3.1.36 \(\int 9765625 e^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 (5+11 x-64 x^2+72 x^3-32 x^4+5 x^5) \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{2} e^{2 (-2+x)^4 x+10 (-x+\log (5 x))} \]

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Rubi [B]  time = 0.21, antiderivative size = 79, normalized size of antiderivative = 3.16, number of steps used = 2, number of rules used = 2, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {12, 2288} \begin {gather*} \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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(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 - 6
4*x + 72*x^2 - 32*x^3 + 5*x^4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=9765625 \int 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^{22 x-64 x^2+48 x^3-16 x^4+2 x^5} x^9 \left (11 x-64 x^2+72 x^3-32 x^4+5 x^5\right )}{2 \left (11-64 x+72 x^2-32 x^3+5 x^4\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 30, normalized size = 1.20 \begin {gather*} \frac {9765625}{2} e^{2 x \left (11-32 x+24 x^2-8 x^3+x^4\right )} x^{10} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(9765625*E^(2*x*(11 - 32*x + 24*x^2 - 8*x^3 + x^4))*x^10)/2

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fricas [A]  time = 0.55, size = 33, normalized size = 1.32 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="f
ricas")

[Out]

1/2*e^(2*x^5 - 16*x^4 + 48*x^3 - 64*x^2 + 22*x + 10*log(5*x))

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giac [A]  time = 0.45, size = 33, normalized size = 1.32 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="g
iac")

[Out]

1/2*e^(2*x^5 - 16*x^4 + 48*x^3 - 64*x^2 + 22*x + 10*log(5*x))

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maple [A]  time = 0.11, size = 28, normalized size = 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	\(\frac {9765625 x^{10} {\mathrm e}^{2 x^{5}-16 x^{4}+48 x^{3}-64 x^{2}+22 x}}{2}\)	\(34\)
meijerg	\(\frac {\sqrt {10}\, \sqrt {\pi }\, \erfi \left (x^{5} \sqrt {10}\, \sqrt {\ln \relax (5)}\right )}{20 \sqrt {\ln \relax (5)}}+\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 \relax (5)^{\frac {2}{5}} \pi \csc \left (\frac {2 \pi }{5}\right )}{\Gamma \left (\frac {3}{5}\right ) \left (-x^{10} \ln \relax (5)\right )^{\frac {2}{5}}}-\frac {x^{4} \left (-1\right )^{\frac {2}{5}} \ln \relax (5)^{\frac {2}{5}} \Gamma \left (\frac {2}{5}, -10 x^{10} \ln \relax (5)\right )}{\left (-x^{10} \ln \relax (5)\right )^{\frac {2}{5}}}\right )}{125 \ln \relax (5)^{\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 \relax (5)^{\frac {3}{10}} \pi \csc \left (\frac {3 \pi }{10}\right )}{\Gamma \left (\frac {7}{10}\right ) \left (-x^{10} \ln \relax (5)\right )^{\frac {3}{10}}}-\frac {x^{3} \left (-1\right )^{\frac {3}{10}} \ln \relax (5)^{\frac {3}{10}} \Gamma \left (\frac {3}{10}, -10 x^{10} \ln \relax (5)\right )}{\left (-x^{10} \ln \relax (5)\right )^{\frac {3}{10}}}\right )}{625 \ln \relax (5)^{\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 \relax (5)^{\frac {1}{5}} \pi \csc \left (\frac {\pi }{5}\right )}{\Gamma \left (\frac {4}{5}\right ) \left (-x^{10} \ln \relax (5)\right )^{\frac {1}{5}}}-\frac {x^{2} \left (-1\right )^{\frac {1}{5}} \ln \relax (5)^{\frac {1}{5}} \Gamma \left (\frac {1}{5}, -10 x^{10} \ln \relax (5)\right )}{\left (-x^{10} \ln \relax (5)\right )^{\frac {1}{5}}}\right )}{25 \ln \relax (5)^{\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 \relax (5)^{\frac {1}{10}} \pi \csc \left (\frac {\pi }{10}\right )}{\Gamma \left (\frac {9}{10}\right ) \left (-x^{10} \ln \relax (5)\right )^{\frac {1}{10}}}-\frac {x \left (-1\right )^{\frac {1}{10}} \ln \relax (5)^{\frac {1}{10}} \Gamma \left (\frac {1}{10}, -10 x^{10} \ln \relax (5)\right )}{\left (-x^{10} \ln \relax (5)\right )^{\frac {1}{10}}}\right )}{100 \ln \relax (5)^{\frac {1}{10}}}-\frac {\ln \left (-10 x^{10} \ln \relax (5)\right )}{2}-\frac {\expIntegralEi \left (1, -10 x^{10} \ln \relax (5)\right )}{2}+5 \ln \relax (x )+\frac {i \pi }{2}+\frac {\ln \relax (2)}{2}+\frac {\ln \relax (5)}{2}+\frac {\ln \left (\ln \relax (5)\right )}{2}\)	\(361\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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=_RETURNVERBOS
E)

[Out]

9765625/2*x^10*exp(2*x*(x^4-8*x^3+24*x^2-32*x+11))

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maxima [A]  time = 0.59, size = 30, normalized size = 1.20 \begin {gather*} \frac {9765625}{2} \, x^{10} e^{\left (2 \, x^{5} - 16 \, x^{4} + 48 \, x^{3} - 64 \, x^{2} + 22 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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="m
axima")

[Out]

9765625/2*x^10*e^(2*x^5 - 16*x^4 + 48*x^3 - 64*x^2 + 22*x)

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mupad [B]  time = 0.27, size = 33, normalized size = 1.32 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

(9765625*x^10*exp(22*x)*exp(2*x^5)*exp(-16*x^4)*exp(48*x^3)*exp(-64*x^2))/2

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sympy [A]  time = 0.14, size = 31, normalized size = 1.24 \begin {gather*} \frac {9765625 x^{10} e^{2 x^{5} - 16 x^{4} + 48 x^{3} - 64 x^{2} + 22 x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

9765625*x**10*exp(2*x**5 - 16*x**4 + 48*x**3 - 64*x**2 + 22*x)/2

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