3.1.12 \(\int x \sinh (\frac {1}{4}+x+x^2) \, dx\) [12]

3.1.12.1 Optimal result

Integrand size = 11, antiderivative size = 52 \[ \int x \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{2} \cosh \left (\frac {1}{4}+x+x^2\right )-\frac {1}{8} \sqrt {\pi } \text {erf}\left (\frac {1}{2} (-1-2 x)\right )-\frac {1}{8} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right ) \]

output

1/2*cosh(1/4+x+x^2)+1/8*erf(1/2+x)*Pi^(1/2)-1/8*erfi(1/2+x)*Pi^(1/2)
 

3.1.12.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.44 \[ \int x \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {2 \left (1+\sqrt {e}\right ) \cosh (x (1+x))+\sqrt [4]{e} \sqrt {\pi } \text {erf}\left (\frac {1}{2}+x\right )-\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )+2 \left (-1+\sqrt {e}\right ) \sinh (x (1+x))}{8 \sqrt [4]{e}} \]

input

Integrate[x*Sinh[1/4 + x + x^2],x]
 

output

(2*(1 + Sqrt[E])*Cosh[x*(1 + x)] + E^(1/4)*Sqrt[Pi]*Erf[1/2 + x] - E^(1/4) 
*Sqrt[Pi]*Erfi[1/2 + x] + 2*(-1 + Sqrt[E])*Sinh[x*(1 + x)])/(8*E^(1/4))
 

3.1.12.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {5905, 5897, 2664, 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.

\(\displaystyle \int x \sinh \left (x^2+x+\frac {1}{4}\right ) \, dx\)

\(\Big \downarrow \) 5905

\(\displaystyle \frac {1}{2} \cosh \left (x^2+x+\frac {1}{4}\right )-\frac {1}{2} \int \sinh \left (x^2+x+\frac {1}{4}\right )dx\)

\(\Big \downarrow \) 5897

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int e^{-x^2-x-\frac {1}{4}}dx-\frac {1}{2} \int e^{x^2+x+\frac {1}{4}}dx\right )+\frac {1}{2} \cosh \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 2664

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int e^{-\frac {1}{4} (2 x+1)^2}dx-\frac {1}{2} \int e^{\frac {1}{4} (2 x+1)^2}dx\right )+\frac {1}{2} \cosh \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int e^{-\frac {1}{4} (2 x+1)^2}dx-\frac {1}{4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 x+1)\right )\right )+\frac {1}{2} \cosh \left (x^2+x+\frac {1}{4}\right )\)

\(\Big \downarrow \) 2634

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

input

Int[x*Sinh[1/4 + x + x^2],x]
 

output

Cosh[1/4 + x + x^2]/2 + ((Sqrt[Pi]*Erf[(1 + 2*x)/2])/4 - (Sqrt[Pi]*Erfi[(1 
 + 2*x)/2])/4)/2
 

3.1.12.3.1 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 2664

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ 
(4*c))   Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
 

rule 5897

Int[Sinh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[1/2   Int[E^ 
(a + b*x + c*x^2), x], x] - Simp[1/2   Int[E^(-a - b*x - c*x^2), x], x] /; 
FreeQ[{a, b, c}, x]
 

rule 5905

Int[((d_.) + (e_.)*(x_))*Sinh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] 
 :> Simp[e*(Cosh[a + b*x + c*x^2]/(2*c)), x] - Simp[(b*e - 2*c*d)/(2*c)   I 
nt[Sinh[a + b*x + c*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*e - 
2*c*d, 0]
 

3.1.12.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {\left (1+2 x \right )^{2}}{4}}}{4}+\frac {\operatorname {erf}\left (\frac {1}{2}+x \right ) \sqrt {\pi }}{8}+\frac {{\mathrm e}^{\frac {\left (1+2 x \right )^{2}}{4}}}{4}+\frac {i \sqrt {\pi }\, \operatorname {erf}\left (i x +\frac {1}{2} i\right )}{8}\)	\(49\)

input

int(x*sinh(1/4+x+x^2),x,method=_RETURNVERBOSE)
 

output

1/4*exp(-1/4*(1+2*x)^2)+1/8*erf(1/2+x)*Pi^(1/2)+1/4*exp(1/4*(1+2*x)^2)+1/8 
*I*Pi^(1/2)*erf(I*x+1/2*I)
 

3.1.12.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (28) = 56\).

Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.08 \[ \int x \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {2 \, \cosh \left (x^{2} + x + \frac {1}{4}\right )^{2} + 4 \, \cosh \left (x^{2} + x + \frac {1}{4}\right ) \sinh \left (x^{2} + x + \frac {1}{4}\right ) + 2 \, \sinh \left (x^{2} + x + \frac {1}{4}\right )^{2} + \sqrt {\pi } {\left (\cosh \left (x^{2} + x + \frac {1}{4}\right ) \operatorname {erf}\left (x + \frac {1}{2}\right ) - \cosh \left (x^{2} + x + \frac {1}{4}\right ) \operatorname {erfi}\left (x + \frac {1}{2}\right ) + {\left (\operatorname {erf}\left (x + \frac {1}{2}\right ) - \operatorname {erfi}\left (x + \frac {1}{2}\right )\right )} \sinh \left (x^{2} + x + \frac {1}{4}\right )\right )} + 2}{8 \, {\left (\cosh \left (x^{2} + x + \frac {1}{4}\right ) + \sinh \left (x^{2} + x + \frac {1}{4}\right )\right )}} \]

input

integrate(x*sinh(1/4+x+x^2),x, algorithm="fricas")
 

output

1/8*(2*cosh(x^2 + x + 1/4)^2 + 4*cosh(x^2 + x + 1/4)*sinh(x^2 + x + 1/4) + 
 2*sinh(x^2 + x + 1/4)^2 + sqrt(pi)*(cosh(x^2 + x + 1/4)*erf(x + 1/2) - co 
sh(x^2 + x + 1/4)*erfi(x + 1/2) + (erf(x + 1/2) - erfi(x + 1/2))*sinh(x^2 
+ x + 1/4)) + 2)/(cosh(x^2 + x + 1/4) + sinh(x^2 + x + 1/4))
 

3.1.12.6 Sympy [F]

\[ \int x \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\int x \sinh {\left (x^{2} + x + \frac {1}{4} \right )}\, dx \]

input

integrate(x*sinh(1/4+x+x**2),x)
 

output

Integral(x*sinh(x**2 + x + 1/4), x)
 

3.1.12.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (28) = 56\).

Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.37 \[ \int x \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{2} \, x^{2} \sinh \left (x^{2} + x + \frac {1}{4}\right ) - \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{4 \, {\left ({\left (2 \, x + 1\right )}^{2}\right )}^{\frac {3}{2}}} - \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{4 \, \left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {1}{16} \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} + \frac {1}{16} \, e^{\left (-\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} + \frac {1}{4} \, \Gamma \left (2, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right ) + \frac {1}{4} \, \Gamma \left (2, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right ) \]

input

integrate(x*sinh(1/4+x+x^2),x, algorithm="maxima")
 

output

1/2*x^2*sinh(x^2 + x + 1/4) - 1/4*(2*x + 1)^3*gamma(3/2, 1/4*(2*x + 1)^2)/ 
((2*x + 1)^2)^(3/2) - 1/4*(2*x + 1)^3*gamma(3/2, -1/4*(2*x + 1)^2)/(-(2*x 
+ 1)^2)^(3/2) - 1/16*e^(1/4*(2*x + 1)^2) + 1/16*e^(-1/4*(2*x + 1)^2) + 1/4 
*gamma(2, 1/4*(2*x + 1)^2) + 1/4*gamma(2, -1/4*(2*x + 1)^2)
 

3.1.12.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int x \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{8} \, \sqrt {\pi } \operatorname {erf}\left (x + \frac {1}{2}\right ) - \frac {1}{8} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, x - \frac {1}{2} i\right ) + \frac {1}{4} \, e^{\left (x^{2} + x + \frac {1}{4}\right )} + \frac {1}{4} \, e^{\left (-x^{2} - x - \frac {1}{4}\right )} \]

input

integrate(x*sinh(1/4+x+x^2),x, algorithm="giac")
 

output

1/8*sqrt(pi)*erf(x + 1/2) - 1/8*I*sqrt(pi)*erf(-I*x - 1/2*I) + 1/4*e^(x^2 
+ x + 1/4) + 1/4*e^(-x^2 - x - 1/4)
 

3.1.12.9 Mupad [F(-1)]

Timed out. \[ \int x \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\int x\,\mathrm {sinh}\left (x^2+x+\frac {1}{4}\right ) \,d x \]

input

int(x*sinh(x + x^2 + 1/4),x)
 

output

int(x*sinh(x + x^2 + 1/4), x)