Integrand size = 79, antiderivative size = 26 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=\frac {\frac {5}{4}+x}{\frac {2 x^2}{3}+\log \left (-1+\frac {x}{1+x}\right )} \] Output:
(x+5/4)/(ln(x/(1+x)-1)+2/3*x^2)
Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=\frac {3 (5+4 x)}{4 \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )} \] Input:
Integrate[(45 - 24*x - 84*x^2 - 24*x^3 + (36 + 36*x)*Log[-(1 + x)^(-1)])/( 16*x^4 + 16*x^5 + (48*x^2 + 48*x^3)*Log[-(1 + x)^(-1)] + (36 + 36*x)*Log[- (1 + x)^(-1)]^2),x]
Output:
(3*(5 + 4*x))/(4*(2*x^2 + 3*Log[-(1 + x)^(-1)]))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-24 x^3-84 x^2-24 x+(36 x+36) \log \left (-\frac {1}{x+1}\right )+45}{16 x^5+16 x^4+\left (48 x^3+48 x^2\right ) \log \left (-\frac {1}{x+1}\right )+(36 x+36) \log ^2\left (-\frac {1}{x+1}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-24 x^3-84 x^2-24 x+(36 x+36) \log \left (-\frac {1}{x+1}\right )+45}{4 (x+1) \left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {3 \left (-8 x^3-28 x^2-8 x+12 (x+1) \log \left (-\frac {1}{x+1}\right )+15\right )}{(x+1) \left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{4} \int \frac {-8 x^3-28 x^2-8 x+12 (x+1) \log \left (-\frac {1}{x+1}\right )+15}{(x+1) \left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {3}{4} \int \left (\frac {-16 x^3-36 x^2-8 x+15}{(x+1) \left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}+\frac {4}{2 x^2+3 \log \left (-\frac {1}{x+1}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{4} \left (12 \int \frac {1}{\left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}dx-20 \int \frac {x}{\left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}dx-16 \int \frac {x^2}{\left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}dx+3 \int \frac {1}{(x+1) \left (2 x^2+3 \log \left (-\frac {1}{x+1}\right )\right )^2}dx+4 \int \frac {1}{2 x^2+3 \log \left (-\frac {1}{x+1}\right )}dx\right )\) |
Input:
Int[(45 - 24*x - 84*x^2 - 24*x^3 + (36 + 36*x)*Log[-(1 + x)^(-1)])/(16*x^4 + 16*x^5 + (48*x^2 + 48*x^3)*Log[-(1 + x)^(-1)] + (36 + 36*x)*Log[-(1 + x )^(-1)]^2),x]
Output:
$Aborted
Time = 0.42 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {\frac {15}{4}+3 x}{2 x^{2}+3 \ln \left (-\frac {1}{1+x}\right )}\) | \(25\) |
risch | \(\frac {\frac {15}{4}+3 x}{2 x^{2}+3 \ln \left (-\frac {1}{1+x}\right )}\) | \(26\) |
parallelrisch | \(\frac {45+36 x}{24 x^{2}+36 \ln \left (-\frac {1}{1+x}\right )}\) | \(26\) |
derivativedivides | \(\frac {\frac {3}{4 \left (1+x \right )^{2}}+\frac {3}{1+x}}{\frac {3 \ln \left (-\frac {1}{1+x}\right )}{\left (1+x \right )^{2}}+\frac {2}{\left (1+x \right )^{2}}-\frac {4}{1+x}+2}\) | \(49\) |
default | \(\frac {\frac {3}{4 \left (1+x \right )^{2}}+\frac {3}{1+x}}{\frac {3 \ln \left (-\frac {1}{1+x}\right )}{\left (1+x \right )^{2}}+\frac {2}{\left (1+x \right )^{2}}-\frac {4}{1+x}+2}\) | \(49\) |
Input:
int(((36*x+36)*ln(-1/(1+x))-24*x^3-84*x^2-24*x+45)/((36*x+36)*ln(-1/(1+x)) ^2+(48*x^3+48*x^2)*ln(-1/(1+x))+16*x^5+16*x^4),x,method=_RETURNVERBOSE)
Output:
(15/4+3*x)/(2*x^2+3*ln(-1/(1+x)))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=\frac {3 \, {\left (4 \, x + 5\right )}}{4 \, {\left (2 \, x^{2} + 3 \, \log \left (-\frac {1}{x + 1}\right )\right )}} \] Input:
integrate(((36*x+36)*log(-1/(1+x))-24*x^3-84*x^2-24*x+45)/((36*x+36)*log(- 1/(1+x))^2+(48*x^3+48*x^2)*log(-1/(1+x))+16*x^5+16*x^4),x, algorithm="fric as")
Output:
3/4*(4*x + 5)/(2*x^2 + 3*log(-1/(x + 1)))
Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=\frac {12 x + 15}{8 x^{2} + 12 \log {\left (- \frac {1}{x + 1} \right )}} \] Input:
integrate(((36*x+36)*ln(-1/(1+x))-24*x**3-84*x**2-24*x+45)/((36*x+36)*ln(- 1/(1+x))**2+(48*x**3+48*x**2)*ln(-1/(1+x))+16*x**5+16*x**4),x)
Output:
(12*x + 15)/(8*x**2 + 12*log(-1/(x + 1)))
Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=\frac {3 \, {\left (4 \, x + 5\right )}}{4 \, {\left (2 \, x^{2} - 3 \, \log \left (-x - 1\right )\right )}} \] Input:
integrate(((36*x+36)*log(-1/(1+x))-24*x^3-84*x^2-24*x+45)/((36*x+36)*log(- 1/(1+x))^2+(48*x^3+48*x^2)*log(-1/(1+x))+16*x^5+16*x^4),x, algorithm="maxi ma")
Output:
3/4*(4*x + 5)/(2*x^2 - 3*log(-x - 1))
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=-\frac {3 \, {\left (\frac {4}{x + 1} + \frac {1}{{\left (x + 1\right )}^{2}}\right )}}{4 \, {\left (\frac {4}{x + 1} - \frac {3 \, \log \left (-\frac {1}{x + 1}\right )}{{\left (x + 1\right )}^{2}} - \frac {2}{{\left (x + 1\right )}^{2}} - 2\right )}} \] Input:
integrate(((36*x+36)*log(-1/(1+x))-24*x^3-84*x^2-24*x+45)/((36*x+36)*log(- 1/(1+x))^2+(48*x^3+48*x^2)*log(-1/(1+x))+16*x^5+16*x^4),x, algorithm="giac ")
Output:
-3/4*(4/(x + 1) + 1/(x + 1)^2)/(4/(x + 1) - 3*log(-1/(x + 1))/(x + 1)^2 - 2/(x + 1)^2 - 2)
Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=\frac {3\,\left (4\,x+5\right )}{4\,\left (3\,\ln \left (-\frac {1}{x+1}\right )+2\,x^2\right )} \] Input:
int(-(24*x - log(-1/(x + 1))*(36*x + 36) + 84*x^2 + 24*x^3 - 45)/(log(-1/( x + 1))*(48*x^2 + 48*x^3) + 16*x^4 + 16*x^5 + log(-1/(x + 1))^2*(36*x + 36 )),x)
Output:
(3*(4*x + 5))/(4*(3*log(-1/(x + 1)) + 2*x^2))
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx=\frac {12 x +15}{12 \,\mathrm {log}\left (-\frac {1}{x +1}\right )+8 x^{2}} \] Input:
int(((36*x+36)*log(-1/(1+x))-24*x^3-84*x^2-24*x+45)/((36*x+36)*log(-1/(1+x ))^2+(48*x^3+48*x^2)*log(-1/(1+x))+16*x^5+16*x^4),x)
Output:
(3*(4*x + 5))/(4*(3*log(( - 1)/(x + 1)) + 2*x**2))