\(\int \sinh (\frac {1}{4}+x+x^2) \, dx\) [13]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [C] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 9, antiderivative size = 39 \[ \int \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{4} \sqrt {\pi } \text {erf}\left (\frac {1}{2} (-1-2 x)\right )+\frac {1}{4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5482, 2266, 2235, 2236} \[ \int \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{4} \sqrt {\pi } \text {erf}\left (\frac {1}{2} (-2 x-1)\right )+\frac {1}{4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (2 x+1)\right ) \]

[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 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]

Rubi steps \begin{align*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.62 \[ \int \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{4} \sqrt {\pi } \left (-\text {erf}\left (\frac {1}{2}+x\right )+\text {erfi}\left (\frac {1}{2}+x\right )\right ) \]

[In]

Integrate[Sinh[1/4 + x + x^2],x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.64

method result size
risch \(-\frac {\operatorname {erf}\left (\frac {1}{2}+x \right ) \sqrt {\pi }}{4}-\frac {i \sqrt {\pi }\, \operatorname {erf}\left (i x +\frac {1}{2} i\right )}{4}\) \(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.41 \[ \int \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=-\frac {1}{4} \, \sqrt {\pi } {\left (\operatorname {erf}\left (x + \frac {1}{2}\right ) - \operatorname {erfi}\left (x + \frac {1}{2}\right )\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sinh(x**2 + x + 1/4), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (19) = 38\).

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.41 \[ \int \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{2 \, {\left ({\left (2 \, x + 1\right )}^{2}\right )}^{\frac {3}{2}}} + \frac {{\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{2 \, \left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} + x \sinh \left (x^{2} + x + \frac {1}{4}\right ) + \frac {1}{4} \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} - \frac {1}{4} \, e^{\left (-\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.54 \[ \int \sinh \left (\frac {1}{4}+x+x^2\right ) \, dx=-\frac {1}{4} \, \sqrt {\pi } \operatorname {erf}\left (x + \frac {1}{2}\right ) + \frac {1}{4} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, x - \frac {1}{2} i\right ) \]

[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)

Mupad [F(-1)]

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

[In]

int(sinh(x + x^2 + 1/4),x)

[Out]

int(sinh(x + x^2 + 1/4), x)