3.17.47 \(\int \frac {-4 x^2 \log ^2(10)+4 x^2 \log (10) \log ^2(3 x) \log (\frac {5 e^3+x^2}{e^3})+(-20 e^3-4 x^2) \log (10) \log (3 x) \log ^2(\frac {5 e^3+x^2}{e^3})+(5 e^3+x^2) \log ^3(\frac {5 e^3+x^2}{e^3})+(20 e^3+4 x^2) \log ^3(3 x) \log ^3(\frac {5 e^3+x^2}{e^3})}{(5 e^3 x+x^3) \log ^3(\frac {5 e^3+x^2}{e^3})} \, dx\)

Optimal. Leaf size=29 \[ \log (x)+\left (-\log ^2(3 x)+\frac {\log (10)}{\log \left (5+\frac {x^2}{e^3}\right )}\right )^2 \]

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Rubi [F]  time = 1.80, 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 {-4 x^2 \log ^2(10)+4 x^2 \log (10) \log ^2(3 x) \log \left (\frac {5 e^3+x^2}{e^3}\right )+\left (-20 e^3-4 x^2\right ) \log (10) \log (3 x) \log ^2\left (\frac {5 e^3+x^2}{e^3}\right )+\left (5 e^3+x^2\right ) \log ^3\left (\frac {5 e^3+x^2}{e^3}\right )+\left (20 e^3+4 x^2\right ) \log ^3(3 x) \log ^3\left (\frac {5 e^3+x^2}{e^3}\right )}{\left (5 e^3 x+x^3\right ) \log ^3\left (\frac {5 e^3+x^2}{e^3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x^2*Log[10]^2 + 4*x^2*Log[10]*Log[3*x]^2*Log[(5*E^3 + x^2)/E^3] + (-20*E^3 - 4*x^2)*Log[10]*Log[3*x]*L
og[(5*E^3 + x^2)/E^3]^2 + (5*E^3 + x^2)*Log[(5*E^3 + x^2)/E^3]^3 + (20*E^3 + 4*x^2)*Log[3*x]^3*Log[(5*E^3 + x^
2)/E^3]^3)/((5*E^3*x + x^3)*Log[(5*E^3 + x^2)/E^3]^3),x]

[Out]

Log[x] + Log[3*x]^4 + Log[10]^2/Log[5 + x^2/E^3]^2 - 2*Log[10]*Defer[Int][Log[3*x]^2/((I*Sqrt[5]*E^(3/2) - x)*
Log[5 + x^2/E^3]^2), x] + 2*Log[10]*Defer[Int][Log[3*x]^2/((I*Sqrt[5]*E^(3/2) + x)*Log[5 + x^2/E^3]^2), x] - 4
*Log[10]*Defer[Int][Log[3*x]/(x*Log[5 + x^2/E^3]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 x^2 \log ^2(10)+4 x^2 \log (10) \log ^2(3 x) \log \left (\frac {5 e^3+x^2}{e^3}\right )+\left (-20 e^3-4 x^2\right ) \log (10) \log (3 x) \log ^2\left (\frac {5 e^3+x^2}{e^3}\right )+\left (5 e^3+x^2\right ) \log ^3\left (\frac {5 e^3+x^2}{e^3}\right )+\left (20 e^3+4 x^2\right ) \log ^3(3 x) \log ^3\left (\frac {5 e^3+x^2}{e^3}\right )}{x \left (5 e^3+x^2\right ) \log ^3\left (\frac {5 e^3+x^2}{e^3}\right )} \, dx\\ &=\int \frac {1+4 \log ^3(3 x)-\frac {4 x^2 \log ^2(10)}{\left (5 e^3+x^2\right ) \log ^3\left (5+\frac {x^2}{e^3}\right )}+\frac {4 x^2 \log (10) \log ^2(3 x)}{\left (5 e^3+x^2\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )}-\frac {4 \log (10) \log (3 x)}{\log \left (5+\frac {x^2}{e^3}\right )}}{x} \, dx\\ &=\int \left (\frac {1+4 \log ^3(3 x)}{x}-\frac {4 x \log ^2(10)}{\left (5 e^3+x^2\right ) \log ^3\left (5+\frac {x^2}{e^3}\right )}+\frac {4 x \log (10) \log ^2(3 x)}{\left (5 e^3+x^2\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )}-\frac {4 \log (10) \log (3 x)}{x \log \left (5+\frac {x^2}{e^3}\right )}\right ) \, dx\\ &=(4 \log (10)) \int \frac {x \log ^2(3 x)}{\left (5 e^3+x^2\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx-(4 \log (10)) \int \frac {\log (3 x)}{x \log \left (5+\frac {x^2}{e^3}\right )} \, dx-\left (4 \log ^2(10)\right ) \int \frac {x}{\left (5 e^3+x^2\right ) \log ^3\left (5+\frac {x^2}{e^3}\right )} \, dx+\int \frac {1+4 \log ^3(3 x)}{x} \, dx\\ &=(4 \log (10)) \int \left (-\frac {\log ^2(3 x)}{2 \left (i \sqrt {5} e^{3/2}-x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )}+\frac {\log ^2(3 x)}{2 \left (i \sqrt {5} e^{3/2}+x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )}\right ) \, dx-(4 \log (10)) \int \frac {\log (3 x)}{x \log \left (5+\frac {x^2}{e^3}\right )} \, dx-\left (2 \log ^2(10)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (5 e^3+x\right ) \log ^3\left (5+\frac {x}{e^3}\right )} \, dx,x,x^2\right )+\int \left (\frac {1}{x}+\frac {4 \log ^3(3 x)}{x}\right ) \, dx\\ &=\log (x)+4 \int \frac {\log ^3(3 x)}{x} \, dx-(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}-x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx+(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}+x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx-(4 \log (10)) \int \frac {\log (3 x)}{x \log \left (5+\frac {x^2}{e^3}\right )} \, dx-\left (2 e^3 \log ^2(10)\right ) \operatorname {Subst}\left (\int \frac {1}{e^3 x \log ^3(x)} \, dx,x,5+\frac {x^2}{e^3}\right )\\ &=\log (x)+4 \operatorname {Subst}\left (\int x^3 \, dx,x,\log (3 x)\right )-(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}-x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx+(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}+x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx-(4 \log (10)) \int \frac {\log (3 x)}{x \log \left (5+\frac {x^2}{e^3}\right )} \, dx-\left (2 \log ^2(10)\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,5+\frac {x^2}{e^3}\right )\\ &=\log (x)+\log ^4(3 x)-(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}-x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx+(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}+x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx-(4 \log (10)) \int \frac {\log (3 x)}{x \log \left (5+\frac {x^2}{e^3}\right )} \, dx-\left (2 \log ^2(10)\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (5+\frac {x^2}{e^3}\right )\right )\\ &=\log (x)+\log ^4(3 x)+\frac {\log ^2(10)}{\log ^2\left (5+\frac {x^2}{e^3}\right )}-(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}-x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx+(2 \log (10)) \int \frac {\log ^2(3 x)}{\left (i \sqrt {5} e^{3/2}+x\right ) \log ^2\left (5+\frac {x^2}{e^3}\right )} \, dx-(4 \log (10)) \int \frac {\log (3 x)}{x \log \left (5+\frac {x^2}{e^3}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 48, normalized size = 1.66 \begin {gather*} \log (x)+\log ^4(3 x)+\frac {\log ^2(10)}{\log ^2\left (5+\frac {x^2}{e^3}\right )}-\frac {2 \log (10) \log ^2(3 x)}{\log \left (5+\frac {x^2}{e^3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^2*Log[10]^2 + 4*x^2*Log[10]*Log[3*x]^2*Log[(5*E^3 + x^2)/E^3] + (-20*E^3 - 4*x^2)*Log[10]*Log[
3*x]*Log[(5*E^3 + x^2)/E^3]^2 + (5*E^3 + x^2)*Log[(5*E^3 + x^2)/E^3]^3 + (20*E^3 + 4*x^2)*Log[3*x]^3*Log[(5*E^
3 + x^2)/E^3]^3)/((5*E^3*x + x^3)*Log[(5*E^3 + x^2)/E^3]^3),x]

[Out]

Log[x] + Log[3*x]^4 + Log[10]^2/Log[5 + x^2/E^3]^2 - (2*Log[10]*Log[3*x]^2)/Log[5 + x^2/E^3]

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fricas [B]  time = 0.83, size = 82, normalized size = 2.83 \begin {gather*} \frac {\log \left ({\left (x^{2} + 5 \, e^{3}\right )} e^{\left (-3\right )}\right )^{2} \log \left (3 \, x\right )^{4} - 2 \, \log \left (10\right ) \log \left ({\left (x^{2} + 5 \, e^{3}\right )} e^{\left (-3\right )}\right ) \log \left (3 \, x\right )^{2} + \log \left ({\left (x^{2} + 5 \, e^{3}\right )} e^{\left (-3\right )}\right )^{2} \log \left (3 \, x\right ) + \log \left (10\right )^{2}}{\log \left ({\left (x^{2} + 5 \, e^{3}\right )} e^{\left (-3\right )}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(3)+4*x^2)*log((5*exp(3)+x^2)/exp(3))^3*log(3*x)^3+4*x^2*log(10)*log((5*exp(3)+x^2)/exp(3))*
log(3*x)^2+(-20*exp(3)-4*x^2)*log(10)*log((5*exp(3)+x^2)/exp(3))^2*log(3*x)+(5*exp(3)+x^2)*log((5*exp(3)+x^2)/
exp(3))^3-4*x^2*log(10)^2)/(5*x*exp(3)+x^3)/log((5*exp(3)+x^2)/exp(3))^3,x, algorithm="fricas")

[Out]

(log((x^2 + 5*e^3)*e^(-3))^2*log(3*x)^4 - 2*log(10)*log((x^2 + 5*e^3)*e^(-3))*log(3*x)^2 + log((x^2 + 5*e^3)*e
^(-3))^2*log(3*x) + log(10)^2)/log((x^2 + 5*e^3)*e^(-3))^2

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giac [B]  time = 0.49, size = 135, normalized size = 4.66 \begin {gather*} \frac {\log \left (x^{2} + 5 \, e^{3}\right )^{2} \log \left (3 \, x\right )^{4} - 6 \, \log \left (x^{2} + 5 \, e^{3}\right ) \log \left (3 \, x\right )^{4} - 2 \, \log \left (10\right ) \log \left (x^{2} + 5 \, e^{3}\right ) \log \left (3 \, x\right )^{2} + 9 \, \log \left (3 \, x\right )^{4} + 6 \, \log \left (10\right ) \log \left (3 \, x\right )^{2} + \log \left (x^{2} + 5 \, e^{3}\right )^{2} \log \relax (x) + \log \left (10\right )^{2} - 6 \, \log \left (x^{2} + 5 \, e^{3}\right ) \log \relax (x) + 9 \, \log \relax (x)}{\log \left (x^{2} + 5 \, e^{3}\right )^{2} - 6 \, \log \left (x^{2} + 5 \, e^{3}\right ) + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(3)+4*x^2)*log((5*exp(3)+x^2)/exp(3))^3*log(3*x)^3+4*x^2*log(10)*log((5*exp(3)+x^2)/exp(3))*
log(3*x)^2+(-20*exp(3)-4*x^2)*log(10)*log((5*exp(3)+x^2)/exp(3))^2*log(3*x)+(5*exp(3)+x^2)*log((5*exp(3)+x^2)/
exp(3))^3-4*x^2*log(10)^2)/(5*x*exp(3)+x^3)/log((5*exp(3)+x^2)/exp(3))^3,x, algorithm="giac")

[Out]

(log(x^2 + 5*e^3)^2*log(3*x)^4 - 6*log(x^2 + 5*e^3)*log(3*x)^4 - 2*log(10)*log(x^2 + 5*e^3)*log(3*x)^2 + 9*log
(3*x)^4 + 6*log(10)*log(3*x)^2 + log(x^2 + 5*e^3)^2*log(x) + log(10)^2 - 6*log(x^2 + 5*e^3)*log(x) + 9*log(x))
/(log(x^2 + 5*e^3)^2 - 6*log(x^2 + 5*e^3) + 9)

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maple [B]  time = 0.28, size = 84, normalized size = 2.90

method	result	size
risch	\(\ln \left (3 x \right )^{4}+\ln \relax (x )+\frac {-2 \ln \left (3 x \right )^{2} \ln \left (\left (5 \,{\mathrm e}^{3}+x^{2}\right ) {\mathrm e}^{-3}\right ) \ln \relax (5)-2 \ln \left (3 x \right )^{2} \ln \left (\left (5 \,{\mathrm e}^{3}+x^{2}\right ) {\mathrm e}^{-3}\right ) \ln \relax (2)+\ln \relax (5)^{2}+2 \ln \relax (2) \ln \relax (5)+\ln \relax (2)^{2}}{\ln \left (\left (5 \,{\mathrm e}^{3}+x^{2}\right ) {\mathrm e}^{-3}\right )^{2}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*exp(3)+4*x^2)*ln((5*exp(3)+x^2)/exp(3))^3*ln(3*x)^3+4*x^2*ln(10)*ln((5*exp(3)+x^2)/exp(3))*ln(3*x)^2+
(-20*exp(3)-4*x^2)*ln(10)*ln((5*exp(3)+x^2)/exp(3))^2*ln(3*x)+(5*exp(3)+x^2)*ln((5*exp(3)+x^2)/exp(3))^3-4*x^2
*ln(10)^2)/(5*x*exp(3)+x^3)/ln((5*exp(3)+x^2)/exp(3))^3,x,method=_RETURNVERBOSE)

[Out]

ln(3*x)^4+ln(x)+(-2*ln(3*x)^2*ln((5*exp(3)+x^2)*exp(-3))*ln(5)-2*ln(3*x)^2*ln((5*exp(3)+x^2)*exp(-3))*ln(2)+ln
(5)^2+2*ln(2)*ln(5)+ln(2)^2)/ln((5*exp(3)+x^2)*exp(-3))^2

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maxima [B]  time = 0.67, size = 163, normalized size = 5.62 \begin {gather*} \frac {\log \left (10\right )^{2}}{\log \left (x^{2} + 5 \, e^{3}\right )^{2} - 6 \, \log \left (x^{2} + 5 \, e^{3}\right ) + 9} - \frac {12 \, \log \relax (3) \log \relax (x)^{3} + 3 \, \log \relax (x)^{4} + 2 \, \log \relax (5) \log \relax (3)^{2} + 2 \, \log \relax (3)^{2} \log \relax (2) + 2 \, {\left (9 \, \log \relax (3)^{2} + \log \relax (5) + \log \relax (2)\right )} \log \relax (x)^{2} - {\left (6 \, \log \relax (3)^{2} \log \relax (x)^{2} + 4 \, \log \relax (3) \log \relax (x)^{3} + \log \relax (x)^{4} + {\left (4 \, \log \relax (3)^{3} + 1\right )} \log \relax (x)\right )} \log \left (x^{2} + 5 \, e^{3}\right ) + {\left (12 \, \log \relax (3)^{3} + 4 \, \log \relax (5) \log \relax (3) + 4 \, \log \relax (3) \log \relax (2) + 3\right )} \log \relax (x)}{\log \left (x^{2} + 5 \, e^{3}\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(3)+4*x^2)*log((5*exp(3)+x^2)/exp(3))^3*log(3*x)^3+4*x^2*log(10)*log((5*exp(3)+x^2)/exp(3))*
log(3*x)^2+(-20*exp(3)-4*x^2)*log(10)*log((5*exp(3)+x^2)/exp(3))^2*log(3*x)+(5*exp(3)+x^2)*log((5*exp(3)+x^2)/
exp(3))^3-4*x^2*log(10)^2)/(5*x*exp(3)+x^3)/log((5*exp(3)+x^2)/exp(3))^3,x, algorithm="maxima")

[Out]

log(10)^2/(log(x^2 + 5*e^3)^2 - 6*log(x^2 + 5*e^3) + 9) - (12*log(3)*log(x)^3 + 3*log(x)^4 + 2*log(5)*log(3)^2
+ 2*log(3)^2*log(2) + 2*(9*log(3)^2 + log(5) + log(2))*log(x)^2 - (6*log(3)^2*log(x)^2 + 4*log(3)*log(x)^3 +
log(x)^4 + (4*log(3)^3 + 1)*log(x))*log(x^2 + 5*e^3) + (12*log(3)^3 + 4*log(5)*log(3) + 4*log(3)*log(2) + 3)*l
og(x))/(log(x^2 + 5*e^3) - 3)

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mupad [B]  time = 1.91, size = 106, normalized size = 3.66 \begin {gather*} \ln \relax (x)+4\,\ln \relax (3)\,{\ln \relax (x)}^3+4\,{\ln \relax (3)}^3\,\ln \relax (x)+{\ln \relax (x)}^4+6\,{\ln \relax (3)}^2\,{\ln \relax (x)}^2+\frac {{\ln \left (10\right )}^2}{{\ln \left ({\mathrm {e}}^{-3}\,x^2+5\right )}^2}-\frac {2\,{\ln \relax (3)}^2\,\ln \left (10\right )}{\ln \left ({\mathrm {e}}^{-3}\,x^2+5\right )}-\frac {2\,\ln \left (10\right )\,{\ln \relax (x)}^2}{\ln \left ({\mathrm {e}}^{-3}\,x^2+5\right )}-\frac {4\,\ln \relax (3)\,\ln \left (10\right )\,\ln \relax (x)}{\ln \left ({\mathrm {e}}^{-3}\,x^2+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(-3)*(5*exp(3) + x^2))^3*(5*exp(3) + x^2) - 4*x^2*log(10)^2 + log(exp(-3)*(5*exp(3) + x^2))^3*log(
3*x)^3*(20*exp(3) + 4*x^2) + 4*x^2*log(exp(-3)*(5*exp(3) + x^2))*log(3*x)^2*log(10) - log(exp(-3)*(5*exp(3) +
x^2))^2*log(3*x)*log(10)*(20*exp(3) + 4*x^2))/(log(exp(-3)*(5*exp(3) + x^2))^3*(5*x*exp(3) + x^3)),x)

[Out]

log(x) + 4*log(3)*log(x)^3 + 4*log(3)^3*log(x) + log(x)^4 + 6*log(3)^2*log(x)^2 + log(10)^2/log(x^2*exp(-3) +
5)^2 - (2*log(3)^2*log(10))/log(x^2*exp(-3) + 5) - (2*log(10)*log(x)^2)/log(x^2*exp(-3) + 5) - (4*log(3)*log(1
0)*log(x))/log(x^2*exp(-3) + 5)

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sympy [B]  time = 0.42, size = 54, normalized size = 1.86 \begin {gather*} \frac {- 2 \log {\left (10 \right )} \log {\left (3 x \right )}^{2} \log {\left (\frac {x^{2} + 5 e^{3}}{e^{3}} \right )} + \log {\left (10 \right )}^{2}}{\log {\left (\frac {x^{2} + 5 e^{3}}{e^{3}} \right )}^{2}} + \log {\relax (x )} + \log {\left (3 x \right )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(3)+4*x**2)*ln((5*exp(3)+x**2)/exp(3))**3*ln(3*x)**3+4*x**2*ln(10)*ln((5*exp(3)+x**2)/exp(3)
)*ln(3*x)**2+(-20*exp(3)-4*x**2)*ln(10)*ln((5*exp(3)+x**2)/exp(3))**2*ln(3*x)+(5*exp(3)+x**2)*ln((5*exp(3)+x**
2)/exp(3))**3-4*x**2*ln(10)**2)/(5*x*exp(3)+x**3)/ln((5*exp(3)+x**2)/exp(3))**3,x)

[Out]

(-2*log(10)*log(3*x)**2*log((x**2 + 5*exp(3))*exp(-3)) + log(10)**2)/log((x**2 + 5*exp(3))*exp(-3))**2 + log(x
) + log(3*x)**4

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