3.94.52 \(\int \frac {-x \log (x)+(5+x+(5+x) \log (x)) \log (5+x)+(-5+e^x (-5-x)-x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx\) [9352]

Optimal. Leaf size=23 \[ -4+e^5-e^x-x+\frac {x \log (x)}{\log (5+x)} \]

[Out]

-x+ln(x)*x/ln(5+x)-exp(x)+exp(5)-4

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Rubi [F]
time = 0.73, 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 {-x \log (x)+(5+x+(5+x) \log (x)) \log (5+x)+\left (-5+e^x (-5-x)-x\right ) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-E^x - x + LogIntegral[5 + x] - Defer[Int][Log[x]/Log[5 + x]^2, x] + Defer[Int][Log[x]/Log[5 + x], x] + 5*Defe
r[Subst][Defer[Int][Log[-5 + x]/(x*Log[x]^2), x], x, 5 + x]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 0.10, size = 30, normalized size = 1.30 \begin {gather*} -e^x-x-\text {ExpIntegralEi}(\log (5+x))+\frac {x \log (x)}{\log (5+x)}+\text {LogIntegral}(5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^x - x - ExpIntegralEi[Log[5 + x]] + (x*Log[x])/Log[5 + x] + LogIntegral[5 + x]

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Maple [A]
time = 3.64, size = 19, normalized size = 0.83

method	result	size
default	\(-x +\frac {\ln \left (x \right ) x}{\ln \left (5+x \right )}-{\mathrm e}^{x}\)	\(19\)
risch	\(-x +\frac {\ln \left (x \right ) x}{\ln \left (5+x \right )}-{\mathrm e}^{x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x+ln(x)*x/ln(5+x)-exp(x)

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Maxima [A]
time = 0.31, size = 23, normalized size = 1.00 \begin {gather*} -\frac {{\left (x + e^{x}\right )} \log \left (x + 5\right ) - x \log \left (x\right )}{\log \left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 23, normalized size = 1.00 \begin {gather*} -\frac {{\left (x + e^{x}\right )} \log \left (x + 5\right ) - x \log \left (x\right )}{\log \left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.16, size = 14, normalized size = 0.61 \begin {gather*} \frac {x \log {\left (x \right )}}{\log {\left (x + 5 \right )}} - x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(x)/log(x + 5) - x - exp(x)

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Giac [A]
time = 0.41, size = 27, normalized size = 1.17 \begin {gather*} -\frac {x \log \left (x + 5\right ) + e^{x} \log \left (x + 5\right ) - x \log \left (x\right )}{\log \left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 7.49, size = 18, normalized size = 0.78 \begin {gather*} \frac {x\,\ln \left (x\right )}{\ln \left (x+5\right )}-{\mathrm {e}}^x-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(x))/log(x + 5) - exp(x) - x

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