Integrand size = 120, antiderivative size = 28 \begin {dmath*} \int \frac {-32+32 x^2+(-32+32 x) \log (3)+\left (16 x^3-16 x^4\right ) \log (1+x+\log (3))+\left (24 x^2-8 x^3-32 x^4+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(1+x+\log (3))+4 x^6 \log ^3(1+x+\log (3))+\left (6 x^5+6 x^6+6 x^5 \log (3)\right ) \log ^4(1+x+\log (3))}{1+x+\log (3)} \, dx=4+x^2 \left (4-\frac {4}{x}-x^2 \log ^2(1+x+\log (3))\right )^2 \end {dmath*}
Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \begin {dmath*} \int \frac {-32+32 x^2+(-32+32 x) \log (3)+\left (16 x^3-16 x^4\right ) \log (1+x+\log (3))+\left (24 x^2-8 x^3-32 x^4+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(1+x+\log (3))+4 x^6 \log ^3(1+x+\log (3))+\left (6 x^5+6 x^6+6 x^5 \log (3)\right ) \log ^4(1+x+\log (3))}{1+x+\log (3)} \, dx=\left (4-4 x+x^3 \log ^2(1+x+\log (3))\right )^2 \end {dmath*}
Integrate[(-32 + 32*x^2 + (-32 + 32*x)*Log[3] + (16*x^3 - 16*x^4)*Log[1 + x + Log[3]] + (24*x^2 - 8*x^3 - 32*x^4 + (24*x^2 - 32*x^3)*Log[3])*Log[1 + x + Log[3]]^2 + 4*x^6*Log[1 + x + Log[3]]^3 + (6*x^5 + 6*x^6 + 6*x^5*Log[ 3])*Log[1 + x + Log[3]]^4)/(1 + x + Log[3]),x]
Time = 0.46 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {4 x^6 \log ^3(x+1+\log (3))+32 x^2+\left (6 x^6+6 x^5+6 x^5 \log (3)\right ) \log ^4(x+1+\log (3))+\left (16 x^3-16 x^4\right ) \log (x+1+\log (3))+\left (-32 x^4-8 x^3+24 x^2+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(x+1+\log (3))+(32 x-32) \log (3)-32}{x+1+\log (3)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (x^3 \log ^2(x+1+\log (3))-4 x+4\right ) \left (2 x^3 \log (x+1+\log (3))+3 x^2 (x+1+\log (3)) \log ^2(x+1+\log (3))-4 (x+1+\log (3))\right )}{x+1+\log (3)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {\left (\log ^2(x+\log (3)+1) x^3-4 x+4\right ) \left (-2 \log (x+\log (3)+1) x^3-3 (x+\log (3)+1) \log ^2(x+\log (3)+1) x^2+4 (x+\log (3)+1)\right )}{x+\log (3)+1}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {\left (\log ^2(x+\log (3)+1) x^3-4 x+4\right ) \left (-2 \log (x+\log (3)+1) x^3-3 (x+\log (3)+1) \log ^2(x+\log (3)+1) x^2+4 (x+\log (3)+1)\right )}{x+\log (3)+1}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \left (x^3 \log ^2(x+1+\log (3))-4 x+4\right )^2\) |
Int[(-32 + 32*x^2 + (-32 + 32*x)*Log[3] + (16*x^3 - 16*x^4)*Log[1 + x + Lo g[3]] + (24*x^2 - 8*x^3 - 32*x^4 + (24*x^2 - 32*x^3)*Log[3])*Log[1 + x + L og[3]]^2 + 4*x^6*Log[1 + x + Log[3]]^3 + (6*x^5 + 6*x^6 + 6*x^5*Log[3])*Lo g[1 + x + Log[3]]^4)/(1 + x + Log[3]),x]
3.3.18.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.85 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50
method | result | size |
risch | \(\ln \left (x +\ln \left (3\right )+1\right )^{4} x^{6}+\left (-8 x^{4}+8 x^{3}\right ) \ln \left (x +\ln \left (3\right )+1\right )^{2}+16 x^{2}-32 x\) | \(42\) |
parallelrisch | \(\ln \left (x +\ln \left (3\right )+1\right )^{4} x^{6}-8 \ln \left (x +\ln \left (3\right )+1\right )^{2} x^{4}+8 \ln \left (x +\ln \left (3\right )+1\right )^{2} x^{3}-16 \ln \left (3\right )^{2}+16 x^{2}+32 \ln \left (3\right )-32 x +48\) | \(59\) |
parts | \(\text {Expression too large to display}\) | \(2855\) |
derivativedivides | \(\text {Expression too large to display}\) | \(2873\) |
default | \(\text {Expression too large to display}\) | \(2873\) |
int(((6*x^5*ln(3)+6*x^6+6*x^5)*ln(x+ln(3)+1)^4+4*x^6*ln(x+ln(3)+1)^3+((-32 *x^3+24*x^2)*ln(3)-32*x^4-8*x^3+24*x^2)*ln(x+ln(3)+1)^2+(-16*x^4+16*x^3)*l n(x+ln(3)+1)+(32*x-32)*ln(3)+32*x^2-32)/(x+ln(3)+1),x,method=_RETURNVERBOS E)
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \begin {dmath*} \int \frac {-32+32 x^2+(-32+32 x) \log (3)+\left (16 x^3-16 x^4\right ) \log (1+x+\log (3))+\left (24 x^2-8 x^3-32 x^4+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(1+x+\log (3))+4 x^6 \log ^3(1+x+\log (3))+\left (6 x^5+6 x^6+6 x^5 \log (3)\right ) \log ^4(1+x+\log (3))}{1+x+\log (3)} \, dx=x^{6} \log \left (x + \log \left (3\right ) + 1\right )^{4} - 8 \, {\left (x^{4} - x^{3}\right )} \log \left (x + \log \left (3\right ) + 1\right )^{2} + 16 \, x^{2} - 32 \, x \end {dmath*}
integrate(((6*x^5*log(3)+6*x^6+6*x^5)*log(x+log(3)+1)^4+4*x^6*log(x+log(3) +1)^3+((-32*x^3+24*x^2)*log(3)-32*x^4-8*x^3+24*x^2)*log(x+log(3)+1)^2+(-16 *x^4+16*x^3)*log(x+log(3)+1)+(32*x-32)*log(3)+32*x^2-32)/(x+log(3)+1),x, a lgorithm=\
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \begin {dmath*} \int \frac {-32+32 x^2+(-32+32 x) \log (3)+\left (16 x^3-16 x^4\right ) \log (1+x+\log (3))+\left (24 x^2-8 x^3-32 x^4+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(1+x+\log (3))+4 x^6 \log ^3(1+x+\log (3))+\left (6 x^5+6 x^6+6 x^5 \log (3)\right ) \log ^4(1+x+\log (3))}{1+x+\log (3)} \, dx=x^{6} \log {\left (x + 1 + \log {\left (3 \right )} \right )}^{4} + 16 x^{2} - 32 x + \left (- 8 x^{4} + 8 x^{3}\right ) \log {\left (x + 1 + \log {\left (3 \right )} \right )}^{2} \end {dmath*}
integrate(((6*x**5*ln(3)+6*x**6+6*x**5)*ln(x+ln(3)+1)**4+4*x**6*ln(x+ln(3) +1)**3+((-32*x**3+24*x**2)*ln(3)-32*x**4-8*x**3+24*x**2)*ln(x+ln(3)+1)**2+ (-16*x**4+16*x**3)*ln(x+ln(3)+1)+(32*x-32)*ln(3)+32*x**2-32)/(x+ln(3)+1),x )
Leaf count of result is larger than twice the leaf count of optimal. 3532 vs. \(2 (27) = 54\).
Time = 0.32 (sec) , antiderivative size = 3532, normalized size of antiderivative = 126.14 \begin {dmath*} \int \frac {-32+32 x^2+(-32+32 x) \log (3)+\left (16 x^3-16 x^4\right ) \log (1+x+\log (3))+\left (24 x^2-8 x^3-32 x^4+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(1+x+\log (3))+4 x^6 \log ^3(1+x+\log (3))+\left (6 x^5+6 x^6+6 x^5 \log (3)\right ) \log ^4(1+x+\log (3))}{1+x+\log (3)} \, dx=\text {Too large to display} \end {dmath*}
integrate(((6*x^5*log(3)+6*x^6+6*x^5)*log(x+log(3)+1)^4+4*x^6*log(x+log(3) +1)^3+((-32*x^3+24*x^2)*log(3)-32*x^4-8*x^3+24*x^2)*log(x+log(3)+1)^2+(-16 *x^4+16*x^3)*log(x+log(3)+1)+(32*x-32)*log(3)+32*x^2-32)/(x+log(3)+1),x, a lgorithm=\
1/54*(54*log(x + log(3) + 1)^4 - 36*log(x + log(3) + 1)^3 + 18*log(x + log (3) + 1)^2 - 6*log(x + log(3) + 1) + 1)*(x + log(3) + 1)^6 + 1/54*(36*log( x + log(3) + 1)^3 - 18*log(x + log(3) + 1)^2 + 6*log(x + log(3) + 1) - 1)* (x + log(3) + 1)^6 - 36/3125*(625*(log(3) + 1)*log(x + log(3) + 1)^4 - 500 *(log(3) + 1)*log(x + log(3) + 1)^3 + 300*(log(3) + 1)*log(x + log(3) + 1) ^2 - 120*(log(3) + 1)*log(x + log(3) + 1) + 24*log(3) + 24)*(x + log(3) + 1)^5 - 24/625*(125*(log(3) + 1)*log(x + log(3) + 1)^3 - 75*(log(3) + 1)*lo g(x + log(3) + 1)^2 + 30*(log(3) + 1)*log(x + log(3) + 1) - 6*log(3) - 6)* (x + log(3) + 1)^5 + 6/3125*(625*log(x + log(3) + 1)^4 - 500*log(x + log(3 ) + 1)^3 + 300*log(x + log(3) + 1)^2 - 120*log(x + log(3) + 1) + 24)*(x + log(3) + 1)^5 + 6/5*(log(3)^6 + 6*log(3)^5 + 15*log(3)^4 + 20*log(3)^3 + 1 5*log(3)^2 + 6*log(3) + 1)*log(x + log(3) + 1)^5 - 6/5*(log(3)^5 + 5*log(3 )^4 + 10*log(3)^3 + 10*log(3)^2 + 5*log(3) + 1)*log(x + log(3) + 1)^5 + 45 /64*(32*(log(3)^2 + 2*log(3) + 1)*log(x + log(3) + 1)^4 - 32*(log(3)^2 + 2 *log(3) + 1)*log(x + log(3) + 1)^3 + 24*(log(3)^2 + 2*log(3) + 1)*log(x + log(3) + 1)^2 + 3*log(3)^2 - 12*(log(3)^2 + 2*log(3) + 1)*log(x + log(3) + 1) + 6*log(3) + 3)*(x + log(3) + 1)^4 - 15/64*(32*(log(3) + 1)*log(x + lo g(3) + 1)^4 - 32*(log(3) + 1)*log(x + log(3) + 1)^3 + 24*(log(3) + 1)*log( x + log(3) + 1)^2 - 12*(log(3) + 1)*log(x + log(3) + 1) + 3*log(3) + 3)*(x + log(3) + 1)^4 + 15/32*(32*(log(3)^2 + 2*log(3) + 1)*log(x + log(3) +...
Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \begin {dmath*} \int \frac {-32+32 x^2+(-32+32 x) \log (3)+\left (16 x^3-16 x^4\right ) \log (1+x+\log (3))+\left (24 x^2-8 x^3-32 x^4+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(1+x+\log (3))+4 x^6 \log ^3(1+x+\log (3))+\left (6 x^5+6 x^6+6 x^5 \log (3)\right ) \log ^4(1+x+\log (3))}{1+x+\log (3)} \, dx=x^{6} \log \left (x + \log \left (3\right ) + 1\right )^{4} - 8 \, {\left (x^{4} - x^{3}\right )} \log \left (x + \log \left (3\right ) + 1\right )^{2} + 16 \, x^{2} - 32 \, x \end {dmath*}
integrate(((6*x^5*log(3)+6*x^6+6*x^5)*log(x+log(3)+1)^4+4*x^6*log(x+log(3) +1)^3+((-32*x^3+24*x^2)*log(3)-32*x^4-8*x^3+24*x^2)*log(x+log(3)+1)^2+(-16 *x^4+16*x^3)*log(x+log(3)+1)+(32*x-32)*log(3)+32*x^2-32)/(x+log(3)+1),x, a lgorithm=\
Timed out. \begin {dmath*} \int \frac {-32+32 x^2+(-32+32 x) \log (3)+\left (16 x^3-16 x^4\right ) \log (1+x+\log (3))+\left (24 x^2-8 x^3-32 x^4+\left (24 x^2-32 x^3\right ) \log (3)\right ) \log ^2(1+x+\log (3))+4 x^6 \log ^3(1+x+\log (3))+\left (6 x^5+6 x^6+6 x^5 \log (3)\right ) \log ^4(1+x+\log (3))}{1+x+\log (3)} \, dx=\text {Hanged} \end {dmath*}