3.7.18 \(\int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+(-60-28 x+28 x^2-4 x^3) \log (2+2 x)+(-60-40 x+20 x^2) \log ^2(2+2 x)} \, dx\)

Optimal. Leaf size=23 \[ \log \left (\frac {-5+\frac {x}{\frac {1}{2}+\log (2+2 x)}}{-3+x}\right ) \]

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Rubi [F]  time = 4.54, 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 {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 - 11*x + 4*x^2 + (-8 - 8*x)*Log[2 + 2*x] + (-20 - 20*x)*Log[2 + 2*x]^2)/(-15 - 4*x + 9*x^2 - 2*x^3 + (-
60 - 28*x + 28*x^2 - 4*x^3)*Log[2 + 2*x] + (-60 - 40*x + 20*x^2)*Log[2 + 2*x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+11 x-4 x^2-(-8-8 x) \log (2+2 x)-(-20-20 x) \log ^2(2+2 x)}{\left (3+2 x-x^2\right ) (1+2 \log (2 (1+x))) (5-2 x+10 \log (2 (1+x)))} \, dx\\ &=\int \left (-\frac {1}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {11 x}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}-\frac {4 x^2}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {8 \log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))}+\frac {20 \log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))}\right ) \, dx\\ &=-\left (4 \int \frac {x^2}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\right )+8 \int \frac {\log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+11 \int \frac {x}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+20 \int \frac {\log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx-\int \frac {1}{(-3+x) (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\\ &=-\left (4 \int \left (\frac {1}{(1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {9}{4 (-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}-\frac {1}{4 (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}\right ) \, dx\right )+8 \int \frac {\log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+11 \int \left (\frac {3}{4 (-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}+\frac {1}{4 (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}\right ) \, dx+20 \int \frac {\log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx-\int \left (\frac {1}{4 (-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}-\frac {1}{4 (1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {1}{(-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\right )+\frac {1}{4} \int \frac {1}{(1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+\frac {11}{4} \int \frac {1}{(1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx-4 \int \frac {1}{(1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+8 \int \frac {\log (2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+\frac {33}{4} \int \frac {1}{(-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx-9 \int \frac {1}{(-3+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx+20 \int \frac {\log ^2(2+2 x)}{(3-x) (1+\log (4)+2 \log (1+x)) (5-2 x+10 \log (2 (1+x)))} \, dx+\int \frac {1}{(1+x) (1+\log (4)+2 \log (1+x)) (-5+2 x-10 \log (2 (1+x)))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.53, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(1 - 11*x + 4*x^2 + (-8 - 8*x)*Log[2 + 2*x] + (-20 - 20*x)*Log[2 + 2*x]^2)/(-15 - 4*x + 9*x^2 - 2*x^
3 + (-60 - 28*x + 28*x^2 - 4*x^3)*Log[2 + 2*x] + (-60 - 40*x + 20*x^2)*Log[2 + 2*x]^2),x]

[Out]

Integrate[(1 - 11*x + 4*x^2 + (-8 - 8*x)*Log[2 + 2*x] + (-20 - 20*x)*Log[2 + 2*x]^2)/(-15 - 4*x + 9*x^2 - 2*x^
3 + (-60 - 28*x + 28*x^2 - 4*x^3)*Log[2 + 2*x] + (-60 - 40*x + 20*x^2)*Log[2 + 2*x]^2), x]

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fricas [A]  time = 0.65, size = 34, normalized size = 1.48 \begin {gather*} -\log \left (x - 3\right ) + \log \left (-2 \, x + 10 \, \log \left (2 \, x + 2\right ) + 5\right ) - \log \left (2 \, \log \left (2 \, x + 2\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-20)*log(2*x+2)^2+(-8*x-8)*log(2*x+2)+4*x^2-11*x+1)/((20*x^2-40*x-60)*log(2*x+2)^2+(-4*x^3+28
*x^2-28*x-60)*log(2*x+2)-2*x^3+9*x^2-4*x-15),x, algorithm="fricas")

[Out]

-log(x - 3) + log(-2*x + 10*log(2*x + 2) + 5) - log(2*log(2*x + 2) + 1)

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giac [A]  time = 0.43, size = 34, normalized size = 1.48 \begin {gather*} -\log \left (x - 3\right ) + \log \left (-2 \, x + 10 \, \log \left (2 \, x + 2\right ) + 5\right ) - \log \left (2 \, \log \left (2 \, x + 2\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-20)*log(2*x+2)^2+(-8*x-8)*log(2*x+2)+4*x^2-11*x+1)/((20*x^2-40*x-60)*log(2*x+2)^2+(-4*x^3+28
*x^2-28*x-60)*log(2*x+2)-2*x^3+9*x^2-4*x-15),x, algorithm="giac")

[Out]

-log(x - 3) + log(-2*x + 10*log(2*x + 2) + 5) - log(2*log(2*x + 2) + 1)

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maple [A]  time = 0.08, size = 31, normalized size = 1.35

method	result	size
risch	\(-\ln \left (x -3\right )+\ln \left (\ln \left (2 x +2\right )-\frac {x}{5}+\frac {1}{2}\right )-\ln \left (\frac {1}{2}+\ln \left (2 x +2\right )\right )\)	\(31\)
norman	\(-\ln \left (x -3\right )-\ln \left (2 \ln \left (2 x +2\right )+1\right )+\ln \left (2 x -10 \ln \left (2 x +2\right )-5\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x-20)*ln(2*x+2)^2+(-8*x-8)*ln(2*x+2)+4*x^2-11*x+1)/((20*x^2-40*x-60)*ln(2*x+2)^2+(-4*x^3+28*x^2-28*x
-60)*ln(2*x+2)-2*x^3+9*x^2-4*x-15),x,method=_RETURNVERBOSE)

[Out]

-ln(x-3)+ln(ln(2*x+2)-1/5*x+1/2)-ln(1/2+ln(2*x+2))

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maxima [A]  time = 0.72, size = 30, normalized size = 1.30 \begin {gather*} -\log \left (x - 3\right ) + \log \left (-\frac {1}{5} \, x + \log \relax (2) + \log \left (x + 1\right ) + \frac {1}{2}\right ) - \log \left (\log \relax (2) + \log \left (x + 1\right ) + \frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-20)*log(2*x+2)^2+(-8*x-8)*log(2*x+2)+4*x^2-11*x+1)/((20*x^2-40*x-60)*log(2*x+2)^2+(-4*x^3+28
*x^2-28*x-60)*log(2*x+2)-2*x^3+9*x^2-4*x-15),x, algorithm="maxima")

[Out]

-log(x - 3) + log(-1/5*x + log(2) + log(x + 1) + 1/2) - log(log(2) + log(x + 1) + 1/2)

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mupad [B]  time = 1.46, size = 51, normalized size = 2.22 \begin {gather*} \ln \left (\frac {10\,\ln \left (2\,x+2\right )-2\,x+5}{x+1}\right )-\ln \left (\frac {\left (2\,\ln \left (2\,x+2\right )+1\right )\,\left (x-4\right )}{x+1}\right )-2\,\mathrm {atanh}\left (2\,x-7\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((11*x + log(2*x + 2)*(8*x + 8) + log(2*x + 2)^2*(20*x + 20) - 4*x^2 - 1)/(4*x + log(2*x + 2)*(28*x - 28*x^
2 + 4*x^3 + 60) + log(2*x + 2)^2*(40*x - 20*x^2 + 60) - 9*x^2 + 2*x^3 + 15),x)

[Out]

log((10*log(2*x + 2) - 2*x + 5)/(x + 1)) - log(((2*log(2*x + 2) + 1)*(x - 4))/(x + 1)) - 2*atanh(2*x - 7)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x-20)*ln(2*x+2)**2+(-8*x-8)*ln(2*x+2)+4*x**2-11*x+1)/((20*x**2-40*x-60)*ln(2*x+2)**2+(-4*x**3+
28*x**2-28*x-60)*ln(2*x+2)-2*x**3+9*x**2-4*x-15),x)

[Out]

Exception raised: PolynomialError

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