3.46.83 \(\int \frac {e^x (50 x \log (5)+450 \log ^2(5))+(100 \log (5)+25 e^x \log (5)) \log (\frac {12}{4+e^x})}{(4+e^x) \log ^3(\frac {12}{4+e^x})} \, dx\)

Optimal. Leaf size=26 \[ \frac {25 \log (5) (x+9 \log (5))}{\log ^2\left (\frac {3}{1+\frac {e^x}{4}}\right )} \]

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Rubi [F]  time = 1.11, 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 {e^x \left (50 x \log (5)+450 \log ^2(5)\right )+\left (100 \log (5)+25 e^x \log (5)\right ) \log \left (\frac {12}{4+e^x}\right )}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(50*x*Log[5] + 450*Log[5]^2) + (100*Log[5] + 25*E^x*Log[5])*Log[12/(4 + E^x)])/((4 + E^x)*Log[12/(4 +
E^x)]^3),x]

[Out]

(225*Log[5]^2)/Log[12/(4 + E^x)]^2 + 50*Log[5]*Defer[Int][x/Log[12/(4 + E^x)]^3, x] - 200*Log[5]*Defer[Int][x/
((4 + E^x)*Log[12/(4 + E^x)]^3), x] + 25*Log[5]*Defer[Subst][Defer[Int][1/(x*Log[12/(4 + x)]^2), x], x, E^x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {200 \log (5) (x+9 \log (5))}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )}+\frac {25 \log (5) \left (2 x+18 \log (5)+\log \left (\frac {12}{4+e^x}\right )\right )}{\log ^3\left (\frac {12}{4+e^x}\right )}\right ) \, dx\\ &=(25 \log (5)) \int \frac {2 x+18 \log (5)+\log \left (\frac {12}{4+e^x}\right )}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-(200 \log (5)) \int \frac {x+9 \log (5)}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx\\ &=(25 \log (5)) \int \left (\frac {2 (x+9 \log (5))}{\log ^3\left (\frac {12}{4+e^x}\right )}+\frac {1}{\log ^2\left (\frac {12}{4+e^x}\right )}\right ) \, dx-(200 \log (5)) \int \left (\frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )}+\frac {9 \log (5)}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )}\right ) \, dx\\ &=(25 \log (5)) \int \frac {1}{\log ^2\left (\frac {12}{4+e^x}\right )} \, dx+(50 \log (5)) \int \frac {x+9 \log (5)}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-(200 \log (5)) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx-\left (1800 \log ^2(5)\right ) \int \frac {1}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx\\ &=(25 \log (5)) \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )+(50 \log (5)) \int \left (\frac {x}{\log ^3\left (\frac {12}{4+e^x}\right )}+\frac {9 \log (5)}{\log ^3\left (\frac {12}{4+e^x}\right )}\right ) \, dx-(200 \log (5)) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx-\left (1800 \log ^2(5)\right ) \operatorname {Subst}\left (\int \frac {1}{x (4+x) \log ^3\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )\\ &=(25 \log (5)) \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )+(50 \log (5)) \int \frac {x}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-(200 \log (5)) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx+\left (450 \log ^2(5)\right ) \int \frac {1}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-\left (1800 \log ^2(5)\right ) \operatorname {Subst}\left (\int \left (\frac {1}{4 x \log ^3\left (\frac {12}{4+x}\right )}-\frac {1}{4 (4+x) \log ^3\left (\frac {12}{4+x}\right )}\right ) \, dx,x,e^x\right )\\ &=(25 \log (5)) \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )+(50 \log (5)) \int \frac {x}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-(200 \log (5)) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx+\left (450 \log ^2(5)\right ) \operatorname {Subst}\left (\int \frac {1}{(4+x) \log ^3\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )\\ &=(25 \log (5)) \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )+(50 \log (5)) \int \frac {x}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-(200 \log (5)) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx+\left (450 \log ^2(5)\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^3\left (\frac {12}{x}\right )} \, dx,x,4+e^x\right )\\ &=(25 \log (5)) \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )+(50 \log (5)) \int \frac {x}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-(200 \log (5)) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx-\left (450 \log ^2(5)\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (\frac {12}{4+e^x}\right )\right )\\ &=\frac {225 \log ^2(5)}{\log ^2\left (\frac {12}{4+e^x}\right )}+(25 \log (5)) \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {12}{4+x}\right )} \, dx,x,e^x\right )+(50 \log (5)) \int \frac {x}{\log ^3\left (\frac {12}{4+e^x}\right )} \, dx-(200 \log (5)) \int \frac {x}{\left (4+e^x\right ) \log ^3\left (\frac {12}{4+e^x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 22, normalized size = 0.85 \begin {gather*} \frac {25 \log (5) (x+9 \log (5))}{\log ^2\left (\frac {12}{4+e^x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(50*x*Log[5] + 450*Log[5]^2) + (100*Log[5] + 25*E^x*Log[5])*Log[12/(4 + E^x)])/((4 + E^x)*Log[1
2/(4 + E^x)]^3),x]

[Out]

(25*Log[5]*(x + 9*Log[5]))/Log[12/(4 + E^x)]^2

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fricas [A]  time = 0.56, size = 24, normalized size = 0.92 \begin {gather*} \frac {25 \, {\left (x \log \relax (5) + 9 \, \log \relax (5)^{2}\right )}}{\log \left (\frac {12}{e^{x} + 4}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)*log(5)+100*log(5))*log(12/(exp(x)+4))+(450*log(5)^2+50*x*log(5))*exp(x))/(exp(x)+4)/log(
12/(exp(x)+4))^3,x, algorithm="fricas")

[Out]

25*(x*log(5) + 9*log(5)^2)/log(12/(e^x + 4))^2

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giac [A]  time = 0.19, size = 36, normalized size = 1.38 \begin {gather*} \frac {25 \, {\left (x \log \relax (5) + 9 \, \log \relax (5)^{2}\right )}}{\log \left (12\right )^{2} - 2 \, \log \left (12\right ) \log \left (e^{x} + 4\right ) + \log \left (e^{x} + 4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)*log(5)+100*log(5))*log(12/(exp(x)+4))+(450*log(5)^2+50*x*log(5))*exp(x))/(exp(x)+4)/log(
12/(exp(x)+4))^3,x, algorithm="giac")

[Out]

25*(x*log(5) + 9*log(5)^2)/(log(12)^2 - 2*log(12)*log(e^x + 4) + log(e^x + 4)^2)

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maple [A]  time = 0.22, size = 25, normalized size = 0.96

method	result	size
default	\(\frac {25 x \ln \relax (5)+225 \ln \relax (5)^{2}}{\ln \left (\frac {12}{{\mathrm e}^{x}+4}\right )^{2}}\)	\(25\)
norman	\(\frac {25 x \ln \relax (5)+225 \ln \relax (5)^{2}}{\ln \left (\frac {12}{{\mathrm e}^{x}+4}\right )^{2}}\)	\(25\)
risch	\(-\frac {100 \ln \relax (5) \left (9 \ln \relax (5)+x \right )}{\left (2 i \ln \relax (3)+4 i \ln \relax (2)-2 i \ln \left ({\mathrm e}^{x}+4\right )\right )^{2}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*exp(x)*ln(5)+100*ln(5))*ln(12/(exp(x)+4))+(450*ln(5)^2+50*x*ln(5))*exp(x))/(exp(x)+4)/ln(12/(exp(x)+4
))^3,x,method=_RETURNVERBOSE)

[Out]

(25*x*ln(5)+225*ln(5)^2)/ln(12/(exp(x)+4))^2

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maxima [B]  time = 0.49, size = 92, normalized size = 3.54 \begin {gather*} \frac {25 \, x \log \relax (5)}{\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \left (e^{x} + 4\right ) + \log \left (e^{x} + 4\right )^{2}} + \frac {225 \, \log \relax (5)^{2}}{\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2} - 2 \, {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \left (e^{x} + 4\right ) + \log \left (e^{x} + 4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)*log(5)+100*log(5))*log(12/(exp(x)+4))+(450*log(5)^2+50*x*log(5))*exp(x))/(exp(x)+4)/log(
12/(exp(x)+4))^3,x, algorithm="maxima")

[Out]

25*x*log(5)/(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2 - 2*(log(3) + 2*log(2))*log(e^x + 4) + log(e^x + 4)^2) +
225*log(5)^2/(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2 - 2*(log(3) + 2*log(2))*log(e^x + 4) + log(e^x + 4)^2)

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mupad [B]  time = 3.26, size = 21, normalized size = 0.81 \begin {gather*} \frac {25\,\ln \relax (5)\,\left (x+9\,\ln \relax (5)\right )}{{\ln \left (\frac {12}{{\mathrm {e}}^x+4}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(50*x*log(5) + 450*log(5)^2) + log(12/(exp(x) + 4))*(100*log(5) + 25*exp(x)*log(5)))/(log(12/(exp(
x) + 4))^3*(exp(x) + 4)),x)

[Out]

(25*log(5)*(x + 9*log(5)))/log(12/(exp(x) + 4))^2

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sympy [A]  time = 0.15, size = 22, normalized size = 0.85 \begin {gather*} \frac {25 x \log {\relax (5 )} + 225 \log {\relax (5 )}^{2}}{\log {\left (\frac {12}{e^{x} + 4} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*exp(x)*ln(5)+100*ln(5))*ln(12/(exp(x)+4))+(450*ln(5)**2+50*x*ln(5))*exp(x))/(exp(x)+4)/ln(12/(e
xp(x)+4))**3,x)

[Out]

(25*x*log(5) + 225*log(5)**2)/log(12/(exp(x) + 4))**2

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