3.85.4 $\int \frac{50x+25x^2+(50x+25x^2)\log (x)+(90+60x+10x^2+(-30-10x)\log (3+x))\log (1+\log (x))+(10x+5x^2+(10x+5x^2)\log (x)){\log }^2(1+\log (x))}{6x+2x^2+(6x+2x^2)\log (x)}dx$

Optimal. Leaf size=22 $$\frac{5}{2}(3+x-\log (3+x))(5+{\log }^2(1+\log (x)))$$

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Rubi [F]  time = 0.91, 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 = {} $$\begin{array}{c}\int \frac{50x+25x^2+(50x+25x^2)\log (x)+(90+60x+10x^2+(-30-10x)\log (3+x))\log (1+\log (x))+(10x+5x^2+(10x+5x^2)\log (x)){\log }^2(1+\log (x))}{6x+2x^2+(6x+2x^2)\log (x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(50*x + 25*x^2 + (50*x + 25*x^2)*Log[x] + (90 + 60*x + 10*x^2 + (-30 - 10*x)*Log[3 + x])*Log[1 + Log[x]] +
(10*x + 5*x^2 + (10*x + 5*x^2)*Log[x])*Log[1 + Log[x]]^2)/(6*x + 2*x^2 + (6*x + 2*x^2)*Log[x]),x]

[Out]

(25*x)/2 - (25*Log[3 + x])/2 + (15*Log[1 + Log[x]]^2)/2 + 5*Defer[Int][Log[1 + Log[x]]/(1 + Log[x]), x] - 5*De
fer[Int][(Log[3 + x]*Log[1 + Log[x]])/(x*(1 + Log[x])), x] + (5*Defer[Int][Log[1 + Log[x]]^2, x])/2 - (5*Defer
[Int][Log[1 + Log[x]]^2/(3 + x), x])/2

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{5(5x(2+x)+2(3+x)(3+x-\log (3+x))\log (1+\log (x))+x(2+x){\log }^2(1+\log (x))+x(2+x)\log (x)(5+{\log }^2(1+\log (x))))}{2x(3+x)(1+\log (x))}dx\\ & =\frac{5}{2}\int \frac{5x(2+x)+2(3+x)(3+x-\log (3+x))\log (1+\log (x))+x(2+x){\log }^2(1+\log (x))+x(2+x)\log (x)(5+{\log }^2(1+\log (x)))}{x(3+x)(1+\log (x))}dx\\ & =\frac{5}{2}\int (\frac{5(2+x)}{3+x}+\frac{2(3+x-\log (3+x))\log (1+\log (x))}{x(1+\log (x))}+\frac{(2+x){\log }^2(1+\log (x))}{3+x})dx\\ & =\frac{5}{2}\int \frac{(2+x){\log }^2(1+\log (x))}{3+x}dx+5\int \frac{(3+x-\log (3+x))\log (1+\log (x))}{x(1+\log (x))}dx+\frac{25}{2}\int \frac{2+x}{3+x}dx\\ & =\frac{5}{2}\int ({\log }^2(1+\log (x))-\frac{{\log }^2(1+\log (x))}{3+x})dx+5\int (\frac{\log (1+\log (x))}{1+\log (x)}+\frac{3\log (1+\log (x))}{x(1+\log (x))}-\frac{\log (3+x)\log (1+\log (x))}{x(1+\log (x))})dx+\frac{25}{2}\int (1+\frac{1}{-3-x})dx\\ & =\frac{25x}{2}-\frac{25}{2}\log (3+x)+\frac{5}{2}\int {\log }^2(1+\log (x))dx-\frac{5}{2}\int \frac{{\log }^2(1+\log (x))}{3+x}dx+5\int \frac{\log (1+\log (x))}{1+\log (x)}dx-5\int \frac{\log (3+x)\log (1+\log (x))}{x(1+\log (x))}dx+15\int \frac{\log (1+\log (x))}{x(1+\log (x))}dx\\ & =\frac{25x}{2}-\frac{25}{2}\log (3+x)+\frac{5}{2}\int {\log }^2(1+\log (x))dx-\frac{5}{2}\int \frac{{\log }^2(1+\log (x))}{3+x}dx+5\int \frac{\log (1+\log (x))}{1+\log (x)}dx-5\int \frac{\log (3+x)\log (1+\log (x))}{x(1+\log (x))}dx+15\mathrm{Subst}(\int \frac{\log (1+x)}{1+x}dx,x,\log (x))\\ & =\frac{25x}{2}-\frac{25}{2}\log (3+x)+\frac{5}{2}\int {\log }^2(1+\log (x))dx-\frac{5}{2}\int \frac{{\log }^2(1+\log (x))}{3+x}dx+5\int \frac{\log (1+\log (x))}{1+\log (x)}dx-5\int \frac{\log (3+x)\log (1+\log (x))}{x(1+\log (x))}dx+15\mathrm{Subst}(\int \frac{\log (x)}{x}dx,x,1+\log (x))\\ & =\frac{25x}{2}-\frac{25}{2}\log (3+x)+\frac{15}{2}{\log }^2(1+\log (x))+\frac{5}{2}\int {\log }^2(1+\log (x))dx-\frac{5}{2}\int \frac{{\log }^2(1+\log (x))}{3+x}dx+5\int \frac{\log (1+\log (x))}{1+\log (x)}dx-5\int \frac{\log (3+x)\log (1+\log (x))}{x(1+\log (x))}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.23, size = 31, normalized size = 1.41 $$\begin{array}{c}\frac{5}{2}(5x-5\log (3+x)+(3+x-\log (3+x)){\log }^2(1+\log (x)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(50*x + 25*x^2 + (50*x + 25*x^2)*Log[x] + (90 + 60*x + 10*x^2 + (-30 - 10*x)*Log[3 + x])*Log[1 + Log
[x]] + (10*x + 5*x^2 + (10*x + 5*x^2)*Log[x])*Log[1 + Log[x]]^2)/(6*x + 2*x^2 + (6*x + 2*x^2)*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.95, size = 28, normalized size = 1.27 $$\begin{array}{c}\frac{5}{2}(x-\log (x+3)+3)\log {(\log (x)+1)}^2+\frac{25}{2}x-\frac{25}{2}\log (x+3)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2+10*x)*log(x)+5*x^2+10*x)*log(log(x)+1)^2+((-10*x-30)*log(3+x)+10*x^2+60*x+90)*log(log(x)+1)
+(25*x^2+50*x)*log(x)+25*x^2+50*x)/((2*x^2+6*x)*log(x)+2*x^2+6*x),x, algorithm="fricas")

[Out]

5/2*(x - log(x + 3) + 3)*log(log(x) + 1)^2 + 25/2*x - 25/2*log(x + 3)

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giac [A]  time = 0.26, size = 36, normalized size = 1.64 $$\begin{array}{c}\frac{5}{2}(x-\log (x+3))\log {(\log (x)+1)}^2+\frac{15}{2}\log {(\log (x)+1)}^2+\frac{25}{2}x-\frac{25}{2}\log (x+3)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2+10*x)*log(x)+5*x^2+10*x)*log(log(x)+1)^2+((-10*x-30)*log(3+x)+10*x^2+60*x+90)*log(log(x)+1)
+(25*x^2+50*x)*log(x)+25*x^2+50*x)/((2*x^2+6*x)*log(x)+2*x^2+6*x),x, algorithm="giac")

[Out]

5/2*(x - log(x + 3))*log(log(x) + 1)^2 + 15/2*log(log(x) + 1)^2 + 25/2*x - 25/2*log(x + 3)

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maple [A]  time = 0.06, size = 30, normalized size = 1.36

method	result	size
risch	$(\frac{5x}{2}-\frac{5\ln (3+x)}{2}+\frac{15}{2})\ln {(\ln (x)+1)}^2+\frac{25x}{2}-\frac{25\ln (3+x)}{2}$	$30$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x^2+10*x)*ln(x)+5*x^2+10*x)*ln(ln(x)+1)^2+((-10*x-30)*ln(3+x)+10*x^2+60*x+90)*ln(ln(x)+1)+(25*x^2+50*
x)*ln(x)+25*x^2+50*x)/((2*x^2+6*x)*ln(x)+2*x^2+6*x),x,method=_RETURNVERBOSE)

[Out]

(5/2*x-5/2*ln(3+x)+15/2)*ln(ln(x)+1)^2+25/2*x-25/2*ln(3+x)

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maxima [A]  time = 0.40, size = 28, normalized size = 1.27 $$\begin{array}{c}\frac{5}{2}(x-\log (x+3)+3)\log {(\log (x)+1)}^2+\frac{25}{2}x-\frac{25}{2}\log (x+3)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2+10*x)*log(x)+5*x^2+10*x)*log(log(x)+1)^2+((-10*x-30)*log(3+x)+10*x^2+60*x+90)*log(log(x)+1)
+(25*x^2+50*x)*log(x)+25*x^2+50*x)/((2*x^2+6*x)*log(x)+2*x^2+6*x),x, algorithm="maxima")

[Out]

5/2*(x - log(x + 3) + 3)*log(log(x) + 1)^2 + 25/2*x - 25/2*log(x + 3)

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mupad [B]  time = 5.47, size = 47, normalized size = 2.14 $$\begin{array}{c}(\frac{5x^3+15x^2}{2x(x+3)}-\frac{5\ln (x+3)}{2}+\frac{15}{2}){\ln (\ln (x)+1)}^2+\frac{25x}{2}-\frac{25\ln (x+3)}{2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x + log(log(x) + 1)*(60*x + 10*x^2 - log(x + 3)*(10*x + 30) + 90) + log(log(x) + 1)^2*(10*x + log(x)*(
10*x + 5*x^2) + 5*x^2) + log(x)*(50*x + 25*x^2) + 25*x^2)/(6*x + log(x)*(6*x + 2*x^2) + 2*x^2),x)

[Out]

(25*x)/2 - (25*log(x + 3))/2 + log(log(x) + 1)^2*((15*x^2 + 5*x^3)/(2*x*(x + 3)) - (5*log(x + 3))/2 + 15/2)

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sympy [A]  time = 0.81, size = 37, normalized size = 1.68 $$\begin{array}{c}\frac{25x}{2}+(\frac{5x}{2}-\frac{5\log (x+3)}{2}+\frac{15}{2})\log {(\log (x)+1)}^2-\frac{25\log (x+3)}{2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x**2+10*x)*ln(x)+5*x**2+10*x)*ln(ln(x)+1)**2+((-10*x-30)*ln(3+x)+10*x**2+60*x+90)*ln(ln(x)+1)+(
25*x**2+50*x)*ln(x)+25*x**2+50*x)/((2*x**2+6*x)*ln(x)+2*x**2+6*x),x)

[Out]

25*x/2 + (5*x/2 - 5*log(x + 3)/2 + 15/2)*log(log(x) + 1)**2 - 25*log(x + 3)/2

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