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

Optimal. Leaf size=22 \[ \log \left (-3+\frac {1}{2} \left (2-\log ^4\left (5+e^x+x+\log (x)\right )\right )\right ) \]

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Rubi [A]  time = 0.59, antiderivative size = 14, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 3, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6688, 12, 6684} \begin {gather*} \log \left (\log ^4\left (x+e^x+\log (x)+5\right )+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[4 + Log[5 + E^x + x + Log[x]]^4]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6684

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (1+x+e^x x\right ) \log ^3\left (5+e^x+x+\log (x)\right )}{x \left (5+e^x+x+\log (x)\right ) \left (4+\log ^4\left (5+e^x+x+\log (x)\right )\right )} \, dx\\ &=4 \int \frac {\left (1+x+e^x x\right ) \log ^3\left (5+e^x+x+\log (x)\right )}{x \left (5+e^x+x+\log (x)\right ) \left (4+\log ^4\left (5+e^x+x+\log (x)\right )\right )} \, dx\\ &=\log \left (4+\log ^4\left (5+e^x+x+\log (x)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 14, normalized size = 0.64 \begin {gather*} \log \left (4+\log ^4\left (5+e^x+x+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[4 + Log[5 + E^x + x + Log[x]]^4]

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fricas [A]  time = 1.09, size = 13, normalized size = 0.59 \begin {gather*} \log \left (\log \left (x + e^{x} + \log \relax (x) + 5\right )^{4} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x + e^x + log(x) + 5)^4 + 4)

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giac [A]  time = 5.24, size = 13, normalized size = 0.59 \begin {gather*} \log \left (\log \left (x + e^{x} + \log \relax (x) + 5\right )^{4} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x + e^x + log(x) + 5)^4 + 4)

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maple [A]  time = 0.03, size = 14, normalized size = 0.64

method	result	size
risch	\(\ln \left (\ln \left (\ln \relax (x )+{\mathrm e}^{x}+5+x \right )^{4}+4\right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(ln(x)+exp(x)+5+x)^4+4)

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maxima [B]  time = 0.57, size = 47, normalized size = 2.14 \begin {gather*} \log \left (\log \left (x + e^{x} + \log \relax (x) + 5\right )^{2} + 2 \, \log \left (x + e^{x} + \log \relax (x) + 5\right ) + 2\right ) + \log \left (\log \left (x + e^{x} + \log \relax (x) + 5\right )^{2} - 2 \, \log \left (x + e^{x} + \log \relax (x) + 5\right ) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x + e^x + log(x) + 5)^2 + 2*log(x + e^x + log(x) + 5) + 2) + log(log(x + e^x + log(x) + 5)^2 - 2*log(x
+ e^x + log(x) + 5) + 2)

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mupad [B]  time = 4.45, size = 13, normalized size = 0.59 \begin {gather*} \ln \left ({\ln \left (x+{\mathrm {e}}^x+\ln \relax (x)+5\right )}^4+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x + exp(x) + log(x) + 5)^4 + 4)

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sympy [A]  time = 11.45, size = 15, normalized size = 0.68 \begin {gather*} \log {\left (\log {\left (x + e^{x} + \log {\relax (x )} + 5 \right )}^{4} + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x + exp(x) + log(x) + 5)**4 + 4)

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