\(\int \frac {-4 x^2-8 x^3+(3 x^2+4 x^3) \log (x)+(16+64 x+64 x^2) \log ^5(x)}{(16+64 x+64 x^2) \log ^5(x)} \, dx\) [9364]

Optimal result

Integrand size = 57, antiderivative size = 22 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=x+\frac {x}{16 \left (\frac {1}{x^2}+\frac {2}{x}\right ) \log ^4(x)} \]

[Out]

x+1/16*x/(2/x+1/x^2)/ln(x)^4

Rubi [F]

\[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx \]

[In]

Int[(-4*x^2 - 8*x^3 + (3*x^2 + 4*x^3)*Log[x] + (16 + 64*x + 64*x^2)*Log[x]^5)/((16 + 64*x + 64*x^2)*Log[x]^5),
x]

[Out]

x - Defer[Int][x^2/((1 + 2*x)*Log[x]^5), x]/4 + Defer[Int][(x^2*(3 + 4*x))/((1 + 2*x)^2*Log[x]^4), x]/16

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{16 (1+2 x)^2 \log ^5(x)} \, dx \\ & = \frac {1}{16} \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{(1+2 x)^2 \log ^5(x)} \, dx \\ & = \frac {1}{16} \int \left (16-\frac {4 x^2}{(1+2 x) \log ^5(x)}+\frac {x^2 (3+4 x)}{(1+2 x)^2 \log ^4(x)}\right ) \, dx \\ & = x+\frac {1}{16} \int \frac {x^2 (3+4 x)}{(1+2 x)^2 \log ^4(x)} \, dx-\frac {1}{4} \int \frac {x^2}{(1+2 x) \log ^5(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.68 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=x+\frac {x^3}{16 (1+2 x) \log ^4(x)} \]

[In]

Integrate[(-4*x^2 - 8*x^3 + (3*x^2 + 4*x^3)*Log[x] + (16 + 64*x + 64*x^2)*Log[x]^5)/((16 + 64*x + 64*x^2)*Log[
x]^5),x]

[Out]

x + x^3/(16*(1 + 2*x)*Log[x]^4)

Maple [A] (verified)

Time = 6.73 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(x +\frac {x^{3}}{16 \left (1+2 x \right ) \ln \left (x \right )^{4}}\)	\(19\)
norman	\(\frac {-\frac {\ln \left (x \right )^{4}}{2}+\frac {x^{3}}{16}+2 x^{2} \ln \left (x \right )^{4}}{\left (1+2 x \right ) \ln \left (x \right )^{4}}\)	\(34\)
parallelrisch	\(\frac {64 x^{2} \ln \left (x \right )^{4}-16 \ln \left (x \right )^{4}+2 x^{3}}{32 \ln \left (x \right )^{4} \left (1+2 x \right )}\)	\(35\)

[In]

int(((64*x^2+64*x+16)*ln(x)^5+(4*x^3+3*x^2)*ln(x)-8*x^3-4*x^2)/(64*x^2+64*x+16)/ln(x)^5,x,method=_RETURNVERBOS
E)

[Out]

x+1/16*x^3/(1+2*x)/ln(x)^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {16 \, {\left (2 \, x^{2} + x\right )} \log \left (x\right )^{4} + x^{3}}{16 \, {\left (2 \, x + 1\right )} \log \left (x\right )^{4}} \]

[In]

integrate(((64*x^2+64*x+16)*log(x)^5+(4*x^3+3*x^2)*log(x)-8*x^3-4*x^2)/(64*x^2+64*x+16)/log(x)^5,x, algorithm=
"fricas")

[Out]

1/16*(16*(2*x^2 + x)*log(x)^4 + x^3)/((2*x + 1)*log(x)^4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {x^{3}}{\left (32 x + 16\right ) \log {\left (x \right )}^{4}} + x \]

[In]

integrate(((64*x**2+64*x+16)*ln(x)**5+(4*x**3+3*x**2)*ln(x)-8*x**3-4*x**2)/(64*x**2+64*x+16)/ln(x)**5,x)

[Out]

x**3/((32*x + 16)*log(x)**4) + x

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {16 \, {\left (2 \, x^{2} + x\right )} \log \left (x\right )^{4} + x^{3}}{16 \, {\left (2 \, x + 1\right )} \log \left (x\right )^{4}} \]

[In]

integrate(((64*x^2+64*x+16)*log(x)^5+(4*x^3+3*x^2)*log(x)-8*x^3-4*x^2)/(64*x^2+64*x+16)/log(x)^5,x, algorithm=
"maxima")

[Out]

1/16*(16*(2*x^2 + x)*log(x)^4 + x^3)/((2*x + 1)*log(x)^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {x^{3}}{16 \, {\left (2 \, x \log \left (x\right )^{4} + \log \left (x\right )^{4}\right )}} + x \]

[In]

integrate(((64*x^2+64*x+16)*log(x)^5+(4*x^3+3*x^2)*log(x)-8*x^3-4*x^2)/(64*x^2+64*x+16)/log(x)^5,x, algorithm=
"giac")

[Out]

1/16*x^3/(2*x*log(x)^4 + log(x)^4) + x

Mupad [B] (verification not implemented)

Time = 14.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {x^3}{16\,{\ln \left (x\right )}^4\,\left (2\,x+1\right )}+\frac {x\,\left (32\,x+16\right )}{16\,\left (2\,x+1\right )} \]

[In]

int((log(x)*(3*x^2 + 4*x^3) + log(x)^5*(64*x + 64*x^2 + 16) - 4*x^2 - 8*x^3)/(log(x)^5*(64*x + 64*x^2 + 16)),x
)

[Out]

x^3/(16*log(x)^4*(2*x + 1)) + (x*(32*x + 16))/(16*(2*x + 1))