\(\int \frac {72 x^2 \log (x)+e^2 (400 x+80 x^2) \log ^2(x)+(-360 x-72 x^2+(720 x+144 x^2) \log (x)) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx\) [9761]

Optimal result

Integrand size = 63, antiderivative size = 25 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=8 \left (e^2 x^2+\frac {9 x^2 \log (5+x)}{5 \log (x)}\right ) \]

[Out]

8*x^2*exp(2)+72/5*x^2/ln(x)*ln(5+x)

Rubi [F]

\[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx \]

[In]

Int[(72*x^2*Log[x] + E^2*(400*x + 80*x^2)*Log[x]^2 + (-360*x - 72*x^2 + (720*x + 144*x^2)*Log[x])*Log[5 + x])/
((25 + 5*x)*Log[x]^2),x]

[Out]

8*E^2*x^2 + (72*Defer[Int][x^2/((5 + x)*Log[x]), x])/5 - (72*Defer[Int][(x*Log[5 + x])/Log[x]^2, x])/5 + (144*
Defer[Int][(x*Log[5 + x])/Log[x], x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {8}{5} x \left (10 e^2-\frac {9 \log (5+x)}{\log ^2(x)}+\frac {9 (x+2 (5+x) \log (5+x))}{(5+x) \log (x)}\right ) \, dx \\ & = \frac {8}{5} \int x \left (10 e^2-\frac {9 \log (5+x)}{\log ^2(x)}+\frac {9 (x+2 (5+x) \log (5+x))}{(5+x) \log (x)}\right ) \, dx \\ & = \frac {8}{5} \int \left (\frac {x \left (9 x+50 e^2 \log (x)+10 e^2 x \log (x)\right )}{(5+x) \log (x)}+\frac {9 x (-1+2 \log (x)) \log (5+x)}{\log ^2(x)}\right ) \, dx \\ & = \frac {8}{5} \int \frac {x \left (9 x+50 e^2 \log (x)+10 e^2 x \log (x)\right )}{(5+x) \log (x)} \, dx+\frac {72}{5} \int \frac {x (-1+2 \log (x)) \log (5+x)}{\log ^2(x)} \, dx \\ & = \frac {8}{5} \int x \left (10 e^2+\frac {9 x}{(5+x) \log (x)}\right ) \, dx+\frac {72}{5} \int \left (-\frac {x \log (5+x)}{\log ^2(x)}+\frac {2 x \log (5+x)}{\log (x)}\right ) \, dx \\ & = \frac {8}{5} \int \left (10 e^2 x+\frac {9 x^2}{(5+x) \log (x)}\right ) \, dx-\frac {72}{5} \int \frac {x \log (5+x)}{\log ^2(x)} \, dx+\frac {144}{5} \int \frac {x \log (5+x)}{\log (x)} \, dx \\ & = 8 e^2 x^2+\frac {72}{5} \int \frac {x^2}{(5+x) \log (x)} \, dx-\frac {72}{5} \int \frac {x \log (5+x)}{\log ^2(x)} \, dx+\frac {144}{5} \int \frac {x \log (5+x)}{\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8}{5} \left (5 e^2 x^2+\frac {9 x^2 \log (5+x)}{\log (x)}\right ) \]

[In]

Integrate[(72*x^2*Log[x] + E^2*(400*x + 80*x^2)*Log[x]^2 + (-360*x - 72*x^2 + (720*x + 144*x^2)*Log[x])*Log[5
+ x])/((25 + 5*x)*Log[x]^2),x]

[Out]

(8*(5*E^2*x^2 + (9*x^2*Log[5 + x])/Log[x]))/5

Maple [A] (verified)

Time = 4.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
risch	\(8 x^{2} {\mathrm e}^{2}+\frac {72 x^{2} \ln \left (5+x \right )}{5 \ln \left (x \right )}\)	\(22\)
parallelrisch	\(\frac {40 x^{2} {\mathrm e}^{2} \ln \left (x \right )+72 \ln \left (5+x \right ) x^{2}-1000 \,{\mathrm e}^{2} \ln \left (x \right )}{5 \ln \left (x \right )}\)	\(32\)

[In]

int((((144*x^2+720*x)*ln(x)-72*x^2-360*x)*ln(5+x)+(80*x^2+400*x)*exp(2)*ln(x)^2+72*x^2*ln(x))/(25+5*x)/ln(x)^2
,x,method=_RETURNVERBOSE)

[Out]

8*x^2*exp(2)+72/5*x^2/ln(x)*ln(5+x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8 \, {\left (5 \, x^{2} e^{2} \log \left (x\right ) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \left (x\right )} \]

[In]

integrate((((144*x^2+720*x)*log(x)-72*x^2-360*x)*log(5+x)+(80*x^2+400*x)*exp(2)*log(x)^2+72*x^2*log(x))/(25+5*
x)/log(x)^2,x, algorithm="fricas")

[Out]

8/5*(5*x^2*e^2*log(x) + 9*x^2*log(x + 5))/log(x)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=8 x^{2} e^{2} + \frac {72 x^{2} \log {\left (x + 5 \right )}}{5 \log {\left (x \right )}} \]

[In]

integrate((((144*x**2+720*x)*ln(x)-72*x**2-360*x)*ln(5+x)+(80*x**2+400*x)*exp(2)*ln(x)**2+72*x**2*ln(x))/(25+5
*x)/ln(x)**2,x)

[Out]

8*x**2*exp(2) + 72*x**2*log(x + 5)/(5*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8 \, {\left (5 \, x^{2} e^{2} \log \left (x\right ) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \left (x\right )} \]

[In]

integrate((((144*x^2+720*x)*log(x)-72*x^2-360*x)*log(5+x)+(80*x^2+400*x)*exp(2)*log(x)^2+72*x^2*log(x))/(25+5*
x)/log(x)^2,x, algorithm="maxima")

[Out]

8/5*(5*x^2*e^2*log(x) + 9*x^2*log(x + 5))/log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=\frac {8 \, {\left (5 \, x^{2} e^{2} \log \left (x\right ) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \left (x\right )} \]

[In]

integrate((((144*x^2+720*x)*log(x)-72*x^2-360*x)*log(5+x)+(80*x^2+400*x)*exp(2)*log(x)^2+72*x^2*log(x))/(25+5*
x)/log(x)^2,x, algorithm="giac")

[Out]

8/5*(5*x^2*e^2*log(x) + 9*x^2*log(x + 5))/log(x)

Mupad [B] (verification not implemented)

Time = 14.74 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx=8\,x^2\,{\mathrm {e}}^2+\frac {72\,x^2\,\ln \left (x+5\right )}{5\,\ln \left (x\right )} \]

[In]

int((72*x^2*log(x) - log(x + 5)*(360*x - log(x)*(720*x + 144*x^2) + 72*x^2) + exp(2)*log(x)^2*(400*x + 80*x^2)
)/(log(x)^2*(5*x + 25)),x)

[Out]

8*x^2*exp(2) + (72*x^2*log(x + 5))/(5*log(x))