3.1.65 \(\int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{(-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+(e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)) \log (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)})) \log (-15+\log (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}))} \, dx\)

Optimal. Leaf size=21 \[ 4 \log \left (\log \left (-15+\log \left (e^{\frac {1}{x^2}}+\frac {e^x}{\log ^2(2)}\right )\right )\right ) \]

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Rubi [A]  time = 0.27, antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 1, number of rules used = 1, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6684} \begin {gather*} 4 \log \left (\log \left (\log \left (\frac {e^{\frac {1}{x^2}} \log ^2(2)+e^x}{\log ^2(2)}\right )-15\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^x*x^3 - 8*E^x^(-2)*Log[2]^2)/((-15*E^x*x^3 - 15*E^x^(-2)*x^3*Log[2]^2 + (E^x*x^3 + E^x^(-2)*x^3*Log[2
]^2)*Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2])*Log[-15 + Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2]]),x]

[Out]

4*Log[Log[-15 + Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2]]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=4 \log \left (\log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 21, normalized size = 1.00 \begin {gather*} 4 \log \left (\log \left (-15+\log \left (e^{\frac {1}{x^2}}+\frac {e^x}{\log ^2(2)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^x*x^3 - 8*E^x^(-2)*Log[2]^2)/((-15*E^x*x^3 - 15*E^x^(-2)*x^3*Log[2]^2 + (E^x*x^3 + E^x^(-2)*x^3
*Log[2]^2)*Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2])*Log[-15 + Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2]]),x]

[Out]

4*Log[Log[-15 + Log[E^x^(-2) + E^x/Log[2]^2]]]

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fricas [A]  time = 0.56, size = 24, normalized size = 1.14 \begin {gather*} 4 \, \log \left (\log \left (\log \left (\frac {e^{\left (\frac {1}{x^{2}}\right )} \log \relax (2)^{2} + e^{x}}{\log \relax (2)^{2}}\right ) - 15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(x)*x^3-8*log(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*log(2)^2*exp(1/x^2))*log((exp(x)+log(2)^2*exp(
1/x^2))/log(2)^2)-15*exp(x)*x^3-15*x^3*log(2)^2*exp(1/x^2))/log(log((exp(x)+log(2)^2*exp(1/x^2))/log(2)^2)-15)
,x, algorithm="fricas")

[Out]

4*log(log(log((e^(x^(-2))*log(2)^2 + e^x)/log(2)^2) - 15))

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giac [A]  time = 0.71, size = 24, normalized size = 1.14 \begin {gather*} 4 \, \log \left (\log \left (\log \left (e^{\left (\frac {1}{x^{2}}\right )} \log \relax (2)^{2} + e^{x}\right ) - 2 \, \log \left (\log \relax (2)\right ) - 15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(x)*x^3-8*log(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*log(2)^2*exp(1/x^2))*log((exp(x)+log(2)^2*exp(
1/x^2))/log(2)^2)-15*exp(x)*x^3-15*x^3*log(2)^2*exp(1/x^2))/log(log((exp(x)+log(2)^2*exp(1/x^2))/log(2)^2)-15)
,x, algorithm="giac")

[Out]

4*log(log(log(e^(x^(-2))*log(2)^2 + e^x) - 2*log(log(2)) - 15))

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maple [A]  time = 0.05, size = 25, normalized size = 1.19

method	result	size
risch	\(4 \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{x}+\ln \relax (2)^{2} {\mathrm e}^{\frac {1}{x^{2}}}}{\ln \relax (2)^{2}}\right )-15\right )\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(x)*x^3-8*ln(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*ln(2)^2*exp(1/x^2))*ln((exp(x)+ln(2)^2*exp(1/x^2))/ln
(2)^2)-15*exp(x)*x^3-15*x^3*ln(2)^2*exp(1/x^2))/ln(ln((exp(x)+ln(2)^2*exp(1/x^2))/ln(2)^2)-15),x,method=_RETUR
NVERBOSE)

[Out]

4*ln(ln(ln((exp(x)+ln(2)^2*exp(1/x^2))/ln(2)^2)-15))

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maxima [A]  time = 0.88, size = 24, normalized size = 1.14 \begin {gather*} 4 \, \log \left (\log \left (\log \left (e^{\left (\frac {1}{x^{2}}\right )} \log \relax (2)^{2} + e^{x}\right ) - 2 \, \log \left (\log \relax (2)\right ) - 15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(x)*x^3-8*log(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*log(2)^2*exp(1/x^2))*log((exp(x)+log(2)^2*exp(
1/x^2))/log(2)^2)-15*exp(x)*x^3-15*x^3*log(2)^2*exp(1/x^2))/log(log((exp(x)+log(2)^2*exp(1/x^2))/log(2)^2)-15)
,x, algorithm="maxima")

[Out]

4*log(log(log(e^(x^(-2))*log(2)^2 + e^x) - 2*log(log(2)) - 15))

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mupad [B]  time = 1.66, size = 24, normalized size = 1.14 \begin {gather*} 4\,\ln \left (\ln \left (\ln \left (\frac {{\mathrm {e}}^x+{\mathrm {e}}^{\frac {1}{x^2}}\,{\ln \relax (2)}^2}{{\ln \relax (2)}^2}\right )-15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x^3*exp(x) - 8*exp(1/x^2)*log(2)^2)/(log(log((exp(x) + exp(1/x^2)*log(2)^2)/log(2)^2) - 15)*(15*x^3*ex
p(x) - log((exp(x) + exp(1/x^2)*log(2)^2)/log(2)^2)*(x^3*exp(x) + x^3*exp(1/x^2)*log(2)^2) + 15*x^3*exp(1/x^2)
*log(2)^2)),x)

[Out]

4*log(log(log((exp(x) + exp(1/x^2)*log(2)^2)/log(2)^2) - 15))

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sympy [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((4*exp(x)*x**3-8*ln(2)**2*exp(1/x**2))/((exp(x)*x**3+x**3*ln(2)**2*exp(1/x**2))*ln((exp(x)+ln(2)**2*
exp(1/x**2))/ln(2)**2)-15*exp(x)*x**3-15*x**3*ln(2)**2*exp(1/x**2))/ln(ln((exp(x)+ln(2)**2*exp(1/x**2))/ln(2)*
*2)-15),x)

[Out]

Timed out

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