3.102.88 \(\int \frac {(-2 x^2+e^2 (12+8 x)+(-10 x^2+e^2 (60+40 x)) \log (5)) \log (\log (25))}{(4 e^4-4 e^2 x+x^2) \log (5)} \, dx\)

Optimal. Leaf size=25 \[ \frac {x (3+x) \left (5+\frac {1}{\log (5)}\right ) \log (\log (25))}{e^2-\frac {x}{2}} \]

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Rubi [B]  time = 0.05, antiderivative size = 53, normalized size of antiderivative = 2.12, number of steps used = 4, number of rules used = 3, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {12, 27, 1850} \begin {gather*} \frac {4 e^2 \left (3+2 e^2\right ) (1+5 \log (5)) \log (\log (25))}{\left (2 e^2-x\right ) \log (5)}-\frac {2 x (1+5 \log (5)) \log (\log (25))}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-2*x^2 + E^2*(12 + 8*x) + (-10*x^2 + E^2*(60 + 40*x))*Log[5])*Log[Log[25]])/((4*E^4 - 4*E^2*x + x^2)*Log
[5]),x]

[Out]

(4*E^2*(3 + 2*E^2)*(1 + 5*Log[5])*Log[Log[25]])/((2*E^2 - x)*Log[5]) - (2*x*(1 + 5*Log[5])*Log[Log[25]])/Log[5
]

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\log (\log (25)) \int \frac {-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)}{4 e^4-4 e^2 x+x^2} \, dx}{\log (5)}\\ &=\frac {\log (\log (25)) \int \frac {-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)}{\left (-2 e^2+x\right )^2} \, dx}{\log (5)}\\ &=\frac {\log (\log (25)) \int \left (-2 (1+5 \log (5))+\frac {4 e^2 \left (3+2 e^2\right ) (1+5 \log (5))}{\left (2 e^2-x\right )^2}\right ) \, dx}{\log (5)}\\ &=\frac {4 e^2 \left (3+2 e^2\right ) (1+5 \log (5)) \log (\log (25))}{\left (2 e^2-x\right ) \log (5)}-\frac {2 x (1+5 \log (5)) \log (\log (25))}{\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 40, normalized size = 1.60 \begin {gather*} \frac {2 \left (-x-\frac {2 e^2 \left (3+2 e^2\right )}{-2 e^2+x}\right ) (1+5 \log (5)) \log (\log (25))}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2*x^2 + E^2*(12 + 8*x) + (-10*x^2 + E^2*(60 + 40*x))*Log[5])*Log[Log[25]])/((4*E^4 - 4*E^2*x + x^
2)*Log[5]),x]

[Out]

(2*(-x - (2*E^2*(3 + 2*E^2))/(-2*E^2 + x))*(1 + 5*Log[5])*Log[Log[25]])/Log[5]

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fricas [B]  time = 0.81, size = 53, normalized size = 2.12 \begin {gather*} -\frac {2 \, {\left (x^{2} - 2 \, {\left (x - 3\right )} e^{2} + 5 \, {\left (x^{2} - 2 \, {\left (x - 3\right )} e^{2} + 4 \, e^{4}\right )} \log \relax (5) + 4 \, e^{4}\right )} \log \left (2 \, \log \relax (5)\right )}{{\left (x - 2 \, e^{2}\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x+60)*exp(2)-10*x^2)*log(5)+(8*x+12)*exp(2)-2*x^2)*log(2*log(5))/(4*exp(2)^2-4*exp(2)*x+x^2)/l
og(5),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x+60)*exp(2)-10*x^2)*log(5)+(8*x+12)*exp(2)-2*x^2)*log(2*log(5))/(4*exp(2)^2-4*exp(2)*x+x^2)/l
og(5),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -ln(5)^-1*ln(2*ln(5))*2*(5*sageVARx*ln(5
)+sageVARx+(-30*exp(2)*ln(5)-6*exp(2)-20*ln(5)*exp(4)-4*exp(4))*1/4/sqrt(exp(2)^2-exp(4))*ln(sqrt((2*sageVARx-
4*exp(2))^2+(-4*sqrt(

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maple [A]  time = 0.35, size = 36, normalized size = 1.44

method	result	size
gosper	\(\frac {2 \left (x^{2}+6 \,{\mathrm e}^{2}\right ) \left (5 \ln \relax (5)+1\right ) \ln \left (2 \ln \relax (5)\right )}{\ln \relax (5) \left (2 \,{\mathrm e}^{2}-x \right )}\)	\(36\)
norman	\(\frac {\frac {2 \left (5 \ln \relax (2) \ln \relax (5)+5 \ln \relax (5) \ln \left (\ln \relax (5)\right )+\ln \relax (2)+\ln \left (\ln \relax (5)\right )\right ) x^{2}}{\ln \relax (5)}+\frac {12 \,{\mathrm e}^{2} \left (5 \ln \relax (2) \ln \relax (5)+5 \ln \relax (5) \ln \left (\ln \relax (5)\right )+\ln \relax (2)+\ln \left (\ln \relax (5)\right )\right )}{\ln \relax (5)}}{2 \,{\mathrm e}^{2}-x}\)	\(68\)
meijerg	\(-\frac {8 \,{\mathrm e}^{2} \left (-\frac {5 \ln \relax (5)}{2}-\frac {1}{2}\right ) \ln \left (2 \ln \relax (5)\right ) \left (-\frac {x \,{\mathrm e}^{-2} \left (-\frac {3 x \,{\mathrm e}^{-2}}{2}+6\right )}{6 \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )}-2 \ln \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )\right )}{\ln \relax (5)}+\frac {4 \left (10 \,{\mathrm e}^{2} \ln \relax (5)+2 \,{\mathrm e}^{2}\right ) \ln \left (2 \ln \relax (5)\right ) \left (\frac {x \,{\mathrm e}^{-2}}{2-x \,{\mathrm e}^{-2}}+\ln \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )\right )}{\ln \relax (5)}+\frac {15 \,{\mathrm e}^{-2} \ln \left (2 \ln \relax (5)\right ) x}{1-\frac {x \,{\mathrm e}^{-2}}{2}}+\frac {3 \,{\mathrm e}^{-2} \ln \left (2 \ln \relax (5)\right ) x}{\ln \relax (5) \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )}\)	\(140\)
risch	\(-10 x \ln \relax (2)-10 x \ln \left (\ln \relax (5)\right )-\frac {2 x \ln \relax (2)}{\ln \relax (5)}-\frac {2 x \ln \left (\ln \relax (5)\right )}{\ln \relax (5)}+\frac {30 \,{\mathrm e}^{2} \ln \relax (2)}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {30 \,{\mathrm e}^{2} \ln \left (\ln \relax (5)\right )}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {20 \ln \relax (2) {\mathrm e}^{4}}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {20 \,{\mathrm e}^{4} \ln \left (\ln \relax (5)\right )}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {6 \,{\mathrm e}^{2} \ln \relax (2)}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {6 \,{\mathrm e}^{2} \ln \left (\ln \relax (5)\right )}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {4 \ln \relax (2) {\mathrm e}^{4}}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {4 \,{\mathrm e}^{4} \ln \left (\ln \relax (5)\right )}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((40*x+60)*exp(2)-10*x^2)*ln(5)+(8*x+12)*exp(2)-2*x^2)*ln(2*ln(5))/(4*exp(2)^2-4*exp(2)*x+x^2)/ln(5),x,me
thod=_RETURNVERBOSE)

[Out]

2*(x^2+6*exp(2))*(5*ln(5)+1)*ln(2*ln(5))/ln(5)/(2*exp(2)-x)

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maxima [B]  time = 0.36, size = 52, normalized size = 2.08 \begin {gather*} -\frac {2 \, {\left (x {\left (5 \, \log \relax (5) + 1\right )} + \frac {2 \, {\left (5 \, {\left (2 \, e^{4} + 3 \, e^{2}\right )} \log \relax (5) + 2 \, e^{4} + 3 \, e^{2}\right )}}{x - 2 \, e^{2}}\right )} \log \left (2 \, \log \relax (5)\right )}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x+60)*exp(2)-10*x^2)*log(5)+(8*x+12)*exp(2)-2*x^2)*log(2*log(5))/(4*exp(2)^2-4*exp(2)*x+x^2)/l
og(5),x, algorithm="maxima")

[Out]

-2*(x*(5*log(5) + 1) + 2*(5*(2*e^4 + 3*e^2)*log(5) + 2*e^4 + 3*e^2)/(x - 2*e^2))*log(2*log(5))/log(5)

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mupad [B]  time = 6.53, size = 42, normalized size = 1.68 \begin {gather*} -\frac {2\,\ln \left (2\,\ln \relax (5)\right )\,\left (5\,\ln \relax (5)+1\right )\,\left (x^2-2\,{\mathrm {e}}^2\,x+6\,{\mathrm {e}}^2+4\,{\mathrm {e}}^4\right )}{\ln \relax (5)\,\left (x-2\,{\mathrm {e}}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*log(5))*(log(5)*(10*x^2 - exp(2)*(40*x + 60)) + 2*x^2 - exp(2)*(8*x + 12)))/(log(5)*(4*exp(4) - 4*
x*exp(2) + x^2)),x)

[Out]

-(2*log(2*log(5))*(5*log(5) + 1)*(6*exp(2) + 4*exp(4) - 2*x*exp(2) + x^2))/(log(5)*(x - 2*exp(2)))

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sympy [B]  time = 0.62, size = 134, normalized size = 5.36 \begin {gather*} - x \left (\frac {2 \log {\left (\log {\relax (5 )} \right )}}{\log {\relax (5 )}} + \frac {2 \log {\relax (2 )}}{\log {\relax (5 )}} + 10 \log {\left (\log {\relax (5 )} \right )} + 10 \log {\relax (2 )}\right ) - \frac {12 e^{2} \log {\left (\log {\relax (5 )} \right )} + 12 e^{2} \log {\relax (2 )} + 8 e^{4} \log {\left (\log {\relax (5 )} \right )} + 8 e^{4} \log {\relax (2 )} + 60 e^{2} \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + 60 e^{2} \log {\relax (2 )} \log {\relax (5 )} + 40 e^{4} \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + 40 e^{4} \log {\relax (2 )} \log {\relax (5 )}}{x \log {\relax (5 )} - 2 e^{2} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x+60)*exp(2)-10*x**2)*ln(5)+(8*x+12)*exp(2)-2*x**2)*ln(2*ln(5))/(4*exp(2)**2-4*exp(2)*x+x**2)/
ln(5),x)

[Out]

-x*(2*log(log(5))/log(5) + 2*log(2)/log(5) + 10*log(log(5)) + 10*log(2)) - (12*exp(2)*log(log(5)) + 12*exp(2)*
log(2) + 8*exp(4)*log(log(5)) + 8*exp(4)*log(2) + 60*exp(2)*log(5)*log(log(5)) + 60*exp(2)*log(2)*log(5) + 40*
exp(4)*log(5)*log(log(5)) + 40*exp(4)*log(2)*log(5))/(x*log(5) - 2*exp(2)*log(5))

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