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

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

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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 = {} $$\begin{array}{c}\int \frac{-e^{e^4+x}x+2x^2+x\log (x)+(-16-x+e^{e^4+x}(-16+15x+x^2)+(16+x)\log (x))\log (16+x)}{64x^2+4x^3+e^{2e^4+2x}(16+x)+e^{e^4+x}(-64x-4x^2)+(e^{e^4+x}(-32-2x)+64x+4x^2)\log (x)+(16+x){\log }^2(x)}dx\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-\left((e^{e^4+x}-2x)x\right)+(-1+e^{e^4+x}(-1+x))(16+x)\log (16+x)+\log (x)(x+(16+x)\log (16+x))}{(16+x){(e^{e^4+x}-2x-\log (x))}^2}dx\\ & =\int (\frac{(-1-2x+2x^2+x\log (x))\log (16+x)}{{(e^{e^4+x}-2x-\log (x))}^2}-\frac{-x-16\log (16+x)+15x\log (16+x)+x^2\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))})dx\\ & =\int \frac{(-1-2x+2x^2+x\log (x))\log (16+x)}{{(e^{e^4+x}-2x-\log (x))}^2}dx-\int \frac{-x-16\log (16+x)+15x\log (16+x)+x^2\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx\\ & =\int (-\frac{\log (16+x)}{{(e^{e^4+x}-2x-\log (x))}^2}-\frac{2x\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}+\frac{2x^2\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}+\frac{x\log (x)\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2})dx-\int (-\frac{x}{(16+x)(-e^{e^4+x}+2x+\log (x))}-\frac{16\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}+\frac{15x\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}+\frac{x^2\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))})dx\\ & =-(2\int \frac{x\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx)+2\int \frac{x^2\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx-15\int \frac{x\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx+16\int \frac{\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx+\int \frac{x}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx-\int \frac{\log (16+x)}{{(e^{e^4+x}-2x-\log (x))}^2}dx+\int \frac{x\log (x)\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx-\int \frac{x^2\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx\\ & =-(2\int \frac{x\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx)+2\int \frac{x^2\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx-15\int (-\frac{\log (16+x)}{e^{e^4+x}-2x-\log (x)}-\frac{16\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))})dx+16\int \frac{\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx+\int (-\frac{1}{e^{e^4+x}-2x-\log (x)}-\frac{16}{(16+x)(-e^{e^4+x}+2x+\log (x))})dx-\int \frac{\log (16+x)}{{(e^{e^4+x}-2x-\log (x))}^2}dx+\int \frac{x\log (x)\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx-\int (\frac{16\log (16+x)}{e^{e^4+x}-2x-\log (x)}+\frac{x\log (16+x)}{-e^{e^4+x}+2x+\log (x)}+\frac{256\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))})dx\\ & =-(2\int \frac{x\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx)+2\int \frac{x^2\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx+15\int \frac{\log (16+x)}{e^{e^4+x}-2x-\log (x)}dx-16\int \frac{1}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx-16\int \frac{\log (16+x)}{e^{e^4+x}-2x-\log (x)}dx+16\int \frac{\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx+240\int \frac{\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx-256\int \frac{\log (16+x)}{(16+x)(-e^{e^4+x}+2x+\log (x))}dx-\int \frac{1}{e^{e^4+x}-2x-\log (x)}dx-\int \frac{\log (16+x)}{{(e^{e^4+x}-2x-\log (x))}^2}dx+\int \frac{x\log (x)\log (16+x)}{{(-e^{e^4+x}+2x+\log (x))}^2}dx-\int \frac{x\log (16+x)}{-e^{e^4+x}+2x+\log (x)}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{x\log (x+16)}{2x-e^{(x+e^4)}+\log (x)}\end{array}$$

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 $$\begin{array}{c}\frac{x\log (x+16)}{2x-e^{(x+e^4)}+\log (x)}\end{array}$$

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	$\frac{x\ln (x+16)}{2x-{\mathrm{e}}^{x+{\mathrm{e}}^4}+\ln (x)}$	$22$

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 $$\begin{array}{c}\frac{x\log (x+16)}{2x-e^{(x+e^4)}+\log (x)}\end{array}$$

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 $$\begin{array}{c}-\frac{30x^4\ln (x+16)-33x^3\ln (x+16)-16x^2\ln (x+16)+2x^5\ln (x+16)+\ln (x)(16x^3\ln (x+16)+x^4\ln (x+16))}{(x+16)(2x-{\mathrm{e}}^{x+{\mathrm{e}}^4}+\ln (x))(x-x^2\ln (x)+2x^2-2x^3)}\end{array}$$

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 $$\begin{array}{c}-\frac{x\log (x+16)}{-2x+e^{x+e^4}-\log (x)}\end{array}$$

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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