Integrand size = 12, antiderivative size = 69 \[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\frac {x \cosh \left (a+b x^2\right )}{2 b}-\frac {e^{-a} \sqrt {\pi } \text {erf}\left (\sqrt {b} x\right )}{8 b^{3/2}}-\frac {e^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x\right )}{8 b^{3/2}} \] Output:
1/2*x*cosh(b*x^2+a)/b-1/8*Pi^(1/2)*erf(b^(1/2)*x)/b^(3/2)/exp(a)-1/8*exp(a )*Pi^(1/2)*erfi(b^(1/2)*x)/b^(3/2)
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97 \[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\frac {4 \sqrt {b} x \cosh \left (a+b x^2\right )+\sqrt {\pi } \text {erf}\left (\sqrt {b} x\right ) (-\cosh (a)+\sinh (a))-\sqrt {\pi } \text {erfi}\left (\sqrt {b} x\right ) (\cosh (a)+\sinh (a))}{8 b^{3/2}} \] Input:
Integrate[x^2*Sinh[a + b*x^2],x]
Output:
(4*Sqrt[b]*x*Cosh[a + b*x^2] + Sqrt[Pi]*Erf[Sqrt[b]*x]*(-Cosh[a] + Sinh[a] ) - Sqrt[Pi]*Erfi[Sqrt[b]*x]*(Cosh[a] + Sinh[a]))/(8*b^(3/2))
Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5847, 5822, 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.
\(\displaystyle \int x^2 \sinh \left (a+b x^2\right ) \, dx\) |
\(\Big \downarrow \) 5847 |
\(\displaystyle \frac {x \cosh \left (a+b x^2\right )}{2 b}-\frac {\int \cosh \left (b x^2+a\right )dx}{2 b}\) |
\(\Big \downarrow \) 5822 |
\(\displaystyle \frac {x \cosh \left (a+b x^2\right )}{2 b}-\frac {\frac {1}{2} \int e^{-b x^2-a}dx+\frac {1}{2} \int e^{b x^2+a}dx}{2 b}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {x \cosh \left (a+b x^2\right )}{2 b}-\frac {\frac {1}{2} \int e^{-b x^2-a}dx+\frac {\sqrt {\pi } e^a \text {erfi}\left (\sqrt {b} x\right )}{4 \sqrt {b}}}{2 b}\) |
\(\Big \downarrow \) 2634 |
\(\displaystyle \frac {x \cosh \left (a+b x^2\right )}{2 b}-\frac {\frac {\sqrt {\pi } e^{-a} \text {erf}\left (\sqrt {b} x\right )}{4 \sqrt {b}}+\frac {\sqrt {\pi } e^a \text {erfi}\left (\sqrt {b} x\right )}{4 \sqrt {b}}}{2 b}\) |
Input:
Int[x^2*Sinh[a + b*x^2],x]
Output:
(x*Cosh[a + b*x^2])/(2*b) - ((Sqrt[Pi]*Erf[Sqrt[b]*x])/(4*Sqrt[b]*E^a) + ( E^a*Sqrt[Pi]*Erfi[Sqrt[b]*x])/(4*Sqrt[b]))/(2*b)
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]
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]
Int[Cosh[(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]
Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[e^( n - 1)*(e*x)^(m - n + 1)*(Cosh[c + d*x^n]/(d*n)), x] - Simp[e^n*((m - n + 1 )/(d*n)) Int[(e*x)^(m - n)*Cosh[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] && LtQ[0, n, m + 1]
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\frac {{\mathrm e}^{-a} x \,{\mathrm e}^{-b \,x^{2}}}{4 b}-\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {b}\, x \right ) {\mathrm e}^{-a}}{8 b^{\frac {3}{2}}}+\frac {{\mathrm e}^{a} {\mathrm e}^{b \,x^{2}} x}{4 b}-\frac {{\mathrm e}^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b}\, x \right )}{8 b \sqrt {-b}}\) | \(74\) |
meijerg | \(-\frac {i \sinh \left (a \right ) \sqrt {\pi }\, \sqrt {2}\, \left (\frac {x \sqrt {2}\, \left (i b \right )^{\frac {3}{2}} {\mathrm e}^{b \,x^{2}}}{4 \sqrt {\pi }\, b}-\frac {x \sqrt {2}\, \left (i b \right )^{\frac {3}{2}} {\mathrm e}^{-b \,x^{2}}}{4 \sqrt {\pi }\, b}+\frac {\left (i b \right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {erf}\left (\sqrt {b}\, x \right )}{8 b^{\frac {3}{2}}}-\frac {\left (i b \right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {erfi}\left (\sqrt {b}\, x \right )}{8 b^{\frac {3}{2}}}\right )}{2 b \sqrt {i b}}-\frac {\cosh \left (a \right ) \sqrt {\pi }\, \sqrt {2}\, \left (\frac {x \sqrt {2}\, \left (i b \right )^{\frac {5}{2}} {\mathrm e}^{-b \,x^{2}}}{4 \sqrt {\pi }\, b^{2}}+\frac {x \sqrt {2}\, \left (i b \right )^{\frac {5}{2}} {\mathrm e}^{b \,x^{2}}}{4 \sqrt {\pi }\, b^{2}}-\frac {\left (i b \right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {erf}\left (\sqrt {b}\, x \right )}{8 b^{\frac {5}{2}}}-\frac {\left (i b \right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {erfi}\left (\sqrt {b}\, x \right )}{8 b^{\frac {5}{2}}}\right )}{2 b \sqrt {i b}}\) | \(221\) |
Input:
int(x^2*sinh(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
1/4/exp(a)/b*x*exp(-b*x^2)-1/8*Pi^(1/2)*erf(b^(1/2)*x)/b^(3/2)/exp(a)+1/4* exp(a)*exp(b*x^2)*x/b-1/8*exp(a)/b*Pi^(1/2)/(-b)^(1/2)*erf((-b)^(1/2)*x)
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (49) = 98\).
Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.75 \[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\frac {2 \, b x \cosh \left (b x^{2} + a\right )^{2} + 4 \, b x \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right ) + 2 \, b x \sinh \left (b x^{2} + a\right )^{2} + \sqrt {\pi } {\left (\cosh \left (b x^{2} + a\right ) \cosh \left (a\right ) + {\left (\cosh \left (a\right ) + \sinh \left (a\right )\right )} \sinh \left (b x^{2} + a\right ) + \cosh \left (b x^{2} + a\right ) \sinh \left (a\right )\right )} \sqrt {-b} \operatorname {erf}\left (\sqrt {-b} x\right ) - \sqrt {\pi } {\left (\cosh \left (b x^{2} + a\right ) \cosh \left (a\right ) + {\left (\cosh \left (a\right ) - \sinh \left (a\right )\right )} \sinh \left (b x^{2} + a\right ) - \cosh \left (b x^{2} + a\right ) \sinh \left (a\right )\right )} \sqrt {b} \operatorname {erf}\left (\sqrt {b} x\right ) + 2 \, b x}{8 \, {\left (b^{2} \cosh \left (b x^{2} + a\right ) + b^{2} \sinh \left (b x^{2} + a\right )\right )}} \] Input:
integrate(x^2*sinh(b*x^2+a),x, algorithm="fricas")
Output:
1/8*(2*b*x*cosh(b*x^2 + a)^2 + 4*b*x*cosh(b*x^2 + a)*sinh(b*x^2 + a) + 2*b *x*sinh(b*x^2 + a)^2 + sqrt(pi)*(cosh(b*x^2 + a)*cosh(a) + (cosh(a) + sinh (a))*sinh(b*x^2 + a) + cosh(b*x^2 + a)*sinh(a))*sqrt(-b)*erf(sqrt(-b)*x) - sqrt(pi)*(cosh(b*x^2 + a)*cosh(a) + (cosh(a) - sinh(a))*sinh(b*x^2 + a) - cosh(b*x^2 + a)*sinh(a))*sqrt(b)*erf(sqrt(b)*x) + 2*b*x)/(b^2*cosh(b*x^2 + a) + b^2*sinh(b*x^2 + a))
\[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\int x^{2} \sinh {\left (a + b x^{2} \right )}\, dx \] Input:
integrate(x**2*sinh(b*x**2+a),x)
Output:
Integral(x**2*sinh(a + b*x**2), x)
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (49) = 98\).
Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\frac {1}{3} \, x^{3} \sinh \left (b x^{2} + a\right ) - \frac {1}{24} \, b {\left (\frac {2 \, {\left (2 \, b x^{3} e^{a} - 3 \, x e^{a}\right )} e^{\left (b x^{2}\right )}}{b^{2}} - \frac {2 \, {\left (2 \, b x^{3} + 3 \, x\right )} e^{\left (-b x^{2} - a\right )}}{b^{2}} + \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (\sqrt {b} x\right ) e^{\left (-a\right )}}{b^{\frac {5}{2}}} + \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (\sqrt {-b} x\right ) e^{a}}{\sqrt {-b} b^{2}}\right )} \] Input:
integrate(x^2*sinh(b*x^2+a),x, algorithm="maxima")
Output:
1/3*x^3*sinh(b*x^2 + a) - 1/24*b*(2*(2*b*x^3*e^a - 3*x*e^a)*e^(b*x^2)/b^2 - 2*(2*b*x^3 + 3*x)*e^(-b*x^2 - a)/b^2 + 3*sqrt(pi)*erf(sqrt(b)*x)*e^(-a)/ b^(5/2) + 3*sqrt(pi)*erf(sqrt(-b)*x)*e^a/(sqrt(-b)*b^2))
Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\frac {x e^{\left (b x^{2} + a\right )}}{4 \, b} + \frac {x e^{\left (-b x^{2} - a\right )}}{4 \, b} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {b} x\right ) e^{\left (-a\right )}}{8 \, b^{\frac {3}{2}}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {-b} x\right ) e^{a}}{8 \, \sqrt {-b} b} \] Input:
integrate(x^2*sinh(b*x^2+a),x, algorithm="giac")
Output:
1/4*x*e^(b*x^2 + a)/b + 1/4*x*e^(-b*x^2 - a)/b + 1/8*sqrt(pi)*erf(-sqrt(b) *x)*e^(-a)/b^(3/2) + 1/8*sqrt(pi)*erf(-sqrt(-b)*x)*e^a/(sqrt(-b)*b)
Timed out. \[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\int x^2\,\mathrm {sinh}\left (b\,x^2+a\right ) \,d x \] Input:
int(x^2*sinh(a + b*x^2),x)
Output:
int(x^2*sinh(a + b*x^2), x)
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.16 \[ \int x^2 \sinh \left (a+b x^2\right ) \, dx=\frac {\sqrt {\pi }\, e^{b \,x^{2}+2 a} \mathrm {erf}\left (\sqrt {b}\, i x \right ) i -\sqrt {\pi }\, e^{b \,x^{2}} \mathrm {erf}\left (\sqrt {b}\, x \right )+2 e^{2 b \,x^{2}+2 a} \sqrt {b}\, x +2 \sqrt {b}\, x}{8 e^{b \,x^{2}+a} \sqrt {b}\, b} \] Input:
int(x^2*sinh(b*x^2+a),x)
Output:
(sqrt(pi)*e**(2*a + b*x**2)*erf(sqrt(b)*i*x)*i - sqrt(pi)*e**(b*x**2)*erf( sqrt(b)*x) + 2*e**(2*a + 2*b*x**2)*sqrt(b)*x + 2*sqrt(b)*x)/(8*e**(a + b*x **2)*sqrt(b)*b)