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

Optimal. Leaf size=18 \[ -x^2 \left (25+\left (3-e^{x^2}\right ) x\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1593, 2226, 2212, 2204} \begin {gather*} -3 x^3-25 x^2+e^{x^2} x^3 \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 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 2212

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 2226

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 15, normalized size = 0.83 \begin {gather*} x^2 \left (-25+\left (-3+e^{x^2}\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.61, size = 19, normalized size = 1.06 \begin {gather*} x^{3} e^{\left (x^{2}\right )} - 3 \, x^{3} - 25 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.15, size = 19, normalized size = 1.06 \begin {gather*} x^{3} e^{\left (x^{2}\right )} - 3 \, x^{3} - 25 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.04, size = 20, normalized size = 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	\(x^{3} {\mathrm e}^{x^{2}}-25 x^{2}-3 x^{3}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.36, size = 19, normalized size = 1.06 \begin {gather*} x^{3} e^{\left (x^{2}\right )} - 3 \, x^{3} - 25 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 5.44, size = 17, normalized size = 0.94 \begin {gather*} -x^2\,\left (3\,x-x\,{\mathrm {e}}^{x^2}+25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.09, size = 17, normalized size = 0.94 \begin {gather*} x^{3} e^{x^{2}} - 3 x^{3} - 25 x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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