3.51.41 \(\int \frac {27 e^{10} x^2+e^{-x^2+x \log (x^2)} (4+8 x-8 x^2+4 x \log (x^2))}{4 e^{10}} \, dx\)

Optimal. Leaf size=24 \[ x \left (e^{-10+x \left (-x+\log \left (x^2\right )\right )}+\frac {9 x^2}{4}\right ) \]

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Rubi [B]  time = 0.07, antiderivative size = 49, normalized size of antiderivative = 2.04, number of steps used = 3, number of rules used = 2, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12, 2288} \begin {gather*} \frac {9 x^3}{4}+\frac {e^{-x^2-10} \left (x^2\right )^x \left (-2 x^2+x \log \left (x^2\right )+2 x\right )}{\log \left (x^2\right )-2 x+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(27*E^10*x^2 + E^(-x^2 + x*Log[x^2])*(4 + 8*x - 8*x^2 + 4*x*Log[x^2]))/(4*E^10),x]

[Out]

(9*x^3)/4 + (E^(-10 - x^2)*(x^2)^x*(2*x - 2*x^2 + x*Log[x^2]))/(2 - 2*x + Log[x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (27 e^{10} x^2+e^{-x^2+x \log \left (x^2\right )} \left (4+8 x-8 x^2+4 x \log \left (x^2\right )\right )\right ) \, dx}{4 e^{10}}\\ &=\frac {9 x^3}{4}+\frac {\int e^{-x^2+x \log \left (x^2\right )} \left (4+8 x-8 x^2+4 x \log \left (x^2\right )\right ) \, dx}{4 e^{10}}\\ &=\frac {9 x^3}{4}+\frac {e^{-10-x^2} \left (x^2\right )^x \left (2 x-2 x^2+x \log \left (x^2\right )\right )}{2-2 x+\log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 1.12 \begin {gather*} \frac {1}{4} \left (9 x^3+4 e^{-10-x^2} x \left (x^2\right )^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(27*E^10*x^2 + E^(-x^2 + x*Log[x^2])*(4 + 8*x - 8*x^2 + 4*x*Log[x^2]))/(4*E^10),x]

[Out]

(9*x^3 + 4*E^(-10 - x^2)*x*(x^2)^x)/4

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fricas [A]  time = 0.96, size = 28, normalized size = 1.17 \begin {gather*} \frac {1}{4} \, {\left (9 \, x^{3} e^{10} + 4 \, x e^{\left (-x^{2} + x \log \left (x^{2}\right )\right )}\right )} e^{\left (-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x*log(x^2)-8*x^2+8*x+4)*exp(x*log(x^2)-x^2)+27*x^2*exp(5)^2)/exp(5)^2,x, algorithm="fricas")

[Out]

1/4*(9*x^3*e^10 + 4*x*e^(-x^2 + x*log(x^2)))*e^(-10)

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giac [A]  time = 0.23, size = 28, normalized size = 1.17 \begin {gather*} \frac {1}{4} \, {\left (9 \, x^{3} e^{10} + 4 \, x e^{\left (-x^{2} + x \log \left (x^{2}\right )\right )}\right )} e^{\left (-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x*log(x^2)-8*x^2+8*x+4)*exp(x*log(x^2)-x^2)+27*x^2*exp(5)^2)/exp(5)^2,x, algorithm="giac")

[Out]

1/4*(9*x^3*e^10 + 4*x*e^(-x^2 + x*log(x^2)))*e^(-10)

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maple [A]  time = 0.09, size = 22, normalized size = 0.92

method	result	size
risch	\(x \left (x^{2}\right )^{x} {\mathrm e}^{-x^{2}-10}+\frac {9 x^{3}}{4}\)	\(22\)
default	\(\frac {{\mathrm e}^{-10} \left (4 \,{\mathrm e}^{x \ln \left (x^{2}\right )-x^{2}} x +9 x^{3} {\mathrm e}^{10}\right )}{4}\)	\(33\)
norman	\(\left (x \,{\mathrm e}^{-5} {\mathrm e}^{x \ln \left (x^{2}\right )-x^{2}}+\frac {9 x^{3} {\mathrm e}^{5}}{4}\right ) {\mathrm e}^{-5}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((4*x*ln(x^2)-8*x^2+8*x+4)*exp(x*ln(x^2)-x^2)+27*x^2*exp(5)^2)/exp(5)^2,x,method=_RETURNVERBOSE)

[Out]

x*(x^2)^x*exp(-x^2-10)+9/4*x^3

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maxima [A]  time = 0.39, size = 27, normalized size = 1.12 \begin {gather*} \frac {1}{4} \, {\left (9 \, x^{3} e^{10} + 4 \, x e^{\left (-x^{2} + 2 \, x \log \relax (x)\right )}\right )} e^{\left (-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x*log(x^2)-8*x^2+8*x+4)*exp(x*log(x^2)-x^2)+27*x^2*exp(5)^2)/exp(5)^2,x, algorithm="maxima")

[Out]

1/4*(9*x^3*e^10 + 4*x*e^(-x^2 + 2*x*log(x)))*e^(-10)

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mupad [B]  time = 3.41, size = 21, normalized size = 0.88 \begin {gather*} \frac {9\,x^3}{4}+x\,{\mathrm {e}}^{-10}\,{\mathrm {e}}^{-x^2}\,{\left (x^2\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-10)*((27*x^2*exp(10))/4 + (exp(x*log(x^2) - x^2)*(8*x + 4*x*log(x^2) - 8*x^2 + 4))/4),x)

[Out]

(9*x^3)/4 + x*exp(-10)*exp(-x^2)*(x^2)^x

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sympy [A]  time = 0.32, size = 22, normalized size = 0.92 \begin {gather*} \frac {9 x^{3}}{4} + \frac {x e^{- x^{2} + x \log {\left (x^{2} \right )}}}{e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x*ln(x**2)-8*x**2+8*x+4)*exp(x*ln(x**2)-x**2)+27*x**2*exp(5)**2)/exp(5)**2,x)

[Out]

9*x**3/4 + x*exp(-10)*exp(-x**2 + x*log(x**2))

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