3.69.100 $\int \frac{-90x^2-18x^3-2x^4-2x^3\log (x)+(-10x^2+18x^3+4x^4+(30x^2+6x^3)\log (x))\log (5+x)}{45+9x+(-30x-6x^2+(-30-6x)\log (x))\log (5+x)+(5x^2+x^3+(10x+2x^2)\log (x)+(5+x){\log }^2(x)){\log }^2(5+x)}dx$

Optimal. Leaf size=20 $$27+\frac{2x^3}{-3+(x+\log (x))\log (5+x)}$$

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Rubi [F]  time = 14.52, 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 = {} $$\begin{array}{c}\int \frac{-90x^2-18x^3-2x^4-2x^3\log (x)+(-10x^2+18x^3+4x^4+(30x^2+6x^3)\log (x))\log (5+x)}{45+9x+(-30x-6x^2+(-30-6x)\log (x))\log (5+x)+(5x^2+x^3+(10x+2x^2)\log (x)+(5+x){\log }^2(x)){\log }^2(5+x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-90*x^2 - 18*x^3 - 2*x^4 - 2*x^3*Log[x] + (-10*x^2 + 18*x^3 + 4*x^4 + (30*x^2 + 6*x^3)*Log[x])*Log[5 + x]
)/(45 + 9*x + (-30*x - 6*x^2 + (-30 - 6*x)*Log[x])*Log[5 + x] + (5*x^2 + x^3 + (10*x + 2*x^2)*Log[x] + (5 + x)
*Log[x]^2)*Log[5 + x]^2),x]

[Out]

-1250*Defer[Int][1/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] + 250*Defer[Int][x/((x + Log[x
])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] - 56*Defer[Int][x^2/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x
]*Log[5 + x])^2), x] + 4*Defer[Int][x^3/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] - 2*Defer
[Int][x^4/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] + 6250*Defer[Int][1/((5 + x)*(x + Log[x
])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] + 500*Defer[Int][Log[x]/((x + Log[x])*(-3 + x*Log[5 + x] + L
og[x]*Log[5 + x])^2), x] - 100*Defer[Int][(x*Log[x])/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2),
x] + 20*Defer[Int][(x^2*Log[x])/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] - 4*Defer[Int][(
x^3*Log[x])/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] - 2500*Defer[Int][Log[x]/((5 + x)*(x
+ Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] - 50*Defer[Int][Log[x]^2/((x + Log[x])*(-3 + x*Log[5
+ x] + Log[x]*Log[5 + x])^2), x] + 10*Defer[Int][(x*Log[x]^2)/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5
+ x])^2), x] - 2*Defer[Int][(x^2*Log[x]^2)/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] + 250*
Defer[Int][Log[x]^2/((5 + x)*(x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])^2), x] - 2*Defer[Int][x^2/((
x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])), x] + 4*Defer[Int][x^3/((x + Log[x])*(-3 + x*Log[5 + x] +
Log[x]*Log[5 + x])), x] + 6*Defer[Int][(x^2*Log[x])/((x + Log[x])*(-3 + x*Log[5 + x] + Log[x]*Log[5 + x])), x
]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2x^2(-45-9x-x^2+(-5+9x+2x^2)\log (5+x)+\log (x)(-x+3(5+x)\log (5+x)))}{(5+x)(3-(x+\log (x))\log (5+x){)}^2}dx\\ & =2\int \frac{x^2(-45-9x-x^2+(-5+9x+2x^2)\log (5+x)+\log (x)(-x+3(5+x)\log (5+x)))}{(5+x)(3-(x+\log (x))\log (5+x){)}^2}dx\\ & =2\int (-\frac{x^2(15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x))}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x^2(-1+2x+3\log (x))}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))})dx\\ & =-(2\int \frac{x^2(15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x))}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx)+2\int \frac{x^2(-1+2x+3\log (x))}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx\\ & =-(2\int (-\frac{5(15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x))}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x(15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x))}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{25(15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x))}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2})dx)+2\int (-\frac{x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}+\frac{2x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}+\frac{3x^2\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))})dx\\ & =-(2\int \frac{x(15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x))}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx)-2\int \frac{x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx+4\int \frac{x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx+6\int \frac{x^2\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx+10\int \frac{15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-50\int \frac{15+18x+3x^2+x^3+2x^2\log (x)+x{\log }^2(x)}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx\\ & =-(2\int \frac{x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx)-2\int (\frac{15x}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{18x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{3x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x^4}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{2x^3\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x^2{\log }^2(x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2})dx+4\int \frac{x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx+6\int \frac{x^2\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx+10\int (\frac{15}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{18x}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{3x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{2x^2\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x{\log }^2(x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2})dx-50\int (\frac{15}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{18x}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{3x^2}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x^3}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{2x^2\log (x)}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}+\frac{x{\log }^2(x)}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2})dx\\ & =-(2\int \frac{x^4}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx)-2\int \frac{x^2{\log }^2(x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-2\int \frac{x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx-4\int \frac{x^3\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx+4\int \frac{x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx-6\int \frac{x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx+6\int \frac{x^2\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x))}dx+10\int \frac{x^3}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx+10\int \frac{x{\log }^2(x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx+20\int \frac{x^2\log (x)}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-30\int \frac{x}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx+30\int \frac{x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-36\int \frac{x^2}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-50\int \frac{x^3}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-50\int \frac{x{\log }^2(x)}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-100\int \frac{x^2\log (x)}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx+150\int \frac{1}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-150\int \frac{x^2}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx+180\int \frac{x}{(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-750\int \frac{1}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx-900\int \frac{x}{(5+x)(x+\log (x))(-3+x\log (5+x)+\log (x)\log (5+x){)}^2}dx\\ & =\text{Rest of rules removed due to large latex content}\end{array}\end{array}$$

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Mathematica [A]  time = 2.38, size = 18, normalized size = 0.90 $$\begin{array}{c}\frac{2x^3}{-3+(x+\log (x))\log (5+x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-90*x^2 - 18*x^3 - 2*x^4 - 2*x^3*Log[x] + (-10*x^2 + 18*x^3 + 4*x^4 + (30*x^2 + 6*x^3)*Log[x])*Log[
5 + x])/(45 + 9*x + (-30*x - 6*x^2 + (-30 - 6*x)*Log[x])*Log[5 + x] + (5*x^2 + x^3 + (10*x + 2*x^2)*Log[x] + (
5 + x)*Log[x]^2)*Log[5 + x]^2),x]

[Out]

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

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fricas [A]  time = 0.78, size = 18, normalized size = 0.90 $$\begin{array}{c}\frac{2x^3}{(x+\log (x))\log (x+5)-3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3+30*x^2)*log(x)+4*x^4+18*x^3-10*x^2)*log(5+x)-2*x^3*log(x)-2*x^4-18*x^3-90*x^2)/(((5+x)*log(
x)^2+(2*x^2+10*x)*log(x)+x^3+5*x^2)*log(5+x)^2+((-6*x-30)*log(x)-6*x^2-30*x)*log(5+x)+9*x+45),x, algorithm="fr
icas")

[Out]

2*x^3/((x + log(x))*log(x + 5) - 3)

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giac [A]  time = 0.30, size = 22, normalized size = 1.10 $$\begin{array}{c}\frac{2x^3}{x\log (x+5)+\log (x+5)\log (x)-3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3+30*x^2)*log(x)+4*x^4+18*x^3-10*x^2)*log(5+x)-2*x^3*log(x)-2*x^4-18*x^3-90*x^2)/(((5+x)*log(
x)^2+(2*x^2+10*x)*log(x)+x^3+5*x^2)*log(5+x)^2+((-6*x-30)*log(x)-6*x^2-30*x)*log(5+x)+9*x+45),x, algorithm="gi
ac")

[Out]

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

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maple [A]  time = 0.03, size = 23, normalized size = 1.15

method	result	size
risch	$\frac{2x^3}{\ln (x)\ln (5+x)+x\ln (5+x)-3}$	$23$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*x^3+30*x^2)*ln(x)+4*x^4+18*x^3-10*x^2)*ln(5+x)-2*x^3*ln(x)-2*x^4-18*x^3-90*x^2)/(((5+x)*ln(x)^2+(2*x^
2+10*x)*ln(x)+x^3+5*x^2)*ln(5+x)^2+((-6*x-30)*ln(x)-6*x^2-30*x)*ln(5+x)+9*x+45),x,method=_RETURNVERBOSE)

[Out]

2*x^3/(ln(x)*ln(5+x)+x*ln(5+x)-3)

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maxima [A]  time = 0.47, size = 18, normalized size = 0.90 $$\begin{array}{c}\frac{2x^3}{(x+\log (x))\log (x+5)-3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3+30*x^2)*log(x)+4*x^4+18*x^3-10*x^2)*log(5+x)-2*x^3*log(x)-2*x^4-18*x^3-90*x^2)/(((5+x)*log(
x)^2+(2*x^2+10*x)*log(x)+x^3+5*x^2)*log(5+x)^2+((-6*x-30)*log(x)-6*x^2-30*x)*log(5+x)+9*x+45),x, algorithm="ma
xima")

[Out]

2*x^3/((x + log(x))*log(x + 5) - 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 $$\begin{array}{c}\int -\frac{2x^3\ln (x)-\ln (x+5)(\ln (x)(6x^3+30x^2)-10x^2+18x^3+4x^4)+90x^2+18x^3+2x^4}{({\ln (x)}^2(x+5)+\ln (x)(2x^2+10x)+5x^2+x^3){\ln (x+5)}^2+(-30x-\ln (x)(6x+30)-6x^2)\ln (x+5)+9x+45}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^3*log(x) - log(x + 5)*(log(x)*(30*x^2 + 6*x^3) - 10*x^2 + 18*x^3 + 4*x^4) + 90*x^2 + 18*x^3 + 2*x^4)
/(9*x - log(x + 5)*(30*x + log(x)*(6*x + 30) + 6*x^2) + log(x + 5)^2*(log(x)^2*(x + 5) + log(x)*(10*x + 2*x^2)
+ 5*x^2 + x^3) + 45),x)

[Out]

int(-(2*x^3*log(x) - log(x + 5)*(log(x)*(30*x^2 + 6*x^3) - 10*x^2 + 18*x^3 + 4*x^4) + 90*x^2 + 18*x^3 + 2*x^4)
/(9*x - log(x + 5)*(30*x + log(x)*(6*x + 30) + 6*x^2) + log(x + 5)^2*(log(x)^2*(x + 5) + log(x)*(10*x + 2*x^2)
+ 5*x^2 + x^3) + 45), x)

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sympy [A]  time = 0.37, size = 15, normalized size = 0.75 $$\begin{array}{c}\frac{2x^3}{(x+\log (x))\log (x+5)-3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x**3+30*x**2)*ln(x)+4*x**4+18*x**3-10*x**2)*ln(5+x)-2*x**3*ln(x)-2*x**4-18*x**3-90*x**2)/(((5+x
)*ln(x)**2+(2*x**2+10*x)*ln(x)+x**3+5*x**2)*ln(5+x)**2+((-6*x-30)*ln(x)-6*x**2-30*x)*ln(5+x)+9*x+45),x)

[Out]

2*x**3/((x + log(x))*log(x + 5) - 3)

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