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

Optimal. Leaf size=39 \[ \frac {1}{4} \sqrt {\pi } \text {erf}\left (\frac {1}{2} (-2 x-1)\right )+\frac {1}{4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 x+1)\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5374, 2234, 2204, 2205} \[ \frac {1}{4} \sqrt {\pi } \text {Erf}\left (\frac {1}{2} (-2 x-1)\right )+\frac {1}{4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (2 x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[1/4 + x + x^2],x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.62 \[ \frac {1}{4} \sqrt {\pi } \left (\text {erfi}\left (x+\frac {1}{2}\right )-\text {erf}\left (x+\frac {1}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[1/4 + x + x^2],x]

[Out]

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

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fricas [A]  time = 0.55, size = 16, normalized size = 0.41 \[ -\frac {1}{4} \, \sqrt {\pi } {\left (\operatorname {erf}\left (x + \frac {1}{2}\right ) - \operatorname {erfi}\left (x + \frac {1}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.13, size = 21, normalized size = 0.54 \[ -\frac {1}{4} \, \sqrt {\pi } \operatorname {erf}\left (x + \frac {1}{2}\right ) + \frac {1}{4} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, x - \frac {1}{2} i\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.04, size = 25, normalized size = 0.64 \[ -\frac {\erf \left (\frac {1}{2}+x \right ) \sqrt {\pi }}{4}-\frac {i \sqrt {\pi }\, \erf \left (i x +\frac {1}{2} i\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.74, size = 94, normalized size = 2.41 \[ \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{2 \, {\left ({\left (2 \, x + 1\right )}^{2}\right )}^{\frac {3}{2}}} + \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{2 \, \left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} + x \sinh \left (x^{2} + x + \frac {1}{4}\right ) + \frac {1}{4} \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} - \frac {1}{4} \, e^{\left (-\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \mathrm {sinh}\left (x^2+x+\frac {1}{4}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x + x^2 + 1/4),x)

[Out]

int(sinh(x + x^2 + 1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(x**2 + x + 1/4), x)

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