Integrand size = 89, antiderivative size = 27 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x-\log (4-x)+\log (x) \left (-e^{7+x}+\log (x)+\log (1+x)\right ) \] Output:
ln(x)*(ln(x)-exp(7+x)+ln(1+x))+x-ln(4-x)
Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x-\log (4-x)-e^{7+x} \log (x)+\log ^2(x)+\log (x) \log (1+x) \] Input:
Integrate[(-5*x - 4*x^2 + x^3 + E^(7 + x)*(4 + 3*x - x^2) + (-8 - 10*x + 3 *x^2 + E^(7 + x)*(4*x + 3*x^2 - x^3))*Log[x] + (-4 - 3*x + x^2)*Log[1 + x] )/(-4*x - 3*x^2 + x^3),x]
Output:
x - Log[4 - x] - E^(7 + x)*Log[x] + Log[x]^2 + Log[x]*Log[1 + x]
Time = 1.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2026, 7279, 2009}
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.
\(\displaystyle \int \frac {x^3-4 x^2+e^{x+7} \left (-x^2+3 x+4\right )+\left (x^2-3 x-4\right ) \log (x+1)+\left (3 x^2+e^{x+7} \left (-x^3+3 x^2+4 x\right )-10 x-8\right ) \log (x)-5 x}{x^3-3 x^2-4 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^3-4 x^2+e^{x+7} \left (-x^2+3 x+4\right )+\left (x^2-3 x-4\right ) \log (x+1)+\left (3 x^2+e^{x+7} \left (-x^3+3 x^2+4 x\right )-10 x-8\right ) \log (x)-5 x}{x \left (x^2-3 x-4\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {x^2}{(x-4) (x+1)}-\frac {4 x}{x^2-3 x-4}-\frac {5}{x^2-3 x-4}-\frac {10 \log (x)}{x^2-3 x-4}-\frac {8 \log (x)}{\left (x^2-3 x-4\right ) x}+\frac {3 x \log (x)}{(x-4) (x+1)}-\frac {e^{x+7} (x \log (x)+1)}{x}+\frac {\log (x+1)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x+\log ^2(x)-e^{x+7} \log (x)+\log (x+1) \log (x)-2 \log (4) \log (8-2 x)-\log (4-x)+2 \log (4) \log (x-4)\) |
Input:
Int[(-5*x - 4*x^2 + x^3 + E^(7 + x)*(4 + 3*x - x^2) + (-8 - 10*x + 3*x^2 + E^(7 + x)*(4*x + 3*x^2 - x^3))*Log[x] + (-4 - 3*x + x^2)*Log[1 + x])/(-4* x - 3*x^2 + x^3),x]
Output:
x - 2*Log[4]*Log[8 - 2*x] - Log[4 - x] + 2*Log[4]*Log[-4 + x] - E^(7 + x)* Log[x] + Log[x]^2 + Log[x]*Log[1 + x]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 62.60 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
default | \(x -\ln \left (x -4\right )-\ln \left (x \right ) {\mathrm e}^{x +7}+\ln \left (x \right ) \ln \left (1+x \right )+\ln \left (x \right )^{2}\) | \(28\) |
risch | \(x -\ln \left (x -4\right )-\ln \left (x \right ) {\mathrm e}^{x +7}+\ln \left (x \right ) \ln \left (1+x \right )+\ln \left (x \right )^{2}\) | \(28\) |
parts | \(x -\ln \left (x -4\right )-\ln \left (x \right ) {\mathrm e}^{x +7}+\ln \left (x \right ) \ln \left (1+x \right )+\ln \left (x \right )^{2}\) | \(28\) |
parallelrisch | \(\ln \left (x \right )^{2}+\ln \left (x \right ) \ln \left (1+x \right )-\ln \left (x \right ) {\mathrm e}^{x +7}-\ln \left (x -4\right )+x -\frac {1}{2}\) | \(29\) |
Input:
int(((x^2-3*x-4)*ln(1+x)+((-x^3+3*x^2+4*x)*exp(x+7)+3*x^2-10*x-8)*ln(x)+(- x^2+3*x+4)*exp(x+7)+x^3-4*x^2-5*x)/(x^3-3*x^2-4*x),x,method=_RETURNVERBOSE )
Output:
x-ln(x-4)-ln(x)*exp(x+7)+ln(x)*ln(1+x)+ln(x)^2
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=-e^{\left (x + 7\right )} \log \left (x\right ) + \log \left (x + 1\right ) \log \left (x\right ) + \log \left (x\right )^{2} + x - \log \left (x - 4\right ) \] Input:
integrate(((x^2-3*x-4)*log(1+x)+((-x^3+3*x^2+4*x)*exp(x+7)+3*x^2-10*x-8)*l og(x)+(-x^2+3*x+4)*exp(x+7)+x^3-4*x^2-5*x)/(x^3-3*x^2-4*x),x, algorithm="f ricas")
Output:
-e^(x + 7)*log(x) + log(x + 1)*log(x) + log(x)^2 + x - log(x - 4)
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x - e^{x + 7} \log {\left (x \right )} + \log {\left (x \right )}^{2} + \log {\left (x \right )} \log {\left (x + 1 \right )} - \log {\left (x - 4 \right )} \] Input:
integrate(((x**2-3*x-4)*ln(1+x)+((-x**3+3*x**2+4*x)*exp(x+7)+3*x**2-10*x-8 )*ln(x)+(-x**2+3*x+4)*exp(x+7)+x**3-4*x**2-5*x)/(x**3-3*x**2-4*x),x)
Output:
x - exp(x + 7)*log(x) + log(x)**2 + log(x)*log(x + 1) - log(x - 4)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=-e^{\left (x + 7\right )} \log \left (x\right ) + \log \left (x + 1\right ) \log \left (x\right ) + \log \left (x\right )^{2} + x - \log \left (x - 4\right ) \] Input:
integrate(((x^2-3*x-4)*log(1+x)+((-x^3+3*x^2+4*x)*exp(x+7)+3*x^2-10*x-8)*l og(x)+(-x^2+3*x+4)*exp(x+7)+x^3-4*x^2-5*x)/(x^3-3*x^2-4*x),x, algorithm="m axima")
Output:
-e^(x + 7)*log(x) + log(x + 1)*log(x) + log(x)^2 + x - log(x - 4)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=-e^{\left (x + 7\right )} \log \left (x\right ) + \log \left (x + 1\right ) \log \left (x\right ) + \log \left (x\right )^{2} + x - \log \left (x - 4\right ) \] Input:
integrate(((x^2-3*x-4)*log(1+x)+((-x^3+3*x^2+4*x)*exp(x+7)+3*x^2-10*x-8)*l og(x)+(-x^2+3*x+4)*exp(x+7)+x^3-4*x^2-5*x)/(x^3-3*x^2-4*x),x, algorithm="g iac")
Output:
-e^(x + 7)*log(x) + log(x + 1)*log(x) + log(x)^2 + x - log(x - 4)
Time = 8.74 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=x-\ln \left (x-4\right )+{\ln \left (x\right )}^2-{\mathrm {e}}^{x+7}\,\ln \left (x\right )+\ln \left (x+1\right )\,\ln \left (x\right ) \] Input:
int((5*x - exp(x + 7)*(3*x - x^2 + 4) + log(x + 1)*(3*x - x^2 + 4) + 4*x^2 - x^3 + log(x)*(10*x - exp(x + 7)*(4*x + 3*x^2 - x^3) - 3*x^2 + 8))/(4*x + 3*x^2 - x^3),x)
Output:
x - log(x - 4) + log(x)^2 - exp(x + 7)*log(x) + log(x + 1)*log(x)
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-5 x-4 x^2+x^3+e^{7+x} \left (4+3 x-x^2\right )+\left (-8-10 x+3 x^2+e^{7+x} \left (4 x+3 x^2-x^3\right )\right ) \log (x)+\left (-4-3 x+x^2\right ) \log (1+x)}{-4 x-3 x^2+x^3} \, dx=-e^{x} \mathrm {log}\left (x \right ) e^{7}-\mathrm {log}\left (x -4\right )+\mathrm {log}\left (x +1\right ) \mathrm {log}\left (x \right )+\mathrm {log}\left (x \right )^{2}+x \] Input:
int(((x^2-3*x-4)*log(1+x)+((-x^3+3*x^2+4*x)*exp(x+7)+3*x^2-10*x-8)*log(x)+ (-x^2+3*x+4)*exp(x+7)+x^3-4*x^2-5*x)/(x^3-3*x^2-4*x),x)
Output:
- e**x*log(x)*e**7 - log(x - 4) + log(x + 1)*log(x) + log(x)**2 + x