Integrand size = 79, antiderivative size = 24 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx=\left (-x^2+4 \log ^4(2)\right ) \log \left (3+e^3 x (3+x)\right ) \] Output:
ln(x*exp(3)*(3+x)+3)*(4*ln(2)^4-x^2)
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx=-\left (\left (x^2-4 \log ^4(2)\right ) \log \left (3+e^3 x (3+x)\right )\right ) \] Input:
Integrate[(E^3*(-3*x^2 - 2*x^3) + E^3*(12 + 8*x)*Log[2]^4 + (-6*x + E^3*(- 6*x^2 - 2*x^3))*Log[3 + E^3*(3*x + x^2)])/(3 + E^3*(3*x + x^2)),x]
Output:
-((x^2 - 4*Log[2]^4)*Log[3 + E^3*x*(3 + x)])
Leaf count is larger than twice the leaf count of optimal. \(90\) vs. \(2(24)=48\).
Time = 0.84 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.75, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {7292, 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.
\(\displaystyle \int \frac {e^3 \left (-2 x^3-3 x^2\right )+\left (e^3 \left (-2 x^3-6 x^2\right )-6 x\right ) \log \left (e^3 \left (x^2+3 x\right )+3\right )+e^3 (8 x+12) \log ^4(2)}{e^3 \left (x^2+3 x\right )+3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^3 \left (-2 x^3-3 x^2\right )+\left (e^3 \left (-2 x^3-6 x^2\right )-6 x\right ) \log \left (e^3 \left (x^2+3 x\right )+3\right )+e^3 (8 x+12) \log ^4(2)}{e^3 x^2+3 e^3 x+3}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (-\frac {e^3 (2 x+3) \left (x^2-4 \log ^4(2)\right )}{e^3 x^2+3 e^3 x+3}-2 x \log \left (e^3 x^2+3 e^3 x+3\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (6-e^3 \left (9-8 \log ^4(2)\right )\right ) \log \left (e^3 x^2+3 e^3 x+3\right )}{2 e^3}+x^2 \left (-\log \left (e^3 x^2+3 e^3 x+3\right )\right )-\frac {3 \left (2-3 e^3\right ) \log \left (e^3 x^2+3 e^3 x+3\right )}{2 e^3}\) |
Input:
Int[(E^3*(-3*x^2 - 2*x^3) + E^3*(12 + 8*x)*Log[2]^4 + (-6*x + E^3*(-6*x^2 - 2*x^3))*Log[3 + E^3*(3*x + x^2)])/(3 + E^3*(3*x + x^2)),x]
Output:
(-3*(2 - 3*E^3)*Log[3 + 3*E^3*x + E^3*x^2])/(2*E^3) - x^2*Log[3 + 3*E^3*x + E^3*x^2] + ((6 - E^3*(9 - 8*Log[2]^4))*Log[3 + 3*E^3*x + E^3*x^2])/(2*E^ 3)
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 = 1.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62
method | result | size |
norman | \(4 \ln \left (2\right )^{4} \ln \left (\left (x^{2}+3 x \right ) {\mathrm e}^{3}+3\right )-x^{2} \ln \left (\left (x^{2}+3 x \right ) {\mathrm e}^{3}+3\right )\) | \(39\) |
risch | \(-x^{2} \ln \left (\left (x^{2}+3 x \right ) {\mathrm e}^{3}+3\right )+4 \ln \left (2\right )^{4} \ln \left (x^{2} {\mathrm e}^{3}+3 x \,{\mathrm e}^{3}+3\right )\) | \(40\) |
default | \(-x^{2} \ln \left (x^{2} {\mathrm e}^{3}+3 x \,{\mathrm e}^{3}+3\right )+4 \ln \left (2\right )^{4} \ln \left (x^{2} {\mathrm e}^{3}+3 x \,{\mathrm e}^{3}+3\right )\) | \(41\) |
parallelrisch | \(4 \ln \left (2\right )^{4} \ln \left (\left (x^{2} {\mathrm e}^{3}+3 x \,{\mathrm e}^{3}+3\right ) {\mathrm e}^{-3}\right )-x^{2} \ln \left (\left (x^{2}+3 x \right ) {\mathrm e}^{3}+3\right )\) | \(45\) |
parts | \({\mathrm e}^{3} \left (-{\mathrm e}^{-3} \left (x^{2}-3 x \right )+{\mathrm e}^{-3} \left (\frac {\left (8 \,{\mathrm e}^{3} \ln \left (2\right )^{4}-9 \,{\mathrm e}^{3}+6\right ) {\mathrm e}^{-3} \ln \left (x^{2} {\mathrm e}^{3}+3 x \,{\mathrm e}^{3}+3\right )}{2}-\frac {2 \left (\frac {27 \,{\mathrm e}^{3}}{2}-18\right ) \operatorname {arctanh}\left (\frac {2 x \,{\mathrm e}^{3}+3 \,{\mathrm e}^{3}}{\sqrt {-12 \,{\mathrm e}^{3}+9 \,{\mathrm e}^{6}}}\right )}{\sqrt {-12 \,{\mathrm e}^{3}+9 \,{\mathrm e}^{6}}}\right )\right )-x^{2} \ln \left (x^{2} {\mathrm e}^{3}+3 x \,{\mathrm e}^{3}+3\right )+{\mathrm e}^{3} \left ({\mathrm e}^{-3} \left (x^{2}-3 x \right )+3 \,{\mathrm e}^{-3} \left (\frac {\left (3 \,{\mathrm e}^{3}-2\right ) {\mathrm e}^{-3} \ln \left (x^{2} {\mathrm e}^{3}+3 x \,{\mathrm e}^{3}+3\right )}{2}-\frac {2 \left (-\frac {9 \,{\mathrm e}^{3}}{2}+6\right ) \operatorname {arctanh}\left (\frac {2 x \,{\mathrm e}^{3}+3 \,{\mathrm e}^{3}}{\sqrt {-12 \,{\mathrm e}^{3}+9 \,{\mathrm e}^{6}}}\right )}{\sqrt {-12 \,{\mathrm e}^{3}+9 \,{\mathrm e}^{6}}}\right )\right )\) | \(215\) |
Input:
int((((-2*x^3-6*x^2)*exp(3)-6*x)*ln((x^2+3*x)*exp(3)+3)+(8*x+12)*exp(3)*ln (2)^4+(-2*x^3-3*x^2)*exp(3))/((x^2+3*x)*exp(3)+3),x,method=_RETURNVERBOSE)
Output:
4*ln(2)^4*ln((x^2+3*x)*exp(3)+3)-x^2*ln((x^2+3*x)*exp(3)+3)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx={\left (4 \, \log \left (2\right )^{4} - x^{2}\right )} \log \left ({\left (x^{2} + 3 \, x\right )} e^{3} + 3\right ) \] Input:
integrate((((-2*x^3-6*x^2)*exp(3)-6*x)*log((x^2+3*x)*exp(3)+3)+(8*x+12)*ex p(3)*log(2)^4+(-2*x^3-3*x^2)*exp(3))/((x^2+3*x)*exp(3)+3),x, algorithm="fr icas")
Output:
(4*log(2)^4 - x^2)*log((x^2 + 3*x)*e^3 + 3)
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx=- x^{2} \log {\left (\left (x^{2} + 3 x\right ) e^{3} + 3 \right )} + 4 \log {\left (2 \right )}^{4} \log {\left (x^{2} e^{3} + 3 x e^{3} + 3 \right )} \] Input:
integrate((((-2*x**3-6*x**2)*exp(3)-6*x)*ln((x**2+3*x)*exp(3)+3)+(8*x+12)* exp(3)*ln(2)**4+(-2*x**3-3*x**2)*exp(3))/((x**2+3*x)*exp(3)+3),x)
Output:
-x**2*log((x**2 + 3*x)*exp(3) + 3) + 4*log(2)**4*log(x**2*exp(3) + 3*x*exp (3) + 3)
Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (23) = 46\).
Time = 0.17 (sec) , antiderivative size = 493, normalized size of antiderivative = 20.54 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx=4 \, {\left (e^{\left (-3\right )} \log \left (x^{2} e^{3} + 3 \, x e^{3} + 3\right ) - \frac {\sqrt {3} e^{\left (-\frac {3}{2}\right )} \log \left (\frac {2 \, x e^{3} - \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}{2 \, x e^{3} + \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}\right )}{\sqrt {3 \, e^{3} - 4}}\right )} e^{3} \log \left (2\right )^{4} + \frac {4 \, \sqrt {3} e^{\frac {3}{2}} \log \left (2\right )^{4} \log \left (\frac {2 \, x e^{3} - \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}{2 \, x e^{3} + \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}\right )}{\sqrt {3 \, e^{3} - 4}} - \frac {3}{2} \, \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\left (-\frac {3}{2}\right )} \log \left (\frac {2 \, x e^{3} - \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}{2 \, x e^{3} + \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}\right ) - {\left (3 \, {\left (3 \, e^{3} - 1\right )} e^{\left (-6\right )} \log \left (x^{2} e^{3} + 3 \, x e^{3} + 3\right ) - \frac {9 \, \sqrt {3} {\left (e^{3} - 1\right )} e^{\left (-\frac {9}{2}\right )} \log \left (\frac {2 \, x e^{3} - \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}{2 \, x e^{3} + \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}\right )}{\sqrt {3 \, e^{3} - 4}} + {\left (x^{2} - 6 \, x\right )} e^{\left (-3\right )}\right )} e^{3} - \frac {3}{2} \, {\left (\frac {\sqrt {3} {\left (3 \, e^{3} - 2\right )} e^{\left (-\frac {9}{2}\right )} \log \left (\frac {2 \, x e^{3} - \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}{2 \, x e^{3} + \sqrt {3} \sqrt {3 \, e^{3} - 4} e^{\frac {3}{2}} + 3 \, e^{3}}\right )}{\sqrt {3 \, e^{3} - 4}} + 2 \, x e^{\left (-3\right )} - 3 \, e^{\left (-3\right )} \log \left (x^{2} e^{3} + 3 \, x e^{3} + 3\right )\right )} e^{3} + \frac {1}{2} \, {\left (2 \, x^{2} e^{3} - 6 \, x e^{3} - {\left (2 \, x^{2} e^{3} - 9 \, e^{3} + 6\right )} \log \left (x^{2} e^{3} + 3 \, x e^{3} + 3\right )\right )} e^{\left (-3\right )} \] Input:
integrate((((-2*x^3-6*x^2)*exp(3)-6*x)*log((x^2+3*x)*exp(3)+3)+(8*x+12)*ex p(3)*log(2)^4+(-2*x^3-3*x^2)*exp(3))/((x^2+3*x)*exp(3)+3),x, algorithm="ma xima")
Output:
4*(e^(-3)*log(x^2*e^3 + 3*x*e^3 + 3) - sqrt(3)*e^(-3/2)*log((2*x*e^3 - sqr t(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3)/(2*x*e^3 + sqrt(3)*sqrt(3*e^3 - 4)*e ^(3/2) + 3*e^3))/sqrt(3*e^3 - 4))*e^3*log(2)^4 + 4*sqrt(3)*e^(3/2)*log(2)^ 4*log((2*x*e^3 - sqrt(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3)/(2*x*e^3 + sqrt( 3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3))/sqrt(3*e^3 - 4) - 3/2*sqrt(3)*sqrt(3* e^3 - 4)*e^(-3/2)*log((2*x*e^3 - sqrt(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3)/ (2*x*e^3 + sqrt(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3)) - (3*(3*e^3 - 1)*e^(- 6)*log(x^2*e^3 + 3*x*e^3 + 3) - 9*sqrt(3)*(e^3 - 1)*e^(-9/2)*log((2*x*e^3 - sqrt(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3)/(2*x*e^3 + sqrt(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3))/sqrt(3*e^3 - 4) + (x^2 - 6*x)*e^(-3))*e^3 - 3/2*(sqr t(3)*(3*e^3 - 2)*e^(-9/2)*log((2*x*e^3 - sqrt(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3)/(2*x*e^3 + sqrt(3)*sqrt(3*e^3 - 4)*e^(3/2) + 3*e^3))/sqrt(3*e^3 - 4) + 2*x*e^(-3) - 3*e^(-3)*log(x^2*e^3 + 3*x*e^3 + 3))*e^3 + 1/2*(2*x^2*e^ 3 - 6*x*e^3 - (2*x^2*e^3 - 9*e^3 + 6)*log(x^2*e^3 + 3*x*e^3 + 3))*e^(-3)
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx=4 \, \log \left (2\right )^{4} \log \left (x^{2} e^{3} + 3 \, x e^{3} + 3\right ) - x^{2} \log \left (x^{2} e^{3} + 3 \, x e^{3} + 3\right ) \] Input:
integrate((((-2*x^3-6*x^2)*exp(3)-6*x)*log((x^2+3*x)*exp(3)+3)+(8*x+12)*ex p(3)*log(2)^4+(-2*x^3-3*x^2)*exp(3))/((x^2+3*x)*exp(3)+3),x, algorithm="gi ac")
Output:
4*log(2)^4*log(x^2*e^3 + 3*x*e^3 + 3) - x^2*log(x^2*e^3 + 3*x*e^3 + 3)
Time = 4.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx=\ln \left ({\mathrm {e}}^3\,x^2+3\,{\mathrm {e}}^3\,x+3\right )\,\left (4\,{\ln \left (2\right )}^4-x^2\right ) \] Input:
int(-(exp(3)*(3*x^2 + 2*x^3) + log(exp(3)*(3*x + x^2) + 3)*(6*x + exp(3)*( 6*x^2 + 2*x^3)) - exp(3)*log(2)^4*(8*x + 12))/(exp(3)*(3*x + x^2) + 3),x)
Output:
log(3*x*exp(3) + x^2*exp(3) + 3)*(4*log(2)^4 - x^2)
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^3 \left (-3 x^2-2 x^3\right )+e^3 (12+8 x) \log ^4(2)+\left (-6 x+e^3 \left (-6 x^2-2 x^3\right )\right ) \log \left (3+e^3 \left (3 x+x^2\right )\right )}{3+e^3 \left (3 x+x^2\right )} \, dx=\mathrm {log}\left (e^{3} x^{2}+3 e^{3} x +3\right ) \left (4 \mathrm {log}\left (2\right )^{4}-x^{2}\right ) \] Input:
int((((-2*x^3-6*x^2)*exp(3)-6*x)*log((x^2+3*x)*exp(3)+3)+(8*x+12)*exp(3)*l og(2)^4+(-2*x^3-3*x^2)*exp(3))/((x^2+3*x)*exp(3)+3),x)
Output:
log(e**3*x**2 + 3*e**3*x + 3)*(4*log(2)**4 - x**2)