3.6.98 \(\int \frac {(12+6 x) \log ^2(\frac {x}{2+x})+e^{\frac {5 x}{\log (\frac {x}{2+x})}} (-20+(20+10 x) \log (\frac {x}{2+x}))}{(2+x) \log ^2(\frac {x}{2+x})} \, dx\)

Optimal. Leaf size=21 \[ 2 \left (e^{\frac {5 x}{\log \left (\frac {x}{2+x}\right )}}+3 x\right ) \]

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Rubi [A]  time = 0.78, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6742, 6688, 6706} \begin {gather*} 6 x+2 e^{\frac {5 x}{\log \left (\frac {x}{x+2}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((12 + 6*x)*Log[x/(2 + x)]^2 + E^((5*x)/Log[x/(2 + x)])*(-20 + (20 + 10*x)*Log[x/(2 + x)]))/((2 + x)*Log[x
/(2 + x)]^2),x]

[Out]

2*E^((5*x)/Log[x/(2 + x)]) + 6*x

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (6+\frac {10 e^{\frac {5 x}{\log \left (\frac {x}{2+x}\right )}} \left (-2+2 \log \left (\frac {x}{2+x}\right )+x \log \left (\frac {x}{2+x}\right )\right )}{(2+x) \log ^2\left (\frac {x}{2+x}\right )}\right ) \, dx\\ &=6 x+10 \int \frac {e^{\frac {5 x}{\log \left (\frac {x}{2+x}\right )}} \left (-2+2 \log \left (\frac {x}{2+x}\right )+x \log \left (\frac {x}{2+x}\right )\right )}{(2+x) \log ^2\left (\frac {x}{2+x}\right )} \, dx\\ &=6 x+10 \int \frac {e^{\frac {5 x}{\log \left (\frac {x}{2+x}\right )}} \left (-2+(2+x) \log \left (\frac {x}{2+x}\right )\right )}{(2+x) \log ^2\left (\frac {x}{2+x}\right )} \, dx\\ &=2 e^{\frac {5 x}{\log \left (\frac {x}{2+x}\right )}}+6 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 21, normalized size = 1.00 \begin {gather*} 2 e^{\frac {5 x}{\log \left (\frac {x}{2+x}\right )}}+6 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((12 + 6*x)*Log[x/(2 + x)]^2 + E^((5*x)/Log[x/(2 + x)])*(-20 + (20 + 10*x)*Log[x/(2 + x)]))/((2 + x)
*Log[x/(2 + x)]^2),x]

[Out]

2*E^((5*x)/Log[x/(2 + x)]) + 6*x

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fricas [A]  time = 0.62, size = 20, normalized size = 0.95 \begin {gather*} 6 \, x + 2 \, e^{\left (\frac {5 \, x}{\log \left (\frac {x}{x + 2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x+20)*log(x/(2+x))-20)*exp(5*x/log(x/(2+x)))+(6*x+12)*log(x/(2+x))^2)/(2+x)/log(x/(2+x))^2,x,
algorithm="fricas")

[Out]

6*x + 2*e^(5*x/log(x/(x + 2)))

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giac [A]  time = 2.01, size = 20, normalized size = 0.95 \begin {gather*} 6 \, x + 2 \, e^{\left (\frac {5 \, x}{\log \left (\frac {x}{x + 2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x+20)*log(x/(2+x))-20)*exp(5*x/log(x/(2+x)))+(6*x+12)*log(x/(2+x))^2)/(2+x)/log(x/(2+x))^2,x,
algorithm="giac")

[Out]

6*x + 2*e^(5*x/log(x/(x + 2)))

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maple [A]  time = 0.23, size = 21, normalized size = 1.00

method	result	size
default	\(2 \,{\mathrm e}^{\frac {5 x}{\ln \left (\frac {x}{2+x}\right )}}+6 x\)	\(21\)
risch	\(2 \,{\mathrm e}^{\frac {5 x}{\ln \left (\frac {x}{2+x}\right )}}+6 x\)	\(21\)
norman	\(\frac {6 \ln \left (\frac {x}{2+x}\right ) x +2 \ln \left (\frac {x}{2+x}\right ) {\mathrm e}^{\frac {5 x}{\ln \left (\frac {x}{2+x}\right )}}}{\ln \left (\frac {x}{2+x}\right )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x+20)*ln(x/(2+x))-20)*exp(5*x/ln(x/(2+x)))+(6*x+12)*ln(x/(2+x))^2)/(2+x)/ln(x/(2+x))^2,x,method=_RET
URNVERBOSE)

[Out]

2*exp(5*x/ln(x/(2+x)))+6*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 6 \, x - 2 \, \int \frac {5 \, {\left ({\left (x + 2\right )} \log \left (x + 2\right ) - {\left (x + 2\right )} \log \relax (x) + 2\right )} e^{\left (-\frac {5 \, x}{\log \left (x + 2\right ) - \log \relax (x)}\right )}}{{\left (x + 2\right )} \log \left (x + 2\right )^{2} - 2 \, {\left (x + 2\right )} \log \left (x + 2\right ) \log \relax (x) + {\left (x + 2\right )} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x+20)*log(x/(2+x))-20)*exp(5*x/log(x/(2+x)))+(6*x+12)*log(x/(2+x))^2)/(2+x)/log(x/(2+x))^2,x,
algorithm="maxima")

[Out]

6*x - 2*integrate(5*((x + 2)*log(x + 2) - (x + 2)*log(x) + 2)*e^(-5*x/(log(x + 2) - log(x)))/((x + 2)*log(x +
2)^2 - 2*(x + 2)*log(x + 2)*log(x) + (x + 2)*log(x)^2), x)

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mupad [B]  time = 0.79, size = 20, normalized size = 0.95 \begin {gather*} 6\,x+2\,{\mathrm {e}}^{\frac {5\,x}{\ln \left (\frac {x}{x+2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x/(x + 2))^2*(6*x + 12) + exp((5*x)/log(x/(x + 2)))*(log(x/(x + 2))*(10*x + 20) - 20))/(log(x/(x + 2)
)^2*(x + 2)),x)

[Out]

6*x + 2*exp((5*x)/log(x/(x + 2)))

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sympy [A]  time = 0.41, size = 15, normalized size = 0.71 \begin {gather*} 6 x + 2 e^{\frac {5 x}{\log {\left (\frac {x}{x + 2} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x+20)*ln(x/(2+x))-20)*exp(5*x/ln(x/(2+x)))+(6*x+12)*ln(x/(2+x))**2)/(2+x)/ln(x/(2+x))**2,x)

[Out]

6*x + 2*exp(5*x/log(x/(x + 2)))

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