3.23.10 \(\int \frac {2+2 x+e^5 (2+2 x)+(2+4 x) \log (x)}{x+2 x^2+e^5 (x+2 x^2)} \, dx\)

Optimal. Leaf size=28 \[ \log (x)+\frac {\log ^2(x)}{1+e^5}+\log \left (\frac {x}{5 (1+2 x)}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {6688, 12, 14, 72, 2301} \begin {gather*} \frac {\log ^2(x)}{1+e^5}+2 \log (x)-\log (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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 (\frac {1+x}{1+2 x}+\frac {\log (x)}{1+e^5}\right )}{x} \, dx\\ &=2 \int \frac {\frac {1+x}{1+2 x}+\frac {\log (x)}{1+e^5}}{x} \, dx\\ &=2 \int \left (\frac {1+x}{x (1+2 x)}+\frac {\log (x)}{\left (1+e^5\right ) x}\right ) \, dx\\ &=2 \int \frac {1+x}{x (1+2 x)} \, dx+\frac {2 \int \frac {\log (x)}{x} \, dx}{1+e^5}\\ &=\frac {\log ^2(x)}{1+e^5}+2 \int \left (\frac {1}{-1-2 x}+\frac {1}{x}\right ) \, dx\\ &=2 \log (x)+\frac {\log ^2(x)}{1+e^5}-\log (1+2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 33, normalized size = 1.18 \begin {gather*} 2 \left (\frac {\left (1+e^5+\log (x)\right )^2}{2 \left (1+e^5\right )}-\frac {1}{2} \log (1+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.94, size = 34, normalized size = 1.21 \begin {gather*} -\frac {{\left (e^{5} + 1\right )} \log \left (2 \, x + 1\right ) - 2 \, {\left (e^{5} + 1\right )} \log \relax (x) - \log \relax (x)^{2}}{e^{5} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.00, size = 40, normalized size = 1.43 \begin {gather*} -\frac {e^{5} \log \left (2 \, x + 1\right ) - 2 \, e^{5} \log \relax (x) - \log \relax (x)^{2} + \log \left (2 \, x + 1\right ) - 2 \, \log \relax (x)}{e^{5} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.38, size = 25, normalized size = 0.89

method	result	size
default	\(-\ln \left (2 x +1\right )+2 \ln \relax (x )+\frac {\ln \relax (x )^{2}}{{\mathrm e}^{5}+1}\)	\(25\)
norman	\(-\ln \left (2 x +1\right )+2 \ln \relax (x )+\frac {\ln \relax (x )^{2}}{{\mathrm e}^{5}+1}\)	\(25\)
risch	\(-\ln \left (2 x +1\right )+2 \ln \relax (x )+\frac {\ln \relax (x )^{2}}{{\mathrm e}^{5}+1}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.71, size = 79, normalized size = 2.82 \begin {gather*} -2 \, {\left (\frac {\log \left (2 \, x + 1\right )}{e^{5} + 1} - \frac {\log \relax (x)}{e^{5} + 1}\right )} e^{5} + \frac {e^{5} \log \left (2 \, x + 1\right )}{e^{5} + 1} + \frac {\log \relax (x)^{2}}{e^{5} + 1} - \frac {\log \left (2 \, x + 1\right )}{e^{5} + 1} + \frac {2 \, \log \relax (x)}{e^{5} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.36, size = 22, normalized size = 0.79 \begin {gather*} \frac {{\ln \relax (x)}^2}{2\,\left (\frac {{\mathrm {e}}^5}{2}+\frac {1}{2}\right )}+2\,\ln \relax (x)-\ln \left (x+\frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log(x)*(4*x + 2) + exp(5)*(2*x + 2) + 2)/(x + exp(5)*(x + 2*x^2) + 2*x^2),x)

[Out]

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

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sympy [A]  time = 0.19, size = 20, normalized size = 0.71 \begin {gather*} \frac {\log {\relax (x )}^{2}}{1 + e^{5}} + 2 \log {\relax (x )} - \log {\left (x + \frac {1}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+2)*ln(x)+(2*x+2)*exp(5)+2*x+2)/((2*x**2+x)*exp(5)+2*x**2+x),x)

[Out]

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

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