3.69.69 $\int \frac{e^5(-10-2x+10x^2+2x^3)+e^{10}(10-8x-2x^2){\log }^2(5+x)+\log (\frac{3e^x}{x})(e^5(-10x-2x^2)+4e^{10}x\log (5+x))}{5x+11x^2+7x^3+x^4+e^5(-10x-12x^2-2x^3){\log }^2(5+x)+e^{10}(5x+x^2){\log }^4(5+x)}dx$

Optimal. Leaf size=29 $$\frac{2\log \left(\frac{3e^x}{x}\right)}{\frac{1+x}{e^5}-{\log }^2(5+x)}$$

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Rubi [F]  time = 2.92, 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 = {} $$\begin{array}{c}\int \frac{e^5(-10-2x+10x^2+2x^3)+e^{10}(10-8x-2x^2){\log }^2(5+x)+\log \left(\frac{3e^x}{x}\right)(e^5(-10x-2x^2)+4e^{10}x\log (5+x))}{5x+11x^2+7x^3+x^4+e^5(-10x-12x^2-2x^3){\log }^2(5+x)+e^{10}(5x+x^2){\log }^4(5+x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^5*(-10 - 2*x + 10*x^2 + 2*x^3) + E^10*(10 - 8*x - 2*x^2)*Log[5 + x]^2 + Log[(3*E^x)/x]*(E^5*(-10*x - 2*
x^2) + 4*E^10*x*Log[5 + x]))/(5*x + 11*x^2 + 7*x^3 + x^4 + E^5*(-10*x - 12*x^2 - 2*x^3)*Log[5 + x]^2 + E^10*(5
*x + x^2)*Log[5 + x]^4),x]

[Out]

-2*E^5*Defer[Int][Log[(3*E^x)/x]/(1 + x - E^5*Log[5 + x]^2)^2, x] + 4*E^10*Defer[Int][(Log[(3*E^x)/x]*Log[5 +
x])/((5 + x)*(1 + x - E^5*Log[5 + x]^2)^2), x] + 2*E^5*Defer[Int][(1 + x - E^5*Log[5 + x]^2)^(-1), x] - 2*E^5*
Defer[Int][1/(x*(1 + x - E^5*Log[5 + x]^2)), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2e^5(-x\log \left(\frac{3e^x}{x}\right)(5+x-2e^5\log (5+x))+(-5+4x+x^2)(1+x-e^5{\log }^2(5+x)))}{x(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx\\ & =\left(2e^5\right)\int \frac{-x\log \left(\frac{3e^x}{x}\right)(5+x-2e^5\log (5+x))+(-5+4x+x^2)(1+x-e^5{\log }^2(5+x))}{x(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx\\ & =\left(2e^5\right)\int (-\frac{\log \left(\frac{3e^x}{x}\right)(5+x-2e^5\log (5+x))}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}+\frac{-1+x}{x(1+x-e^5{\log }^2(5+x))})dx\\ & =-(\left(2e^5\right)\int \frac{\log \left(\frac{3e^x}{x}\right)(5+x-2e^5\log (5+x))}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx)+\left(2e^5\right)\int \frac{-1+x}{x(1+x-e^5{\log }^2(5+x))}dx\\ & =-(\left(2e^5\right)\int (\frac{5\log \left(\frac{3e^x}{x}\right)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}+\frac{x\log \left(\frac{3e^x}{x}\right)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}-\frac{2e^5\log \left(\frac{3e^x}{x}\right)\log (5+x)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2})dx)+\left(2e^5\right)\int (\frac{1}{1+x-e^5{\log }^2(5+x)}-\frac{1}{x(1+x-e^5{\log }^2(5+x))})dx\\ & =-(\left(2e^5\right)\int \frac{x\log \left(\frac{3e^x}{x}\right)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx)+\left(2e^5\right)\int \frac{1}{1+x-e^5{\log }^2(5+x)}dx-\left(2e^5\right)\int \frac{1}{x(1+x-e^5{\log }^2(5+x))}dx-\left(10e^5\right)\int \frac{\log \left(\frac{3e^x}{x}\right)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx+\left(4e^{10}\right)\int \frac{\log \left(\frac{3e^x}{x}\right)\log (5+x)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx\\ & =\left(2e^5\right)\int \frac{1}{1+x-e^5{\log }^2(5+x)}dx-\left(2e^5\right)\int \frac{1}{x(1+x-e^5{\log }^2(5+x))}dx-\left(2e^5\right)\int (\frac{\log \left(\frac{3e^x}{x}\right)}{{(1+x-e^5{\log }^2(5+x))}^2}-\frac{5\log \left(\frac{3e^x}{x}\right)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2})dx-\left(10e^5\right)\int \frac{\log \left(\frac{3e^x}{x}\right)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx+\left(4e^{10}\right)\int \frac{\log \left(\frac{3e^x}{x}\right)\log (5+x)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx\\ & =-(\left(2e^5\right)\int \frac{\log \left(\frac{3e^x}{x}\right)}{{(1+x-e^5{\log }^2(5+x))}^2}dx)+\left(2e^5\right)\int \frac{1}{1+x-e^5{\log }^2(5+x)}dx-\left(2e^5\right)\int \frac{1}{x(1+x-e^5{\log }^2(5+x))}dx+\left(4e^{10}\right)\int \frac{\log \left(\frac{3e^x}{x}\right)\log (5+x)}{(5+x){(1+x-e^5{\log }^2(5+x))}^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.86, size = 31, normalized size = 1.07 $$\begin{array}{c}-\frac{2e^5\log \left(\frac{3e^x}{x}\right)}{-1-x+e^5{\log }^2(5+x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-10 - 2*x + 10*x^2 + 2*x^3) + E^10*(10 - 8*x - 2*x^2)*Log[5 + x]^2 + Log[(3*E^x)/x]*(E^5*(-10*
x - 2*x^2) + 4*E^10*x*Log[5 + x]))/(5*x + 11*x^2 + 7*x^3 + x^4 + E^5*(-10*x - 12*x^2 - 2*x^3)*Log[5 + x]^2 + E
^10*(5*x + x^2)*Log[5 + x]^4),x]

[Out]

(-2*E^5*Log[(3*E^x)/x])/(-1 - x + E^5*Log[5 + x]^2)

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fricas [A]  time = 0.84, size = 30, normalized size = 1.03 $$\begin{array}{c}-\frac{2e^5\log \left(\frac{e^{(x+\log (3))}}{x}\right)}{e^5\log {(x+5)}^2-x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)
^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x
^3+11*x^2+5*x),x, algorithm="fricas")

[Out]

-2*e^5*log(e^(x + log(3))/x)/(e^5*log(x + 5)^2 - x - 1)

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giac [A]  time = 0.80, size = 34, normalized size = 1.17 $$\begin{array}{c}-\frac{4(x{e}^5+e^5\log (3)-e^5\log (x))}{e^5\log {(x+5)}^2-x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)
^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x
^3+11*x^2+5*x),x, algorithm="giac")

[Out]

-4*(x*e^5 + e^5*log(3) - e^5*log(x))/(e^5*log(x + 5)^2 - x - 1)

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maple [C]  time = 2.36, size = 135, normalized size = 4.66

method	result	size
risch	$-\frac{2{\mathrm{e}}^5\ln \left(3{\mathrm{e}}^x\right)}{{\mathrm{e}}^5\ln {(5+x)}^2-x-1}+\frac{{\mathrm{e}}^5(i\pi \mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}\left(i{\mathrm{e}}^x\right)\mathrm{csgn}\left(\frac{i{\mathrm{e}}^x}{x}\right)-i\pi \mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^x}{x}\right)}^2-i\pi \mathrm{csgn}\left(i{\mathrm{e}}^x\right)\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^x}{x}\right)}^2+i\pi \mathrm{csgn}{\left(\frac{i{\mathrm{e}}^x}{x}\right)}^3+2\ln (x))}{{\mathrm{e}}^5\ln {(5+x)}^2-x-1}$	$135$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x*exp(5)^2*ln(5+x)+(-2*x^2-10*x)*exp(5))*ln(exp(ln(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*ln(5+x)^2+(2*x^3+
10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*ln(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*ln(5+x)^2+x^4+7*x^3+11*x^2+5*
x),x,method=_RETURNVERBOSE)

[Out]

-2*exp(5)/(exp(5)*ln(5+x)^2-x-1)*ln(3*exp(x))+exp(5)*(I*Pi*csgn(I/x)*csgn(I*exp(x))*csgn(I/x*exp(x))-I*Pi*csgn
(I/x)*csgn(I/x*exp(x))^2-I*Pi*csgn(I*exp(x))*csgn(I/x*exp(x))^2+I*Pi*csgn(I/x*exp(x))^3+2*ln(x))/(exp(5)*ln(5+
x)^2-x-1)

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maxima [A]  time = 0.50, size = 34, normalized size = 1.17 $$\begin{array}{c}-\frac{2(x{e}^5+e^5\log (3)-e^5\log (x))}{e^5\log {(x+5)}^2-x-1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)
^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x
^3+11*x^2+5*x),x, algorithm="maxima")

[Out]

-2*(x*e^5 + e^5*log(3) - e^5*log(x))/(e^5*log(x + 5)^2 - x - 1)

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mupad [B]  time = 4.62, size = 27, normalized size = 0.93 $$\begin{array}{c}\frac{2{\mathrm{e}}^5\ln \left(\frac{3{\mathrm{e}}^x}{x}\right)}{-{\mathrm{e}}^5{\ln (x+5)}^2+x+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*(2*x - 10*x^2 - 2*x^3 + 10) + log(exp(x + log(3))/x)*(exp(5)*(10*x + 2*x^2) - 4*x*log(x + 5)*exp(
10)) + log(x + 5)^2*exp(10)*(8*x + 2*x^2 - 10))/(5*x + 11*x^2 + 7*x^3 + x^4 - log(x + 5)^2*exp(5)*(10*x + 12*x
^2 + 2*x^3) + log(x + 5)^4*exp(10)*(5*x + x^2)),x)

[Out]

(2*exp(5)*log((3*exp(x))/x))/(x - log(x + 5)^2*exp(5) + 1)

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sympy [A]  time = 0.64, size = 26, normalized size = 0.90 $$\begin{array}{c}\frac{2e^5\log \left(\frac{3e^x}{x}\right)}{x-e^5\log {(x+5)}^2+1}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(5)**2*ln(5+x)+(-2*x**2-10*x)*exp(5))*ln(exp(ln(3)+x)/x)+(-2*x**2-8*x+10)*exp(5)**2*ln(5+x)
**2+(2*x**3+10*x**2-2*x-10)*exp(5))/((x**2+5*x)*exp(5)**2*ln(5+x)**4+(-2*x**3-12*x**2-10*x)*exp(5)*ln(5+x)**2+
x**4+7*x**3+11*x**2+5*x),x)

[Out]

2*exp(5)*log(3*exp(x)/x)/(x - exp(5)*log(x + 5)**2 + 1)

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