3.37.7 \(\int \frac {e^3 (8 x^2-8 e x^2)+(e^3 (-4 x+4 e x)+(-2+2 e) \log (2)) \log (5)}{16 e^3 x^2-8 e^3 x \log (5)+e^3 \log ^2(5)} \, dx\)

Optimal. Leaf size=27 \[ \frac {(1-e) \left (x+\frac {\log (2)}{e^3}\right )}{2-\frac {\log (5)}{2 x}} \]

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Rubi [A]  time = 0.06, antiderivative size = 44, normalized size of antiderivative = 1.63, number of steps used = 4, number of rules used = 3, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {27, 12, 1850} \begin {gather*} \frac {1}{2} (1-e) x+\frac {(1-e) \log (5) \left (e^3 \log (5)+\log (16)\right )}{8 e^3 (4 x-\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(8*x^2 - 8*E*x^2) + (E^3*(-4*x + 4*E*x) + (-2 + 2*E)*Log[2])*Log[5])/(16*E^3*x^2 - 8*E^3*x*Log[5] + E
^3*Log[5]^2),x]

[Out]

((1 - E)*x)/2 + ((1 - E)*Log[5]*(E^3*Log[5] + Log[16]))/(8*E^3*(4*x - 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} &=\int \frac {e^3 \left (8 x^2-8 e x^2\right )+\left (e^3 (-4 x+4 e x)+(-2+2 e) \log (2)\right ) \log (5)}{e^3 (4 x-\log (5))^2} \, dx\\ &=\frac {\int \frac {e^3 \left (8 x^2-8 e x^2\right )+\left (e^3 (-4 x+4 e x)+(-2+2 e) \log (2)\right ) \log (5)}{(4 x-\log (5))^2} \, dx}{e^3}\\ &=\frac {\int \left (-\frac {1}{2} (-1+e) e^3+\frac {(-1+e) \log (5) \left (e^3 \log (5)+\log (16)\right )}{2 (-4 x+\log (5))^2}\right ) \, dx}{e^3}\\ &=\frac {1}{2} (1-e) x+\frac {(1-e) \log (5) \left (e^3 \log (5)+\log (16)\right )}{8 e^3 (4 x-\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 1.59 \begin {gather*} -\frac {2 (-1+e) \left (\log (4) \log (5)+e^3 \left (8 x^2-4 x \log (5)+\log ^2(5)\right )\right )}{e^3 (32 x-8 \log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(8*x^2 - 8*E*x^2) + (E^3*(-4*x + 4*E*x) + (-2 + 2*E)*Log[2])*Log[5])/(16*E^3*x^2 - 8*E^3*x*Log[
5] + E^3*Log[5]^2),x]

[Out]

(-2*(-1 + E)*(Log[4]*Log[5] + E^3*(8*x^2 - 4*x*Log[5] + Log[5]^2)))/(E^3*(32*x - 8*Log[5]))

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fricas [B]  time = 0.72, size = 65, normalized size = 2.41 \begin {gather*} -\frac {16 \, x^{2} e^{4} - 16 \, x^{2} e^{3} + {\left (e^{4} - e^{3}\right )} \log \relax (5)^{2} - 4 \, {\left (x e^{4} - x e^{3} - {\left (e - 1\right )} \log \relax (2)\right )} \log \relax (5)}{8 \, {\left (4 \, x e^{3} - e^{3} \log \relax (5)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(1)-2)*log(2)+(4*x*exp(1)-4*x)*exp(3))*log(5)+(-8*x^2*exp(1)+8*x^2)*exp(3))/(exp(3)*log(5)^2
-8*x*exp(3)*log(5)+16*x^2*exp(3)),x, algorithm="fricas")

[Out]

-1/8*(16*x^2*e^4 - 16*x^2*e^3 + (e^4 - e^3)*log(5)^2 - 4*(x*e^4 - x*e^3 - (e - 1)*log(2))*log(5))/(4*x*e^3 - e
^3*log(5))

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giac [B]  time = 0.16, size = 59, normalized size = 2.19 \begin {gather*} -\frac {1}{2} \, {\left (x e^{4} - x e^{3}\right )} e^{\left (-3\right )} - \frac {{\left (e^{4} \log \relax (5)^{2} - e^{3} \log \relax (5)^{2} + 4 \, e \log \relax (5) \log \relax (2) - 4 \, \log \relax (5) \log \relax (2)\right )} e^{\left (-3\right )}}{8 \, {\left (4 \, x - \log \relax (5)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(1)-2)*log(2)+(4*x*exp(1)-4*x)*exp(3))*log(5)+(-8*x^2*exp(1)+8*x^2)*exp(3))/(exp(3)*log(5)^2
-8*x*exp(3)*log(5)+16*x^2*exp(3)),x, algorithm="giac")

[Out]

-1/2*(x*e^4 - x*e^3)*e^(-3) - 1/8*(e^4*log(5)^2 - e^3*log(5)^2 + 4*e*log(5)*log(2) - 4*log(5)*log(2))*e^(-3)/(
4*x - log(5))

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maple [A]  time = 0.16, size = 32, normalized size = 1.19

method	result	size
gosper	\(\frac {\left (4 x^{2} {\mathrm e}^{3}+\ln \relax (2) \ln \relax (5)\right ) \left ({\mathrm e}-1\right ) {\mathrm e}^{-3}}{2 \ln \relax (5)-8 x}\)	\(32\)
norman	\(\frac {\left (2 \,{\mathrm e}-2\right ) x^{2}+\frac {\ln \relax (5) \ln \relax (2) \left ({\mathrm e}-1\right ) {\mathrm e}^{-3}}{2}}{\ln \relax (5)-4 x}\)	\(35\)
default	\(\left (2 \,{\mathrm e}-2\right ) {\mathrm e}^{-3} \left (-\frac {x \,{\mathrm e}^{3}}{4}-\frac {\ln \relax (5) \left ({\mathrm e}^{3} \ln \relax (5)+4 \ln \relax (2)\right )}{16 \left (4 x -\ln \relax (5)\right )}\right )\)	\(42\)
risch	\(-\frac {x \,{\mathrm e}}{2}+\frac {x}{2}+\frac {{\mathrm e}^{-3} \ln \relax (5)^{2} {\mathrm e}^{4}}{8 \ln \relax (5)-32 x}-\frac {{\mathrm e}^{-3} {\mathrm e}^{3} \ln \relax (5)^{2}}{8 \left (\ln \relax (5)-4 x \right )}+\frac {{\mathrm e}^{-3} \ln \relax (5) {\mathrm e} \ln \relax (2)}{2 \ln \relax (5)-8 x}-\frac {{\mathrm e}^{-3} \ln \relax (5) \ln \relax (2)}{2 \left (\ln \relax (5)-4 x \right )}\)	\(80\)
meijerg	\(\frac {2 \ln \relax (2) \left ({\mathrm e}-1\right ) {\mathrm e}^{-3} x}{\ln \relax (5) \left (1-\frac {4 x}{\ln \relax (5)}\right )}-\frac {\left (-\frac {{\mathrm e}}{2}+\frac {1}{2}\right ) \ln \relax (5) \left (-\frac {4 x \left (-\frac {12 x}{\ln \relax (5)}+6\right )}{3 \ln \relax (5) \left (1-\frac {4 x}{\ln \relax (5)}\right )}-2 \ln \left (1-\frac {4 x}{\ln \relax (5)}\right )\right )}{4}-\frac {\left (1-{\mathrm e}\right ) \ln \relax (5) \left (\frac {4 x}{\ln \relax (5) \left (1-\frac {4 x}{\ln \relax (5)}\right )}+\ln \left (1-\frac {4 x}{\ln \relax (5)}\right )\right )}{4}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*exp(1)-2)*ln(2)+(4*x*exp(1)-4*x)*exp(3))*ln(5)+(-8*x^2*exp(1)+8*x^2)*exp(3))/(exp(3)*ln(5)^2-8*x*exp(
3)*ln(5)+16*x^2*exp(3)),x,method=_RETURNVERBOSE)

[Out]

1/2*(4*x^2*exp(3)+ln(2)*ln(5))*(exp(1)-1)/exp(3)/(ln(5)-4*x)

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maxima [B]  time = 0.65, size = 47, normalized size = 1.74 \begin {gather*} -\frac {1}{2} \, x {\left (e - 1\right )} - \frac {{\left (e^{4} - e^{3}\right )} \log \relax (5)^{2} + 4 \, {\left (e - 1\right )} \log \relax (5) \log \relax (2)}{8 \, {\left (4 \, x e^{3} - e^{3} \log \relax (5)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(1)-2)*log(2)+(4*x*exp(1)-4*x)*exp(3))*log(5)+(-8*x^2*exp(1)+8*x^2)*exp(3))/(exp(3)*log(5)^2
-8*x*exp(3)*log(5)+16*x^2*exp(3)),x, algorithm="maxima")

[Out]

-1/2*x*(e - 1) - 1/8*((e^4 - e^3)*log(5)^2 + 4*(e - 1)*log(5)*log(2))/(4*x*e^3 - e^3*log(5))

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mupad [B]  time = 2.26, size = 46, normalized size = 1.70 \begin {gather*} -\frac {{\mathrm {e}}^{-3}\,\left (\mathrm {e}-1\right )\,\left (16\,{\mathrm {e}}^3\,x^2-4\,{\mathrm {e}}^3\,\ln \relax (5)\,x+4\,\ln \relax (2)\,\ln \relax (5)+{\mathrm {e}}^3\,{\ln \relax (5)}^2\right )}{8\,\left (4\,x-\ln \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(exp(3)*(4*x - 4*x*exp(1)) - log(2)*(2*exp(1) - 2)) + exp(3)*(8*x^2*exp(1) - 8*x^2))/(exp(3)*log(
5)^2 + 16*x^2*exp(3) - 8*x*exp(3)*log(5)),x)

[Out]

-(exp(-3)*(exp(1) - 1)*(4*log(2)*log(5) + exp(3)*log(5)^2 + 16*x^2*exp(3) - 4*x*exp(3)*log(5)))/(8*(4*x - log(
5)))

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sympy [B]  time = 0.35, size = 61, normalized size = 2.26 \begin {gather*} x \left (\frac {1}{2} - \frac {e}{2}\right ) + \frac {- e^{4} \log {\relax (5 )}^{2} - 4 e \log {\relax (2 )} \log {\relax (5 )} + 4 \log {\relax (2 )} \log {\relax (5 )} + e^{3} \log {\relax (5 )}^{2}}{32 x e^{3} - 8 e^{3} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(1)-2)*ln(2)+(4*x*exp(1)-4*x)*exp(3))*ln(5)+(-8*x**2*exp(1)+8*x**2)*exp(3))/(exp(3)*ln(5)**2
-8*x*exp(3)*ln(5)+16*x**2*exp(3)),x)

[Out]

x*(1/2 - E/2) + (-exp(4)*log(5)**2 - 4*E*log(2)*log(5) + 4*log(2)*log(5) + exp(3)*log(5)**2)/(32*x*exp(3) - 8*
exp(3)*log(5))

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