3.41.18 \(\int \frac {-e^{e^4+x} x+2 x^2+x \log (x)+(-16-x+e^{e^4+x} (-16+15 x+x^2)+(16+x) \log (x)) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} (-64 x-4 x^2)+(e^{e^4+x} (-32-2 x)+64 x+4 x^2) \log (x)+(16+x) \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ 2+\frac {x \log (16+x)}{-e^{e^4+x}+2 x+\log (x)} \]

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Rubi [F]  time = 7.35, 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^{e^4+x} x+2 x^2+x \log (x)+\left (-16-x+e^{e^4+x} \left (-16+15 x+x^2\right )+(16+x) \log (x)\right ) \log (16+x)}{64 x^2+4 x^3+e^{2 e^4+2 x} (16+x)+e^{e^4+x} \left (-64 x-4 x^2\right )+\left (e^{e^4+x} (-32-2 x)+64 x+4 x^2\right ) \log (x)+(16+x) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^(E^4 + x)*x) + 2*x^2 + x*Log[x] + (-16 - x + E^(E^4 + x)*(-16 + 15*x + x^2) + (16 + x)*Log[x])*Log[16
+ x])/(64*x^2 + 4*x^3 + E^(2*E^4 + 2*x)*(16 + x) + E^(E^4 + x)*(-64*x - 4*x^2) + (E^(E^4 + x)*(-32 - 2*x) + 6
4*x + 4*x^2)*Log[x] + (16 + x)*Log[x]^2),x]

[Out]

-Defer[Int][(E^(E^4 + x) - 2*x - Log[x])^(-1), x] - 16*Defer[Int][1/((16 + x)*(-E^(E^4 + x) + 2*x + Log[x])),
x] - Defer[Int][Log[16 + x]/(E^(E^4 + x) - 2*x - Log[x])^2, x] - Defer[Int][Log[16 + x]/(E^(E^4 + x) - 2*x - L
og[x]), x] - 2*Defer[Int][(x*Log[16 + x])/(-E^(E^4 + x) + 2*x + Log[x])^2, x] + 2*Defer[Int][(x^2*Log[16 + x])
/(-E^(E^4 + x) + 2*x + Log[x])^2, x] + Defer[Int][(x*Log[x]*Log[16 + x])/(-E^(E^4 + x) + 2*x + Log[x])^2, x] -
Defer[Int][(x*Log[16 + x])/(-E^(E^4 + x) + 2*x + Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\left (\left (e^{e^4+x}-2 x\right ) x\right )+\left (-1+e^{e^4+x} (-1+x)\right ) (16+x) \log (16+x)+\log (x) (x+(16+x) \log (16+x))}{(16+x) \left (e^{e^4+x}-2 x-\log (x)\right )^2} \, dx\\ &=\int \left (\frac {\left (-1-2 x+2 x^2+x \log (x)\right ) \log (16+x)}{\left (e^{e^4+x}-2 x-\log (x)\right )^2}-\frac {-x-16 \log (16+x)+15 x \log (16+x)+x^2 \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}\right ) \, dx\\ &=\int \frac {\left (-1-2 x+2 x^2+x \log (x)\right ) \log (16+x)}{\left (e^{e^4+x}-2 x-\log (x)\right )^2} \, dx-\int \frac {-x-16 \log (16+x)+15 x \log (16+x)+x^2 \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx\\ &=\int \left (-\frac {\log (16+x)}{\left (e^{e^4+x}-2 x-\log (x)\right )^2}-\frac {2 x \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2}+\frac {2 x^2 \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2}+\frac {x \log (x) \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2}\right ) \, dx-\int \left (-\frac {x}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}-\frac {16 \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}+\frac {15 x \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}+\frac {x^2 \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {x \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx\right )+2 \int \frac {x^2 \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx-15 \int \frac {x \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx+16 \int \frac {\log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx+\int \frac {x}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx-\int \frac {\log (16+x)}{\left (e^{e^4+x}-2 x-\log (x)\right )^2} \, dx+\int \frac {x \log (x) \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx-\int \frac {x^2 \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx\\ &=-\left (2 \int \frac {x \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx\right )+2 \int \frac {x^2 \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx-15 \int \left (-\frac {\log (16+x)}{e^{e^4+x}-2 x-\log (x)}-\frac {16 \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}\right ) \, dx+16 \int \frac {\log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx+\int \left (-\frac {1}{e^{e^4+x}-2 x-\log (x)}-\frac {16}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}\right ) \, dx-\int \frac {\log (16+x)}{\left (e^{e^4+x}-2 x-\log (x)\right )^2} \, dx+\int \frac {x \log (x) \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx-\int \left (\frac {16 \log (16+x)}{e^{e^4+x}-2 x-\log (x)}+\frac {x \log (16+x)}{-e^{e^4+x}+2 x+\log (x)}+\frac {256 \log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {x \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx\right )+2 \int \frac {x^2 \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx+15 \int \frac {\log (16+x)}{e^{e^4+x}-2 x-\log (x)} \, dx-16 \int \frac {1}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx-16 \int \frac {\log (16+x)}{e^{e^4+x}-2 x-\log (x)} \, dx+16 \int \frac {\log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx+240 \int \frac {\log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx-256 \int \frac {\log (16+x)}{(16+x) \left (-e^{e^4+x}+2 x+\log (x)\right )} \, dx-\int \frac {1}{e^{e^4+x}-2 x-\log (x)} \, dx-\int \frac {\log (16+x)}{\left (e^{e^4+x}-2 x-\log (x)\right )^2} \, dx+\int \frac {x \log (x) \log (16+x)}{\left (-e^{e^4+x}+2 x+\log (x)\right )^2} \, dx-\int \frac {x \log (16+x)}{-e^{e^4+x}+2 x+\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.85, size = 23, normalized size = 0.92 \begin {gather*} \frac {x \log (16+x)}{-e^{e^4+x}+2 x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(E^4 + x)*x) + 2*x^2 + x*Log[x] + (-16 - x + E^(E^4 + x)*(-16 + 15*x + x^2) + (16 + x)*Log[x])*
Log[16 + x])/(64*x^2 + 4*x^3 + E^(2*E^4 + 2*x)*(16 + x) + E^(E^4 + x)*(-64*x - 4*x^2) + (E^(E^4 + x)*(-32 - 2*
x) + 64*x + 4*x^2)*Log[x] + (16 + x)*Log[x]^2),x]

[Out]

(x*Log[16 + x])/(-E^(E^4 + x) + 2*x + Log[x])

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fricas [A]  time = 0.65, size = 21, normalized size = 0.84 \begin {gather*} \frac {x \log \left (x + 16\right )}{2 \, x - e^{\left (x + e^{4}\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+16)*log(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*log(x+16)+x*log(x)-x*exp(x+exp(4))+2*x^2)/((x+16)*
log(x)^2+((-2*x-32)*exp(x+exp(4))+4*x^2+64*x)*log(x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+
64*x^2),x, algorithm="fricas")

[Out]

x*log(x + 16)/(2*x - e^(x + e^4) + log(x))

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giac [A]  time = 0.18, size = 21, normalized size = 0.84 \begin {gather*} \frac {x \log \left (x + 16\right )}{2 \, x - e^{\left (x + e^{4}\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+16)*log(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*log(x+16)+x*log(x)-x*exp(x+exp(4))+2*x^2)/((x+16)*
log(x)^2+((-2*x-32)*exp(x+exp(4))+4*x^2+64*x)*log(x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+
64*x^2),x, algorithm="giac")

[Out]

x*log(x + 16)/(2*x - e^(x + e^4) + log(x))

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maple [A]  time = 0.04, size = 22, normalized size = 0.88

method	result	size
risch	\(\frac {x \ln \left (x +16\right )}{2 x -{\mathrm e}^{x +{\mathrm e}^{4}}+\ln \relax (x )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+16)*ln(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*ln(x+16)+x*ln(x)-x*exp(x+exp(4))+2*x^2)/((x+16)*ln(x)^2+(
(-2*x-32)*exp(x+exp(4))+4*x^2+64*x)*ln(x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+64*x^2),x,m
ethod=_RETURNVERBOSE)

[Out]

x*ln(x+16)/(2*x-exp(x+exp(4))+ln(x))

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maxima [A]  time = 0.45, size = 21, normalized size = 0.84 \begin {gather*} \frac {x \log \left (x + 16\right )}{2 \, x - e^{\left (x + e^{4}\right )} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+16)*log(x)+(x^2+15*x-16)*exp(x+exp(4))-x-16)*log(x+16)+x*log(x)-x*exp(x+exp(4))+2*x^2)/((x+16)*
log(x)^2+((-2*x-32)*exp(x+exp(4))+4*x^2+64*x)*log(x)+(x+16)*exp(x+exp(4))^2+(-4*x^2-64*x)*exp(x+exp(4))+4*x^3+
64*x^2),x, algorithm="maxima")

[Out]

x*log(x + 16)/(2*x - e^(x + e^4) + log(x))

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mupad [B]  time = 3.32, size = 101, normalized size = 4.04 \begin {gather*} -\frac {30\,x^4\,\ln \left (x+16\right )-33\,x^3\,\ln \left (x+16\right )-16\,x^2\,\ln \left (x+16\right )+2\,x^5\,\ln \left (x+16\right )+\ln \relax (x)\,\left (16\,x^3\,\ln \left (x+16\right )+x^4\,\ln \left (x+16\right )\right )}{\left (x+16\right )\,\left (2\,x-{\mathrm {e}}^{x+{\mathrm {e}}^4}+\ln \relax (x)\right )\,\left (x-x^2\,\ln \relax (x)+2\,x^2-2\,x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*exp(x + exp(4)) + log(x + 16)*(x - exp(x + exp(4))*(15*x + x^2 - 16) - log(x)*(x + 16) + 16) - x*log(x
) - 2*x^2)/(log(x)^2*(x + 16) - exp(x + exp(4))*(64*x + 4*x^2) + exp(2*x + 2*exp(4))*(x + 16) + 64*x^2 + 4*x^3
+ log(x)*(64*x - exp(x + exp(4))*(2*x + 32) + 4*x^2)),x)

[Out]

-(30*x^4*log(x + 16) - 33*x^3*log(x + 16) - 16*x^2*log(x + 16) + 2*x^5*log(x + 16) + log(x)*(16*x^3*log(x + 16
) + x^4*log(x + 16)))/((x + 16)*(2*x - exp(x + exp(4)) + log(x))*(x - x^2*log(x) + 2*x^2 - 2*x^3))

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sympy [A]  time = 0.36, size = 20, normalized size = 0.80 \begin {gather*} - \frac {x \log {\left (x + 16 \right )}}{- 2 x + e^{x + e^{4}} - \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+16)*ln(x)+(x**2+15*x-16)*exp(x+exp(4))-x-16)*ln(x+16)+x*ln(x)-x*exp(x+exp(4))+2*x**2)/((x+16)*l
n(x)**2+((-2*x-32)*exp(x+exp(4))+4*x**2+64*x)*ln(x)+(x+16)*exp(x+exp(4))**2+(-4*x**2-64*x)*exp(x+exp(4))+4*x**
3+64*x**2),x)

[Out]

-x*log(x + 16)/(-2*x + exp(x + exp(4)) - log(x))

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