3.48.21 \(\int \frac {-8 x-2 x^2+(16+12 x+2 x^2-48 x^3-36 x^4-6 x^5) \log (2+x)+(8+6 x+x^2-24 x^3-18 x^4-3 x^5) \log ^2(2+x)+((-16 x^3-8 x^4) \log (2+x)+(-8 x^3-4 x^4) \log ^2(2+x)+((16+8 x) \log (2+x)+(8+4 x) \log ^2(2+x)) \log (\frac {2 x+x \log (2+x)}{\log (2+x)})) \log (-x^3+\log (\frac {2 x+x \log (2+x)}{\log (2+x)}))}{(-64 x^3-64 x^4-20 x^5-2 x^6) \log (2+x)+(-32 x^3-32 x^4-10 x^5-x^6) \log ^2(2+x)+((64+64 x+20 x^2+2 x^3) \log (2+x)+(32+32 x+10 x^2+x^3) \log ^2(2+x)) \log (\frac {2 x+x \log (2+x)}{\log (2+x)})} \, dx\)

Optimal. Leaf size=26 \[ \frac {x \log \left (-x^3+\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}{4+x} \]

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Rubi [F]  time = 14.78, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-8 x-2 x^2+\left (16+12 x+2 x^2-48 x^3-36 x^4-6 x^5\right ) \log (2+x)+\left (8+6 x+x^2-24 x^3-18 x^4-3 x^5\right ) \log ^2(2+x)+\left (\left (-16 x^3-8 x^4\right ) \log (2+x)+\left (-8 x^3-4 x^4\right ) \log ^2(2+x)+\left ((16+8 x) \log (2+x)+(8+4 x) \log ^2(2+x)\right ) \log \left (\frac {2 x+x \log (2+x)}{\log (2+x)}\right )\right ) \log \left (-x^3+\log \left (\frac {2 x+x \log (2+x)}{\log (2+x)}\right )\right )}{\left (-64 x^3-64 x^4-20 x^5-2 x^6\right ) \log (2+x)+\left (-32 x^3-32 x^4-10 x^5-x^6\right ) \log ^2(2+x)+\left (\left (64+64 x+20 x^2+2 x^3\right ) \log (2+x)+\left (32+32 x+10 x^2+x^3\right ) \log ^2(2+x)\right ) \log \left (\frac {2 x+x \log (2+x)}{\log (2+x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8*x - 2*x^2 + (16 + 12*x + 2*x^2 - 48*x^3 - 36*x^4 - 6*x^5)*Log[2 + x] + (8 + 6*x + x^2 - 24*x^3 - 18*x^
4 - 3*x^5)*Log[2 + x]^2 + ((-16*x^3 - 8*x^4)*Log[2 + x] + (-8*x^3 - 4*x^4)*Log[2 + x]^2 + ((16 + 8*x)*Log[2 +
x] + (8 + 4*x)*Log[2 + x]^2)*Log[(2*x + x*Log[2 + x])/Log[2 + x]])*Log[-x^3 + Log[(2*x + x*Log[2 + x])/Log[2 +
x]]])/((-64*x^3 - 64*x^4 - 20*x^5 - 2*x^6)*Log[2 + x] + (-32*x^3 - 32*x^4 - 10*x^5 - x^6)*Log[2 + x]^2 + ((64
+ 64*x + 20*x^2 + 2*x^3)*Log[2 + x] + (32 + 32*x + 10*x^2 + x^3)*Log[2 + x]^2)*Log[(2*x + x*Log[2 + x])/Log[2
+ x]]),x]

[Out]

96*Defer[Int][1/((2 + Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])), x] - 24*Defer[Int][x/((2 + Log[2 + x])*(
x^3 - Log[x + (2*x)/Log[2 + x]])), x] + 6*Defer[Int][x^2/((2 + Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])),
x] - 386*Defer[Int][1/((4 + x)*(2 + Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])), x] - 2*Defer[Int][1/((2 +
x)*Log[2 + x]*(2 + Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])), x] + 4*Defer[Int][1/((4 + x)*Log[2 + x]*(2
+ Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])), x] + 48*Defer[Int][Log[2 + x]/((2 + Log[2 + x])*(x^3 - Log[
x + (2*x)/Log[2 + x]])), x] - 12*Defer[Int][(x*Log[2 + x])/((2 + Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])
), x] + 3*Defer[Int][(x^2*Log[2 + x])/((2 + Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])), x] - 193*Defer[Int
][Log[2 + x]/((4 + x)*(2 + Log[2 + x])*(x^3 - Log[x + (2*x)/Log[2 + x]])), x] + 4*Defer[Int][Log[-x^3 + Log[x
+ (2*x)/Log[2 + x]]]/(4 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x (4+x)+2 (2+x) \log (2+x) \left (-4-x+12 x^3+3 x^4+4 \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right ) \log \left (-x^3+\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )\right )+(2+x) \log ^2(2+x) \left (-4-x+12 x^3+3 x^4+4 \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right ) \log \left (-x^3+\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )\right )}{(2+x) (4+x)^2 \log (2+x) (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx\\ &=\int \left (-\frac {8}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}-\frac {2 x}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}+\frac {24 x^3}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}+\frac {6 x^4}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}+\frac {2 x}{(2+x) (4+x) \log (2+x) (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}-\frac {4 \log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}-\frac {x \log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}+\frac {12 x^3 \log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}+\frac {3 x^4 \log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}+\frac {4 \log \left (-x^3+\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}{(4+x)^2}\right ) \, dx\\ &=-\left (2 \int \frac {x}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx\right )+2 \int \frac {x}{(2+x) (4+x) \log (2+x) (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx+3 \int \frac {x^4 \log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx-4 \int \frac {\log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx+4 \int \frac {\log \left (-x^3+\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}{(4+x)^2} \, dx+6 \int \frac {x^4}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx-8 \int \frac {1}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx+12 \int \frac {x^3 \log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx+24 \int \frac {x^3}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx-\int \frac {x \log (2+x)}{(4+x)^2 (2+\log (2+x)) \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 46, normalized size = 1.77 \begin {gather*} \log \left (x^3-\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )-\frac {4 \log \left (-x^3+\log \left (x+\frac {2 x}{\log (2+x)}\right )\right )}{4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x - 2*x^2 + (16 + 12*x + 2*x^2 - 48*x^3 - 36*x^4 - 6*x^5)*Log[2 + x] + (8 + 6*x + x^2 - 24*x^3 -
18*x^4 - 3*x^5)*Log[2 + x]^2 + ((-16*x^3 - 8*x^4)*Log[2 + x] + (-8*x^3 - 4*x^4)*Log[2 + x]^2 + ((16 + 8*x)*Lo
g[2 + x] + (8 + 4*x)*Log[2 + x]^2)*Log[(2*x + x*Log[2 + x])/Log[2 + x]])*Log[-x^3 + Log[(2*x + x*Log[2 + x])/L
og[2 + x]]])/((-64*x^3 - 64*x^4 - 20*x^5 - 2*x^6)*Log[2 + x] + (-32*x^3 - 32*x^4 - 10*x^5 - x^6)*Log[2 + x]^2
+ ((64 + 64*x + 20*x^2 + 2*x^3)*Log[2 + x] + (32 + 32*x + 10*x^2 + x^3)*Log[2 + x]^2)*Log[(2*x + x*Log[2 + x])
/Log[2 + x]]),x]

[Out]

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

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fricas [A]  time = 1.06, size = 32, normalized size = 1.23 \begin {gather*} \frac {x \log \left (-x^{3} + \log \left (\frac {x \log \left (x + 2\right ) + 2 \, x}{\log \left (x + 2\right )}\right )\right )}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x+8)*log(2+x)^2+(8*x+16)*log(2+x))*log((x*log(2+x)+2*x)/log(2+x))+(-4*x^4-8*x^3)*log(2+x)^2+(-
8*x^4-16*x^3)*log(2+x))*log(log((x*log(2+x)+2*x)/log(2+x))-x^3)+(-3*x^5-18*x^4-24*x^3+x^2+6*x+8)*log(2+x)^2+(-
6*x^5-36*x^4-48*x^3+2*x^2+12*x+16)*log(2+x)-2*x^2-8*x)/(((x^3+10*x^2+32*x+32)*log(2+x)^2+(2*x^3+20*x^2+64*x+64
)*log(2+x))*log((x*log(2+x)+2*x)/log(2+x))+(-x^6-10*x^5-32*x^4-32*x^3)*log(2+x)^2+(-2*x^6-20*x^5-64*x^4-64*x^3
)*log(2+x)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.42, size = 56, normalized size = 2.15 \begin {gather*} -\frac {4 \, \log \left (-x^{3} + \log \left (x \log \left (x + 2\right ) + 2 \, x\right ) - \log \left (\log \left (x + 2\right )\right )\right )}{x + 4} + \log \left (x^{3} - \log \left (x \log \left (x + 2\right ) + 2 \, x\right ) + \log \left (\log \left (x + 2\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x+8)*log(2+x)^2+(8*x+16)*log(2+x))*log((x*log(2+x)+2*x)/log(2+x))+(-4*x^4-8*x^3)*log(2+x)^2+(-
8*x^4-16*x^3)*log(2+x))*log(log((x*log(2+x)+2*x)/log(2+x))-x^3)+(-3*x^5-18*x^4-24*x^3+x^2+6*x+8)*log(2+x)^2+(-
6*x^5-36*x^4-48*x^3+2*x^2+12*x+16)*log(2+x)-2*x^2-8*x)/(((x^3+10*x^2+32*x+32)*log(2+x)^2+(2*x^3+20*x^2+64*x+64
)*log(2+x))*log((x*log(2+x)+2*x)/log(2+x))+(-x^6-10*x^5-32*x^4-32*x^3)*log(2+x)^2+(-2*x^6-20*x^5-64*x^4-64*x^3
)*log(2+x)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.31, size = 470, normalized size = 18.08

method	result	size
risch	\(-\frac {4 \ln \left (\ln \relax (x )-\ln \left (\ln \left (2+x \right )\right )+\ln \left (\ln \left (2+x \right )+2\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )+\mathrm {csgn}\left (\frac {i}{\ln \left (2+x \right )}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )+\mathrm {csgn}\left (i \left (\ln \left (2+x \right )+2\right )\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right ) \left (-\mathrm {csgn}\left (\frac {i x \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )+\mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )\right )}{2}-x^{3}\right )}{4+x}+\ln \left (\ln \left (\ln \left (2+x \right )+2\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right ) \mathrm {csgn}\left (\frac {i x \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (\ln \left (2+x \right )+2\right )\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (2+x \right )}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )-\pi \,\mathrm {csgn}\left (i \left (\ln \left (2+x \right )+2\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (2+x \right )}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right ) \mathrm {csgn}\left (\frac {i x \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i x \left (\ln \left (2+x \right )+2\right )}{\ln \left (2+x \right )}\right )^{3}-2 i x^{3}+2 i \ln \relax (x )-2 i \ln \left (\ln \left (2+x \right )\right )\right )}{2}\right )\)	\(470\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((4*x+8)*ln(2+x)^2+(8*x+16)*ln(2+x))*ln((x*ln(2+x)+2*x)/ln(2+x))+(-4*x^4-8*x^3)*ln(2+x)^2+(-8*x^4-16*x^3
)*ln(2+x))*ln(ln((x*ln(2+x)+2*x)/ln(2+x))-x^3)+(-3*x^5-18*x^4-24*x^3+x^2+6*x+8)*ln(2+x)^2+(-6*x^5-36*x^4-48*x^
3+2*x^2+12*x+16)*ln(2+x)-2*x^2-8*x)/(((x^3+10*x^2+32*x+32)*ln(2+x)^2+(2*x^3+20*x^2+64*x+64)*ln(2+x))*ln((x*ln(
2+x)+2*x)/ln(2+x))+(-x^6-10*x^5-32*x^4-32*x^3)*ln(2+x)^2+(-2*x^6-20*x^5-64*x^4-64*x^3)*ln(2+x)),x,method=_RETU
RNVERBOSE)

[Out]

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

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maxima [A]  time = 0.46, size = 30, normalized size = 1.15 \begin {gather*} \frac {x \log \left (-x^{3} + \log \relax (x) + \log \left (\log \left (x + 2\right ) + 2\right ) - \log \left (\log \left (x + 2\right )\right )\right )}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x+8)*log(2+x)^2+(8*x+16)*log(2+x))*log((x*log(2+x)+2*x)/log(2+x))+(-4*x^4-8*x^3)*log(2+x)^2+(-
8*x^4-16*x^3)*log(2+x))*log(log((x*log(2+x)+2*x)/log(2+x))-x^3)+(-3*x^5-18*x^4-24*x^3+x^2+6*x+8)*log(2+x)^2+(-
6*x^5-36*x^4-48*x^3+2*x^2+12*x+16)*log(2+x)-2*x^2-8*x)/(((x^3+10*x^2+32*x+32)*log(2+x)^2+(2*x^3+20*x^2+64*x+64
)*log(2+x))*log((x*log(2+x)+2*x)/log(2+x))+(-x^6-10*x^5-32*x^4-32*x^3)*log(2+x)^2+(-2*x^6-20*x^5-64*x^4-64*x^3
)*log(2+x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.98, size = 32, normalized size = 1.23 \begin {gather*} \frac {x\,\ln \left (\ln \left (\frac {2\,x+x\,\ln \left (x+2\right )}{\ln \left (x+2\right )}\right )-x^3\right )}{x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + log(log((2*x + x*log(x + 2))/log(x + 2)) - x^3)*(log(x + 2)*(16*x^3 + 8*x^4) - log((2*x + x*log(x +
2))/log(x + 2))*(log(x + 2)^2*(4*x + 8) + log(x + 2)*(8*x + 16)) + log(x + 2)^2*(8*x^3 + 4*x^4)) - log(x + 2)
*(12*x + 2*x^2 - 48*x^3 - 36*x^4 - 6*x^5 + 16) - log(x + 2)^2*(6*x + x^2 - 24*x^3 - 18*x^4 - 3*x^5 + 8) + 2*x^
2)/(log(x + 2)*(64*x^3 + 64*x^4 + 20*x^5 + 2*x^6) - log((2*x + x*log(x + 2))/log(x + 2))*(log(x + 2)*(64*x + 2
0*x^2 + 2*x^3 + 64) + log(x + 2)^2*(32*x + 10*x^2 + x^3 + 32)) + log(x + 2)^2*(32*x^3 + 32*x^4 + 10*x^5 + x^6)
),x)

[Out]

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

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sympy [B]  time = 44.28, size = 48, normalized size = 1.85 \begin {gather*} \log {\left (- x^{3} + \log {\left (\frac {x \log {\left (x + 2 \right )} + 2 x}{\log {\left (x + 2 \right )}} \right )} \right )} - \frac {4 \log {\left (- x^{3} + \log {\left (\frac {x \log {\left (x + 2 \right )} + 2 x}{\log {\left (x + 2 \right )}} \right )} \right )}}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x+8)*ln(2+x)**2+(8*x+16)*ln(2+x))*ln((x*ln(2+x)+2*x)/ln(2+x))+(-4*x**4-8*x**3)*ln(2+x)**2+(-8*
x**4-16*x**3)*ln(2+x))*ln(ln((x*ln(2+x)+2*x)/ln(2+x))-x**3)+(-3*x**5-18*x**4-24*x**3+x**2+6*x+8)*ln(2+x)**2+(-
6*x**5-36*x**4-48*x**3+2*x**2+12*x+16)*ln(2+x)-2*x**2-8*x)/(((x**3+10*x**2+32*x+32)*ln(2+x)**2+(2*x**3+20*x**2
+64*x+64)*ln(2+x))*ln((x*ln(2+x)+2*x)/ln(2+x))+(-x**6-10*x**5-32*x**4-32*x**3)*ln(2+x)**2+(-2*x**6-20*x**5-64*
x**4-64*x**3)*ln(2+x)),x)

[Out]

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

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