3.39.86 \(\int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{((-3-x+2 x^2) \log ^2(16)+(-6-2 x+4 x^2) \log (16) \log (1+x)+(-3-x+2 x^2) \log ^2(1+x)) \log ^2(-3+2 x)} \, dx\)

Optimal. Leaf size=19 \[ \frac {3}{(\log (16)+\log (1+x)) \log (-3+2 x)} \]

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Rubi [F]  time = 3.39, 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 {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-6 - 6*x)*Log[16] + (-6 - 6*x)*Log[1 + x] + (9 - 6*x)*Log[-3 + 2*x])/(((-3 - x + 2*x^2)*Log[16]^2 + (-6
- 2*x + 4*x^2)*Log[16]*Log[1 + x] + (-3 - x + 2*x^2)*Log[1 + x]^2)*Log[-3 + 2*x]^2),x]

[Out]

3*Log[256]*Defer[Int][1/((3 - 2*x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x] + (3*Log[256]*Defer[Int][1/((-1 - x)*
Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x])/5 + (3*Log[256]*Defer[Int][1/((1 + x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2)
, x])/5 + 6*Defer[Int][Log[1 + x]/((3 - 2*x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x] + (6*Defer[Int][Log[1 + x]/
((-1 - x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x])/5 + (6*Defer[Int][Log[1 + x]/((1 + x)*Log[-3 + 2*x]^2*Log[16
+ 16*x]^2), x])/5 + (18*Defer[Int][1/((3 - 2*x)*Log[-3 + 2*x]*Log[16 + 16*x]^2), x])/5 + 3*Defer[Int][1/((-1 -
x)*Log[-3 + 2*x]*Log[16 + 16*x]^2), x] + (18*Defer[Int][1/((-3 + 2*x)*Log[-3 + 2*x]*Log[16 + 16*x]^2), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 ((1+x) \log (256)+2 (1+x) \log (1+x)+(-3+2 x) \log (-3+2 x))}{\left (3+x-2 x^2\right ) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx\\ &=3 \int \frac {(1+x) \log (256)+2 (1+x) \log (1+x)+(-3+2 x) \log (-3+2 x)}{\left (3+x-2 x^2\right ) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx\\ &=3 \int \left (\frac {2 (\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x))}{5 (3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x)}{5 (1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx\\ &=\frac {3}{5} \int \frac {\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx\\ &=\frac {3}{5} \int \left (\frac {\log (256)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {x \log (256)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 \log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 x \log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {3}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)}+\frac {2 x}{(1+x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {6}{5} \int \left (\frac {\log (256)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {x \log (256)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 \log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 x \log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 x}{(3-2 x) \log (-3+2 x) \log ^2(16+16 x)}+\frac {3}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx\\ &=\frac {6}{5} \int \frac {\log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {x \log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {x}{(1+x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {9}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {x \log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {x}{(3-2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {x}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \frac {x}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx\\ &=\frac {6}{5} \int \left (\frac {\log (1+x)}{\log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {\log (1+x)}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {6}{5} \int \left (\frac {1}{\log (-3+2 x) \log ^2(16+16 x)}+\frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {6}{5} \int \frac {\log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {9}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \left (-\frac {\log (1+x)}{2 \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {3 \log (1+x)}{2 (3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {12}{5} \int \left (-\frac {1}{2 \log (-3+2 x) \log ^2(16+16 x)}+\frac {3}{2 (3-2 x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {12}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \left (\frac {1}{\log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {1}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \left (-\frac {1}{2 \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {3}{2 (3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {1}{5} (6 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx\\ &=\frac {6}{5} \int \frac {\log (1+x)}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {\log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {9}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(3-2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (9 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 29, normalized size = 1.53 \begin {gather*} \frac {3 (\log (256)+2 \log (1+x))}{2 \log ^2(16 (1+x)) \log (-3+2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-6 - 6*x)*Log[16] + (-6 - 6*x)*Log[1 + x] + (9 - 6*x)*Log[-3 + 2*x])/(((-3 - x + 2*x^2)*Log[16]^2
+ (-6 - 2*x + 4*x^2)*Log[16]*Log[1 + x] + (-3 - x + 2*x^2)*Log[1 + x]^2)*Log[-3 + 2*x]^2),x]

[Out]

(3*(Log[256] + 2*Log[1 + x]))/(2*Log[16*(1 + x)]^2*Log[-3 + 2*x])

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fricas [A]  time = 0.65, size = 21, normalized size = 1.11 \begin {gather*} \frac {3}{{\left (4 \, \log \relax (2) + \log \left (x + 1\right )\right )} \log \left (2 \, x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+9)*log(2*x-3)+(-6*x-6)*log(x+1)+4*(-6*x-6)*log(2))/((2*x^2-x-3)*log(x+1)^2+4*(4*x^2-2*x-6)*lo
g(2)*log(x+1)+16*(2*x^2-x-3)*log(2)^2)/log(2*x-3)^2,x, algorithm="fricas")

[Out]

3/((4*log(2) + log(x + 1))*log(2*x - 3))

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giac [A]  time = 0.21, size = 26, normalized size = 1.37 \begin {gather*} \frac {3}{4 \, \log \relax (2) \log \left (2 \, x - 3\right ) + \log \left (2 \, x - 3\right ) \log \left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+9)*log(2*x-3)+(-6*x-6)*log(x+1)+4*(-6*x-6)*log(2))/((2*x^2-x-3)*log(x+1)^2+4*(4*x^2-2*x-6)*lo
g(2)*log(x+1)+16*(2*x^2-x-3)*log(2)^2)/log(2*x-3)^2,x, algorithm="giac")

[Out]

3/(4*log(2)*log(2*x - 3) + log(2*x - 3)*log(x + 1))

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maple [A]  time = 0.25, size = 22, normalized size = 1.16

method	result	size
risch	\(\frac {3}{\left (\ln \left (x +1\right )+4 \ln \relax (2)\right ) \ln \left (2 x -3\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x+9)*ln(2*x-3)+(-6*x-6)*ln(x+1)+4*(-6*x-6)*ln(2))/((2*x^2-x-3)*ln(x+1)^2+4*(4*x^2-2*x-6)*ln(2)*ln(x+1
)+16*(2*x^2-x-3)*ln(2)^2)/ln(2*x-3)^2,x,method=_RETURNVERBOSE)

[Out]

3/(ln(x+1)+4*ln(2))/ln(2*x-3)

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maxima [A]  time = 0.49, size = 21, normalized size = 1.11 \begin {gather*} \frac {3}{{\left (4 \, \log \relax (2) + \log \left (x + 1\right )\right )} \log \left (2 \, x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+9)*log(2*x-3)+(-6*x-6)*log(x+1)+4*(-6*x-6)*log(2))/((2*x^2-x-3)*log(x+1)^2+4*(4*x^2-2*x-6)*lo
g(2)*log(x+1)+16*(2*x^2-x-3)*log(2)^2)/log(2*x-3)^2,x, algorithm="maxima")

[Out]

3/((4*log(2) + log(x + 1))*log(2*x - 3))

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mupad [B]  time = 2.69, size = 194, normalized size = 10.21 \begin {gather*} \frac {15}{4\,\left (x+1\right )}-\frac {\frac {15\,\left (4\,\ln \relax (2)+1\right )}{4\,\left (x+1\right )}+\frac {15\,\ln \left (x+1\right )}{4\,\left (x+1\right )}}{\ln \left (x+1\right )+4\,\ln \relax (2)}+\frac {\frac {3}{\ln \left (x+1\right )+4\,\ln \relax (2)}+\frac {3\,\ln \left (2\,x-3\right )\,\left (2\,x-3\right )}{2\,\left (8\,\ln \left (x+1\right )\,\ln \relax (2)+x\,{\ln \left (x+1\right )}^2+16\,x\,{\ln \relax (2)}^2+{\ln \left (x+1\right )}^2+16\,{\ln \relax (2)}^2+8\,x\,\ln \left (x+1\right )\,\ln \relax (2)\right )}}{\ln \left (2\,x-3\right )}+\frac {\frac {3\,\left (10\,\ln \relax (2)-2\,x+3\right )}{2\,\left (x+1\right )}+\frac {15\,\ln \left (x+1\right )}{4\,\left (x+1\right )}}{{\ln \left (x+1\right )}^2+8\,\ln \relax (2)\,\ln \left (x+1\right )+16\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(2)*(6*x + 6) + log(2*x - 3)*(6*x - 9) + log(x + 1)*(6*x + 6))/(log(2*x - 3)^2*(log(x + 1)^2*(x - 2*
x^2 + 3) + 16*log(2)^2*(x - 2*x^2 + 3) + 4*log(x + 1)*log(2)*(2*x - 4*x^2 + 6))),x)

[Out]

15/(4*(x + 1)) - ((15*(4*log(2) + 1))/(4*(x + 1)) + (15*log(x + 1))/(4*(x + 1)))/(log(x + 1) + 4*log(2)) + (3/
(log(x + 1) + 4*log(2)) + (3*log(2*x - 3)*(2*x - 3))/(2*(8*log(x + 1)*log(2) + x*log(x + 1)^2 + 16*x*log(2)^2
+ log(x + 1)^2 + 16*log(2)^2 + 8*x*log(x + 1)*log(2))))/log(2*x - 3) + ((3*(10*log(2) - 2*x + 3))/(2*(x + 1))
+ (15*log(x + 1))/(4*(x + 1)))/(8*log(x + 1)*log(2) + log(x + 1)^2 + 16*log(2)^2)

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sympy [A]  time = 0.36, size = 17, normalized size = 0.89 \begin {gather*} \frac {3}{\left (\log {\left (x + 1 \right )} + 4 \log {\relax (2 )}\right ) \log {\left (2 x - 3 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+9)*ln(2*x-3)+(-6*x-6)*ln(x+1)+4*(-6*x-6)*ln(2))/((2*x**2-x-3)*ln(x+1)**2+4*(4*x**2-2*x-6)*ln(
2)*ln(x+1)+16*(2*x**2-x-3)*ln(2)**2)/ln(2*x-3)**2,x)

[Out]

3/((log(x + 1) + 4*log(2))*log(2*x - 3))

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