\(\int (-50 x-9 x^2+e^{x^2} (3 x^2+2 x^4)) \, dx\) [8700]

Optimal result

Integrand size = 26, antiderivative size = 18 \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=-x^2 \left (25+\left (3-e^{x^2}\right ) x\right ) \]

[Out]

-x^2*(25+(3-exp(x^2))*x)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1607, 2258, 2243, 2235} \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=-3 x^3-25 x^2+e^{x^2} x^3 \]

[In]

Int[-50*x - 9*x^2 + E^x^2*(3*x^2 + 2*x^4),x]

[Out]

-25*x^2 - 3*x^3 + E^x^2*x^3

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps \begin{align*} \text {integral}& = -25 x^2-3 x^3+\int e^{x^2} \left (3 x^2+2 x^4\right ) \, dx \\ & = -25 x^2-3 x^3+\int e^{x^2} x^2 \left (3+2 x^2\right ) \, dx \\ & = -25 x^2-3 x^3+\int \left (3 e^{x^2} x^2+2 e^{x^2} x^4\right ) \, dx \\ & = -25 x^2-3 x^3+2 \int e^{x^2} x^4 \, dx+3 \int e^{x^2} x^2 \, dx \\ & = \frac {3 e^{x^2} x}{2}-25 x^2-3 x^3+e^{x^2} x^3-\frac {3}{2} \int e^{x^2} \, dx-3 \int e^{x^2} x^2 \, dx \\ & = -25 x^2-3 x^3+e^{x^2} x^3-\frac {3}{4} \sqrt {\pi } \text {erfi}(x)+\frac {3}{2} \int e^{x^2} \, dx \\ & = -25 x^2-3 x^3+e^{x^2} x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=x^2 \left (-25+\left (-3+e^{x^2}\right ) x\right ) \]

[In]

Integrate[-50*x - 9*x^2 + E^x^2*(3*x^2 + 2*x^4),x]

[Out]

x^2*(-25 + (-3 + E^x^2)*x)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method	result	size
default	\(x^{3} {\mathrm e}^{x^{2}}-25 x^{2}-3 x^{3}\)	\(20\)
norman	\(x^{3} {\mathrm e}^{x^{2}}-25 x^{2}-3 x^{3}\)	\(20\)
risch	\(x^{3} {\mathrm e}^{x^{2}}-25 x^{2}-3 x^{3}\)	\(20\)
parallelrisch	\(x^{3} {\mathrm e}^{x^{2}}-25 x^{2}-3 x^{3}\)	\(20\)
parts	\(x^{3} {\mathrm e}^{x^{2}}-25 x^{2}-3 x^{3}\)	\(20\)

[In]

int((2*x^4+3*x^2)*exp(x^2)-9*x^2-50*x,x,method=_RETURNVERBOSE)

[Out]

x^3*exp(x^2)-25*x^2-3*x^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=x^{3} e^{\left (x^{2}\right )} - 3 \, x^{3} - 25 \, x^{2} \]

[In]

integrate((2*x^4+3*x^2)*exp(x^2)-9*x^2-50*x,x, algorithm="fricas")

[Out]

x^3*e^(x^2) - 3*x^3 - 25*x^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=x^{3} e^{x^{2}} - 3 x^{3} - 25 x^{2} \]

[In]

integrate((2*x**4+3*x**2)*exp(x**2)-9*x**2-50*x,x)

[Out]

x**3*exp(x**2) - 3*x**3 - 25*x**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=x^{3} e^{\left (x^{2}\right )} - 3 \, x^{3} - 25 \, x^{2} \]

[In]

integrate((2*x^4+3*x^2)*exp(x^2)-9*x^2-50*x,x, algorithm="maxima")

[Out]

x^3*e^(x^2) - 3*x^3 - 25*x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=x^{3} e^{\left (x^{2}\right )} - 3 \, x^{3} - 25 \, x^{2} \]

[In]

integrate((2*x^4+3*x^2)*exp(x^2)-9*x^2-50*x,x, algorithm="giac")

[Out]

x^3*e^(x^2) - 3*x^3 - 25*x^2

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \left (-50 x-9 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=-x^2\,\left (3\,x-x\,{\mathrm {e}}^{x^2}+25\right ) \]

[In]

int(exp(x^2)*(3*x^2 + 2*x^4) - 50*x - 9*x^2,x)

[Out]

-x^2*(3*x - x*exp(x^2) + 25)