3.84.61 \(\int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 (x^3+x^3 \log (2))}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} (5+12 x^2) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+(-12 x^2+15 x^4) \log (2))}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 (x^3+x^3 \log (2))}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2))} \, dx\)

Optimal. Leaf size=33 \[ \frac {5}{3 \left (2+e^{\frac {1}{x}}-e^{\frac {x^2 (x+x \log (2))}{\log (2)}}\right )}+x \]

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Rubi [F]  time = 5.33, 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 {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(12*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2]))/Log[2])*x^2*Log[2] + E^x^(-1)*(5 + 12*
x^2)*Log[2] + E^((x^3 + x^3*Log[2])/Log[2])*(15*x^4 - 6*E^x^(-1)*x^2*Log[2] + (-12*x^2 + 15*x^4)*Log[2]))/(12*
x^2*Log[2] + 12*E^x^(-1)*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2]))/Log[2])*x^2*Log[2] +
E^((x^3 + x^3*Log[2])/Log[2])*(-12*x^2*Log[2] - 6*E^x^(-1)*x^2*Log[2])),x]

[Out]

(x*Log[8])/(3*Log[2]) + (Log[32]*Defer[Int][E^x^(-1)/((2 + E^x^(-1) - E^(x^3*(1 + Log[2]^(-1))))^2*x^2), x])/(
3*Log[2]) + (10*(1 + Log[2])*Defer[Int][x^2/(2 + E^x^(-1) - E^(x^3*(1 + Log[2]^(-1))))^2, x])/Log[2] + (5*(1 +
Log[2])*Defer[Int][(E^x^(-1)*x^2)/(2 + E^x^(-1) - E^(x^3*(1 + Log[2]^(-1))))^2, x])/Log[2] - (5*(1 + Log[2])*
Defer[Int][x^2/(2 + E^x^(-1) - E^(x^3*(1 + Log[2]^(-1)))), x])/Log[2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12 x^2 \log (2)-6 e^{\frac {1}{x}+x^3 \left (1+\frac {1}{\log (2)}\right )} x^2 \log (2)+3 e^{2 x^3 \left (1+\frac {1}{\log (2)}\right )} x^2 \log (2)+3 e^{x^3 \left (1+\frac {1}{\log (2)}\right )} x^2 \left (-4 \log (2)+5 x^2 (1+\log (2))\right )+e^{2/x} x^2 \log (8)+e^{\frac {1}{x}} \left (12 x^2 \log (2)+\log (32)\right )}{3 \left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2 \log (2)} \, dx\\ &=\frac {\int \frac {12 x^2 \log (2)-6 e^{\frac {1}{x}+x^3 \left (1+\frac {1}{\log (2)}\right )} x^2 \log (2)+3 e^{2 x^3 \left (1+\frac {1}{\log (2)}\right )} x^2 \log (2)+3 e^{x^3 \left (1+\frac {1}{\log (2)}\right )} x^2 \left (-4 \log (2)+5 x^2 (1+\log (2))\right )+e^{2/x} x^2 \log (8)+e^{\frac {1}{x}} \left (12 x^2 \log (2)+\log (32)\right )}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2} \, dx}{3 \log (2)}\\ &=\frac {\int \left (-\frac {15 x^2 (1+\log (2))}{2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}}+\log (8)+\frac {30 x^4 (1+\log (2))+15 e^{\frac {1}{x}} x^4 (1+\log (2))+e^{\frac {1}{x}} \log (32)}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2}\right ) \, dx}{3 \log (2)}\\ &=\frac {x \log (8)}{3 \log (2)}+\frac {\int \frac {30 x^4 (1+\log (2))+15 e^{\frac {1}{x}} x^4 (1+\log (2))+e^{\frac {1}{x}} \log (32)}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2} \, dx}{3 \log (2)}-\frac {(5 (1+\log (2))) \int \frac {x^2}{2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}} \, dx}{\log (2)}\\ &=\frac {x \log (8)}{3 \log (2)}+\frac {\int \left (\frac {30 x^2 (1+\log (2))}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}+\frac {15 e^{\frac {1}{x}} x^2 (1+\log (2))}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}+\frac {e^{\frac {1}{x}} \log (32)}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2}\right ) \, dx}{3 \log (2)}-\frac {(5 (1+\log (2))) \int \frac {x^2}{2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}} \, dx}{\log (2)}\\ &=\frac {x \log (8)}{3 \log (2)}+\frac {(5 (1+\log (2))) \int \frac {e^{\frac {1}{x}} x^2}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2} \, dx}{\log (2)}-\frac {(5 (1+\log (2))) \int \frac {x^2}{2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}} \, dx}{\log (2)}+\frac {(10 (1+\log (2))) \int \frac {x^2}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2} \, dx}{\log (2)}+\frac {\log (32) \int \frac {e^{\frac {1}{x}}}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2} \, dx}{3 \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.34, size = 86, normalized size = 2.61 \begin {gather*} \frac {1}{3} \left (3 x+\frac {15 \left (2+e^{\frac {1}{x}}\right ) x^4 (1+\log (2))+e^{\frac {1}{x}} \log (32)}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right ) \left (e^{\frac {1}{x}} \log (2)+x^4 \left (6+e^{\frac {1}{x}} (3+\log (8))+\log (64)\right )\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2]))/Log[2])*x^2*Log[2] + E^x^(-1)*(5
+ 12*x^2)*Log[2] + E^((x^3 + x^3*Log[2])/Log[2])*(15*x^4 - 6*E^x^(-1)*x^2*Log[2] + (-12*x^2 + 15*x^4)*Log[2])
)/(12*x^2*Log[2] + 12*E^x^(-1)*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2]))/Log[2])*x^2*Log
[2] + E^((x^3 + x^3*Log[2])/Log[2])*(-12*x^2*Log[2] - 6*E^x^(-1)*x^2*Log[2])),x]

[Out]

(3*x + (15*(2 + E^x^(-1))*x^4*(1 + Log[2]) + E^x^(-1)*Log[32])/((2 + E^x^(-1) - E^(x^3*(1 + Log[2]^(-1))))*(E^
x^(-1)*Log[2] + x^4*(6 + E^x^(-1)*(3 + Log[8]) + Log[64]))))/3

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fricas [B]  time = 0.64, size = 59, normalized size = 1.79 \begin {gather*} \frac {3 \, x e^{\left (\frac {x^{3} \log \relax (2) + x^{3}}{\log \relax (2)}\right )} - 3 \, x e^{\frac {1}{x}} - 6 \, x - 5}{3 \, {\left (e^{\left (\frac {x^{3} \log \relax (2) + x^{3}}{\log \relax (2)}\right )} - e^{\frac {1}{x}} - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(2)*exp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)+(15*x^4-12*x^2)*log(2)+15*x^4)*
exp((x^3*log(2)+x^3)/log(2))+3*x^2*log(2)*exp(1/x)^2+(12*x^2+5)*log(2)*exp(1/x)+12*x^2*log(2))/(3*x^2*log(2)*e
xp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)-12*x^2*log(2))*exp((x^3*log(2)+x^3)/log(2))+3*x^2*log(2)
*exp(1/x)^2+12*x^2*log(2)*exp(1/x)+12*x^2*log(2)),x, algorithm="fricas")

[Out]

1/3*(3*x*e^((x^3*log(2) + x^3)/log(2)) - 3*x*e^(1/x) - 6*x - 5)/(e^((x^3*log(2) + x^3)/log(2)) - e^(1/x) - 2)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(2)*exp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)+(15*x^4-12*x^2)*log(2)+15*x^4)*
exp((x^3*log(2)+x^3)/log(2))+3*x^2*log(2)*exp(1/x)^2+(12*x^2+5)*log(2)*exp(1/x)+12*x^2*log(2))/(3*x^2*log(2)*e
xp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)-12*x^2*log(2))*exp((x^3*log(2)+x^3)/log(2))+3*x^2*log(2)
*exp(1/x)^2+12*x^2*log(2)*exp(1/x)+12*x^2*log(2)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.29, size = 28, normalized size = 0.85

method	result	size
risch	\(x +\frac {5}{3 \left ({\mathrm e}^{\frac {1}{x}}-{\mathrm e}^{\frac {x^{3} \left (1+\ln \relax (2)\right )}{\ln \relax (2)}}+2\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2*ln(2)*exp((x^3*ln(2)+x^3)/ln(2))^2+(-6*x^2*ln(2)*exp(1/x)+(15*x^4-12*x^2)*ln(2)+15*x^4)*exp((x^3*ln
(2)+x^3)/ln(2))+3*x^2*ln(2)*exp(1/x)^2+(12*x^2+5)*ln(2)*exp(1/x)+12*x^2*ln(2))/(3*x^2*ln(2)*exp((x^3*ln(2)+x^3
)/ln(2))^2+(-6*x^2*ln(2)*exp(1/x)-12*x^2*ln(2))*exp((x^3*ln(2)+x^3)/ln(2))+3*x^2*ln(2)*exp(1/x)^2+12*x^2*ln(2)
*exp(1/x)+12*x^2*ln(2)),x,method=_RETURNVERBOSE)

[Out]

x+5/3/(exp(1/x)-exp(x^3*(1+ln(2))/ln(2))+2)

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maxima [A]  time = 0.50, size = 53, normalized size = 1.61 \begin {gather*} \frac {3 \, x e^{\left (x^{3} + \frac {x^{3}}{\log \relax (2)}\right )} - 3 \, x e^{\frac {1}{x}} - 6 \, x - 5}{3 \, {\left (e^{\left (x^{3} + \frac {x^{3}}{\log \relax (2)}\right )} - e^{\frac {1}{x}} - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(2)*exp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)+(15*x^4-12*x^2)*log(2)+15*x^4)*
exp((x^3*log(2)+x^3)/log(2))+3*x^2*log(2)*exp(1/x)^2+(12*x^2+5)*log(2)*exp(1/x)+12*x^2*log(2))/(3*x^2*log(2)*e
xp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)-12*x^2*log(2))*exp((x^3*log(2)+x^3)/log(2))+3*x^2*log(2)
*exp(1/x)^2+12*x^2*log(2)*exp(1/x)+12*x^2*log(2)),x, algorithm="maxima")

[Out]

1/3*(3*x*e^(x^3 + x^3/log(2)) - 3*x*e^(1/x) - 6*x - 5)/(e^(x^3 + x^3/log(2)) - e^(1/x) - 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {12\,x^2\,\ln \relax (2)-{\mathrm {e}}^{\frac {x^3\,\ln \relax (2)+x^3}{\ln \relax (2)}}\,\left (\ln \relax (2)\,\left (12\,x^2-15\,x^4\right )-15\,x^4+6\,x^2\,{\mathrm {e}}^{1/x}\,\ln \relax (2)\right )+3\,x^2\,{\mathrm {e}}^{\frac {2\,\left (x^3\,\ln \relax (2)+x^3\right )}{\ln \relax (2)}}\,\ln \relax (2)+3\,x^2\,{\mathrm {e}}^{2/x}\,\ln \relax (2)+{\mathrm {e}}^{1/x}\,\ln \relax (2)\,\left (12\,x^2+5\right )}{12\,x^2\,\ln \relax (2)-{\mathrm {e}}^{\frac {x^3\,\ln \relax (2)+x^3}{\ln \relax (2)}}\,\left (12\,x^2\,\ln \relax (2)+6\,x^2\,{\mathrm {e}}^{1/x}\,\ln \relax (2)\right )+3\,x^2\,{\mathrm {e}}^{\frac {2\,\left (x^3\,\ln \relax (2)+x^3\right )}{\ln \relax (2)}}\,\ln \relax (2)+3\,x^2\,{\mathrm {e}}^{2/x}\,\ln \relax (2)+12\,x^2\,{\mathrm {e}}^{1/x}\,\ln \relax (2)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(log(2)*(12*x^2 - 15*x^4) - 15*x^4 + 6*x^2*exp(1/x)*log(2)
) + 3*x^2*exp((2*(x^3*log(2) + x^3))/log(2))*log(2) + 3*x^2*exp(2/x)*log(2) + exp(1/x)*log(2)*(12*x^2 + 5))/(1
2*x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(12*x^2*log(2) + 6*x^2*exp(1/x)*log(2)) + 3*x^2*exp((2*(x^3*log(
2) + x^3))/log(2))*log(2) + 3*x^2*exp(2/x)*log(2) + 12*x^2*exp(1/x)*log(2)),x)

[Out]

int((12*x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(log(2)*(12*x^2 - 15*x^4) - 15*x^4 + 6*x^2*exp(1/x)*log(2)
) + 3*x^2*exp((2*(x^3*log(2) + x^3))/log(2))*log(2) + 3*x^2*exp(2/x)*log(2) + exp(1/x)*log(2)*(12*x^2 + 5))/(1
2*x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(12*x^2*log(2) + 6*x^2*exp(1/x)*log(2)) + 3*x^2*exp((2*(x^3*log(
2) + x^3))/log(2))*log(2) + 3*x^2*exp(2/x)*log(2) + 12*x^2*exp(1/x)*log(2)), x)

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sympy [A]  time = 0.37, size = 27, normalized size = 0.82 \begin {gather*} x - \frac {5}{- 3 e^{\frac {1}{x}} + 3 e^{\frac {x^{3} \log {\relax (2 )} + x^{3}}{\log {\relax (2 )}}} - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2*ln(2)*exp((x**3*ln(2)+x**3)/ln(2))**2+(-6*x**2*ln(2)*exp(1/x)+(15*x**4-12*x**2)*ln(2)+15*x**
4)*exp((x**3*ln(2)+x**3)/ln(2))+3*x**2*ln(2)*exp(1/x)**2+(12*x**2+5)*ln(2)*exp(1/x)+12*x**2*ln(2))/(3*x**2*ln(
2)*exp((x**3*ln(2)+x**3)/ln(2))**2+(-6*x**2*ln(2)*exp(1/x)-12*x**2*ln(2))*exp((x**3*ln(2)+x**3)/ln(2))+3*x**2*
ln(2)*exp(1/x)**2+12*x**2*ln(2)*exp(1/x)+12*x**2*ln(2)),x)

[Out]

x - 5/(-3*exp(1/x) + 3*exp((x**3*log(2) + x**3)/log(2)) - 6)

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