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

Optimal. Leaf size=66 \[ -\frac {3}{16} \sqrt {\pi } \text {erf}\left (\frac {1}{2} (-2 x-1)\right )-\frac {1}{16} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 x+1)\right )+\frac {1}{2} x \sinh \left (x^2+x+\frac {1}{4}\right )-\frac {1}{4} \sinh \left (x^2+x+\frac {1}{4}\right ) \]

[Out]

-1/4*sinh(1/4+x+x^2)+1/2*x*sinh(1/4+x+x^2)+3/16*erf(1/2+x)*Pi^(1/2)-1/16*erfi(1/2+x)*Pi^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {5387, 5374, 2234, 2204, 2205, 5383, 5375} \[ -\frac {3}{16} \sqrt {\pi } \text {Erf}\left (\frac {1}{2} (-2 x-1)\right )-\frac {1}{16} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (2 x+1)\right )+\frac {1}{2} x \sinh \left (x^2+x+\frac {1}{4}\right )-\frac {1}{4} \sinh \left (x^2+x+\frac {1}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cosh[1/4 + x + x^2],x]

[Out]

(-3*Sqrt[Pi]*Erf[(-1 - 2*x)/2])/16 - (Sqrt[Pi]*Erfi[(1 + 2*x)/2])/16 - Sinh[1/4 + x + x^2]/4 + (x*Sinh[1/4 + x
+ x^2])/2

Rule 2204

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 2205

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 2234

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 5374

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 5375

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 5383

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 5387

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[(e^2*(m - 1))/(2*c), Int[(d + e*x)^(m - 2)*Sinh[a + b*x + c*x^2], x]
, x] - Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Cosh[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]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 72, normalized size = 1.09 \[ \frac {1}{16} \left (3 \sqrt {\pi } \text {erf}\left (x+\frac {1}{2}\right )-\sqrt {\pi } \text {erfi}\left (x+\frac {1}{2}\right )+\frac {2 (2 x-1) \left (\left (1+\sqrt {e}\right ) \sinh (x (x+1))+\left (\sqrt {e}-1\right ) \cosh (x (x+1))\right )}{\sqrt [4]{e}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cosh[1/4 + x + x^2],x]

[Out]

(3*Sqrt[Pi]*Erf[1/2 + x] - Sqrt[Pi]*Erfi[1/2 + x] + (2*(-1 + 2*x)*((-1 + Sqrt[E])*Cosh[x*(1 + x)] + (1 + Sqrt[
E])*Sinh[x*(1 + x)]))/E^(1/4))/16

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fricas [A]  time = 0.53, size = 60, normalized size = 0.91 \[ \frac {1}{16} \, {\left (\sqrt {\pi } {\left (3 \, \operatorname {erf}\left (x + \frac {1}{2}\right ) - \operatorname {erfi}\left (x + \frac {1}{2}\right )\right )} e^{\left (x^{2} + x + \frac {1}{4}\right )} + 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x^{2} + 2 \, x + \frac {1}{2}\right )} - 4 \, x + 2\right )} e^{\left (-x^{2} - x - \frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(sqrt(pi)*(3*erf(x + 1/2) - erfi(x + 1/2))*e^(x^2 + x + 1/4) + 2*(2*x - 1)*e^(2*x^2 + 2*x + 1/2) - 4*x +
2)*e^(-x^2 - x - 1/4)

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giac [C]  time = 0.13, size = 53, normalized size = 0.80 \[ \frac {1}{8} \, {\left (2 \, x - 1\right )} e^{\left (x^{2} + x + \frac {1}{4}\right )} - \frac {1}{8} \, {\left (2 \, x - 1\right )} e^{\left (-x^{2} - x - \frac {1}{4}\right )} + \frac {3}{16} \, \sqrt {\pi } \operatorname {erf}\left (x + \frac {1}{2}\right ) - \frac {1}{16} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, x - \frac {1}{2} i\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*x - 1)*e^(x^2 + x + 1/4) - 1/8*(2*x - 1)*e^(-x^2 - x - 1/4) + 3/16*sqrt(pi)*erf(x + 1/2) - 1/16*I*sqrt(
pi)*erf(-I*x - 1/2*I)

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maple [C]  time = 0.13, size = 75, normalized size = 1.14 \[ -\frac {x \,{\mathrm e}^{-\frac {\left (1+2 x \right )^{2}}{4}}}{4}+\frac {{\mathrm e}^{-\frac {\left (1+2 x \right )^{2}}{4}}}{8}+\frac {3 \erf \left (\frac {1}{2}+x \right ) \sqrt {\pi }}{16}+\frac {x \,{\mathrm e}^{\frac {\left (1+2 x \right )^{2}}{4}}}{4}-\frac {{\mathrm e}^{\frac {\left (1+2 x \right )^{2}}{4}}}{8}+\frac {i \sqrt {\pi }\, \erf \left (i x +\frac {1}{2} i\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x*exp(-1/4*(1+2*x)^2)+1/8*exp(-1/4*(1+2*x)^2)+3/16*erf(1/2+x)*Pi^(1/2)+1/4*x*exp(1/4*(1+2*x)^2)-1/8*exp(1
/4*(1+2*x)^2)+1/16*I*Pi^(1/2)*erf(I*x+1/2*I)

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maxima [B]  time = 0.45, size = 183, normalized size = 2.77 \[ \frac {1}{3} \, x^{3} \cosh \left (x^{2} + x + \frac {1}{4}\right ) - \frac {{\left (2 \, x + 1\right )}^{5} \Gamma \left (\frac {5}{2}, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{6 \, {\left ({\left (2 \, x + 1\right )}^{2}\right )}^{\frac {5}{2}}} + \frac {{\left (2 \, x + 1\right )}^{5} \Gamma \left (\frac {5}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{6 \, \left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {5}{2}}} - \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{8 \, {\left ({\left (2 \, x + 1\right )}^{2}\right )}^{\frac {3}{2}}} + \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{8 \, \left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} + \frac {1}{48} \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} + \frac {1}{48} \, e^{\left (-\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} + \frac {1}{4} \, \Gamma \left (2, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right ) - \frac {1}{4} \, \Gamma \left (2, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*cosh(x^2 + x + 1/4) - 1/6*(2*x + 1)^5*gamma(5/2, 1/4*(2*x + 1)^2)/((2*x + 1)^2)^(5/2) + 1/6*(2*x + 1)^
5*gamma(5/2, -1/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(5/2) - 1/8*(2*x + 1)^3*gamma(3/2, 1/4*(2*x + 1)^2)/((2*x + 1)^2
)^(3/2) + 1/8*(2*x + 1)^3*gamma(3/2, -1/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3/2) + 1/48*e^(1/4*(2*x + 1)^2) + 1/48*
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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^2\,\mathrm {cosh}\left (x^2+x+\frac {1}{4}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(x + x^2 + 1/4),x)

[Out]

int(x^2*cosh(x + x^2 + 1/4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cosh {\left (x^{2} + x + \frac {1}{4} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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