3.59.10 \(\int \frac {e^{2 x} (4+16 x)+e^{2 x} (x+2 x^2) \log ^2(3 x+6 x^2)+(e^{2 x} (-4-16 x-16 x^2) \log (3 x+6 x^2)+e^{2 x} (4+17 x+20 x^2+4 x^3) \log ^2(3 x+6 x^2)) \log (\frac {-4+(4+x) \log (3 x+6 x^2)}{\log (3 x+6 x^2)})}{(-4-8 x) \log (3 x+6 x^2)+(4+9 x+2 x^2) \log ^2(3 x+6 x^2)} \, dx\)

Optimal. Leaf size=24 \[ e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \]

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Rubi [F]  time = 7.97, 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 {e^{2 x} (4+16 x)+e^{2 x} \left (x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )+\left (e^{2 x} \left (-4-16 x-16 x^2\right ) \log \left (3 x+6 x^2\right )+e^{2 x} \left (4+17 x+20 x^2+4 x^3\right ) \log ^2\left (3 x+6 x^2\right )\right ) \log \left (\frac {-4+(4+x) \log \left (3 x+6 x^2\right )}{\log \left (3 x+6 x^2\right )}\right )}{(-4-8 x) \log \left (3 x+6 x^2\right )+\left (4+9 x+2 x^2\right ) \log ^2\left (3 x+6 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(4 + 16*x) + E^(2*x)*(x + 2*x^2)*Log[3*x + 6*x^2]^2 + (E^(2*x)*(-4 - 16*x - 16*x^2)*Log[3*x + 6*x
^2] + E^(2*x)*(4 + 17*x + 20*x^2 + 4*x^3)*Log[3*x + 6*x^2]^2)*Log[(-4 + (4 + x)*Log[3*x + 6*x^2])/Log[3*x + 6*
x^2]])/((-4 - 8*x)*Log[3*x + 6*x^2] + (4 + 9*x + 2*x^2)*Log[3*x + 6*x^2]^2),x]

[Out]

E^(2*x)/2 - (4*ExpIntegralEi[2*(4 + x)])/E^8 - 2*Defer[Int][E^(2*x)/Log[3*x*(1 + 2*x)], x] + Defer[Int][E^(2*x
)/((1 + 2*x)*Log[3*x*(1 + 2*x)]), x] - 16*Defer[Int][E^(2*x)/((4 + x)*(-4 + 4*Log[3*x*(1 + 2*x)] + x*Log[3*x*(
1 + 2*x)])), x] - (7*Defer[Int][E^(2*x)/((1 + 2*x)*(-4 + 4*Log[3*x*(1 + 2*x)] + x*Log[3*x*(1 + 2*x)])), x])/2
+ (23*Defer[Int][E^(2*x)/(-4 + (4 + x)*Log[3*x*(1 + 2*x)]), x])/2 + 2*Defer[Int][(E^(2*x)*x)/(-4 + (4 + x)*Log
[3*x*(1 + 2*x)]), x] + Defer[Int][E^(2*x)*Log[4 + x - 4/Log[3*x*(1 + 2*x)]], x] + 2*Defer[Int][E^(2*x)*x*Log[4
+ x - 4/Log[3*x*(1 + 2*x)]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (-4-16 x-x (1+2 x) \log ^2(3 x (1+2 x))-(1+2 x)^2 \log (3 x (1+2 x)) (-4+(4+x) \log (3 x (1+2 x))) \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )\right )}{(1+2 x) \log (3 x (1+2 x)) (4-(4+x) \log (3 x (1+2 x)))} \, dx\\ &=\int \left (\frac {e^{2 x} \left (4+16 x+x \log ^2(3 x (1+2 x))+2 x^2 \log ^2(3 x (1+2 x))\right )}{(1+2 x) \log (3 x (1+2 x)) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}+e^{2 x} (1+2 x) \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )\right ) \, dx\\ &=\int \frac {e^{2 x} \left (4+16 x+x \log ^2(3 x (1+2 x))+2 x^2 \log ^2(3 x (1+2 x))\right )}{(1+2 x) \log (3 x (1+2 x)) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int e^{2 x} (1+2 x) \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx\\ &=\int \frac {e^{2 x} \left (-4-16 x-x (1+2 x) \log ^2(3 x (1+2 x))\right )}{(1+2 x) \log (3 x (1+2 x)) (4-(4+x) \log (3 x (1+2 x)))} \, dx+\int \left (e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )+2 e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right )\right ) \, dx\\ &=2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx+\int \left (\frac {e^{2 x} x}{4+x}+\frac {e^{2 x} (-1-4 x)}{(1+2 x) \log (3 x (1+2 x))}+\frac {4 e^{2 x} x}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}+\frac {e^{2 x} (4+x) (1+4 x)}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}\right ) \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx\\ &=2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx+4 \int \frac {e^{2 x} x}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int \frac {e^{2 x} x}{4+x} \, dx+\int \frac {e^{2 x} (-1-4 x)}{(1+2 x) \log (3 x (1+2 x))} \, dx+\int \frac {e^{2 x} (4+x) (1+4 x)}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx\\ &=2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx+4 \int \left (\frac {e^{2 x}}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))}-\frac {4 e^{2 x}}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}\right ) \, dx+\int \left (e^{2 x}-\frac {4 e^{2 x}}{4+x}\right ) \, dx+\int \left (-\frac {2 e^{2 x}}{\log (3 x (1+2 x))}+\frac {e^{2 x}}{(1+2 x) \log (3 x (1+2 x))}\right ) \, dx+\int \left (\frac {15 e^{2 x}}{2 (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}+\frac {2 e^{2 x} x}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))}-\frac {7 e^{2 x}}{2 (1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))}\right ) \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 x}}{\log (3 x (1+2 x))} \, dx\right )+2 \int \frac {e^{2 x} x}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))} \, dx+2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx-\frac {7}{2} \int \frac {e^{2 x}}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx-4 \int \frac {e^{2 x}}{4+x} \, dx+4 \int \frac {e^{2 x}}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))} \, dx+\frac {15}{2} \int \frac {e^{2 x}}{-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x))} \, dx-16 \int \frac {e^{2 x}}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int e^{2 x} \, dx+\int \frac {e^{2 x}}{(1+2 x) \log (3 x (1+2 x))} \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx\\ &=\frac {e^{2 x}}{2}-\frac {4 \text {Ei}(2 (4+x))}{e^8}-2 \int \frac {e^{2 x}}{\log (3 x (1+2 x))} \, dx+2 \int \frac {e^{2 x} x}{-4+(4+x) \log (3 x (1+2 x))} \, dx+2 \int e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx-\frac {7}{2} \int \frac {e^{2 x}}{(1+2 x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+4 \int \frac {e^{2 x}}{-4+(4+x) \log (3 x (1+2 x))} \, dx+\frac {15}{2} \int \frac {e^{2 x}}{-4+(4+x) \log (3 x (1+2 x))} \, dx-16 \int \frac {e^{2 x}}{(4+x) (-4+4 \log (3 x (1+2 x))+x \log (3 x (1+2 x)))} \, dx+\int \frac {e^{2 x}}{(1+2 x) \log (3 x (1+2 x))} \, dx+\int e^{2 x} \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 24, normalized size = 1.00 \begin {gather*} e^{2 x} x \log \left (4+x-\frac {4}{\log (3 x (1+2 x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(4 + 16*x) + E^(2*x)*(x + 2*x^2)*Log[3*x + 6*x^2]^2 + (E^(2*x)*(-4 - 16*x - 16*x^2)*Log[3*x
+ 6*x^2] + E^(2*x)*(4 + 17*x + 20*x^2 + 4*x^3)*Log[3*x + 6*x^2]^2)*Log[(-4 + (4 + x)*Log[3*x + 6*x^2])/Log[3*
x + 6*x^2]])/((-4 - 8*x)*Log[3*x + 6*x^2] + (4 + 9*x + 2*x^2)*Log[3*x + 6*x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.65, size = 36, normalized size = 1.50 \begin {gather*} x e^{\left (2 \, x\right )} \log \left (\frac {{\left (x + 4\right )} \log \left (6 \, x^{2} + 3 \, x\right ) - 4}{\log \left (6 \, x^{2} + 3 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3+20*x^2+17*x+4)*exp(x)^2*log(6*x^2+3*x)^2+(-16*x^2-16*x-4)*exp(x)^2*log(6*x^2+3*x))*log(((4+
x)*log(6*x^2+3*x)-4)/log(6*x^2+3*x))+(2*x^2+x)*exp(x)^2*log(6*x^2+3*x)^2+(16*x+4)*exp(x)^2)/((2*x^2+9*x+4)*log
(6*x^2+3*x)^2+(-8*x-4)*log(6*x^2+3*x)),x, algorithm="fricas")

[Out]

x*e^(2*x)*log(((x + 4)*log(6*x^2 + 3*x) - 4)/log(6*x^2 + 3*x))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3+20*x^2+17*x+4)*exp(x)^2*log(6*x^2+3*x)^2+(-16*x^2-16*x-4)*exp(x)^2*log(6*x^2+3*x))*log(((4+
x)*log(6*x^2+3*x)-4)/log(6*x^2+3*x))+(2*x^2+x)*exp(x)^2*log(6*x^2+3*x)^2+(16*x+4)*exp(x)^2)/((2*x^2+9*x+4)*log
(6*x^2+3*x)^2+(-8*x-4)*log(6*x^2+3*x)),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 1.63, size = 2063, normalized size = 85.96

method	result	size
risch	\(\text {Expression too large to display}\)	\(2063\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^3+20*x^2+17*x+4)*exp(x)^2*ln(6*x^2+3*x)^2+(-16*x^2-16*x-4)*exp(x)^2*ln(6*x^2+3*x))*ln(((4+x)*ln(6*x
^2+3*x)-4)/ln(6*x^2+3*x))+(2*x^2+x)*exp(x)^2*ln(6*x^2+3*x)^2+(16*x+4)*exp(x)^2)/((2*x^2+9*x+4)*ln(6*x^2+3*x)^2
+(-8*x-4)*ln(6*x^2+3*x)),x,method=_RETURNVERBOSE)

[Out]

x*exp(2*x)*ln(-8*I+8*I*ln(x)+x*Pi*csgn(I*x*(1/2+x))^3-x*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-x*Pi*csgn(I*(1/2+x))*
csgn(I*x*(1/2+x))^2+4*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+4*Pi*csgn(I*x*(1/2+x))^3+8*I*ln(3)+8*I*ln
(1/2+x)+2*I*x*ln(3)+2*I*x*ln(x)+2*I*x*ln(1/2+x)-4*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-4*Pi*csgn(I*(1/2+x))*csgn(I
*x*(1/2+x))^2+x*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x)))-x*exp(2*x)*ln(Pi*csgn(I*x)*csgn(I*(1/2+x))*csg
n(I*x*(1/2+x))+Pi*csgn(I*x*(1/2+x))^3+2*I*ln(x)+2*I*ln(1/2+x)+2*I*ln(3)-Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-Pi*cs
gn(I*(1/2+x))*csgn(I*x*(1/2+x))^2)-1/2*I*Pi*x*csgn(I/(Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+Pi*csgn(I
*x*(1/2+x))^3+2*I*ln(x)+2*I*ln(1/2+x)+2*I*ln(3)-Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-Pi*csgn(I*(1/2+x))*csgn(I*x*(
1/2+x))^2))*csgn(I*(-8*I+8*I*ln(x)+x*Pi*csgn(I*x*(1/2+x))^3-x*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-x*Pi*csgn(I*(1/
2+x))*csgn(I*x*(1/2+x))^2+4*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+4*Pi*csgn(I*x*(1/2+x))^3+8*I*ln(3)+
8*I*ln(1/2+x)+2*I*x*ln(3)+2*I*x*ln(x)+2*I*x*ln(1/2+x)-4*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-4*Pi*csgn(I*(1/2+x))*
csgn(I*x*(1/2+x))^2+x*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))))*csgn(I/(Pi*csgn(I*x)*csgn(I*(1/2+x))*cs
gn(I*x*(1/2+x))+Pi*csgn(I*x*(1/2+x))^3+2*I*ln(x)+2*I*ln(1/2+x)+2*I*ln(3)-Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-Pi*c
sgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2)*(-8*I+8*I*ln(x)+x*Pi*csgn(I*x*(1/2+x))^3-x*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^
2-x*Pi*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2+4*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+4*Pi*csgn(I*x*(1/2
+x))^3+8*I*ln(3)+8*I*ln(1/2+x)+2*I*x*ln(3)+2*I*x*ln(x)+2*I*x*ln(1/2+x)-4*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-4*Pi
*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2+x*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))))*exp(2*x)+1/2*I*Pi*x*cs
gn(I/(Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+Pi*csgn(I*x*(1/2+x))^3+2*I*ln(x)+2*I*ln(1/2+x)+2*I*ln(3)-
Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-Pi*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2))*csgn(I/(Pi*csgn(I*x)*csgn(I*(1/2+x))
*csgn(I*x*(1/2+x))+Pi*csgn(I*x*(1/2+x))^3+2*I*ln(x)+2*I*ln(1/2+x)+2*I*ln(3)-Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-P
i*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2)*(-8*I+8*I*ln(x)+x*Pi*csgn(I*x*(1/2+x))^3-x*Pi*csgn(I*x)*csgn(I*x*(1/2+x
))^2-x*Pi*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2+4*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+4*Pi*csgn(I*x*(
1/2+x))^3+8*I*ln(3)+8*I*ln(1/2+x)+2*I*x*ln(3)+2*I*x*ln(x)+2*I*x*ln(1/2+x)-4*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-4
*Pi*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2+x*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))))^2*exp(2*x)+1/2*I*Pi
*x*csgn(I*(-8*I+8*I*ln(x)+x*Pi*csgn(I*x*(1/2+x))^3-x*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-x*Pi*csgn(I*(1/2+x))*csg
n(I*x*(1/2+x))^2+4*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+4*Pi*csgn(I*x*(1/2+x))^3+8*I*ln(3)+8*I*ln(1/
2+x)+2*I*x*ln(3)+2*I*x*ln(x)+2*I*x*ln(1/2+x)-4*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-4*Pi*csgn(I*(1/2+x))*csgn(I*x*
(1/2+x))^2+x*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))))*csgn(I/(Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1
/2+x))+Pi*csgn(I*x*(1/2+x))^3+2*I*ln(x)+2*I*ln(1/2+x)+2*I*ln(3)-Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-Pi*csgn(I*(1/
2+x))*csgn(I*x*(1/2+x))^2)*(-8*I+8*I*ln(x)+x*Pi*csgn(I*x*(1/2+x))^3-x*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-x*Pi*cs
gn(I*(1/2+x))*csgn(I*x*(1/2+x))^2+4*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+4*Pi*csgn(I*x*(1/2+x))^3+8*
I*ln(3)+8*I*ln(1/2+x)+2*I*x*ln(3)+2*I*x*ln(x)+2*I*x*ln(1/2+x)-4*Pi*csgn(I*x)*csgn(I*x*(1/2+x))^2-4*Pi*csgn(I*(
1/2+x))*csgn(I*x*(1/2+x))^2+x*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))))^2*exp(2*x)-1/2*I*Pi*x*csgn(I/(P
i*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))+Pi*csgn(I*x*(1/2+x))^3+2*I*ln(x)+2*I*ln(1/2+x)+2*I*ln(3)-Pi*csgn
(I*x)*csgn(I*x*(1/2+x))^2-Pi*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2)*(-8*I+8*I*ln(x)+x*Pi*csgn(I*x*(1/2+x))^3-x*P
i*csgn(I*x)*csgn(I*x*(1/2+x))^2-x*Pi*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2+4*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I
*x*(1/2+x))+4*Pi*csgn(I*x*(1/2+x))^3+8*I*ln(3)+8*I*ln(1/2+x)+2*I*x*ln(3)+2*I*x*ln(x)+2*I*x*ln(1/2+x)-4*Pi*csgn
(I*x)*csgn(I*x*(1/2+x))^2-4*Pi*csgn(I*(1/2+x))*csgn(I*x*(1/2+x))^2+x*Pi*csgn(I*x)*csgn(I*(1/2+x))*csgn(I*x*(1/
2+x))))^3*exp(2*x)

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maxima [B]  time = 0.51, size = 58, normalized size = 2.42 \begin {gather*} x e^{\left (2 \, x\right )} \log \left (x {\left (\log \relax (3) + \log \left (2 \, x + 1\right )\right )} + {\left (x + 4\right )} \log \relax (x) + 4 \, \log \relax (3) + 4 \, \log \left (2 \, x + 1\right ) - 4\right ) - x e^{\left (2 \, x\right )} \log \left (\log \relax (3) + \log \left (2 \, x + 1\right ) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3+20*x^2+17*x+4)*exp(x)^2*log(6*x^2+3*x)^2+(-16*x^2-16*x-4)*exp(x)^2*log(6*x^2+3*x))*log(((4+
x)*log(6*x^2+3*x)-4)/log(6*x^2+3*x))+(2*x^2+x)*exp(x)^2*log(6*x^2+3*x)^2+(16*x+4)*exp(x)^2)/((2*x^2+9*x+4)*log
(6*x^2+3*x)^2+(-8*x-4)*log(6*x^2+3*x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.79, size = 36, normalized size = 1.50 \begin {gather*} x\,{\mathrm {e}}^{2\,x}\,\ln \left (\frac {\ln \left (6\,x^2+3\,x\right )\,\left (x+4\right )-4}{\ln \left (6\,x^2+3\,x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((log(3*x + 6*x^2)*(x + 4) - 4)/log(3*x + 6*x^2))*(exp(2*x)*log(3*x + 6*x^2)^2*(17*x + 20*x^2 + 4*x^3
+ 4) - exp(2*x)*log(3*x + 6*x^2)*(16*x + 16*x^2 + 4)) + exp(2*x)*(16*x + 4) + exp(2*x)*log(3*x + 6*x^2)^2*(x +
2*x^2))/(log(3*x + 6*x^2)^2*(9*x + 2*x^2 + 4) - log(3*x + 6*x^2)*(8*x + 4)),x)

[Out]

x*exp(2*x)*log((log(3*x + 6*x^2)*(x + 4) - 4)/log(3*x + 6*x^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**3+20*x**2+17*x+4)*exp(x)**2*ln(6*x**2+3*x)**2+(-16*x**2-16*x-4)*exp(x)**2*ln(6*x**2+3*x))*ln
(((4+x)*ln(6*x**2+3*x)-4)/ln(6*x**2+3*x))+(2*x**2+x)*exp(x)**2*ln(6*x**2+3*x)**2+(16*x+4)*exp(x)**2)/((2*x**2+
9*x+4)*ln(6*x**2+3*x)**2+(-8*x-4)*ln(6*x**2+3*x)),x)

[Out]

Timed out

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