3.86.36 \(\int \frac {e^{\frac {-3 x^3+(x-4 x^2) \log (3)}{3 x+4 \log (3)}} (90 x^2-180 x^4+(240 x-480 x^3) \log (3)+(160+40 x-320 x^2) \log ^2(3))}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx\)

Optimal. Leaf size=25 \[ 10 e^{-2-x^2+\frac {x}{4+\frac {3 x}{\log (3)}}} x \]

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Rubi [F]  time = 1.91, 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 {\exp \left (\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-3*x^3 + (x - 4*x^2)*Log[3])/(3*x + 4*Log[3]))*(90*x^2 - 180*x^4 + (240*x - 480*x^3)*Log[3] + (160 +
40*x - 320*x^2)*Log[3]^2))/(9*E^2*x^2 + 24*E^2*x*Log[3] + 16*E^2*Log[3]^2),x]

[Out]

10*Defer[Int][E^(-2 + (-3*x^3 + (x - 4*x^2)*Log[3])/(3*x + 4*Log[3])), x] - 20*Defer[Int][E^(-2 + (-3*x^3 + (x
- 4*x^2)*Log[3])/(3*x + 4*Log[3]))*x^2, x] + (40*Log[3]^2*Defer[Int][E^(-2 + (-3*x^3 + (x - 4*x^2)*Log[3])/(3
*x + 4*Log[3]))/(3*x + 4*Log[3]), x])/3 - (160*Log[3]^3*Defer[Int][E^(-2 + (-3*x^3 + (x - 4*x^2)*Log[3])/(3*x
+ 4*Log[3]))/(3*x + Log[81])^2, x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{(3 x+4 \log (3))^2} \, dx\\ &=\int \frac {\exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) \left (-180 x^4-480 x^3 \log (3)+160 \log ^2(3)+40 x \log (3) (6+\log (3))+10 x^2 \left (9-32 \log ^2(3)\right )\right )}{(3 x+4 \log (3))^2} \, dx\\ &=\int \left (10 \exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right )-20 \exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) x^2+\frac {40 \exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) \log ^2(3)}{3 (3 x+4 \log (3))}-\frac {160 \exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) \log ^3(3)}{3 (3 x+\log (81))^2}\right ) \, dx\\ &=10 \int \exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) \, dx-20 \int \exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right ) x^2 \, dx+\frac {1}{3} \left (40 \log ^2(3)\right ) \int \frac {\exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right )}{3 x+4 \log (3)} \, dx-\frac {1}{3} \left (160 \log ^3(3)\right ) \int \frac {\exp \left (-2+\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}\right )}{(3 x+\log (81))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((-3*x^3 + (x - 4*x^2)*Log[3])/(3*x + 4*Log[3]))*(90*x^2 - 180*x^4 + (240*x - 480*x^3)*Log[3] + (
160 + 40*x - 320*x^2)*Log[3]^2))/(9*E^2*x^2 + 24*E^2*x*Log[3] + 16*E^2*Log[3]^2),x]

[Out]

Integrate[(E^((-3*x^3 + (x - 4*x^2)*Log[3])/(3*x + 4*Log[3]))*(90*x^2 - 180*x^4 + (240*x - 480*x^3)*Log[3] + (
160 + 40*x - 320*x^2)*Log[3]^2))/(9*E^2*x^2 + 24*E^2*x*Log[3] + 16*E^2*Log[3]^2), x]

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fricas [A]  time = 0.52, size = 36, normalized size = 1.44 \begin {gather*} 10 \, x e^{\left (-\frac {3 \, x^{3} + {\left (4 \, x^{2} - x\right )} \log \relax (3)}{3 \, x + 4 \, \log \relax (3)} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-320*x^2+40*x+160)*log(3)^2+(-480*x^3+240*x)*log(3)-180*x^4+90*x^2)*exp(((-4*x^2+x)*log(3)-3*x^3)/
(4*log(3)+3*x))/(16*exp(2)*log(3)^2+24*x*exp(2)*log(3)+9*x^2*exp(2)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 40, normalized size = 1.60 \begin {gather*} 10 \cdot 3^{\frac {1}{3}} x e^{\left (-\frac {9 \, x^{3} + 12 \, x^{2} \log \relax (3) + 4 \, \log \relax (3)^{2}}{3 \, {\left (3 \, x + 4 \, \log \relax (3)\right )}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-320*x^2+40*x+160)*log(3)^2+(-480*x^3+240*x)*log(3)-180*x^4+90*x^2)*exp(((-4*x^2+x)*log(3)-3*x^3)/
(4*log(3)+3*x))/(16*exp(2)*log(3)^2+24*x*exp(2)*log(3)+9*x^2*exp(2)),x, algorithm="giac")

[Out]

10*3^(1/3)*x*e^(-1/3*(9*x^3 + 12*x^2*log(3) + 4*log(3)^2)/(3*x + 4*log(3)) - 2)

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maple [A]  time = 0.42, size = 37, normalized size = 1.48

method	result	size
gosper	\(10 \,{\mathrm e}^{-2} x \,{\mathrm e}^{-\frac {x \left (4 x \ln \relax (3)+3 x^{2}-\ln \relax (3)\right )}{4 \ln \relax (3)+3 x}}\)	\(37\)
risch	\(10 x \,{\mathrm e}^{-\frac {4 x^{2} \ln \relax (3)+3 x^{3}-x \ln \relax (3)+8 \ln \relax (3)+6 x}{4 \ln \relax (3)+3 x}}\)	\(42\)
norman	\(\frac {30 x^{2} {\mathrm e}^{-2} {\mathrm e}^{\frac {\left (-4 x^{2}+x \right ) \ln \relax (3)-3 x^{3}}{4 \ln \relax (3)+3 x}}+40 \,{\mathrm e}^{-2} \ln \relax (3) x \,{\mathrm e}^{\frac {\left (-4 x^{2}+x \right ) \ln \relax (3)-3 x^{3}}{4 \ln \relax (3)+3 x}}}{4 \ln \relax (3)+3 x}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-320*x^2+40*x+160)*ln(3)^2+(-480*x^3+240*x)*ln(3)-180*x^4+90*x^2)*exp(((-4*x^2+x)*ln(3)-3*x^3)/(4*ln(3)+
3*x))/(16*exp(2)*ln(3)^2+24*x*exp(2)*ln(3)+9*x^2*exp(2)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.61, size = 30, normalized size = 1.20 \begin {gather*} 10 \cdot 3^{\frac {1}{3}} x e^{\left (-x^{2} - \frac {4 \, \log \relax (3)^{2}}{3 \, {\left (3 \, x + 4 \, \log \relax (3)\right )}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-320*x^2+40*x+160)*log(3)^2+(-480*x^3+240*x)*log(3)-180*x^4+90*x^2)*exp(((-4*x^2+x)*log(3)-3*x^3)/
(4*log(3)+3*x))/(16*exp(2)*log(3)^2+24*x*exp(2)*log(3)+9*x^2*exp(2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.63, size = 37, normalized size = 1.48 \begin {gather*} 10\,3^{\frac {x-4\,x^2}{3\,x+\ln \left (81\right )}}\,x\,{\mathrm {e}}^{-\frac {3\,x^3}{3\,x+\ln \left (81\right )}-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(3*x^3 - log(3)*(x - 4*x^2))/(3*x + 4*log(3)))*(log(3)*(240*x - 480*x^3) + log(3)^2*(40*x - 320*x^2
+ 160) + 90*x^2 - 180*x^4))/(16*exp(2)*log(3)^2 + 9*x^2*exp(2) + 24*x*exp(2)*log(3)),x)

[Out]

10*3^((x - 4*x^2)/(3*x + log(81)))*x*exp(- (3*x^3)/(3*x + log(81)) - 2)

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sympy [A]  time = 0.44, size = 31, normalized size = 1.24 \begin {gather*} \frac {10 x e^{\frac {- 3 x^{3} + \left (- 4 x^{2} + x\right ) \log {\relax (3 )}}{3 x + 4 \log {\relax (3 )}}}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-320*x**2+40*x+160)*ln(3)**2+(-480*x**3+240*x)*ln(3)-180*x**4+90*x**2)*exp(((-4*x**2+x)*ln(3)-3*x*
*3)/(4*ln(3)+3*x))/(16*exp(2)*ln(3)**2+24*x*exp(2)*ln(3)+9*x**2*exp(2)),x)

[Out]

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

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