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

Optimal. Leaf size=52 \[ \frac {1}{8} \sqrt {\pi } \text {erf}\left (\frac {1}{2} (-2 x-1)\right )-\frac {1}{8} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 x+1)\right )+\frac {1}{2} \sinh \left (x^2+x+\frac {1}{4}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {5383, 5375, 2234, 2204, 2205} \[ \frac {1}{8} \sqrt {\pi } \text {Erf}\left (\frac {1}{2} (-2 x-1)\right )-\frac {1}{8} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (2 x+1)\right )+\frac {1}{2} \sinh \left (x^2+x+\frac {1}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[1/4 + x + x^2],x]

[Out]

(Sqrt[Pi]*Erf[(-1 - 2*x)/2])/8 - (Sqrt[Pi]*Erfi[(1 + 2*x)/2])/8 + 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 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]

Rubi steps

\begin {align*} \int x \cosh \left (\frac {1}{4}+x+x^2\right ) \, dx &=\frac {1}{2} \sinh \left (\frac {1}{4}+x+x^2\right )-\frac {1}{2} \int \cosh \left (\frac {1}{4}+x+x^2\right ) \, dx\\ &=\frac {1}{2} \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}{2} \sinh \left (\frac {1}{4}+x+x^2\right )-\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}{2} \sinh \left (\frac {1}{4}+x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 76, normalized size = 1.46 \[ \frac {-\sqrt [4]{e} \sqrt {\pi } \text {erf}\left (x+\frac {1}{2}\right )-\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (x+\frac {1}{2}\right )+2 \left (1+\sqrt {e}\right ) \sinh (x (x+1))+2 \left (\sqrt {e}-1\right ) \cosh (x (x+1))}{8 \sqrt [4]{e}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[1/4 + x + x^2],x]

[Out]

(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 + Sq
rt[E])*Sinh[x*(1 + x)])/(8*E^(1/4))

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fricas [A]  time = 0.69, size = 48, normalized size = 0.92 \[ -\frac {1}{8} \, {\left (\sqrt {\pi } {\left (\operatorname {erf}\left (x + \frac {1}{2}\right ) + \operatorname {erfi}\left (x + \frac {1}{2}\right )\right )} e^{\left (x^{2} + x + \frac {1}{4}\right )} - 2 \, e^{\left (2 \, x^{2} + 2 \, x + \frac {1}{2}\right )} + 2\right )} e^{\left (-x^{2} - x - \frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.14, size = 43, normalized size = 0.83 \[ -\frac {1}{8} \, \sqrt {\pi } \operatorname {erf}\left (x + \frac {1}{2}\right ) - \frac {1}{8} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, x - \frac {1}{2} i\right ) + \frac {1}{4} \, e^{\left (x^{2} + x + \frac {1}{4}\right )} - \frac {1}{4} \, e^{\left (-x^{2} - x - \frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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maple [C]  time = 0.12, size = 49, normalized size = 0.94 \[ -\frac {{\mathrm e}^{-\frac {\left (1+2 x \right )^{2}}{4}}}{4}-\frac {\erf \left (\frac {1}{2}+x \right ) \sqrt {\pi }}{8}+\frac {{\mathrm e}^{\frac {\left (1+2 x \right )^{2}}{4}}}{4}+\frac {i \sqrt {\pi }\, \erf \left (i x +\frac {1}{2} i\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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maxima [B]  time = 0.47, size = 123, normalized size = 2.37 \[ \frac {1}{2} \, x^{2} \cosh \left (x^{2} + x + \frac {1}{4}\right ) + \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{4 \, {\left ({\left (2 \, x + 1\right )}^{2}\right )}^{\frac {3}{2}}} - \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{4 \, \left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {1}{16} \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} - \frac {1}{16} \, e^{\left (-\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} - \frac {1}{4} \, \Gamma \left (2, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right ) + \frac {1}{4} \, \Gamma \left (2, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*cosh(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(x + x^2 + 1/4),x)

[Out]

int(x*cosh(x + x^2 + 1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*cosh(x**2 + x + 1/4), x)

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