Integrand size = 131, antiderivative size = 24 \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=\left (-e^{\frac {1}{4} x^3 (4+\log (5))}+\log \left (5+x^2\right )\right )^2 \] Output:
(ln(x^2+5)-exp(1/4*(ln(5)+4)*x^3))^2
Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=\left (-5^{\frac {x^3}{4}} e^{x^3}+\log \left (5+x^2\right )\right )^2 \] Input:
Integrate[(-8*E^((4*x^3 + x^3*Log[5])/4)*x + E^((4*x^3 + x^3*Log[5])/2)*(6 0*x^2 + 12*x^4 + (15*x^2 + 3*x^4)*Log[5]) + (8*x + E^((4*x^3 + x^3*Log[5]) /4)*(-60*x^2 - 12*x^4 + (-15*x^2 - 3*x^4)*Log[5]))*Log[5 + x^2])/(10 + 2*x ^2),x]
Output:
(-(5^(x^3/4)*E^x^3) + Log[5 + x^2])^2
Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(24)=48\).
Time = 1.90 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.29, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {7239, 27, 25, 7276, 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.
\(\displaystyle \int \frac {-8 x e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )}+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (12 x^4+60 x^2+\left (3 x^4+15 x^2\right ) \log (5)\right )+\left (e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-12 x^4-60 x^2+\left (-3 x^4-15 x^2\right ) \log (5)\right )+8 x\right ) \log \left (x^2+5\right )}{2 x^2+10} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x \left (8-3\ 5^{\frac {x^3}{4}} e^{x^3} x \left (x^2+5\right ) (4+\log (5))\right ) \left (\log \left (x^2+5\right )-5^{\frac {x^3}{4}} e^{x^3}\right )}{2 \left (x^2+5\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {x \left (8-3\ 5^{\frac {x^3}{4}} e^{x^3} x \left (x^2+5\right ) (4+\log (5))\right ) \left (5^{\frac {x^3}{4}} e^{x^3}-\log \left (x^2+5\right )\right )}{x^2+5}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {x \left (8-3\ 5^{\frac {x^3}{4}} e^{x^3} x \left (x^2+5\right ) (4+\log (5))\right ) \left (5^{\frac {x^3}{4}} e^{x^3}-\log \left (x^2+5\right )\right )}{x^2+5}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {1}{2} \int \left (-3 5^{\frac {x^3}{2}} e^{2 x^3} (4+\log (5)) x^2-\frac {8 \log \left (x^2+5\right ) x}{x^2+5}+\frac {5^{\frac {x^3}{4}} e^{x^3} \left (12 \left (1+\frac {\log (5)}{4}\right ) \log \left (x^2+5\right ) x^3+60 \left (1+\frac {\log (5)}{4}\right ) \log \left (x^2+5\right ) x+8\right ) x}{x^2+5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (2\ 5^{\frac {x^3}{2}} e^{2 x^3}+2 \log ^2\left (x^2+5\right )-\frac {4\ 5^{\frac {x^3}{4}} e^{x^3} \left (5 x \log \left (x^2+5\right )+x^3 \log \left (x^2+5\right )\right )}{x \left (x^2+5\right )}\right )\) |
Input:
Int[(-8*E^((4*x^3 + x^3*Log[5])/4)*x + E^((4*x^3 + x^3*Log[5])/2)*(60*x^2 + 12*x^4 + (15*x^2 + 3*x^4)*Log[5]) + (8*x + E^((4*x^3 + x^3*Log[5])/4)*(- 60*x^2 - 12*x^4 + (-15*x^2 - 3*x^4)*Log[5]))*Log[5 + x^2])/(10 + 2*x^2),x]
Output:
(2*5^(x^3/2)*E^(2*x^3) + 2*Log[5 + x^2]^2 - (4*5^(x^3/4)*E^x^3*(5*x*Log[5 + x^2] + x^3*Log[5 + x^2]))/(x*(5 + x^2)))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 22.59 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67
method | result | size |
parallelrisch | \(\ln \left (x^{2}+5\right )^{2}-2 \ln \left (x^{2}+5\right ) {\mathrm e}^{\frac {\left (\ln \left (5\right )+4\right ) x^{3}}{4}}+{\mathrm e}^{\frac {\left (\ln \left (5\right )+4\right ) x^{3}}{2}}\) | \(40\) |
risch | \(5^{\frac {x^{3}}{2}} {\mathrm e}^{2 x^{3}}-2 \,5^{\frac {x^{3}}{4}} {\mathrm e}^{x^{3}} \ln \left (x^{2}+5\right )+\ln \left (x^{2}+5\right )^{2}\) | \(45\) |
default | \(-2 \,{\mathrm e}^{\frac {x^{3} \ln \left (5\right )}{4}+x^{3}} \ln \left (x^{2}+5\right )+\ln \left (x^{2}+5\right )^{2}+{\mathrm e}^{\frac {x^{3} \ln \left (5\right )}{2}+2 x^{3}}\) | \(58\) |
parts | \(-2 \,{\mathrm e}^{\frac {x^{3} \ln \left (5\right )}{4}+x^{3}} \ln \left (x^{2}+5\right )+\ln \left (x^{2}+5\right )^{2}+\frac {2 \left (6+\frac {3 \ln \left (5\right )}{2}\right ) {\mathrm e}^{\frac {x^{3} \ln \left (5\right )}{2}+2 x^{3}}}{3 \left (\ln \left (5\right )+4\right )}\) | \(58\) |
Input:
int(((((-3*x^4-15*x^2)*ln(5)-12*x^4-60*x^2)*exp(1/4*x^3*ln(5)+x^3)+8*x)*ln (x^2+5)+((3*x^4+15*x^2)*ln(5)+12*x^4+60*x^2)*exp(1/4*x^3*ln(5)+x^3)^2-8*x* exp(1/4*x^3*ln(5)+x^3))/(2*x^2+10),x,method=_RETURNVERBOSE)
Output:
ln(x^2+5)^2-2*ln(x^2+5)*exp(1/4*(ln(5)+4)*x^3)+exp(1/4*(ln(5)+4)*x^3)^2
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).
Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=-2 \, e^{\left (\frac {1}{4} \, x^{3} \log \left (5\right ) + x^{3}\right )} \log \left (x^{2} + 5\right ) + \log \left (x^{2} + 5\right )^{2} + e^{\left (\frac {1}{2} \, x^{3} \log \left (5\right ) + 2 \, x^{3}\right )} \] Input:
integrate(((((-3*x^4-15*x^2)*log(5)-12*x^4-60*x^2)*exp(1/4*x^3*log(5)+x^3) +8*x)*log(x^2+5)+((3*x^4+15*x^2)*log(5)+12*x^4+60*x^2)*exp(1/4*x^3*log(5)+ x^3)^2-8*x*exp(1/4*x^3*log(5)+x^3))/(2*x^2+10),x, algorithm="fricas")
Output:
-2*e^(1/4*x^3*log(5) + x^3)*log(x^2 + 5) + log(x^2 + 5)^2 + e^(1/2*x^3*log (5) + 2*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
Time = 0.38 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=- 2 e^{\frac {x^{3} \log {\left (5 \right )}}{4} + x^{3}} \log {\left (x^{2} + 5 \right )} + e^{\frac {x^{3} \log {\left (5 \right )}}{2} + 2 x^{3}} + \log {\left (x^{2} + 5 \right )}^{2} \] Input:
integrate(((((-3*x**4-15*x**2)*ln(5)-12*x**4-60*x**2)*exp(1/4*x**3*ln(5)+x **3)+8*x)*ln(x**2+5)+((3*x**4+15*x**2)*ln(5)+12*x**4+60*x**2)*exp(1/4*x**3 *ln(5)+x**3)**2-8*x*exp(1/4*x**3*ln(5)+x**3))/(2*x**2+10),x)
Output:
-2*exp(x**3*log(5)/4 + x**3)*log(x**2 + 5) + exp(x**3*log(5)/2 + 2*x**3) + log(x**2 + 5)**2
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=-2 \, e^{\left (\frac {1}{4} \, x^{3} \log \left (5\right ) + x^{3}\right )} \log \left (x^{2} + 5\right ) + \log \left (x^{2} + 5\right )^{2} + e^{\left (\frac {1}{2} \, x^{3} \log \left (5\right ) + 2 \, x^{3}\right )} \] Input:
integrate(((((-3*x^4-15*x^2)*log(5)-12*x^4-60*x^2)*exp(1/4*x^3*log(5)+x^3) +8*x)*log(x^2+5)+((3*x^4+15*x^2)*log(5)+12*x^4+60*x^2)*exp(1/4*x^3*log(5)+ x^3)^2-8*x*exp(1/4*x^3*log(5)+x^3))/(2*x^2+10),x, algorithm="maxima")
Output:
-2*e^(1/4*x^3*log(5) + x^3)*log(x^2 + 5) + log(x^2 + 5)^2 + e^(1/2*x^3*log (5) + 2*x^3)
Timed out. \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=\text {Timed out} \] Input:
integrate(((((-3*x^4-15*x^2)*log(5)-12*x^4-60*x^2)*exp(1/4*x^3*log(5)+x^3) +8*x)*log(x^2+5)+((3*x^4+15*x^2)*log(5)+12*x^4+60*x^2)*exp(1/4*x^3*log(5)+ x^3)^2-8*x*exp(1/4*x^3*log(5)+x^3))/(2*x^2+10),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=\int \frac {\ln \left (x^2+5\right )\,\left (8\,x-{\mathrm {e}}^{\frac {x^3\,\ln \left (5\right )}{4}+x^3}\,\left (\ln \left (5\right )\,\left (3\,x^4+15\,x^2\right )+60\,x^2+12\,x^4\right )\right )-8\,x\,{\mathrm {e}}^{\frac {x^3\,\ln \left (5\right )}{4}+x^3}+{\mathrm {e}}^{\frac {x^3\,\ln \left (5\right )}{2}+2\,x^3}\,\left (\ln \left (5\right )\,\left (3\,x^4+15\,x^2\right )+60\,x^2+12\,x^4\right )}{2\,x^2+10} \,d x \] Input:
int((log(x^2 + 5)*(8*x - exp((x^3*log(5))/4 + x^3)*(log(5)*(15*x^2 + 3*x^4 ) + 60*x^2 + 12*x^4)) - 8*x*exp((x^3*log(5))/4 + x^3) + exp((x^3*log(5))/2 + 2*x^3)*(log(5)*(15*x^2 + 3*x^4) + 60*x^2 + 12*x^4))/(2*x^2 + 10),x)
Output:
int((log(x^2 + 5)*(8*x - exp((x^3*log(5))/4 + x^3)*(log(5)*(15*x^2 + 3*x^4 ) + 60*x^2 + 12*x^4)) - 8*x*exp((x^3*log(5))/4 + x^3) + exp((x^3*log(5))/2 + 2*x^3)*(log(5)*(15*x^2 + 3*x^4) + 60*x^2 + 12*x^4))/(2*x^2 + 10), x)
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {-8 e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} x+e^{\frac {1}{2} \left (4 x^3+x^3 \log (5)\right )} \left (60 x^2+12 x^4+\left (15 x^2+3 x^4\right ) \log (5)\right )+\left (8 x+e^{\frac {1}{4} \left (4 x^3+x^3 \log (5)\right )} \left (-60 x^2-12 x^4+\left (-15 x^2-3 x^4\right ) \log (5)\right )\right ) \log \left (5+x^2\right )}{10+2 x^2} \, dx=e^{2 x^{3}} 5^{\frac {x^{3}}{2}}-2 e^{x^{3}} 5^{\frac {x^{3}}{4}} \mathrm {log}\left (x^{2}+5\right )+\mathrm {log}\left (x^{2}+5\right )^{2} \] Input:
int(((((-3*x^4-15*x^2)*log(5)-12*x^4-60*x^2)*exp(1/4*x^3*log(5)+x^3)+8*x)* log(x^2+5)+((3*x^4+15*x^2)*log(5)+12*x^4+60*x^2)*exp(1/4*x^3*log(5)+x^3)^2 -8*x*exp(1/4*x^3*log(5)+x^3))/(2*x^2+10),x)
Output:
e**(2*x**3)*5**(x**3/2) - 2*e**(x**3)*5**(x**3/4)*log(x**2 + 5) + log(x**2 + 5)**2