\(\int x^2 \sinh (a+\frac {b}{x^2}) \, dx\) [43]

Optimal result

Integrand size = 12, antiderivative size = 86 \[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {2}{3} b x \cosh \left (a+\frac {b}{x^2}\right )+\frac {1}{3} b^{3/2} e^{-a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {1}{3} b^{3/2} e^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+\frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98 \[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{3} \left (2 b x \cosh \left (a+\frac {b}{x^2}\right )+b^{3/2} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)-\sinh (a))-b^{3/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)+\sinh (a))+x^3 \sinh \left (a+\frac {b}{x^2}\right )\right ) \] Input:

Integrate[x^2*Sinh[a + b/x^2],x]
 

Output:

(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
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5869, 5849, 5850, 5821, 2633, 2634}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5869

\(\displaystyle -\int x^4 \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x}\)

\(\Big \downarrow \) 5849

\(\displaystyle \frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )-\frac {2}{3} b \int x^2 \cosh \left (a+\frac {b}{x^2}\right )d\frac {1}{x}\)

\(\Big \downarrow \) 5850

\(\displaystyle \frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )-\frac {2}{3} b \left (2 b \int \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x}-x \cosh \left (a+\frac {b}{x^2}\right )\right )\)

\(\Big \downarrow \) 5821

\(\displaystyle \frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )-\frac {2}{3} b \left (2 b \left (\frac {1}{2} \int e^{a+\frac {b}{x^2}}d\frac {1}{x}-\frac {1}{2} \int e^{-a-\frac {b}{x^2}}d\frac {1}{x}\right )-x \cosh \left (a+\frac {b}{x^2}\right )\right )\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )-\frac {2}{3} b \left (2 b \left (\frac {\sqrt {\pi } e^a \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{4 \sqrt {b}}-\frac {1}{2} \int e^{-a-\frac {b}{x^2}}d\frac {1}{x}\right )-x \cosh \left (a+\frac {b}{x^2}\right )\right )\)

\(\Big \downarrow \) 2634

\(\displaystyle \frac {1}{3} x^3 \sinh \left (a+\frac {b}{x^2}\right )-\frac {2}{3} b \left (2 b \left (\frac {\sqrt {\pi } e^a \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{4 \sqrt {b}}-\frac {\sqrt {\pi } e^{-a} \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{4 \sqrt {b}}\right )-x \cosh \left (a+\frac {b}{x^2}\right )\right )\)

Input:

Int[x^2*Sinh[a + b/x^2],x]
 

Output:

(-2*b*(-(x*Cosh[a + b/x^2]) + 2*b*(-1/4*(Sqrt[Pi]*Erf[Sqrt[b]/x])/(Sqrt[b] 
*E^a) + (E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/(4*Sqrt[b]))))/3 + (x^3*Sinh[a + b/ 
x^2])/3
 

Defintions of rubi rules used

rule 2633

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 2634

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] /; Fr 
eeQ[{F, a, b, c, d}, x] && NegQ[b]
 

rule 5821

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

rule 5849

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] - Simp[d*(n/(e^n*(m + 1)))   In 
t[(e*x)^(m + n)*Cosh[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0 
] && LtQ[m, -1]
 

rule 5850

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] - Simp[d*(n/(e^n*(m + 1)))   In 
t[(e*x)^(m + n)*Sinh[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0 
] && LtQ[m, -1]
 

rule 5869

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo 
l] :> -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]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.20

method	result	size
risch	\(-\frac {{\mathrm e}^{-a} x^{3} {\mathrm e}^{-\frac {b}{x^{2}}}}{6}+\frac {b^{\frac {3}{2}} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) {\mathrm e}^{-a}}{3}+\frac {{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}} b x}{3}+\frac {{\mathrm e}^{a} x^{3} {\mathrm e}^{\frac {b}{x^{2}}}}{6}+\frac {{\mathrm e}^{a} b \,{\mathrm e}^{\frac {b}{x^{2}}} x}{3}-\frac {{\mathrm e}^{a} b^{2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right )}{3 \sqrt {-b}}\)	\(103\)
meijerg	\(-\frac {b \sqrt {\pi }\, \cosh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (-\frac {4 x^{3} \sqrt {2}\, \left (\frac {2 b}{x^{2}}+1\right ) {\mathrm e}^{\frac {b}{x^{2}}}}{3 \sqrt {\pi }\, \sqrt {i b}\, b}+\frac {4 x^{3} \sqrt {2}\, \left (-\frac {2 b}{x^{2}}+1\right ) {\mathrm e}^{-\frac {b}{x^{2}}}}{3 \sqrt {\pi }\, \sqrt {i b}\, b}-\frac {8 \sqrt {2}\, \sqrt {b}\, \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right )}{3 \sqrt {i b}}+\frac {8 \sqrt {2}\, \sqrt {b}\, \operatorname {erfi}\left (\frac {\sqrt {b}}{x}\right )}{3 \sqrt {i b}}\right )}{16}-\frac {i b \sqrt {\pi }\, \sinh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (-\frac {8 x^{3} \sqrt {2}\, \left (-\frac {b}{x^{2}}+\frac {1}{2}\right ) {\mathrm e}^{-\frac {b}{x^{2}}}}{3 \sqrt {\pi }\, \left (i b \right )^{\frac {3}{2}}}-\frac {8 x^{3} \sqrt {2}\, \left (\frac {b}{x^{2}}+\frac {1}{2}\right ) {\mathrm e}^{\frac {b}{x^{2}}}}{3 \sqrt {\pi }\, \left (i b \right )^{\frac {3}{2}}}+\frac {8 \sqrt {2}\, b^{\frac {3}{2}} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right )}{3 \left (i b \right )^{\frac {3}{2}}}+\frac {8 \sqrt {2}\, b^{\frac {3}{2}} \operatorname {erfi}\left (\frac {\sqrt {b}}{x}\right )}{3 \left (i b \right )^{\frac {3}{2}}}\right )}{16}\)	\(258\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (64) = 128\).

Time = 0.10 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.10 \[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=-\frac {x^{3} - {\left (x^{3} + 2 \, b x\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 2 \, \sqrt {\pi } {\left (b \cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (b \cosh \left (a\right ) + b \sinh \left (a\right )\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {-b} \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right ) - 2 \, \sqrt {\pi } {\left (b \cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (b \cosh \left (a\right ) - b \sinh \left (a\right )\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {b} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) - 2 \, {\left (x^{3} + 2 \, b x\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) - {\left (x^{3} + 2 \, b x\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 2 \, b x}{6 \, {\left (\cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )}} \] Input:

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

Output:

-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*s 
inh(a))*sinh((a*x^2 + b)/x^2))*sqrt(-b)*erf(sqrt(-b)/x) - 2*sqrt(pi)*(b*co 
sh(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))
 

Sympy [F]

\[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int x^{2} \sinh {\left (a + \frac {b}{x^{2}} \right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{3} \, x^{3} \sinh \left (a + \frac {b}{x^{2}}\right ) + \frac {1}{6} \, {\left (x \sqrt {\frac {b}{x^{2}}} e^{\left (-a\right )} \Gamma \left (-\frac {1}{2}, \frac {b}{x^{2}}\right ) + x \sqrt {-\frac {b}{x^{2}}} e^{a} \Gamma \left (-\frac {1}{2}, -\frac {b}{x^{2}}\right )\right )} b \] Input:

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

Output:

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
 

Giac [F]

\[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int { x^{2} \sinh \left (a + \frac {b}{x^{2}}\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int x^2\,\mathrm {sinh}\left (a+\frac {b}{x^2}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^2 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {-4 e^{\frac {2 a \,x^{2}+2 b}{x^{2}}} b^{2}+2 e^{\frac {2 a \,x^{2}+2 b}{x^{2}}} b \,x^{2}+e^{\frac {2 a \,x^{2}+2 b}{x^{2}}} x^{4}-8 e^{\frac {a \,x^{2}+b}{x^{2}}} \left (\int \frac {e^{\frac {a \,x^{2}+b}{x^{2}}}}{x^{4}}d x \right ) b^{3} x -8 e^{\frac {a \,x^{2}+b}{x^{2}}} \left (\int \frac {1}{e^{\frac {a \,x^{2}+b}{x^{2}}} x^{4}}d x \right ) b^{3} x +4 b^{2}+2 b \,x^{2}-x^{4}}{6 e^{\frac {a \,x^{2}+b}{x^{2}}} x} \] Input:

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

Output:

( - 4*e**((2*a*x**2 + 2*b)/x**2)*b**2 + 2*e**((2*a*x**2 + 2*b)/x**2)*b*x** 
2 + e**((2*a*x**2 + 2*b)/x**2)*x**4 - 8*e**((a*x**2 + b)/x**2)*int(e**((a* 
x**2 + b)/x**2)/x**4,x)*b**3*x - 8*e**((a*x**2 + b)/x**2)*int(1/(e**((a*x* 
*2 + b)/x**2)*x**4),x)*b**3*x + 4*b**2 + 2*b*x**2 - x**4)/(6*e**((a*x**2 + 
 b)/x**2)*x)