\(\int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+(528 x+84 x^2-12 x^3) \log (4+x)+(12 x^2+3 x^3) \log ^2(4+x)} \, dx\) [4862]

Optimal result

Integrand size = 63, antiderivative size = 27 \[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\frac {x}{3 (-7+3 x-5 (3+x)+x (4-\log (4+x)))} \]

[Out]

1/3*x/(-2*x-22+x*(4-ln(4+x)))

Rubi [F]

\[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=\int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx \]

[In]

Int[(-88 - 22*x + x^2)/(5808 + 396*x - 216*x^2 + 12*x^3 + (528*x + 84*x^2 - 12*x^3)*Log[4 + x] + (12*x^2 + 3*x
^3)*Log[4 + x]^2),x]

[Out]

(-26*Defer[Int][(22 - 2*x + x*Log[4 + x])^(-2), x])/3 + Defer[Int][x/(22 - 2*x + x*Log[4 + x])^2, x]/3 + (16*D
efer[Int][1/((4 + x)*(22 - 2*x + x*Log[4 + x])^2), x])/3

Rubi steps \begin{align*} \text {integral}& = \int \frac {-88-22 x+x^2}{3 (4+x) (22-2 x+x \log (4+x))^2} \, dx \\ & = \frac {1}{3} \int \frac {-88-22 x+x^2}{(4+x) (22-2 x+x \log (4+x))^2} \, dx \\ & = \frac {1}{3} \int \left (-\frac {26}{(22-2 x+x \log (4+x))^2}+\frac {x}{(22-2 x+x \log (4+x))^2}+\frac {16}{(4+x) (22-2 x+x \log (4+x))^2}\right ) \, dx \\ & = \frac {1}{3} \int \frac {x}{(22-2 x+x \log (4+x))^2} \, dx+\frac {16}{3} \int \frac {1}{(4+x) (22-2 x+x \log (4+x))^2} \, dx-\frac {26}{3} \int \frac {1}{(22-2 x+x \log (4+x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {x}{3 (22-2 x+x \log (4+x))} \]

[In]

Integrate[(-88 - 22*x + x^2)/(5808 + 396*x - 216*x^2 + 12*x^3 + (528*x + 84*x^2 - 12*x^3)*Log[4 + x] + (12*x^2
+ 3*x^3)*Log[4 + x]^2),x]

[Out]

-1/3*x/(22 - 2*x + x*Log[4 + x])

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63

method	result	size
norman	\(-\frac {x}{3 \left (x \ln \left (4+x \right )-2 x +22\right )}\)	\(17\)
risch	\(-\frac {x}{3 \left (x \ln \left (4+x \right )-2 x +22\right )}\)	\(17\)
parallelrisch	\(-\frac {x}{3 \left (x \ln \left (4+x \right )-2 x +22\right )}\)	\(17\)
derivativedivides	\(-\frac {x}{3 \left (\left (4+x \right ) \ln \left (4+x \right )+22-2 x -4 \ln \left (4+x \right )\right )}\)	\(25\)
default	\(-\frac {x}{3 \left (\left (4+x \right ) \ln \left (4+x \right )+22-2 x -4 \ln \left (4+x \right )\right )}\)	\(25\)

[In]

int((x^2-22*x-88)/((3*x^3+12*x^2)*ln(4+x)^2+(-12*x^3+84*x^2+528*x)*ln(4+x)+12*x^3-216*x^2+396*x+5808),x,method
=_RETURNVERBOSE)

[Out]

-1/3*x/(x*ln(4+x)-2*x+22)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59 \[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {x}{3 \, {\left (x \log \left (x + 4\right ) - 2 \, x + 22\right )}} \]

[In]

integrate((x^2-22*x-88)/((3*x^3+12*x^2)*log(4+x)^2+(-12*x^3+84*x^2+528*x)*log(4+x)+12*x^3-216*x^2+396*x+5808),
x, algorithm="fricas")

[Out]

-1/3*x/(x*log(x + 4) - 2*x + 22)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=- \frac {x}{3 x \log {\left (x + 4 \right )} - 6 x + 66} \]

[In]

integrate((x**2-22*x-88)/((3*x**3+12*x**2)*ln(4+x)**2+(-12*x**3+84*x**2+528*x)*ln(4+x)+12*x**3-216*x**2+396*x+
5808),x)

[Out]

-x/(3*x*log(x + 4) - 6*x + 66)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59 \[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {x}{3 \, {\left (x \log \left (x + 4\right ) - 2 \, x + 22\right )}} \]

[In]

integrate((x^2-22*x-88)/((3*x^3+12*x^2)*log(4+x)^2+(-12*x^3+84*x^2+528*x)*log(4+x)+12*x^3-216*x^2+396*x+5808),
x, algorithm="maxima")

[Out]

-1/3*x/(x*log(x + 4) - 2*x + 22)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59 \[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {x}{3 \, {\left (x \log \left (x + 4\right ) - 2 \, x + 22\right )}} \]

[In]

integrate((x^2-22*x-88)/((3*x^3+12*x^2)*log(4+x)^2+(-12*x^3+84*x^2+528*x)*log(4+x)+12*x^3-216*x^2+396*x+5808),
x, algorithm="giac")

[Out]

-1/3*x/(x*log(x + 4) - 2*x + 22)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {-88-22 x+x^2}{5808+396 x-216 x^2+12 x^3+\left (528 x+84 x^2-12 x^3\right ) \log (4+x)+\left (12 x^2+3 x^3\right ) \log ^2(4+x)} \, dx=-\frac {x}{3\,\left (x\,\ln \left (x+4\right )-2\,x+22\right )} \]

[In]

int(-(22*x - x^2 + 88)/(396*x + log(x + 4)*(528*x + 84*x^2 - 12*x^3) + log(x + 4)^2*(12*x^2 + 3*x^3) - 216*x^2
+ 12*x^3 + 5808),x)

[Out]

-x/(3*(x*log(x + 4) - 2*x + 22))