Integrand size = 56, antiderivative size = 25 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {1}{2} e^{2 (-2+x)^4 x+10 (-x+\log (5 x))} \] Output:
1/2*exp(5*ln(5*x)-5*x+x*(-2+x)^4)^2
Time = 0.78 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {9765625}{2} e^{2 x \left (11-32 x+24 x^2-8 x^3+x^4\right )} x^{10} \] Input:
Integrate[9765625*E^(22*x - 64*x^2 + 48*x^3 - 16*x^4 + 2*x^5)*x^9*(5 + 11* x - 64*x^2 + 72*x^3 - 32*x^4 + 5*x^5),x]
Output:
(9765625*E^(2*x*(11 - 32*x + 24*x^2 - 8*x^3 + x^4))*x^10)/2
Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(25)=50\).
Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {27, 2726}
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.
\(\displaystyle \int 9765625 e^{2 x^5-16 x^4+48 x^3-64 x^2+22 x} x^9 \left (5 x^5-32 x^4+72 x^3-64 x^2+11 x+5\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 9765625 \int e^{2 x^5-16 x^4+48 x^3-64 x^2+22 x} x^9 \left (5 x^5-32 x^4+72 x^3-64 x^2+11 x+5\right )dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {9765625 e^{2 x^5-16 x^4+48 x^3-64 x^2+22 x} x^9 \left (5 x^5-32 x^4+72 x^3-64 x^2+11 x\right )}{2 \left (5 x^4-32 x^3+72 x^2-64 x+11\right )}\) |
Input:
Int[9765625*E^(22*x - 64*x^2 + 48*x^3 - 16*x^4 + 2*x^5)*x^9*(5 + 11*x - 64 *x^2 + 72*x^3 - 32*x^4 + 5*x^5),x]
Output:
(9765625*E^(22*x - 64*x^2 + 48*x^3 - 16*x^4 + 2*x^5)*x^9*(11*x - 64*x^2 + 72*x^3 - 32*x^4 + 5*x^5))/(2*(11 - 64*x + 72*x^2 - 32*x^3 + 5*x^4))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {9765625 x^{10} {\mathrm e}^{2 x \left (x^{4}-8 x^{3}+24 x^{2}-32 x +11\right )}}{2}\) | \(28\) |
gosper | \(\frac {9765625 x^{10} {\mathrm e}^{2 x^{5}-16 x^{4}+48 x^{3}-64 x^{2}+22 x}}{2}\) | \(34\) |
default | \(\frac {9765625 x^{10} {\mathrm e}^{2 x^{5}-16 x^{4}+48 x^{3}-64 x^{2}+22 x}}{2}\) | \(34\) |
parallelrisch | \(\frac {9765625 x^{10} {\mathrm e}^{2 x^{5}-16 x^{4}+48 x^{3}-64 x^{2}+22 x}}{2}\) | \(34\) |
orering | \(\frac {9765625 x^{10} {\mathrm e}^{2 x^{5}-16 x^{4}+48 x^{3}-64 x^{2}+22 x}}{2}\) | \(34\) |
meijerg | \(\frac {\sqrt {10}\, \sqrt {\pi }\, \operatorname {erfi}\left (x^{5} \sqrt {10}\, \sqrt {\ln \left (5\right )}\right )}{20 \sqrt {\ln \left (5\right )}}+\frac {8 \,8^{\frac {1}{5}} 25^{\frac {4}{5}} \left (-1\right )^{\frac {3}{5}} \left (\frac {x^{4} \left (-1\right )^{\frac {2}{5}} \ln \left (5\right )^{\frac {2}{5}} \pi \csc \left (\frac {2 \pi }{5}\right )}{\Gamma \left (\frac {3}{5}\right ) \left (-x^{10} \ln \left (5\right )\right )^{\frac {2}{5}}}-\frac {x^{4} \left (-1\right )^{\frac {2}{5}} \ln \left (5\right )^{\frac {2}{5}} \Gamma \left (\frac {2}{5}, -10 x^{10} \ln \left (5\right )\right )}{\left (-x^{10} \ln \left (5\right )\right )^{\frac {2}{5}}}\right )}{125 \ln \left (5\right )^{\frac {2}{5}}}-\frac {18 \,128^{\frac {1}{10}} 125^{\frac {9}{10}} \left (-1\right )^{\frac {7}{10}} \left (\frac {x^{3} \left (-1\right )^{\frac {3}{10}} \ln \left (5\right )^{\frac {3}{10}} \pi \csc \left (\frac {3 \pi }{10}\right )}{\Gamma \left (\frac {7}{10}\right ) \left (-x^{10} \ln \left (5\right )\right )^{\frac {3}{10}}}-\frac {x^{3} \left (-1\right )^{\frac {3}{10}} \ln \left (5\right )^{\frac {3}{10}} \Gamma \left (\frac {3}{10}, -10 x^{10} \ln \left (5\right )\right )}{\left (-x^{10} \ln \left (5\right )\right )^{\frac {3}{10}}}\right )}{625 \ln \left (5\right )^{\frac {3}{10}}}+\frac {16 \,16^{\frac {1}{5}} 5^{\frac {4}{5}} \left (-1\right )^{\frac {4}{5}} \left (\frac {x^{2} \left (-1\right )^{\frac {1}{5}} \ln \left (5\right )^{\frac {1}{5}} \pi \csc \left (\frac {\pi }{5}\right )}{\Gamma \left (\frac {4}{5}\right ) \left (-x^{10} \ln \left (5\right )\right )^{\frac {1}{5}}}-\frac {x^{2} \left (-1\right )^{\frac {1}{5}} \ln \left (5\right )^{\frac {1}{5}} \Gamma \left (\frac {1}{5}, -10 x^{10} \ln \left (5\right )\right )}{\left (-x^{10} \ln \left (5\right )\right )^{\frac {1}{5}}}\right )}{25 \ln \left (5\right )^{\frac {1}{5}}}-\frac {11 \,10^{\frac {9}{10}} \left (-1\right )^{\frac {9}{10}} \left (\frac {x \left (-1\right )^{\frac {1}{10}} \ln \left (5\right )^{\frac {1}{10}} \pi \csc \left (\frac {\pi }{10}\right )}{\Gamma \left (\frac {9}{10}\right ) \left (-x^{10} \ln \left (5\right )\right )^{\frac {1}{10}}}-\frac {x \left (-1\right )^{\frac {1}{10}} \ln \left (5\right )^{\frac {1}{10}} \Gamma \left (\frac {1}{10}, -10 x^{10} \ln \left (5\right )\right )}{\left (-x^{10} \ln \left (5\right )\right )^{\frac {1}{10}}}\right )}{100 \ln \left (5\right )^{\frac {1}{10}}}+5 \ln \left (x \right )+\frac {i \pi }{2}+\frac {\ln \left (2\right )}{2}+\frac {\ln \left (5\right )}{2}+\frac {\ln \left (\ln \left (5\right )\right )}{2}-\frac {\ln \left (-10 x^{10} \ln \left (5\right )\right )}{2}-\frac {\operatorname {expIntegral}_{1}\left (-10 x^{10} \ln \left (5\right )\right )}{2}\) | \(361\) |
Input:
int((5*x^5-32*x^4+72*x^3-64*x^2+11*x+5)*exp(5*ln(5*x)+x^5-8*x^4+24*x^3-32* x^2+11*x)^2/x,x,method=_RETURNVERBOSE)
Output:
9765625/2*x^10*exp(2*x*(x^4-8*x^3+24*x^2-32*x+11))
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {1}{2} \, e^{\left (2 \, x^{5} - 16 \, x^{4} + 48 \, x^{3} - 64 \, x^{2} + 22 \, x + 10 \, \log \left (5 \, x\right )\right )} \] Input:
integrate((5*x^5-32*x^4+72*x^3-64*x^2+11*x+5)*exp(5*log(5*x)+x^5-8*x^4+24* x^3-32*x^2+11*x)^2/x,x, algorithm="fricas")
Output:
1/2*e^(2*x^5 - 16*x^4 + 48*x^3 - 64*x^2 + 22*x + 10*log(5*x))
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {9765625 x^{10} e^{2 x^{5} - 16 x^{4} + 48 x^{3} - 64 x^{2} + 22 x}}{2} \] Input:
integrate((5*x**5-32*x**4+72*x**3-64*x**2+11*x+5)*exp(5*ln(5*x)+x**5-8*x** 4+24*x**3-32*x**2+11*x)**2/x,x)
Output:
9765625*x**10*exp(2*x**5 - 16*x**4 + 48*x**3 - 64*x**2 + 22*x)/2
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {9765625}{2} \, x^{10} e^{\left (2 \, x^{5} - 16 \, x^{4} + 48 \, x^{3} - 64 \, x^{2} + 22 \, x\right )} \] Input:
integrate((5*x^5-32*x^4+72*x^3-64*x^2+11*x+5)*exp(5*log(5*x)+x^5-8*x^4+24* x^3-32*x^2+11*x)^2/x,x, algorithm="maxima")
Output:
9765625/2*x^10*e^(2*x^5 - 16*x^4 + 48*x^3 - 64*x^2 + 22*x)
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {1}{2} \, e^{\left (2 \, x^{5} - 16 \, x^{4} + 48 \, x^{3} - 64 \, x^{2} + 22 \, x + 10 \, \log \left (5 \, x\right )\right )} \] Input:
integrate((5*x^5-32*x^4+72*x^3-64*x^2+11*x+5)*exp(5*log(5*x)+x^5-8*x^4+24* x^3-32*x^2+11*x)^2/x,x, algorithm="giac")
Output:
1/2*e^(2*x^5 - 16*x^4 + 48*x^3 - 64*x^2 + 22*x + 10*log(5*x))
Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {9765625\,x^{10}\,{\mathrm {e}}^{22\,x}\,{\mathrm {e}}^{2\,x^5}\,{\mathrm {e}}^{-16\,x^4}\,{\mathrm {e}}^{48\,x^3}\,{\mathrm {e}}^{-64\,x^2}}{2} \] Input:
int((exp(22*x + 10*log(5*x) - 64*x^2 + 48*x^3 - 16*x^4 + 2*x^5)*(11*x - 64 *x^2 + 72*x^3 - 32*x^4 + 5*x^5 + 5))/x,x)
Output:
(9765625*x^10*exp(22*x)*exp(2*x^5)*exp(-16*x^4)*exp(48*x^3)*exp(-64*x^2))/ 2
Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5\right ) \, dx=\frac {9765625 e^{2 x^{5}+48 x^{3}+22 x} x^{10}}{2 e^{16 x^{4}+64 x^{2}}} \] Input:
int((5*x^5-32*x^4+72*x^3-64*x^2+11*x+5)*exp(5*log(5*x)+x^5-8*x^4+24*x^3-32 *x^2+11*x)^2/x,x)
Output:
(9765625*e**(2*x**5 + 48*x**3 + 22*x)*x**10)/(2*e**(16*x**4 + 64*x**2))