3.18.22 \(\int \frac {1}{12} e^{\frac {1}{12} (-3 x+6 e^x x+2 e^x x \log (x^2))} (-3+e^x (10+6 x)+e^x (2+2 x) \log (x^2)) \, dx\)

Optimal. Leaf size=21 \[ e^{x \left (-\frac {1}{4}+\frac {1}{6} e^x \left (3+\log \left (x^2\right )\right )\right )} \]

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Rubi [F]  time = 2.03, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-3*x + 6*E^x*x + 2*E^x*x*Log[x^2])/12)*(-3 + E^x*(10 + 6*x) + E^x*(2 + 2*x)*Log[x^2]))/12,x]

[Out]

(5*Defer[Int][E^((x*(9 + 6*E^x + 2*E^x*Log[x^2]))/12), x])/6 - Defer[Int][E^((-3*x + 6*E^x*x + 2*E^x*x*Log[x^2
])/12), x]/4 + Defer[Int][E^((x*(9 + 6*E^x + 2*E^x*Log[x^2]))/12)*x, x]/2 + Defer[Int][E^((x*(9 + 6*E^x + 2*E^
x*Log[x^2]))/12)*Log[x^2], x]/6 + Defer[Int][E^((x*(9 + 6*E^x + 2*E^x*Log[x^2]))/12)*x*Log[x^2], x]/6

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{12} \int e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx\\ &=\frac {1}{12} \int \left (-3 e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )}+2 \exp \left (x+\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )\right ) (5+3 x)+2 \exp \left (x+\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )\right ) (1+x) \log \left (x^2\right )\right ) \, dx\\ &=\frac {1}{6} \int \exp \left (x+\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )\right ) (5+3 x) \, dx+\frac {1}{6} \int \exp \left (x+\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )\right ) (1+x) \log \left (x^2\right ) \, dx-\frac {1}{4} \int e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \, dx\\ &=\frac {1}{6} \int e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} (5+3 x) \, dx+\frac {1}{6} \int e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} (1+x) \log \left (x^2\right ) \, dx-\frac {1}{4} \int e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \, dx\\ &=\frac {1}{6} \int \left (5 e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )}+3 e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} x\right ) \, dx+\frac {1}{6} \int \left (e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} \log \left (x^2\right )+e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} x \log \left (x^2\right )\right ) \, dx-\frac {1}{4} \int e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \, dx\\ &=\frac {1}{6} \int e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} \log \left (x^2\right ) \, dx+\frac {1}{6} \int e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} x \log \left (x^2\right ) \, dx-\frac {1}{4} \int e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \, dx+\frac {1}{2} \int e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} x \, dx+\frac {5}{6} \int e^{\frac {1}{12} x \left (9+6 e^x+2 e^x \log \left (x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.37, size = 29, normalized size = 1.38 \begin {gather*} e^{-\frac {x}{4}+\frac {e^x x}{2}} \left (x^2\right )^{\frac {e^x x}{6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-3*x + 6*E^x*x + 2*E^x*x*Log[x^2])/12)*(-3 + E^x*(10 + 6*x) + E^x*(2 + 2*x)*Log[x^2]))/12,x]

[Out]

E^(-1/4*x + (E^x*x)/2)*(x^2)^((E^x*x)/6)

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fricas [A]  time = 0.55, size = 19, normalized size = 0.90 \begin {gather*} e^{\left (\frac {1}{6} \, x e^{x} \log \left (x^{2}\right ) + \frac {1}{2} \, x e^{x} - \frac {1}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((2*x+2)*exp(x)*log(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x)*log(x^2)+1/2*exp(x)*x-1/4*x),x, al
gorithm="fricas")

[Out]

e^(1/6*x*e^x*log(x^2) + 1/2*x*e^x - 1/4*x)

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giac [A]  time = 0.25, size = 19, normalized size = 0.90 \begin {gather*} e^{\left (\frac {1}{6} \, x e^{x} \log \left (x^{2}\right ) + \frac {1}{2} \, x e^{x} - \frac {1}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((2*x+2)*exp(x)*log(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x)*log(x^2)+1/2*exp(x)*x-1/4*x),x, al
gorithm="giac")

[Out]

e^(1/6*x*e^x*log(x^2) + 1/2*x*e^x - 1/4*x)

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maple [A]  time = 0.35, size = 20, normalized size = 0.95

method	result	size
default	\({\mathrm e}^{\frac {x \,{\mathrm e}^{x} \ln \left (x^{2}\right )}{6}+\frac {{\mathrm e}^{x} x}{2}-\frac {x}{4}}\)	\(20\)
norman	\({\mathrm e}^{\frac {x \,{\mathrm e}^{x} \ln \left (x^{2}\right )}{6}+\frac {{\mathrm e}^{x} x}{2}-\frac {x}{4}}\)	\(20\)
risch	\({\mathrm e}^{\frac {x \left (-i {\mathrm e}^{x} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i {\mathrm e}^{x} \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i {\mathrm e}^{x} \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+4 \,{\mathrm e}^{x} \ln \relax (x )+6 \,{\mathrm e}^{x}-3\right )}{12}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*((2*x+2)*exp(x)*ln(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x)*ln(x^2)+1/2*exp(x)*x-1/4*x),x,method=_RET
URNVERBOSE)

[Out]

exp(1/6*x*exp(x)*ln(x^2)+1/2*exp(x)*x-1/4*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{12} \, \int {\left (2 \, {\left (x + 1\right )} e^{x} \log \left (x^{2}\right ) + 2 \, {\left (3 \, x + 5\right )} e^{x} - 3\right )} e^{\left (\frac {1}{6} \, x e^{x} \log \left (x^{2}\right ) + \frac {1}{2} \, x e^{x} - \frac {1}{4} \, x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((2*x+2)*exp(x)*log(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x)*log(x^2)+1/2*exp(x)*x-1/4*x),x, al
gorithm="maxima")

[Out]

1/12*integrate((2*(x + 1)*e^x*log(x^2) + 2*(3*x + 5)*e^x - 3)*e^(1/6*x*e^x*log(x^2) + 1/2*x*e^x - 1/4*x), x)

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mupad [B]  time = 1.26, size = 20, normalized size = 0.95 \begin {gather*} {\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{2}-\frac {x}{4}}\,{\left (x^2\right )}^{\frac {x\,{\mathrm {e}}^x}{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x*exp(x))/2 - x/4 + (x*log(x^2)*exp(x))/6)*(exp(x)*(6*x + 10) + log(x^2)*exp(x)*(2*x + 2) - 3))/12,x
)

[Out]

exp((x*exp(x))/2 - x/4)*(x^2)^((x*exp(x))/6)

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sympy [A]  time = 0.50, size = 22, normalized size = 1.05 \begin {gather*} e^{\frac {x e^{x} \log {\left (x^{2} \right )}}{6} + \frac {x e^{x}}{2} - \frac {x}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((2*x+2)*exp(x)*ln(x**2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x)*ln(x**2)+1/2*exp(x)*x-1/4*x),x)

[Out]

exp(x*exp(x)*log(x**2)/6 + x*exp(x)/2 - x/4)

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