3.40.18 \(\int \frac {e^{\log ^2(4+2 x)} (8+4 x+(4+4 x) \log (4+2 x) \log (4+8 x+4 x^2))}{18+27 x+9 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {2}{9} e^{\log ^2(4+2 x)} \log \left (\left (5-\left (-2+\frac {3}{x}\right ) x\right )^2\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 41, normalized size of antiderivative = 1.41, number of steps used = 1, number of rules used = 1, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {2288} \begin {gather*} \frac {2 (x+1) (x+2) e^{\log ^2(2 x+4)} \log \left (4 x^2+8 x+4\right )}{9 \left (x^2+3 x+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^Log[4 + 2*x]^2*(8 + 4*x + (4 + 4*x)*Log[4 + 2*x]*Log[4 + 8*x + 4*x^2]))/(18 + 27*x + 9*x^2),x]

[Out]

(2*E^Log[4 + 2*x]^2*(1 + x)*(2 + x)*Log[4 + 8*x + 4*x^2])/(9*(2 + 3*x + x^2))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {2 e^{\log ^2(4+2 x)} (1+x) (2+x) \log \left (4+8 x+4 x^2\right )}{9 \left (2+3 x+x^2\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 22, normalized size = 0.76 \begin {gather*} \frac {2}{9} e^{\log ^2(2 (2+x))} \log \left (4 (1+x)^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[4 + 2*x]^2*(8 + 4*x + (4 + 4*x)*Log[4 + 2*x]*Log[4 + 8*x + 4*x^2]))/(18 + 27*x + 9*x^2),x]

[Out]

(2*E^Log[2*(2 + x)]^2*Log[4*(1 + x)^2])/9

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 22, normalized size = 0.76 \begin {gather*} \frac {2}{9} \, e^{\left (\log \left (2 \, x + 4\right )^{2}\right )} \log \left (4 \, x^{2} + 8 \, x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(2*x+4)*log(4*x^2+8*x+4)+4*x+8)*exp(log(2*x+4)^2)/(9*x^2+27*x+18),x, algorithm="fricas")

[Out]

2/9*e^(log(2*x + 4)^2)*log(4*x^2 + 8*x + 4)

________________________________________________________________________________________

giac [B]  time = 0.20, size = 56, normalized size = 1.93 \begin {gather*} \frac {4}{9} \, e^{\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (x + 2\right ) + \log \left (x + 2\right )^{2}\right )} \log \relax (2) + \frac {2}{9} \, e^{\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (x + 2\right ) + \log \left (x + 2\right )^{2}\right )} \log \left (x^{2} + 2 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(2*x+4)*log(4*x^2+8*x+4)+4*x+8)*exp(log(2*x+4)^2)/(9*x^2+27*x+18),x, algorithm="giac")

[Out]

4/9*e^(log(2)^2 + 2*log(2)*log(x + 2) + log(x + 2)^2)*log(2) + 2/9*e^(log(2)^2 + 2*log(2)*log(x + 2) + log(x +
2)^2)*log(x^2 + 2*x + 1)

________________________________________________________________________________________

maple [C]  time = 0.12, size = 81, normalized size = 2.79

method	result	size
risch	\(\left (-\frac {i \pi \mathrm {csgn}\left (i \left (x +1\right )\right )^{2} \mathrm {csgn}\left (i \left (x +1\right )^{2}\right )}{9}+\frac {2 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (i \left (x +1\right )^{2}\right )^{2}}{9}-\frac {i \pi \mathrm {csgn}\left (i \left (x +1\right )^{2}\right )^{3}}{9}+\frac {4 \ln \relax (2)}{9}+\frac {4 \ln \left (x +1\right )}{9}\right ) {\mathrm e}^{\ln \left (2 x +4\right )^{2}}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+4)*ln(2*x+4)*ln(4*x^2+8*x+4)+4*x+8)*exp(ln(2*x+4)^2)/(9*x^2+27*x+18),x,method=_RETURNVERBOSE)

[Out]

(-1/9*I*Pi*csgn(I*(x+1))^2*csgn(I*(x+1)^2)+2/9*I*Pi*csgn(I*(x+1))*csgn(I*(x+1)^2)^2-1/9*I*Pi*csgn(I*(x+1)^2)^3
+4/9*ln(2)+4/9*ln(x+1))*exp(ln(2*x+4)^2)

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 37, normalized size = 1.28 \begin {gather*} \frac {4}{9} \, {\left (e^{\left (\log \relax (2)^{2}\right )} \log \relax (2) + e^{\left (\log \relax (2)^{2}\right )} \log \left (x + 1\right )\right )} e^{\left (2 \, \log \relax (2) \log \left (x + 2\right ) + \log \left (x + 2\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(2*x+4)*log(4*x^2+8*x+4)+4*x+8)*exp(log(2*x+4)^2)/(9*x^2+27*x+18),x, algorithm="maxima")

[Out]

4/9*(e^(log(2)^2)*log(2) + e^(log(2)^2)*log(x + 1))*e^(2*log(2)*log(x + 2) + log(x + 2)^2)

________________________________________________________________________________________

mupad [B]  time = 0.24, size = 22, normalized size = 0.76 \begin {gather*} \frac {2\,{\mathrm {e}}^{{\ln \left (2\,x+4\right )}^2}\,\ln \left (4\,x^2+8\,x+4\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(2*x + 4)^2)*(4*x + log(2*x + 4)*log(8*x + 4*x^2 + 4)*(4*x + 4) + 8))/(27*x + 9*x^2 + 18),x)

[Out]

(2*exp(log(2*x + 4)^2)*log(8*x + 4*x^2 + 4))/9

________________________________________________________________________________________

sympy [A]  time = 6.86, size = 24, normalized size = 0.83 \begin {gather*} \frac {2 e^{\log {\left (2 x + 4 \right )}^{2}} \log {\left (4 x^{2} + 8 x + 4 \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*ln(2*x+4)*ln(4*x**2+8*x+4)+4*x+8)*exp(ln(2*x+4)**2)/(9*x**2+27*x+18),x)

[Out]

2*exp(log(2*x + 4)**2)*log(4*x**2 + 8*x + 4)/9

________________________________________________________________________________________