3.76.51 \(\int \frac {e^5+4 \log (\frac {-1+7 \log (2)}{\log (2)})}{8 e^{10}-8 e^5 x+2 x^2+(32 e^5-16 x) \log (\frac {-1+7 \log (2)}{\log (2)})+32 \log ^2(\frac {-1+7 \log (2)}{\log (2)})} \, dx\)

Optimal. Leaf size=34 \[ \frac {1}{2 \left (2-\frac {-e^5+x}{\frac {x}{4}+\log \left (7-\frac {1}{\log (2)}\right )}\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 1981, 27, 32} \begin {gather*} -\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (x-2 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5 + 4*Log[(-1 + 7*Log[2])/Log[2]])/(8*E^10 - 8*E^5*x + 2*x^2 + (32*E^5 - 16*x)*Log[(-1 + 7*Log[2])/Log[
2]] + 32*Log[(-1 + 7*Log[2])/Log[2]]^2),x]

[Out]

-1/2*(E^5 + 4*Log[7 - Log[2]^(-1)])/(x - 2*(E^5 + 2*Log[7 - Log[2]^(-1)]))

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx\\ &=\left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{2 x^2-8 x \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )+8 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )^2} \, dx\\ &=\left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{2 \left (2 e^5-x+4 \log \left (7-\frac {1}{\log (2)}\right )\right )^2} \, dx\\ &=\frac {1}{2} \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{\left (2 e^5-x+4 \log \left (7-\frac {1}{\log (2)}\right )\right )^2} \, dx\\ &=-\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (x-2 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 1.15 \begin {gather*} -\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (-2 e^5+x-4 \log \left (7-\frac {1}{\log (2)}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5 + 4*Log[(-1 + 7*Log[2])/Log[2]])/(8*E^10 - 8*E^5*x + 2*x^2 + (32*E^5 - 16*x)*Log[(-1 + 7*Log[2]
)/Log[2]] + 32*Log[(-1 + 7*Log[2])/Log[2]]^2),x]

[Out]

-1/2*(E^5 + 4*Log[7 - Log[2]^(-1)])/(-2*E^5 + x - 4*Log[7 - Log[2]^(-1)])

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fricas [A]  time = 0.85, size = 41, normalized size = 1.21 \begin {gather*} -\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \relax (2) - 1}{\log \relax (2)}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \relax (2) - 1}{\log \relax (2)}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log((7*log(2)-1)/log(2))+exp(5))/(32*log((7*log(2)-1)/log(2))^2+(32*exp(5)-16*x)*log((7*log(2)-1)
/log(2))+8*exp(5)^2-8*x*exp(5)+2*x^2),x, algorithm="fricas")

[Out]

-1/2*(e^5 + 4*log((7*log(2) - 1)/log(2)))/(x - 2*e^5 - 4*log((7*log(2) - 1)/log(2)))

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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((4*log((7*log(2)-1)/log(2))+exp(5))/(32*log((7*log(2)-1)/log(2))^2+(32*exp(5)-16*x)*log((7*log(2)-1)
/log(2))+8*exp(5)^2-8*x*exp(5)+2*x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (exp(5)+4*ln((7*ln(2)-1)*ln(2)^-1))/2*1/
4/sqrt(-exp(10)+exp(5)^2)*ln(abs(2*sageVARx-4*exp(5)-8*ln((7*ln(2)-1)*ln(2)^-1)-4*sqrt(-exp(10)+exp(5)^2))/abs
(2*sageVARx-4*exp(5)-

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maple [A]  time = 0.36, size = 44, normalized size = 1.29

method	result	size
gosper	\(\frac {4 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )+{\mathrm e}^{5}}{8 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )+4 \,{\mathrm e}^{5}-2 x}\)	\(44\)
norman	\(\frac {\frac {{\mathrm e}^{5}}{2}+2 \ln \left (7 \ln \relax (2)-1\right )-2 \ln \left (\ln \relax (2)\right )}{4 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )+2 \,{\mathrm e}^{5}-x}\)	\(45\)
risch	\(\frac {\ln \left (7 \ln \relax (2)-1\right )}{{\mathrm e}^{5}+2 \ln \left (7 \ln \relax (2)-1\right )-2 \ln \left (\ln \relax (2)\right )-\frac {x}{2}}-\frac {\ln \left (\ln \relax (2)\right )}{{\mathrm e}^{5}+2 \ln \left (7 \ln \relax (2)-1\right )-2 \ln \left (\ln \relax (2)\right )-\frac {x}{2}}+\frac {{\mathrm e}^{5}}{4 \,{\mathrm e}^{5}+8 \ln \left (7 \ln \relax (2)-1\right )-8 \ln \left (\ln \relax (2)\right )-2 x}\)	\(85\)
meijerg	\(-\frac {\ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right ) x}{\left (-2 \,{\mathrm e}^{5}-4 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )\right ) \left (1-\frac {x}{2 \left (2 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )+{\mathrm e}^{5}\right )}\right ) \left (2 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )+{\mathrm e}^{5}\right )}-\frac {{\mathrm e}^{5} x}{4 \left (-2 \,{\mathrm e}^{5}-4 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )\right ) \left (1-\frac {x}{2 \left (2 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )+{\mathrm e}^{5}\right )}\right ) \left (2 \ln \left (\frac {7 \ln \relax (2)-1}{\ln \relax (2)}\right )+{\mathrm e}^{5}\right )}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln((7*ln(2)-1)/ln(2))+exp(5))/(32*ln((7*ln(2)-1)/ln(2))^2+(32*exp(5)-16*x)*ln((7*ln(2)-1)/ln(2))+8*exp(
5)^2-8*x*exp(5)+2*x^2),x,method=_RETURNVERBOSE)

[Out]

1/2*(4*ln((7*ln(2)-1)/ln(2))+exp(5))/(4*ln((7*ln(2)-1)/ln(2))+2*exp(5)-x)

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maxima [A]  time = 0.36, size = 41, normalized size = 1.21 \begin {gather*} -\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \relax (2) - 1}{\log \relax (2)}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \relax (2) - 1}{\log \relax (2)}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log((7*log(2)-1)/log(2))+exp(5))/(32*log((7*log(2)-1)/log(2))^2+(32*exp(5)-16*x)*log((7*log(2)-1)
/log(2))+8*exp(5)^2-8*x*exp(5)+2*x^2),x, algorithm="maxima")

[Out]

-1/2*(e^5 + 4*log((7*log(2) - 1)/log(2)))/(x - 2*e^5 - 4*log((7*log(2) - 1)/log(2)))

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mupad [B]  time = 0.36, size = 39, normalized size = 1.15 \begin {gather*} -\frac {2\,\ln \left (\ln \left (128\right )-1\right )+\frac {{\mathrm {e}}^5}{2}-2\,\ln \left (\ln \relax (2)\right )}{x-4\,\ln \left (\ln \left (128\right )-1\right )-2\,{\mathrm {e}}^5+4\,\ln \left (\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5) + 4*log((7*log(2) - 1)/log(2)))/(8*exp(10) - log((7*log(2) - 1)/log(2))*(16*x - 32*exp(5)) + 32*lo
g((7*log(2) - 1)/log(2))^2 - 8*x*exp(5) + 2*x^2),x)

[Out]

-(2*log(log(128) - 1) + exp(5)/2 - 2*log(log(2)))/(x - 4*log(log(128) - 1) - 2*exp(5) + 4*log(log(2)))

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sympy [B]  time = 0.37, size = 46, normalized size = 1.35 \begin {gather*} - \frac {- 4 \log {\left (\log {\relax (2 )} \right )} + 4 \log {\left (-1 + 7 \log {\relax (2 )} \right )} + e^{5}}{2 x - 4 e^{5} - 8 \log {\left (-1 + 7 \log {\relax (2 )} \right )} + 8 \log {\left (\log {\relax (2 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln((7*ln(2)-1)/ln(2))+exp(5))/(32*ln((7*ln(2)-1)/ln(2))**2+(32*exp(5)-16*x)*ln((7*ln(2)-1)/ln(2))
+8*exp(5)**2-8*x*exp(5)+2*x**2),x)

[Out]

-(-4*log(log(2)) + 4*log(-1 + 7*log(2)) + exp(5))/(2*x - 4*exp(5) - 8*log(-1 + 7*log(2)) + 8*log(log(2)))

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