3.72.64 \(\int \frac {(-20 x+e^{3+2 x} (16 x+16 x^2+4 x^3)) \log (x) \log (\frac {-5 x+e^{3+2 x} (2+x)}{2+x})+(20 x+10 x^2+e^{3+2 x} (-8-8 x-2 x^2)+(-10 x-5 x^2+e^{3+2 x} (4+4 x+x^2)) \log (x)) \log ^2(\frac {-5 x+e^{3+2 x} (2+x)}{2+x})}{(-10 x-5 x^2+e^{3+2 x} (4+4 x+x^2)) \log ^3(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {x \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \]

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Rubi [F]  time = 11.93, 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 {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-20*x + E^(3 + 2*x)*(16*x + 16*x^2 + 4*x^3))*Log[x]*Log[(-5*x + E^(3 + 2*x)*(2 + x))/(2 + x)] + (20*x +
10*x^2 + E^(3 + 2*x)*(-8 - 8*x - 2*x^2) + (-10*x - 5*x^2 + E^(3 + 2*x)*(4 + 4*x + x^2))*Log[x])*Log[(-5*x + E^
(3 + 2*x)*(2 + x))/(2 + x)]^2)/((-10*x - 5*x^2 + E^(3 + 2*x)*(4 + 4*x + x^2))*Log[x]^3),x]

[Out]

4*Defer[Int][(x*Log[E^(3 + 2*x) - (5*x)/(2 + x)])/Log[x]^2, x] + 40*Defer[Int][Log[E^(3 + 2*x) - (5*x)/(2 + x)
]/((2 + x)*(2*E^(3 + 2*x) - 5*x + E^(3 + 2*x)*x)*Log[x]^2), x] - 20*Defer[Int][Log[E^(3 + 2*x) - (5*x)/(2 + x)
]/((-5*x + E^(3 + 2*x)*(2 + x))*Log[x]^2), x] + 20*Defer[Int][(x^2*Log[E^(3 + 2*x) - (5*x)/(2 + x)])/((-5*x +
E^(3 + 2*x)*(2 + x))*Log[x]^2), x] - 2*Defer[Int][Log[E^(3 + 2*x) - (5*x)/(2 + x)]^2/Log[x]^3, x] + Defer[Int]
[Log[E^(3 + 2*x) - (5*x)/(2 + x)]^2/Log[x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right ) \left (\frac {4 x \left (-5+e^{3+2 x} (2+x)^2\right ) \log (x)}{(2+x) \left (-5 x+e^{3+2 x} (2+x)\right )}+(-2+\log (x)) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )\right )}{\log ^3(x)} \, dx\\ &=\int \left (\frac {20 x \left (-1+2 x+x^2\right ) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}+\frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right ) \left (4 x \log (x)-2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )+\log (x) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )\right )}{\log ^3(x)}\right ) \, dx\\ &=20 \int \frac {x \left (-1+2 x+x^2\right ) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right ) \left (4 x \log (x)-2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )+\log (x) \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )\right )}{\log ^3(x)} \, dx\\ &=20 \int \left (-\frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}+\frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}+\frac {2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)}\right ) \, dx+\int \left (\frac {4 x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)}+\frac {(-2+\log (x)) \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)}\right ) \, dx\\ &=4 \int \frac {x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx-20 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+20 \int \frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+40 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \frac {(-2+\log (x)) \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)} \, dx\\ &=4 \int \frac {x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx-20 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+20 \int \frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+40 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \left (-\frac {2 \log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)}+\frac {\log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {\log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^3(x)} \, dx\right )+4 \int \frac {x \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx-20 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+20 \int \frac {x^2 \log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\left (-5 x+e^{3+2 x} (2+x)\right ) \log ^2(x)} \, dx+40 \int \frac {\log \left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{(2+x) \left (2 e^{3+2 x}-5 x+e^{3+2 x} x\right ) \log ^2(x)} \, dx+\int \frac {\log ^2\left (e^{3+2 x}-\frac {5 x}{2+x}\right )}{\log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.40, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-20 x+e^{3+2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log (x) \log \left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )+\left (20 x+10 x^2+e^{3+2 x} \left (-8-8 x-2 x^2\right )+\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log (x)\right ) \log ^2\left (\frac {-5 x+e^{3+2 x} (2+x)}{2+x}\right )}{\left (-10 x-5 x^2+e^{3+2 x} \left (4+4 x+x^2\right )\right ) \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-20*x + E^(3 + 2*x)*(16*x + 16*x^2 + 4*x^3))*Log[x]*Log[(-5*x + E^(3 + 2*x)*(2 + x))/(2 + x)] + (2
0*x + 10*x^2 + E^(3 + 2*x)*(-8 - 8*x - 2*x^2) + (-10*x - 5*x^2 + E^(3 + 2*x)*(4 + 4*x + x^2))*Log[x])*Log[(-5*
x + E^(3 + 2*x)*(2 + x))/(2 + x)]^2)/((-10*x - 5*x^2 + E^(3 + 2*x)*(4 + 4*x + x^2))*Log[x]^3),x]

[Out]

Integrate[((-20*x + E^(3 + 2*x)*(16*x + 16*x^2 + 4*x^3))*Log[x]*Log[(-5*x + E^(3 + 2*x)*(2 + x))/(2 + x)] + (2
0*x + 10*x^2 + E^(3 + 2*x)*(-8 - 8*x - 2*x^2) + (-10*x - 5*x^2 + E^(3 + 2*x)*(4 + 4*x + x^2))*Log[x])*Log[(-5*
x + E^(3 + 2*x)*(2 + x))/(2 + x)]^2)/((-10*x - 5*x^2 + E^(3 + 2*x)*(4 + 4*x + x^2))*Log[x]^3), x]

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fricas [A]  time = 0.57, size = 29, normalized size = 1.16 \begin {gather*} \frac {x \log \left (\frac {{\left (x + 2\right )} e^{\left (2 \, x + 3\right )} - 5 \, x}{x + 2}\right )^{2}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2+4*x+4)*exp(2*x+3)-5*x^2-10*x)*log(x)+(-2*x^2-8*x-8)*exp(2*x+3)+10*x^2+20*x)*log(((2+x)*exp(2
*x+3)-5*x)/(2+x))^2+((4*x^3+16*x^2+16*x)*exp(2*x+3)-20*x)*log(x)*log(((2+x)*exp(2*x+3)-5*x)/(2+x)))/((x^2+4*x+
4)*exp(2*x+3)-5*x^2-10*x)/log(x)^3,x, algorithm="fricas")

[Out]

x*log(((x + 2)*e^(2*x + 3) - 5*x)/(x + 2))^2/log(x)^2

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giac [B]  time = 0.87, size = 67, normalized size = 2.68 \begin {gather*} \frac {x \log \left (x e^{\left (2 \, x + 3\right )} - 5 \, x + 2 \, e^{\left (2 \, x + 3\right )}\right )^{2} - 2 \, x \log \left (x e^{\left (2 \, x + 3\right )} - 5 \, x + 2 \, e^{\left (2 \, x + 3\right )}\right ) \log \left (x + 2\right ) + x \log \left (x + 2\right )^{2}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2+4*x+4)*exp(2*x+3)-5*x^2-10*x)*log(x)+(-2*x^2-8*x-8)*exp(2*x+3)+10*x^2+20*x)*log(((2+x)*exp(2
*x+3)-5*x)/(2+x))^2+((4*x^3+16*x^2+16*x)*exp(2*x+3)-20*x)*log(x)*log(((2+x)*exp(2*x+3)-5*x)/(2+x)))/((x^2+4*x+
4)*exp(2*x+3)-5*x^2-10*x)/log(x)^3,x, algorithm="giac")

[Out]

(x*log(x*e^(2*x + 3) - 5*x + 2*e^(2*x + 3))^2 - 2*x*log(x*e^(2*x + 3) - 5*x + 2*e^(2*x + 3))*log(x + 2) + x*lo
g(x + 2)^2)/log(x)^2

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maple [C]  time = 0.26, size = 1011, normalized size = 40.44

method	result	size
risch	\(\frac {x \ln \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )^{2}}{\ln \relax (x )^{2}}-\frac {x \left (i \pi \,\mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{3}+2 \ln \left (2+x \right )\right ) \ln \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{\ln \relax (x )^{2}}+\frac {x \left (-\pi ^{2} \mathrm {csgn}\left (\frac {i}{2+x}\right )^{2} \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right )^{2} \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{2}+2 \pi ^{2} \mathrm {csgn}\left (\frac {i}{2+x}\right )^{2} \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{3}-\pi ^{2} \mathrm {csgn}\left (\frac {i}{2+x}\right )^{2} \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{4}+2 \pi ^{2} \mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right )^{2} \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{3}-4 \pi ^{2} \mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{4}+2 \pi ^{2} \mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{5}-\pi ^{2} \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right )^{2} \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{4}+2 \pi ^{2} \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{5}-\pi ^{2} \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{6}+4 i \ln \left (2+x \right ) \pi \,\mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )-4 i \ln \left (2+x \right ) \pi \,\mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{2}-4 i \ln \left (2+x \right ) \pi \,\mathrm {csgn}\left (i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{2}+4 i \ln \left (2+x \right ) \pi \mathrm {csgn}\left (\frac {i \left (\left ({\mathrm e}^{2 x +3}-5\right ) x +2 \,{\mathrm e}^{2 x +3}\right )}{2+x}\right )^{3}+4 \ln \left (2+x \right )^{2}\right )}{4 \ln \relax (x )^{2}}\)	\(1011\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x^2+4*x+4)*exp(2*x+3)-5*x^2-10*x)*ln(x)+(-2*x^2-8*x-8)*exp(2*x+3)+10*x^2+20*x)*ln(((2+x)*exp(2*x+3)-5*
x)/(2+x))^2+((4*x^3+16*x^2+16*x)*exp(2*x+3)-20*x)*ln(x)*ln(((2+x)*exp(2*x+3)-5*x)/(2+x)))/((x^2+4*x+4)*exp(2*x
+3)-5*x^2-10*x)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

x/ln(x)^2*ln((exp(2*x+3)-5)*x+2*exp(2*x+3))^2-x*(I*Pi*csgn(I/(2+x))*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))*cs
gn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))-I*Pi*csgn(I/(2+x))*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^2
-I*Pi*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^2+I*Pi*csgn(I/(2+x
)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^3+2*ln(2+x))/ln(x)^2*ln((exp(2*x+3)-5)*x+2*exp(2*x+3))+1/4*x*(-Pi^2*csgn(I/
(2+x))^2*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^2*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^2+2*Pi^2*csgn
(I/(2+x))^2*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^3-Pi^2*csgn(
I/(2+x))^2*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^4+2*Pi^2*csgn(I/(2+x))*csgn(I*((exp(2*x+3)-5)*x+2*exp
(2*x+3)))^2*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^3-4*Pi^2*csgn(I/(2+x))*csgn(I*((exp(2*x+3)-5)*x+2*ex
p(2*x+3)))*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^4+2*Pi^2*csgn(I/(2+x))*csgn(I/(2+x)*((exp(2*x+3)-5)*x
+2*exp(2*x+3)))^5-Pi^2*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^2*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))
^4+2*Pi^2*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^5-Pi^2*csgn(I/
(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^6+4*I*ln(2+x)*Pi*csgn(I/(2+x))*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))*
csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))-4*I*ln(2+x)*Pi*csgn(I/(2+x))*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*ex
p(2*x+3)))^2-4*I*ln(2+x)*Pi*csgn(I*((exp(2*x+3)-5)*x+2*exp(2*x+3)))*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3
)))^2+4*I*ln(2+x)*Pi*csgn(I/(2+x)*((exp(2*x+3)-5)*x+2*exp(2*x+3)))^3+4*ln(2+x)^2)/ln(x)^2

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maxima [B]  time = 0.46, size = 63, normalized size = 2.52 \begin {gather*} \frac {x \log \left ({\left (x e^{3} + 2 \, e^{3}\right )} e^{\left (2 \, x\right )} - 5 \, x\right )^{2} - 2 \, x \log \left ({\left (x e^{3} + 2 \, e^{3}\right )} e^{\left (2 \, x\right )} - 5 \, x\right ) \log \left (x + 2\right ) + x \log \left (x + 2\right )^{2}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2+4*x+4)*exp(2*x+3)-5*x^2-10*x)*log(x)+(-2*x^2-8*x-8)*exp(2*x+3)+10*x^2+20*x)*log(((2+x)*exp(2
*x+3)-5*x)/(2+x))^2+((4*x^3+16*x^2+16*x)*exp(2*x+3)-20*x)*log(x)*log(((2+x)*exp(2*x+3)-5*x)/(2+x)))/((x^2+4*x+
4)*exp(2*x+3)-5*x^2-10*x)/log(x)^3,x, algorithm="maxima")

[Out]

(x*log((x*e^3 + 2*e^3)*e^(2*x) - 5*x)^2 - 2*x*log((x*e^3 + 2*e^3)*e^(2*x) - 5*x)*log(x + 2) + x*log(x + 2)^2)/
log(x)^2

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mupad [B]  time = 4.62, size = 31, normalized size = 1.24 \begin {gather*} \frac {x\,{\ln \left (-\frac {5\,x-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3\,\left (x+2\right )}{x+2}\right )}^2}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(-(5*x - exp(2*x + 3)*(x + 2))/(x + 2))^2*(20*x - exp(2*x + 3)*(8*x + 2*x^2 + 8) - log(x)*(10*x - exp
(2*x + 3)*(4*x + x^2 + 4) + 5*x^2) + 10*x^2) - log(-(5*x - exp(2*x + 3)*(x + 2))/(x + 2))*log(x)*(20*x - exp(2
*x + 3)*(16*x + 16*x^2 + 4*x^3)))/(log(x)^3*(10*x - exp(2*x + 3)*(4*x + x^2 + 4) + 5*x^2)),x)

[Out]

(x*log(-(5*x - exp(2*x)*exp(3)*(x + 2))/(x + 2))^2)/log(x)^2

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sympy [A]  time = 0.96, size = 26, normalized size = 1.04 \begin {gather*} \frac {x \log {\left (\frac {- 5 x + \left (x + 2\right ) e^{2 x + 3}}{x + 2} \right )}^{2}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x**2+4*x+4)*exp(2*x+3)-5*x**2-10*x)*ln(x)+(-2*x**2-8*x-8)*exp(2*x+3)+10*x**2+20*x)*ln(((2+x)*exp
(2*x+3)-5*x)/(2+x))**2+((4*x**3+16*x**2+16*x)*exp(2*x+3)-20*x)*ln(x)*ln(((2+x)*exp(2*x+3)-5*x)/(2+x)))/((x**2+
4*x+4)*exp(2*x+3)-5*x**2-10*x)/ln(x)**3,x)

[Out]

x*log((-5*x + (x + 2)*exp(2*x + 3))/(x + 2))**2/log(x)**2

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