3.16.100 \(\int (6 x^5+e^{\frac {2 (e^2+x^3+2 x^2 \log (5)+x \log ^2(5))}{x}} (-2 e^2+2 x+4 x^3+4 x^2 \log (5))+e^{\frac {e^2+x^3+2 x^2 \log (5)+x \log ^2(5)}{x}} (-2 e^2 x^2+8 x^3+4 x^5+4 x^4 \log (5))) \, dx\)

Optimal. Leaf size=35 \[ x^2 \left (e^{\frac {e^2}{x}+(x+\log (5))^2}-x+\left (1+\frac {1}{x}\right ) x^2\right )^2 \]

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Rubi [B]  time = 0.34, antiderivative size = 187, normalized size of antiderivative = 5.34, number of steps used = 3, number of rules used = 1, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {2288} \begin {gather*} x^6-\frac {5^{4 x} e^{\frac {2 \left (x^3+x \log ^2(5)+e^2\right )}{x}} \left (-2 x^3-2 x^2 \log (5)+e^2\right )}{\frac {3 x^2+4 x \log (5)+\log ^2(5)}{x}-\frac {x^3+2 x^2 \log (5)+x \log ^2(5)+e^2}{x^2}}-\frac {2\ 5^{2 x} e^{\frac {x^3+x \log ^2(5)+e^2}{x}} \left (-2 x^5-2 x^4 \log (5)+e^2 x^2\right )}{\frac {3 x^2+4 x \log (5)+\log ^2(5)}{x}-\frac {x^3+2 x^2 \log (5)+x \log ^2(5)+e^2}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[6*x^5 + E^((2*(E^2 + x^3 + 2*x^2*Log[5] + x*Log[5]^2))/x)*(-2*E^2 + 2*x + 4*x^3 + 4*x^2*Log[5]) + E^((E^2
+ x^3 + 2*x^2*Log[5] + x*Log[5]^2)/x)*(-2*E^2*x^2 + 8*x^3 + 4*x^5 + 4*x^4*Log[5]),x]

[Out]

x^6 - (5^(4*x)*E^((2*(E^2 + x^3 + x*Log[5]^2))/x)*(E^2 - 2*x^3 - 2*x^2*Log[5]))/((3*x^2 + 4*x*Log[5] + Log[5]^
2)/x - (E^2 + x^3 + 2*x^2*Log[5] + x*Log[5]^2)/x^2) - (2*5^(2*x)*E^((E^2 + x^3 + x*Log[5]^2)/x)*(E^2*x^2 - 2*x
^5 - 2*x^4*Log[5]))/((3*x^2 + 4*x*Log[5] + Log[5]^2)/x - (E^2 + x^3 + 2*x^2*Log[5] + x*Log[5]^2)/x^2)

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} &=x^6+\int e^{\frac {2 \left (e^2+x^3+2 x^2 \log (5)+x \log ^2(5)\right )}{x}} \left (-2 e^2+2 x+4 x^3+4 x^2 \log (5)\right ) \, dx+\int e^{\frac {e^2+x^3+2 x^2 \log (5)+x \log ^2(5)}{x}} \left (-2 e^2 x^2+8 x^3+4 x^5+4 x^4 \log (5)\right ) \, dx\\ &=x^6-\frac {5^{4 x} e^{\frac {2 \left (e^2+x^3+x \log ^2(5)\right )}{x}} \left (e^2-2 x^3-2 x^2 \log (5)\right )}{\frac {3 x^2+4 x \log (5)+\log ^2(5)}{x}-\frac {e^2+x^3+2 x^2 \log (5)+x \log ^2(5)}{x^2}}-\frac {2\ 5^{2 x} e^{\frac {e^2+x^3+x \log ^2(5)}{x}} \left (e^2 x^2-2 x^5-2 x^4 \log (5)\right )}{\frac {3 x^2+4 x \log (5)+\log ^2(5)}{x}-\frac {e^2+x^3+2 x^2 \log (5)+x \log ^2(5)}{x^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 69, normalized size = 1.97 \begin {gather*} x^6+\frac {4\ 25^x e^{\frac {e^2}{x}+x^2+\log ^2(5)} x^4 \log (5)}{\log (25)}+\frac {2\ 625^x e^{\frac {2 \left (e^2+x^3+x \log ^2(5)\right )}{x}} x^2 \log (25)}{\log (625)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[6*x^5 + E^((2*(E^2 + x^3 + 2*x^2*Log[5] + x*Log[5]^2))/x)*(-2*E^2 + 2*x + 4*x^3 + 4*x^2*Log[5]) + E^
((E^2 + x^3 + 2*x^2*Log[5] + x*Log[5]^2)/x)*(-2*E^2*x^2 + 8*x^3 + 4*x^5 + 4*x^4*Log[5]),x]

[Out]

x^6 + (4*25^x*E^(E^2/x + x^2 + Log[5]^2)*x^4*Log[5])/Log[25] + (2*625^x*E^((2*(E^2 + x^3 + x*Log[5]^2))/x)*x^2
*Log[25])/Log[625]

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fricas [A]  time = 0.88, size = 62, normalized size = 1.77 \begin {gather*} x^{6} + 2 \, x^{4} e^{\left (\frac {x^{3} + 2 \, x^{2} \log \relax (5) + x \log \relax (5)^{2} + e^{2}}{x}\right )} + x^{2} e^{\left (\frac {2 \, {\left (x^{3} + 2 \, x^{2} \log \relax (5) + x \log \relax (5)^{2} + e^{2}\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(5)-2*exp(2)+4*x^3+2*x)*exp((x*log(5)^2+2*x^2*log(5)+exp(2)+x^3)/x)^2+(4*x^4*log(5)-2*x^2*
exp(2)+4*x^5+8*x^3)*exp((x*log(5)^2+2*x^2*log(5)+exp(2)+x^3)/x)+6*x^5,x, algorithm="fricas")

[Out]

x^6 + 2*x^4*e^((x^3 + 2*x^2*log(5) + x*log(5)^2 + e^2)/x) + x^2*e^(2*(x^3 + 2*x^2*log(5) + x*log(5)^2 + e^2)/x
)

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giac [A]  time = 0.40, size = 62, normalized size = 1.77 \begin {gather*} x^{6} + 2 \, x^{4} e^{\left (\frac {x^{3} + 2 \, x^{2} \log \relax (5) + x \log \relax (5)^{2} + e^{2}}{x}\right )} + x^{2} e^{\left (\frac {2 \, {\left (x^{3} + 2 \, x^{2} \log \relax (5) + x \log \relax (5)^{2} + e^{2}\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(5)-2*exp(2)+4*x^3+2*x)*exp((x*log(5)^2+2*x^2*log(5)+exp(2)+x^3)/x)^2+(4*x^4*log(5)-2*x^2*
exp(2)+4*x^5+8*x^3)*exp((x*log(5)^2+2*x^2*log(5)+exp(2)+x^3)/x)+6*x^5,x, algorithm="giac")

[Out]

x^6 + 2*x^4*e^((x^3 + 2*x^2*log(5) + x*log(5)^2 + e^2)/x) + x^2*e^(2*(x^3 + 2*x^2*log(5) + x*log(5)^2 + e^2)/x
)

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maple [A]  time = 0.16, size = 57, normalized size = 1.63

method	result	size
risch	\(25^{2 x} {\mathrm e}^{\frac {2 x \ln \relax (5)^{2}+2 x^{3}+2 \,{\mathrm e}^{2}}{x}} x^{2}+2 \,25^{x} {\mathrm e}^{\frac {x \ln \relax (5)^{2}+x^{3}+{\mathrm e}^{2}}{x}} x^{4}+x^{6}\)	\(57\)
default	\({\mathrm e}^{\frac {2 x \ln \relax (5)^{2}+2 x^{2} \ln \left (25\right )+2 \,{\mathrm e}^{2}+2 x^{3}}{x}} x^{2}+2 \,{\mathrm e}^{\frac {x \ln \relax (5)^{2}+2 x^{2} \ln \relax (5)+{\mathrm e}^{2}+x^{3}}{x}} x^{4}+x^{6}\)	\(64\)
norman	\({\mathrm e}^{\frac {2 x \ln \relax (5)^{2}+2 x^{2} \ln \left (25\right )+2 \,{\mathrm e}^{2}+2 x^{3}}{x}} x^{2}+2 \,{\mathrm e}^{\frac {x \ln \relax (5)^{2}+2 x^{2} \ln \relax (5)+{\mathrm e}^{2}+x^{3}}{x}} x^{4}+x^{6}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2*ln(5)-2*exp(2)+4*x^3+2*x)*exp((x*ln(5)^2+2*x^2*ln(5)+exp(2)+x^3)/x)^2+(4*x^4*ln(5)-2*x^2*exp(2)+4*x
^5+8*x^3)*exp((x*ln(5)^2+2*x^2*ln(5)+exp(2)+x^3)/x)+6*x^5,x,method=_RETURNVERBOSE)

[Out]

(25^x)^2*exp(2*(x*ln(5)^2+x^3+exp(2))/x)*x^2+2*25^x*exp((x*ln(5)^2+x^3+exp(2))/x)*x^4+x^6

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maxima [A]  time = 1.04, size = 58, normalized size = 1.66 \begin {gather*} x^{6} + 2 \, x^{4} e^{\left (x^{2} + 2 \, x \log \relax (5) + \log \relax (5)^{2} + \frac {e^{2}}{x}\right )} + x^{2} e^{\left (2 \, x^{2} + 4 \, x \log \relax (5) + 2 \, \log \relax (5)^{2} + \frac {2 \, e^{2}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(5)-2*exp(2)+4*x^3+2*x)*exp((x*log(5)^2+2*x^2*log(5)+exp(2)+x^3)/x)^2+(4*x^4*log(5)-2*x^2*
exp(2)+4*x^5+8*x^3)*exp((x*log(5)^2+2*x^2*log(5)+exp(2)+x^3)/x)+6*x^5,x, algorithm="maxima")

[Out]

x^6 + 2*x^4*e^(x^2 + 2*x*log(5) + log(5)^2 + e^2/x) + x^2*e^(2*x^2 + 4*x*log(5) + 2*log(5)^2 + 2*e^2/x)

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mupad [B]  time = 1.19, size = 31, normalized size = 0.89 \begin {gather*} x^2\,{\left (5^{2\,x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x}+{\ln \relax (5)}^2+x^2}+x^2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp((2*(exp(2) + x*log(5)^2 + 2*x^2*log(5) + x^3))/x)*(2*x - 2*exp(2) + 4*x^2*log(5) + 4*x^3) + 6*x^5 + ex
p((exp(2) + x*log(5)^2 + 2*x^2*log(5) + x^3)/x)*(4*x^4*log(5) - 2*x^2*exp(2) + 8*x^3 + 4*x^5),x)

[Out]

x^2*(5^(2*x)*exp(exp(2)/x + log(5)^2 + x^2) + x^2)^2

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sympy [B]  time = 0.24, size = 66, normalized size = 1.89 \begin {gather*} x^{6} + 2 x^{4} e^{\frac {x^{3} + 2 x^{2} \log {\relax (5 )} + x \log {\relax (5 )}^{2} + e^{2}}{x}} + x^{2} e^{\frac {2 \left (x^{3} + 2 x^{2} \log {\relax (5 )} + x \log {\relax (5 )}^{2} + e^{2}\right )}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2*ln(5)-2*exp(2)+4*x**3+2*x)*exp((x*ln(5)**2+2*x**2*ln(5)+exp(2)+x**3)/x)**2+(4*x**4*ln(5)-2*x
**2*exp(2)+4*x**5+8*x**3)*exp((x*ln(5)**2+2*x**2*ln(5)+exp(2)+x**3)/x)+6*x**5,x)

[Out]

x**6 + 2*x**4*exp((x**3 + 2*x**2*log(5) + x*log(5)**2 + exp(2))/x) + x**2*exp(2*(x**3 + 2*x**2*log(5) + x*log(
5)**2 + exp(2))/x)

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