3.94.94 \(\int \frac {1+2 x}{(x+x^2) \log ^2(4)} \, dx\)

Optimal. Leaf size=13 \[ \frac {\log \left (12 \left (x+x^2\right )\right )}{\log ^2(4)} \]

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 628} \begin {gather*} \frac {\log \left (x^2+x\right )}{\log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)/((x + x^2)*Log[4]^2),x]

[Out]

Log[x + x^2]/Log[4]^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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {1+2 x}{x+x^2} \, dx}{\log ^2(4)}\\ &=\frac {\log \left (x+x^2\right )}{\log ^2(4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.92 \begin {gather*} \frac {\log (x)+\log (1+x)}{\log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)/((x + x^2)*Log[4]^2),x]

[Out]

(Log[x] + Log[1 + x])/Log[4]^2

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fricas [A]  time = 0.85, size = 12, normalized size = 0.92 \begin {gather*} \frac {\log \left (x^{2} + x\right )}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)/(x^2+x)/log(2)^2,x, algorithm="fricas")

[Out]

1/4*log(x^2 + x)/log(2)^2

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giac [A]  time = 0.21, size = 13, normalized size = 1.00 \begin {gather*} \frac {\log \left ({\left | x^{2} + x \right |}\right )}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)/(x^2+x)/log(2)^2,x, algorithm="giac")

[Out]

1/4*log(abs(x^2 + x))/log(2)^2

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maple [A]  time = 0.56, size = 13, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\ln \left (x^{2}+x \right )}{4 \ln \relax (2)^{2}}\)	\(13\)
default	\(\frac {\ln \left (\left (x +1\right ) x \right )}{4 \ln \relax (2)^{2}}\)	\(13\)
risch	\(\frac {\ln \left (x^{2}+x \right )}{4 \ln \relax (2)^{2}}\)	\(13\)
norman	\(\frac {\ln \relax (x )}{4 \ln \relax (2)^{2}}+\frac {\ln \left (x +1\right )}{4 \ln \relax (2)^{2}}\)	\(20\)
meijerg	\(\frac {\ln \relax (x )-\ln \left (x +1\right )}{4 \ln \relax (2)^{2}}+\frac {\ln \left (x +1\right )}{2 \ln \relax (2)^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(2*x+1)/(x^2+x)/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

1/4/ln(2)^2*ln(x^2+x)

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maxima [A]  time = 0.35, size = 12, normalized size = 0.92 \begin {gather*} \frac {\log \left (x^{2} + x\right )}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)/(x^2+x)/log(2)^2,x, algorithm="maxima")

[Out]

1/4*log(x^2 + x)/log(2)^2

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mupad [B]  time = 0.11, size = 12, normalized size = 0.92 \begin {gather*} \frac {\ln \left (x\,\left (x+1\right )\right )}{4\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x/2 + 1/4)/(log(2)^2*(x + x^2)),x)

[Out]

log(x*(x + 1))/(4*log(2)^2)

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sympy [A]  time = 0.17, size = 12, normalized size = 0.92 \begin {gather*} \frac {\log {\left (x^{2} + x \right )}}{4 \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)/(x**2+x)/ln(2)**2,x)

[Out]

log(x**2 + x)/(4*log(2)**2)

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