\(\int x^2 \cosh ^2(\frac {1}{4}+x+x^2) \, dx\) [24]

Optimal result

Integrand size = 15, antiderivative size = 68 \[ \int x^2 \cosh ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x^3}{6}+\frac {1}{16} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {1+2 x}{\sqrt {2}}\right )-\frac {1}{16} \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{8} x \sinh \left (\frac {1}{2}+2 x+2 x^2\right ) \]

[Out]

1/6*x^3-1/16*sinh(1/2+2*x+2*x^2)+1/8*x*sinh(1/2+2*x+2*x^2)+1/32*erf(1/2*(1+2*x)*2^(1/2))*2^(1/2)*Pi^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {5503, 5495, 5491, 5483, 2266, 2235, 2236, 5482} \[ \int x^2 \cosh ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{16} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {2 x+1}{\sqrt {2}}\right )+\frac {x^3}{6}+\frac {1}{8} x \sinh \left (2 x^2+2 x+\frac {1}{2}\right )-\frac {1}{16} \sinh \left (2 x^2+2 x+\frac {1}{2}\right ) \]

[In]

Int[x^2*Cosh[1/4 + x + x^2]^2,x]

[Out]

x^3/6 + (Sqrt[Pi/2]*Erf[(1 + 2*x)/Sqrt[2]])/16 - Sinh[1/2 + 2*x + 2*x^2]/16 + (x*Sinh[1/2 + 2*x + 2*x^2])/8

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 2266

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

Rule 5482

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

Rule 5483

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

Rule 5491

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

Rule 5495

Int[Cosh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*(
Sinh[a + b*x + c*x^2]/(2*c)), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Cosh[a + b*x + c*x^2], x]
, x] - Dist[e^2*((m - 1)/(2*c)), Int[(d + e*x)^(m - 2)*Sinh[a + b*x + c*x^2], x], x]) /; FreeQ[{a, b, c, d, e}
, x] && GtQ[m, 1] && NeQ[b*e - 2*c*d, 0]

Rule 5503

Int[Cosh[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]^(n_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce
[(d + e*x)^m, Cosh[a + b*x + c*x^2]^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x^2}{2}+\frac {1}{2} x^2 \cosh \left (\frac {1}{2}+2 x+2 x^2\right )\right ) \, dx \\ & = \frac {x^3}{6}+\frac {1}{2} \int x^2 \cosh \left (\frac {1}{2}+2 x+2 x^2\right ) \, dx \\ & = \frac {x^3}{6}+\frac {1}{8} x \sinh \left (\frac {1}{2}+2 x+2 x^2\right )-\frac {1}{8} \int \sinh \left (\frac {1}{2}+2 x+2 x^2\right ) \, dx-\frac {1}{4} \int x \cosh \left (\frac {1}{2}+2 x+2 x^2\right ) \, dx \\ & = \frac {x^3}{6}-\frac {1}{16} \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{8} x \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{16} \int e^{-\frac {1}{2}-2 x-2 x^2} \, dx-\frac {1}{16} \int e^{\frac {1}{2}+2 x+2 x^2} \, dx+\frac {1}{8} \int \cosh \left (\frac {1}{2}+2 x+2 x^2\right ) \, dx \\ & = \frac {x^3}{6}-\frac {1}{16} \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{8} x \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{16} \int e^{-\frac {1}{8} (-2-4 x)^2} \, dx-\frac {1}{16} \int e^{\frac {1}{8} (2+4 x)^2} \, dx+\frac {1}{16} \int e^{-\frac {1}{2}-2 x-2 x^2} \, dx+\frac {1}{16} \int e^{\frac {1}{2}+2 x+2 x^2} \, dx \\ & = \frac {x^3}{6}+\frac {1}{32} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {1+2 x}{\sqrt {2}}\right )-\frac {1}{32} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {1+2 x}{\sqrt {2}}\right )-\frac {1}{16} \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{8} x \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{16} \int e^{-\frac {1}{8} (-2-4 x)^2} \, dx+\frac {1}{16} \int e^{\frac {1}{8} (2+4 x)^2} \, dx \\ & = \frac {x^3}{6}+\frac {1}{16} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {1+2 x}{\sqrt {2}}\right )-\frac {1}{16} \sinh \left (\frac {1}{2}+2 x+2 x^2\right )+\frac {1}{8} x \sinh \left (\frac {1}{2}+2 x+2 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.46 \[ \int x^2 \cosh ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {16 \sqrt {e} x^3+3 (-1+e) (-1+2 x) \cosh (2 x (1+x))+3 \sqrt {2 e \pi } \text {erf}\left (\frac {1+2 x}{\sqrt {2}}\right )-3 \sinh (2 x (1+x))-3 e \sinh (2 x (1+x))+6 x \sinh (2 x (1+x))+6 e x \sinh (2 x (1+x))}{96 \sqrt {e}} \]

[In]

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

[Out]

(16*Sqrt[E]*x^3 + 3*(-1 + E)*(-1 + 2*x)*Cosh[2*x*(1 + x)] + 3*Sqrt[2*E*Pi]*Erf[(1 + 2*x)/Sqrt[2]] - 3*Sinh[2*x
*(1 + x)] - 3*E*Sinh[2*x*(1 + x)] + 6*x*Sinh[2*x*(1 + x)] + 6*E*x*Sinh[2*x*(1 + x)])/(96*Sqrt[E])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.13

method	result	size
risch	\(\frac {x^{3}}{6}-\frac {x \,{\mathrm e}^{-\frac {\left (1+2 x \right )^{2}}{2}}}{16}+\frac {{\mathrm e}^{-\frac {\left (1+2 x \right )^{2}}{2}}}{32}+\frac {\sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, x +\frac {\sqrt {2}}{2}\right )}{32}+\frac {x \,{\mathrm e}^{\frac {\left (1+2 x \right )^{2}}{2}}}{16}-\frac {{\mathrm e}^{\frac {\left (1+2 x \right )^{2}}{2}}}{32}\)	\(77\)

[In]

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

[Out]

1/6*x^3-1/16*x*exp(-1/2*(1+2*x)^2)+1/32*exp(-1/2*(1+2*x)^2)+1/32*Pi^(1/2)*2^(1/2)*erf(2^(1/2)*x+1/2*2^(1/2))+1
/16*x*exp(1/2*(1+2*x)^2)-1/32*exp(1/2*(1+2*x)^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (52) = 104\).

Time = 0.25 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.94 \[ \int x^2 \cosh ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {16 \, x^{3} \cosh \left (x^{2} + x + \frac {1}{4}\right )^{2} + 3 \, {\left (2 \, x - 1\right )} \cosh \left (x^{2} + x + \frac {1}{4}\right )^{4} + 12 \, {\left (2 \, x - 1\right )} \cosh \left (x^{2} + x + \frac {1}{4}\right ) \sinh \left (x^{2} + x + \frac {1}{4}\right )^{3} + 3 \, {\left (2 \, x - 1\right )} \sinh \left (x^{2} + x + \frac {1}{4}\right )^{4} + 2 \, {\left (8 \, x^{3} + 9 \, {\left (2 \, x - 1\right )} \cosh \left (x^{2} + x + \frac {1}{4}\right )^{2}\right )} \sinh \left (x^{2} + x + \frac {1}{4}\right )^{2} + 4 \, {\left (8 \, x^{3} \cosh \left (x^{2} + x + \frac {1}{4}\right ) + 3 \, {\left (2 \, x - 1\right )} \cosh \left (x^{2} + x + \frac {1}{4}\right )^{3}\right )} \sinh \left (x^{2} + x + \frac {1}{4}\right ) + 3 \, \sqrt {\pi } {\left (\sqrt {2} \cosh \left (x^{2} + x + \frac {1}{4}\right )^{2} \operatorname {erf}\left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) + 2 \, \sqrt {2} \cosh \left (x^{2} + x + \frac {1}{4}\right ) \operatorname {erf}\left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) \sinh \left (x^{2} + x + \frac {1}{4}\right ) + \sqrt {2} \operatorname {erf}\left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) \sinh \left (x^{2} + x + \frac {1}{4}\right )^{2}\right )} - 6 \, x + 3}{96 \, {\left (\cosh \left (x^{2} + x + \frac {1}{4}\right )^{2} + 2 \, \cosh \left (x^{2} + x + \frac {1}{4}\right ) \sinh \left (x^{2} + x + \frac {1}{4}\right ) + \sinh \left (x^{2} + x + \frac {1}{4}\right )^{2}\right )}} \]

[In]

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

[Out]

1/96*(16*x^3*cosh(x^2 + x + 1/4)^2 + 3*(2*x - 1)*cosh(x^2 + x + 1/4)^4 + 12*(2*x - 1)*cosh(x^2 + x + 1/4)*sinh
(x^2 + x + 1/4)^3 + 3*(2*x - 1)*sinh(x^2 + x + 1/4)^4 + 2*(8*x^3 + 9*(2*x - 1)*cosh(x^2 + x + 1/4)^2)*sinh(x^2
+ x + 1/4)^2 + 4*(8*x^3*cosh(x^2 + x + 1/4) + 3*(2*x - 1)*cosh(x^2 + x + 1/4)^3)*sinh(x^2 + x + 1/4) + 3*sqrt
(pi)*(sqrt(2)*cosh(x^2 + x + 1/4)^2*erf(1/2*sqrt(2)*(2*x + 1)) + 2*sqrt(2)*cosh(x^2 + x + 1/4)*erf(1/2*sqrt(2)
*(2*x + 1))*sinh(x^2 + x + 1/4) + sqrt(2)*erf(1/2*sqrt(2)*(2*x + 1))*sinh(x^2 + x + 1/4)^2) - 6*x + 3)/(cosh(x
^2 + x + 1/4)^2 + 2*cosh(x^2 + x + 1/4)*sinh(x^2 + x + 1/4) + sinh(x^2 + x + 1/4)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.66 \[ \int x^2 \cosh ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{6} \, x^{3} + \frac {1}{32} \, {\left (2 \, x e^{\frac {1}{2}} - e^{\frac {1}{2}}\right )} e^{\left (2 \, x^{2} + 2 \, x\right )} - \frac {1}{64} i \, \sqrt {2} {\left (-\frac {2 i \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {1}{2} \, {\left (2 \, x + 1\right )}^{2}\right )}{{\left ({\left (2 \, x + 1\right )}^{2}\right )}^{\frac {3}{2}}} + \frac {i \, \sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {{\left (2 \, x + 1\right )}^{2}}} + 2 i \, \sqrt {2} e^{\left (-\frac {1}{2} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} \]

[In]

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

[Out]

1/6*x^3 + 1/32*(2*x*e^(1/2) - e^(1/2))*e^(2*x^2 + 2*x) - 1/64*I*sqrt(2)*(-2*I*(2*x + 1)^3*gamma(3/2, 1/2*(2*x
+ 1)^2)/((2*x + 1)^2)^(3/2) + I*sqrt(pi)*(2*x + 1)*(erf(sqrt(1/2)*sqrt((2*x + 1)^2)) - 1)/sqrt((2*x + 1)^2) +
2*I*sqrt(2)*e^(-1/2*(2*x + 1)^2))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int x^2 \cosh ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{6} \, x^{3} + \frac {1}{32} \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) + \frac {1}{32} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x^{2} + 2 \, x + \frac {1}{2}\right )} - \frac {1}{32} \, {\left (2 \, x - 1\right )} e^{\left (-2 \, x^{2} - 2 \, x - \frac {1}{2}\right )} \]

[In]

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

[Out]

1/6*x^3 + 1/32*sqrt(2)*sqrt(pi)*erf(1/2*sqrt(2)*(2*x + 1)) + 1/32*(2*x - 1)*e^(2*x^2 + 2*x + 1/2) - 1/32*(2*x
- 1)*e^(-2*x^2 - 2*x - 1/2)

Mupad [F(-1)]

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

[In]

int(x^2*cosh(x + x^2 + 1/4)^2,x)

[Out]

int(x^2*cosh(x + x^2 + 1/4)^2, x)