 ) \text {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 ) \text {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 ) \text {Subst}\left (\int e^{-a-b x^2} \, dx,x,\frac {1}{x}\right )-\frac {1}{15} \left (4 b^3\right ) \text {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*}

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Mathematica [A]
time = 0.08, size = 102, normalized size = 0.98 \begin {gather*} \frac {1}{15} \left (2 b x^3 \cosh \left (a+\frac {b}{x^2}\right )+2 b^{5/2} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {b}}{x}\right ) (-\cosh (a)+\sinh (a))-2 b^{5/2} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)+\sinh (a))+4 b^2 x \sinh \left (a+\frac {b}{x^2}\right )+3 x^5 \sinh \left (a+\frac {b}{x^2}\right )\right ) \end {gather*}

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 = 0.48, size = 138, normalized size = 1.33

method	result	size
risch	\(-\frac {{\mathrm e}^{-a} x^{5} {\mathrm e}^{-\frac {b}{x^{2}}}}{10}+\frac {{\mathrm e}^{-a} b \,x^{3} {\mathrm e}^{-\frac {b}{x^{2}}}}{15}-\frac {2 \,{\mathrm e}^{-a} b^{\frac {5}{2}} \sqrt {\pi }\, \erf \left (\frac {\sqrt {b}}{x}\right )}{15}-\frac {2 \,{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}} b^{2} x}{15}+\frac {{\mathrm e}^{a} x^{5} {\mathrm e}^{\frac {b}{x^{2}}}}{10}+\frac {{\mathrm e}^{a} b \,x^{3} {\mathrm e}^{\frac {b}{x^{2}}}}{15}+\frac {2 \,{\mathrm e}^{a} b^{2} {\mathrm e}^{\frac {b}{x^{2}}} x}{15}-\frac {2 \,{\mathrm e}^{a} b^{3} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b}}{x}\right )}{15 \sqrt {-b}}\)	\(138\)
meijerg	\(-\frac {i b^{2} \sqrt {\pi }\, \cosh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (\frac {8 x^{5} \sqrt {2}\, \left (\frac {4 b^{2}}{x^{4}}-\frac {2 b}{x^{2}}+3\right ) {\mathrm e}^{-\frac {b}{x^{2}}}}{15 \sqrt {\pi }\, \left (i b \right )^{\frac {3}{2}} b}-\frac {8 x^{5} \sqrt {2}\, \left (\frac {4 b^{2}}{x^{4}}+\frac {2 b}{x^{2}}+3\right ) {\mathrm e}^{\frac {b}{x^{2}}}}{15 \sqrt {\pi }\, \left (i b \right )^{\frac {3}{2}} b}+\frac {32 \sqrt {2}\, b^{\frac {3}{2}} \erf \left (\frac {\sqrt {b}}{x}\right )}{15 \left (i b \right )^{\frac {3}{2}}}+\frac {32 \sqrt {2}\, b^{\frac {3}{2}} \erfi \left (\frac {\sqrt {b}}{x}\right )}{15 \left (i b \right )^{\frac {3}{2}}}\right )}{32}+\frac {b^{2} \sqrt {\pi }\, \sinh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (-\frac {16 x^{5} \sqrt {2}\, \left (\frac {2 b^{2}}{3 x^{4}}+\frac {b}{3 x^{2}}+\frac {1}{2}\right ) {\mathrm e}^{\frac {b}{x^{2}}}}{5 \sqrt {\pi }\, \left (i b \right )^{\frac {5}{2}}}-\frac {16 x^{5} \sqrt {2}\, \left (\frac {2 b^{2}}{3 x^{4}}-\frac {b}{3 x^{2}}+\frac {1}{2}\right ) {\mathrm e}^{-\frac {b}{x^{2}}}}{5 \sqrt {\pi }\, \left (i b \right )^{\frac {5}{2}}}-\frac {32 \sqrt {2}\, b^{\frac {5}{2}} \erf \left (\frac {\sqrt {b}}{x}\right )}{15 \left (i b \right )^{\frac {5}{2}}}+\frac {32 \sqrt {2}\, b^{\frac {5}{2}} \erfi \left (\frac {\sqrt {b}}{x}\right )}{15 \left (i b \right )^{\frac {5}{2}}}\right )}{32}\)	\(295\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*Pi^(1/2)*erf(b^(1/2)/x)-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*x^3*exp(b/x^2)+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 = 0.34, size = 62, normalized size = 0.60 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (80) = 160\).
time = 0.35, size = 323, normalized size = 3.11 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \sinh {\left (a + \frac {b}{x^{2}} \right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,\mathrm {sinh}\left (a+\frac {b}{x^2}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*sinh(a + b/x^2),x)

[Out]

int(x^4*sinh(a + b/x^2), x)

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