3.39.99 \(\int \frac {-20-10 x-5 x^2+(8 x+8 x^2-2 x^3-2 x^4+e^{12 x} (-48-48 x+12 x^2+12 x^3)) \log ^2(\frac {-4+x^2}{1+x})}{(-20-20 x+5 x^2+5 x^3) \log (\frac {-4+x^2}{1+x})+(4 x^2+4 x^3-x^4-x^5+e^{12 x} (-4-4 x+x^2+x^3)) \log ^2(\frac {-4+x^2}{1+x})} \, dx\)

Optimal. Leaf size=32 \[ \log \left (\log (4) \left (e^{12 x}-x^2+\frac {5}{\log \left (x-\frac {4+x}{1+x}\right )}\right )\right ) \]

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Rubi [F]  time = 85.94, 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 {-20-10 x-5 x^2+\left (8 x+8 x^2-2 x^3-2 x^4+e^{12 x} \left (-48-48 x+12 x^2+12 x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (-20-20 x+5 x^2+5 x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+\left (4 x^2+4 x^3-x^4-x^5+e^{12 x} \left (-4-4 x+x^2+x^3\right )\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20 - 10*x - 5*x^2 + (8*x + 8*x^2 - 2*x^3 - 2*x^4 + E^(12*x)*(-48 - 48*x + 12*x^2 + 12*x^3))*Log[(-4 + x^
2)/(1 + x)]^2)/((-20 - 20*x + 5*x^2 + 5*x^3)*Log[(-4 + x^2)/(1 + x)] + (4*x^2 + 4*x^3 - x^4 - x^5 + E^(12*x)*(
-4 - 4*x + x^2 + x^3))*Log[(-4 + x^2)/(1 + x)]^2),x]

[Out]

12*x + 5*Defer[Int][1/((-2 + x)*Log[(-4 + x^2)/(1 + x)]*(-5 - E^(12*x)*Log[(-4 + x^2)/(1 + x)] + x^2*Log[(-4 +
x^2)/(1 + x)])), x] - 5*Defer[Int][1/((1 + x)*Log[(-4 + x^2)/(1 + x)]*(-5 - E^(12*x)*Log[(-4 + x^2)/(1 + x)]
+ x^2*Log[(-4 + x^2)/(1 + x)])), x] + 5*Defer[Int][1/((2 + x)*Log[(-4 + x^2)/(1 + x)]*(-5 - E^(12*x)*Log[(-4 +
x^2)/(1 + x)] + x^2*Log[(-4 + x^2)/(1 + x)])), x] - 60*Defer[Int][(5 + (E^(12*x) - x^2)*Log[(-4 + x^2)/(1 + x
)])^(-1), x] + 2*Defer[Int][(x*Log[(-4 + x^2)/(1 + x)])/(-5 + (-E^(12*x) + x^2)*Log[(-4 + x^2)/(1 + x)]), x] -
12*Defer[Int][(x^2*Log[(-4 + x^2)/(1 + x)])/(-5 + (-E^(12*x) + x^2)*Log[(-4 + x^2)/(1 + x)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (4+2 x+x^2\right )-2 \left (6 e^{12 x}-x\right ) \left (-4-4 x+x^2+x^3\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{\left (4+4 x-x^2-x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx\\ &=\int \left (12-\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(-2+x) (1+x) (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}\right ) \, dx\\ &=12 x-\int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(-2+x) (1+x) (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx\\ &=12 x-\int \frac {-5 \left (4+2 x+x^2\right )-60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2-x) (1+x) (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx\\ &=12 x-\int \left (\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{12 (-2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}-\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{3 (1+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}+\frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{4 (2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )}\right ) \, dx\\ &=12 x-\frac {1}{12} \int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(-2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx-\frac {1}{4} \int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx+\frac {1}{3} \int \frac {-20-10 x-5 x^2+240 \log \left (\frac {-4+x^2}{1+x}\right )+240 x \log \left (\frac {-4+x^2}{1+x}\right )-60 x^2 \log \left (\frac {-4+x^2}{1+x}\right )-60 x^3 \log \left (\frac {-4+x^2}{1+x}\right )+8 x \log ^2\left (\frac {-4+x^2}{1+x}\right )-40 x^2 \log ^2\left (\frac {-4+x^2}{1+x}\right )-50 x^3 \log ^2\left (\frac {-4+x^2}{1+x}\right )+10 x^4 \log ^2\left (\frac {-4+x^2}{1+x}\right )+12 x^5 \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(1+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (-5-e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )+x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx\\ &=12 x-\frac {1}{12} \int \frac {-5 \left (4+2 x+x^2\right )-60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )+2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2-x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx-\frac {1}{4} \int \frac {5 \left (4+2 x+x^2\right )+60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )-2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(2+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+x}\right )\right )} \, dx+\frac {1}{3} \int \frac {5 \left (4+2 x+x^2\right )+60 \left (-4-4 x+x^2+x^3\right ) \log \left (\frac {-4+x^2}{1+x}\right )-2 x \left (4-20 x-25 x^2+5 x^3+6 x^4\right ) \log ^2\left (\frac {-4+x^2}{1+x}\right )}{(1+x) \log \left (\frac {-4+x^2}{1+x}\right ) \left (5-\left (-e^{12 x}+x^2\right ) \log \left (\frac {-4+x^2}{1+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 = 0.94, size = 54, normalized size = 1.69 \begin {gather*} -\log \left (\log \left (\frac {-4+x^2}{1+x}\right )\right )+\log \left (5+e^{12 x} \log \left (\frac {-4+x^2}{1+x}\right )-x^2 \log \left (\frac {-4+x^2}{1+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 10*x - 5*x^2 + (8*x + 8*x^2 - 2*x^3 - 2*x^4 + E^(12*x)*(-48 - 48*x + 12*x^2 + 12*x^3))*Log[(-
4 + x^2)/(1 + x)]^2)/((-20 - 20*x + 5*x^2 + 5*x^3)*Log[(-4 + x^2)/(1 + x)] + (4*x^2 + 4*x^3 - x^4 - x^5 + E^(1
2*x)*(-4 - 4*x + x^2 + x^3))*Log[(-4 + x^2)/(1 + x)]^2),x]

[Out]

-Log[Log[(-4 + x^2)/(1 + x)]] + Log[5 + E^(12*x)*Log[(-4 + x^2)/(1 + x)] - x^2*Log[(-4 + x^2)/(1 + x)]]

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fricas [B]  time = 0.50, size = 66, normalized size = 2.06 \begin {gather*} \log \left (-x^{2} + e^{\left (12 \, x\right )}\right ) + \log \left (\frac {{\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (\frac {x^{2} - 4}{x + 1}\right ) - 5}{x^{2} - e^{\left (12 \, x\right )}}\right ) - \log \left (\log \left (\frac {x^{2} - 4}{x + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^3+12*x^2-48*x-48)*exp(12*x)-2*x^4-2*x^3+8*x^2+8*x)*log((x^2-4)/(x+1))^2-5*x^2-10*x-20)/(((x^
3+x^2-4*x-4)*exp(12*x)-x^5-x^4+4*x^3+4*x^2)*log((x^2-4)/(x+1))^2+(5*x^3+5*x^2-20*x-20)*log((x^2-4)/(x+1))),x,
algorithm="fricas")

[Out]

log(-x^2 + e^(12*x)) + log(((x^2 - e^(12*x))*log((x^2 - 4)/(x + 1)) - 5)/(x^2 - e^(12*x))) - log(log((x^2 - 4)
/(x + 1)))

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giac [A]  time = 1.00, size = 53, normalized size = 1.66 \begin {gather*} \log \left (-x^{2} \log \left (\frac {x^{2} - 4}{x + 1}\right ) + e^{\left (12 \, x\right )} \log \left (\frac {x^{2} - 4}{x + 1}\right ) + 5\right ) - \log \left (\log \left (\frac {x^{2} - 4}{x + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^3+12*x^2-48*x-48)*exp(12*x)-2*x^4-2*x^3+8*x^2+8*x)*log((x^2-4)/(x+1))^2-5*x^2-10*x-20)/(((x^
3+x^2-4*x-4)*exp(12*x)-x^5-x^4+4*x^3+4*x^2)*log((x^2-4)/(x+1))^2+(5*x^3+5*x^2-20*x-20)*log((x^2-4)/(x+1))),x,
algorithm="giac")

[Out]

log(-x^2*log((x^2 - 4)/(x + 1)) + e^(12*x)*log((x^2 - 4)/(x + 1)) + 5) - log(log((x^2 - 4)/(x + 1)))

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maple [C]  time = 0.43, size = 433, normalized size = 13.53

method	result	size
risch	\(\ln \left (-x^{2}+{\mathrm e}^{12 x}\right )+\ln \left (\ln \left (x^{2}-4\right )-\frac {i \left (\pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )-\pi \,x^{2} \mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{2}-\pi \,x^{2} \mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{2}+\pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right ) {\mathrm e}^{12 x}+\pi \,\mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{2} {\mathrm e}^{12 x}+\pi \,\mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{2} {\mathrm e}^{12 x}-\pi \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{3} {\mathrm e}^{12 x}-2 i x^{2} \ln \left (x +1\right )+2 i {\mathrm e}^{12 x} \ln \left (x +1\right )-10 i\right )}{2 \left (x^{2}-{\mathrm e}^{12 x}\right )}\right )-\ln \left (\ln \left (x^{2}-4\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x +1}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x +1}\right )^{3}-2 i \ln \left (x +1\right )\right )}{2}\right )\)	\(433\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*x^3+12*x^2-48*x-48)*exp(12*x)-2*x^4-2*x^3+8*x^2+8*x)*ln((x^2-4)/(x+1))^2-5*x^2-10*x-20)/(((x^3+x^2-4
*x-4)*exp(12*x)-x^5-x^4+4*x^3+4*x^2)*ln((x^2-4)/(x+1))^2+(5*x^3+5*x^2-20*x-20)*ln((x^2-4)/(x+1))),x,method=_RE
TURNVERBOSE)

[Out]

ln(-x^2+exp(12*x))+ln(ln(x^2-4)-1/2*I*(Pi*x^2*csgn(I/(x+1))*csgn(I*(x^2-4))*csgn(I/(x+1)*(x^2-4))-Pi*x^2*csgn(
I/(x+1))*csgn(I/(x+1)*(x^2-4))^2-Pi*x^2*csgn(I*(x^2-4))*csgn(I/(x+1)*(x^2-4))^2+Pi*x^2*csgn(I/(x+1)*(x^2-4))^3
-Pi*csgn(I/(x+1))*csgn(I*(x^2-4))*csgn(I/(x+1)*(x^2-4))*exp(12*x)+Pi*csgn(I/(x+1))*csgn(I/(x+1)*(x^2-4))^2*exp
(12*x)+Pi*csgn(I*(x^2-4))*csgn(I/(x+1)*(x^2-4))^2*exp(12*x)-Pi*csgn(I/(x+1)*(x^2-4))^3*exp(12*x)-2*I*x^2*ln(x+
1)+2*I*exp(12*x)*ln(x+1)-10*I)/(x^2-exp(12*x)))-ln(ln(x^2-4)-1/2*I*(Pi*csgn(I/(x+1))*csgn(I*(x^2-4))*csgn(I/(x
+1)*(x^2-4))-Pi*csgn(I/(x+1))*csgn(I/(x+1)*(x^2-4))^2-Pi*csgn(I*(x^2-4))*csgn(I/(x+1)*(x^2-4))^2+Pi*csgn(I/(x+
1)*(x^2-4))^3-2*I*ln(x+1)))

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maxima [B]  time = 0.44, size = 97, normalized size = 3.03 \begin {gather*} \log \left (x + e^{\left (6 \, x\right )}\right ) + \log \left (-x + e^{\left (6 \, x\right )}\right ) + \log \left (\frac {{\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (x + 2\right ) - {\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (x + 1\right ) + {\left (x^{2} - e^{\left (12 \, x\right )}\right )} \log \left (x - 2\right ) - 5}{x^{2} - e^{\left (12 \, x\right )}}\right ) - \log \left (\log \left (x + 2\right ) - \log \left (x + 1\right ) + \log \left (x - 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^3+12*x^2-48*x-48)*exp(12*x)-2*x^4-2*x^3+8*x^2+8*x)*log((x^2-4)/(x+1))^2-5*x^2-10*x-20)/(((x^
3+x^2-4*x-4)*exp(12*x)-x^5-x^4+4*x^3+4*x^2)*log((x^2-4)/(x+1))^2+(5*x^3+5*x^2-20*x-20)*log((x^2-4)/(x+1))),x,
algorithm="maxima")

[Out]

log(x + e^(6*x)) + log(-x + e^(6*x)) + log(((x^2 - e^(12*x))*log(x + 2) - (x^2 - e^(12*x))*log(x + 1) + (x^2 -
e^(12*x))*log(x - 2) - 5)/(x^2 - e^(12*x))) - log(log(x + 2) - log(x + 1) + log(x - 2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {10\,x+{\ln \left (\frac {x^2-4}{x+1}\right )}^2\,\left ({\mathrm {e}}^{12\,x}\,\left (-12\,x^3-12\,x^2+48\,x+48\right )-8\,x-8\,x^2+2\,x^3+2\,x^4\right )+5\,x^2+20}{\left ({\mathrm {e}}^{12\,x}\,\left (-x^3-x^2+4\,x+4\right )-4\,x^2-4\,x^3+x^4+x^5\right )\,{\ln \left (\frac {x^2-4}{x+1}\right )}^2+\left (-5\,x^3-5\,x^2+20\,x+20\right )\,\ln \left (\frac {x^2-4}{x+1}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + log((x^2 - 4)/(x + 1))^2*(exp(12*x)*(48*x - 12*x^2 - 12*x^3 + 48) - 8*x - 8*x^2 + 2*x^3 + 2*x^4) +
5*x^2 + 20)/(log((x^2 - 4)/(x + 1))*(20*x - 5*x^2 - 5*x^3 + 20) + log((x^2 - 4)/(x + 1))^2*(exp(12*x)*(4*x -
x^2 - x^3 + 4) - 4*x^2 - 4*x^3 + x^4 + x^5)),x)

[Out]

int((10*x + log((x^2 - 4)/(x + 1))^2*(exp(12*x)*(48*x - 12*x^2 - 12*x^3 + 48) - 8*x - 8*x^2 + 2*x^3 + 2*x^4) +
5*x^2 + 20)/(log((x^2 - 4)/(x + 1))*(20*x - 5*x^2 - 5*x^3 + 20) + log((x^2 - 4)/(x + 1))^2*(exp(12*x)*(4*x -
x^2 - x^3 + 4) - 4*x^2 - 4*x^3 + x^4 + x^5)), x)

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sympy [A]  time = 1.38, size = 31, normalized size = 0.97 \begin {gather*} \log {\left (\frac {- x^{2} \log {\left (\frac {x^{2} - 4}{x + 1} \right )} + 5}{\log {\left (\frac {x^{2} - 4}{x + 1} \right )}} + e^{12 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x**3+12*x**2-48*x-48)*exp(12*x)-2*x**4-2*x**3+8*x**2+8*x)*ln((x**2-4)/(x+1))**2-5*x**2-10*x-20
)/(((x**3+x**2-4*x-4)*exp(12*x)-x**5-x**4+4*x**3+4*x**2)*ln((x**2-4)/(x+1))**2+(5*x**3+5*x**2-20*x-20)*ln((x**
2-4)/(x+1))),x)

[Out]

log((-x**2*log((x**2 - 4)/(x + 1)) + 5)/log((x**2 - 4)/(x + 1)) + exp(12*x))

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