3.20.49 \(\int \frac {-1+e^2+5 e^4+(e^2+5 e^4) \log (x)+(1+10 e^2+(1+10 e^2) \log (x)) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+(2 e^2+2 e^2 \log (x)) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx\)

Optimal. Leaf size=17 \[ 5 x+\frac {x}{e^2+\log (1+\log (x))} \]

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Rubi [F]  time = 0.47, 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 {-1+e^2+5 e^4+\left (e^2+5 e^4\right ) \log (x)+\left (1+10 e^2+\left (1+10 e^2\right ) \log (x)\right ) \log (1+\log (x))+(5+5 \log (x)) \log ^2(1+\log (x))}{e^4+e^4 \log (x)+\left (2 e^2+2 e^2 \log (x)\right ) \log (1+\log (x))+(1+\log (x)) \log ^2(1+\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + E^2 + 5*E^4 + (E^2 + 5*E^4)*Log[x] + (1 + 10*E^2 + (1 + 10*E^2)*Log[x])*Log[1 + Log[x]] + (5 + 5*Log
[x])*Log[1 + Log[x]]^2)/(E^4 + E^4*Log[x] + (2*E^2 + 2*E^2*Log[x])*Log[1 + Log[x]] + (1 + Log[x])*Log[1 + Log[
x]]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 17, normalized size = 1.00 \begin {gather*} 5 x+\frac {x}{e^2+\log (1+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^2 + 5*E^4 + (E^2 + 5*E^4)*Log[x] + (1 + 10*E^2 + (1 + 10*E^2)*Log[x])*Log[1 + Log[x]] + (5 +
5*Log[x])*Log[1 + Log[x]]^2)/(E^4 + E^4*Log[x] + (2*E^2 + 2*E^2*Log[x])*Log[1 + Log[x]] + (1 + Log[x])*Log[1
+ Log[x]]^2),x]

[Out]

5*x + x/(E^2 + Log[1 + Log[x]])

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fricas [A]  time = 0.74, size = 26, normalized size = 1.53 \begin {gather*} \frac {5 \, x e^{2} + 5 \, x \log \left (\log \relax (x) + 1\right ) + x}{e^{2} + \log \left (\log \relax (x) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(x)+5)*log(log(x)+1)^2+((10*exp(2)+1)*log(x)+10*exp(2)+1)*log(log(x)+1)+(5*exp(2)^2+exp(2))*l
og(x)+5*exp(2)^2+exp(2)-1)/((log(x)+1)*log(log(x)+1)^2+(2*exp(2)*log(x)+2*exp(2))*log(log(x)+1)+exp(2)^2*log(x
)+exp(2)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.34, size = 46, normalized size = 2.71 \begin {gather*} \frac {5 \, x e^{2}}{e^{2} + \log \left (\log \relax (x) + 1\right )} + \frac {5 \, x \log \left (\log \relax (x) + 1\right )}{e^{2} + \log \left (\log \relax (x) + 1\right )} + \frac {x}{e^{2} + \log \left (\log \relax (x) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(x)+5)*log(log(x)+1)^2+((10*exp(2)+1)*log(x)+10*exp(2)+1)*log(log(x)+1)+(5*exp(2)^2+exp(2))*l
og(x)+5*exp(2)^2+exp(2)-1)/((log(x)+1)*log(log(x)+1)^2+(2*exp(2)*log(x)+2*exp(2))*log(log(x)+1)+exp(2)^2*log(x
)+exp(2)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 17, normalized size = 1.00

method	result	size
risch	\(\frac {x}{{\mathrm e}^{2}+\ln \left (\ln \relax (x )+1\right )}+5 x\)	\(17\)
norman	\(\frac {\left (5 \,{\mathrm e}^{2}+1\right ) x +5 x \ln \left (\ln \relax (x )+1\right )}{{\mathrm e}^{2}+\ln \left (\ln \relax (x )+1\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*ln(x)+5)*ln(ln(x)+1)^2+((10*exp(2)+1)*ln(x)+10*exp(2)+1)*ln(ln(x)+1)+(5*exp(2)^2+exp(2))*ln(x)+5*exp(2
)^2+exp(2)-1)/((ln(x)+1)*ln(ln(x)+1)^2+(2*exp(2)*ln(x)+2*exp(2))*ln(ln(x)+1)+exp(2)^2*ln(x)+exp(2)^2),x,method
=_RETURNVERBOSE)

[Out]

x/(exp(2)+ln(ln(x)+1))+5*x

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maxima [A]  time = 0.56, size = 28, normalized size = 1.65 \begin {gather*} \frac {x {\left (5 \, e^{2} + 1\right )} + 5 \, x \log \left (\log \relax (x) + 1\right )}{e^{2} + \log \left (\log \relax (x) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(x)+5)*log(log(x)+1)^2+((10*exp(2)+1)*log(x)+10*exp(2)+1)*log(log(x)+1)+(5*exp(2)^2+exp(2))*l
og(x)+5*exp(2)^2+exp(2)-1)/((log(x)+1)*log(log(x)+1)^2+(2*exp(2)*log(x)+2*exp(2))*log(log(x)+1)+exp(2)^2*log(x
)+exp(2)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.57, size = 25, normalized size = 1.47 \begin {gather*} \frac {x\,\left (5\,{\mathrm {e}}^2+5\,\ln \left (\ln \relax (x)+1\right )+1\right )}{{\mathrm {e}}^2+\ln \left (\ln \relax (x)+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2) + 5*exp(4) + log(log(x) + 1)^2*(5*log(x) + 5) + log(log(x) + 1)*(10*exp(2) + log(x)*(10*exp(2) + 1
) + 1) + log(x)*(exp(2) + 5*exp(4)) - 1)/(exp(4) + exp(4)*log(x) + log(log(x) + 1)^2*(log(x) + 1) + log(log(x)
+ 1)*(2*exp(2) + 2*exp(2)*log(x))),x)

[Out]

(x*(5*exp(2) + 5*log(log(x) + 1) + 1))/(exp(2) + log(log(x) + 1))

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sympy [A]  time = 0.28, size = 14, normalized size = 0.82 \begin {gather*} 5 x + \frac {x}{\log {\left (\log {\relax (x )} + 1 \right )} + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*ln(x)+5)*ln(ln(x)+1)**2+((10*exp(2)+1)*ln(x)+10*exp(2)+1)*ln(ln(x)+1)+(5*exp(2)**2+exp(2))*ln(x)
+5*exp(2)**2+exp(2)-1)/((ln(x)+1)*ln(ln(x)+1)**2+(2*exp(2)*ln(x)+2*exp(2))*ln(ln(x)+1)+exp(2)**2*ln(x)+exp(2)*
*2),x)

[Out]

5*x + x/(log(log(x) + 1) + exp(2))

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