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

Optimal. Leaf size=23 \[ \left (-x+\log \left (-1+x+\frac {\left (2+e^4\right )^2 \log (5)}{x^2}\right )\right )^2 \]

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Rubi [A]  time = 0.37, antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 3, number of rules used = 3, integrand size = 144, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6688, 12, 6686} \begin {gather*} \left (x-\log \left (-\frac {-x^3+x^2-\left (2+e^4\right )^2 \log (5)}{x^2}\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 30, normalized size = 1.30 \begin {gather*} \left (x-\log \left (\frac {-x^2+x^3+\left (2+e^4\right )^2 \log (5)}{x^2}\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 59, normalized size = 2.57 \begin {gather*} x^{2} - 2 \, x \log \left (\frac {x^{3} - x^{2} + {\left (e^{8} + 4 \, e^{4} + 4\right )} \log \relax (5)}{x^{2}}\right ) + \log \left (\frac {x^{3} - x^{2} + {\left (e^{8} + 4 \, e^{4} + 4\right )} \log \relax (5)}{x^{2}}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.33, size = 72, normalized size = 3.13 \begin {gather*} x^{2} - 2 \, x \log \left (x^{3} - x^{2} + e^{8} \log \relax (5) + 4 \, e^{4} \log \relax (5) + 4 \, \log \relax (5)\right ) + 4 \, x \log \relax (x) - 4 \, \log \left (x^{3} - x^{2} + e^{8} \log \relax (5) + 4 \, e^{4} \log \relax (5) + 4 \, \log \relax (5)\right ) \log \relax (x) + 4 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.44, size = 64, normalized size = 2.78

method	result	size
norman	\(x^{2}+\ln \left (\frac {\left ({\mathrm e}^{8}+4 \,{\mathrm e}^{4}+4\right ) \ln \relax (5)+x^{3}-x^{2}}{x^{2}}\right )^{2}-2 x \ln \left (\frac {\left ({\mathrm e}^{8}+4 \,{\mathrm e}^{4}+4\right ) \ln \relax (5)+x^{3}-x^{2}}{x^{2}}\right )\)	\(64\)
default	error in gcdex: invalid arguments\	N/A

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 67, normalized size = 2.91 \begin {gather*} x^{2} - 2 \, {\left (x + 2 \, \log \relax (x)\right )} \log \left (x^{3} - x^{2} + {\left (e^{8} + 4 \, e^{4} + 4\right )} \log \relax (5)\right ) + \log \left (x^{3} - x^{2} + {\left (e^{8} + 4 \, e^{4} + 4\right )} \log \relax (5)\right )^{2} + 4 \, x \log \relax (x) + 4 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 10.41, size = 31, normalized size = 1.35 \begin {gather*} {\left (x-\ln \left (\frac {x^3-x^2+\ln \relax (5)\,\left (4\,{\mathrm {e}}^4+{\mathrm {e}}^8+4\right )}{x^2}\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.34, size = 58, normalized size = 2.52 \begin {gather*} x^{2} - 2 x \log {\left (\frac {x^{3} - x^{2} + \left (4 + 4 e^{4} + e^{8}\right ) \log {\relax (5 )}}{x^{2}} \right )} + \log {\left (\frac {x^{3} - x^{2} + \left (4 + 4 e^{4} + e^{8}\right ) \log {\relax (5 )}}{x^{2}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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