3.93.68 \(\int \frac {-3+6 x+6 x^2+e (-1+2 x+2 x^2)+(3 x+e x) \log (2 x)}{8 x+4 x^2+2 x \log (2 x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{2} \left (6+e^5+(3+e) (x-\log (4+2 x+\log (2 x)))\right ) \]

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Rubi [A]  time = 0.27, antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6688, 12, 6742, 6684} \begin {gather*} \frac {1}{2} (3+e) x-\frac {1}{2} (3+e) \log (2 x+\log (2 x)+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + 6*x + 6*x^2 + E*(-1 + 2*x + 2*x^2) + (3*x + E*x)*Log[2*x])/(8*x + 4*x^2 + 2*x*Log[2*x]),x]

[Out]

((3 + E)*x)/2 - ((3 + E)*Log[4 + 2*x + Log[2*x]])/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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 \frac {(3+e) \left (-1+2 x+2 x^2+x \log (2 x)\right )}{2 x (4+2 x+\log (2 x))} \, dx\\ &=\frac {1}{2} (3+e) \int \frac {-1+2 x+2 x^2+x \log (2 x)}{x (4+2 x+\log (2 x))} \, dx\\ &=\frac {1}{2} (3+e) \int \left (1+\frac {-1-2 x}{x (4+2 x+\log (2 x))}\right ) \, dx\\ &=\frac {1}{2} (3+e) x+\frac {1}{2} (3+e) \int \frac {-1-2 x}{x (4+2 x+\log (2 x))} \, dx\\ &=\frac {1}{2} (3+e) x-\frac {1}{2} (3+e) \log (4+2 x+\log (2 x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 21, normalized size = 0.78 \begin {gather*} \frac {1}{2} (3+e) (x-\log (4+2 x+\log (2 x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + 6*x + 6*x^2 + E*(-1 + 2*x + 2*x^2) + (3*x + E*x)*Log[2*x])/(8*x + 4*x^2 + 2*x*Log[2*x]),x]

[Out]

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

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fricas [A]  time = 0.53, size = 25, normalized size = 0.93 \begin {gather*} \frac {1}{2} \, x e - \frac {1}{2} \, {\left (e + 3\right )} \log \left (2 \, x + \log \left (2 \, x\right ) + 4\right ) + \frac {3}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(1)+3*x)*log(2*x)+(2*x^2+2*x-1)*exp(1)+6*x^2+6*x-3)/(2*x*log(2*x)+4*x^2+8*x),x, algorithm="fr
icas")

[Out]

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

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giac [A]  time = 0.13, size = 37, normalized size = 1.37 \begin {gather*} \frac {1}{2} \, x e - \frac {1}{2} \, e \log \left (2 \, x + \log \left (2 \, x\right ) + 4\right ) + \frac {3}{2} \, x - \frac {3}{2} \, \log \left (-2 \, x - \log \left (2 \, x\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(1)+3*x)*log(2*x)+(2*x^2+2*x-1)*exp(1)+6*x^2+6*x-3)/(2*x*log(2*x)+4*x^2+8*x),x, algorithm="gi
ac")

[Out]

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

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maple [A]  time = 0.06, size = 27, normalized size = 1.00

method	result	size
norman	\(\left (\frac {3}{2}+\frac {{\mathrm e}}{2}\right ) x +\left (-\frac {3}{2}-\frac {{\mathrm e}}{2}\right ) \ln \left (\ln \left (2 x \right )+2 x +4\right )\)	\(27\)
risch	\(\frac {3 x}{2}+\frac {x \,{\mathrm e}}{2}-\frac {3 \ln \left (\ln \left (2 x \right )+2 x +4\right )}{2}-\frac {\ln \left (\ln \left (2 x \right )+2 x +4\right ) {\mathrm e}}{2}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(1)+3*x)*ln(2*x)+(2*x^2+2*x-1)*exp(1)+6*x^2+6*x-3)/(2*x*ln(2*x)+4*x^2+8*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.47, size = 24, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, x {\left (e + 3\right )} - \frac {1}{2} \, {\left (e + 3\right )} \log \left (2 \, x + \log \relax (2) + \log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(1)+3*x)*log(2*x)+(2*x^2+2*x-1)*exp(1)+6*x^2+6*x-3)/(2*x*log(2*x)+4*x^2+8*x),x, algorithm="ma
xima")

[Out]

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

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mupad [B]  time = 7.27, size = 21, normalized size = 0.78 \begin {gather*} \left (x-\ln \left (2\,x+\ln \left (2\,x\right )+4\right )\right )\,\left (\frac {\mathrm {e}}{2}+\frac {3}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + log(2*x)*(3*x + x*exp(1)) + exp(1)*(2*x + 2*x^2 - 1) + 6*x^2 - 3)/(8*x + 2*x*log(2*x) + 4*x^2),x)

[Out]

(x - log(2*x + log(2*x) + 4))*(exp(1)/2 + 3/2)

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sympy [A]  time = 0.18, size = 27, normalized size = 1.00 \begin {gather*} x \left (\frac {e}{2} + \frac {3}{2}\right ) - \frac {\left (e + 3\right ) \log {\left (2 x + \log {\left (2 x \right )} + 4 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(1)+3*x)*ln(2*x)+(2*x**2+2*x-1)*exp(1)+6*x**2+6*x-3)/(2*x*ln(2*x)+4*x**2+8*x),x)

[Out]

x*(E/2 + 3/2) - (E + 3)*log(2*x + log(2*x) + 4)/2

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