3.74.43 \(\int \frac {(15-17 x-4 x^2) \log (\frac {9-24 x+16 x^2}{x^4})+(-120+92 x-8 x^2+(60-40 x) \log (x)) \log (10-x-5 \log (x))+(30 x-43 x^2+4 x^3+(-15 x+20 x^2) \log (x)) \log ^2(10-x-5 \log (x))}{(30 x-43 x^2+4 x^3+(-15 x+20 x^2) \log (x)) \log ^2(10-x-5 \log (x))} \, dx\)

Optimal. Leaf size=32 \[ x+\frac {\log \left (\frac {\left (-4+\frac {3}{x}\right )^2}{x^2}\right )}{\log (-x+5 (2-\log (x)))} \]

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Rubi [F]  time = 2.62, 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 (15-17 x-4 x^2\right ) \log \left (\frac {9-24 x+16 x^2}{x^4}\right )+\left (-120+92 x-8 x^2+(60-40 x) \log (x)\right ) \log (10-x-5 \log (x))+\left (30 x-43 x^2+4 x^3+\left (-15 x+20 x^2\right ) \log (x)\right ) \log ^2(10-x-5 \log (x))}{\left (30 x-43 x^2+4 x^3+\left (-15 x+20 x^2\right ) \log (x)\right ) \log ^2(10-x-5 \log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((15 - 17*x - 4*x^2)*Log[(9 - 24*x + 16*x^2)/x^4] + (-120 + 92*x - 8*x^2 + (60 - 40*x)*Log[x])*Log[10 - x
- 5*Log[x]] + (30*x - 43*x^2 + 4*x^3 + (-15*x + 20*x^2)*Log[x])*Log[10 - x - 5*Log[x]]^2)/((30*x - 43*x^2 + 4*
x^3 + (-15*x + 20*x^2)*Log[x])*Log[10 - x - 5*Log[x]]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (15-17 x-4 x^2\right ) \log \left (\frac {(-3+4 x)^2}{x^4}\right )+(-10+x+5 \log (x)) \log (10-x-5 \log (x)) (12-8 x+x (-3+4 x) \log (10-x-5 \log (x)))}{(3-4 x) x (10-x-5 \log (x)) \log ^2(10-x-5 \log (x))} \, dx\\ &=\int \left (1-\frac {(5+x) \log \left (\frac {(-3+4 x)^2}{x^4}\right )}{x (-10+x+5 \log (x)) \log ^2(10-x-5 \log (x))}-\frac {4 (-3+2 x)}{x (-3+4 x) \log (10-x-5 \log (x))}\right ) \, dx\\ &=x-4 \int \frac {-3+2 x}{x (-3+4 x) \log (10-x-5 \log (x))} \, dx-\int \frac {(5+x) \log \left (\frac {(-3+4 x)^2}{x^4}\right )}{x (-10+x+5 \log (x)) \log ^2(10-x-5 \log (x))} \, dx\\ &=x-4 \int \left (\frac {1}{x \log (10-x-5 \log (x))}-\frac {2}{(-3+4 x) \log (10-x-5 \log (x))}\right ) \, dx-\int \left (\frac {\log \left (\frac {(-3+4 x)^2}{x^4}\right )}{(-10+x+5 \log (x)) \log ^2(10-x-5 \log (x))}+\frac {5 \log \left (\frac {(-3+4 x)^2}{x^4}\right )}{x (-10+x+5 \log (x)) \log ^2(10-x-5 \log (x))}\right ) \, dx\\ &=x-4 \int \frac {1}{x \log (10-x-5 \log (x))} \, dx-5 \int \frac {\log \left (\frac {(-3+4 x)^2}{x^4}\right )}{x (-10+x+5 \log (x)) \log ^2(10-x-5 \log (x))} \, dx+8 \int \frac {1}{(-3+4 x) \log (10-x-5 \log (x))} \, dx-\int \frac {\log \left (\frac {(-3+4 x)^2}{x^4}\right )}{(-10+x+5 \log (x)) \log ^2(10-x-5 \log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 27, normalized size = 0.84 \begin {gather*} x+\frac {\log \left (\frac {(-3+4 x)^2}{x^4}\right )}{\log (10-x-5 \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((15 - 17*x - 4*x^2)*Log[(9 - 24*x + 16*x^2)/x^4] + (-120 + 92*x - 8*x^2 + (60 - 40*x)*Log[x])*Log[1
0 - x - 5*Log[x]] + (30*x - 43*x^2 + 4*x^3 + (-15*x + 20*x^2)*Log[x])*Log[10 - x - 5*Log[x]]^2)/((30*x - 43*x^
2 + 4*x^3 + (-15*x + 20*x^2)*Log[x])*Log[10 - x - 5*Log[x]]^2),x]

[Out]

x + Log[(-3 + 4*x)^2/x^4]/Log[10 - x - 5*Log[x]]

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fricas [A]  time = 0.81, size = 41, normalized size = 1.28 \begin {gather*} \frac {x \log \left (-x - 5 \, \log \relax (x) + 10\right ) + \log \left (\frac {16 \, x^{2} - 24 \, x + 9}{x^{4}}\right )}{\log \left (-x - 5 \, \log \relax (x) + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2-15*x)*log(x)+4*x^3-43*x^2+30*x)*log(-5*log(x)+10-x)^2+((-40*x+60)*log(x)-8*x^2+92*x-120)*l
og(-5*log(x)+10-x)+(-4*x^2-17*x+15)*log((16*x^2-24*x+9)/x^4))/((20*x^2-15*x)*log(x)+4*x^3-43*x^2+30*x)/log(-5*
log(x)+10-x)^2,x, algorithm="fricas")

[Out]

(x*log(-x - 5*log(x) + 10) + log((16*x^2 - 24*x + 9)/x^4))/log(-x - 5*log(x) + 10)

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giac [A]  time = 0.26, size = 42, normalized size = 1.31 \begin {gather*} x + \frac {\log \left (16 \, x^{2} - 24 \, x + 9\right )}{\log \left (-x - 5 \, \log \relax (x) + 10\right )} - \frac {4 \, \log \relax (x)}{\log \left (-x - 5 \, \log \relax (x) + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2-15*x)*log(x)+4*x^3-43*x^2+30*x)*log(-5*log(x)+10-x)^2+((-40*x+60)*log(x)-8*x^2+92*x-120)*l
og(-5*log(x)+10-x)+(-4*x^2-17*x+15)*log((16*x^2-24*x+9)/x^4))/((20*x^2-15*x)*log(x)+4*x^3-43*x^2+30*x)/log(-5*
log(x)+10-x)^2,x, algorithm="giac")

[Out]

x + log(16*x^2 - 24*x + 9)/log(-x - 5*log(x) + 10) - 4*log(x)/log(-x - 5*log(x) + 10)

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maple [C]  time = 0.14, size = 390, normalized size = 12.19

method	result	size
risch	\(x +\frac {8 \ln \relax (2)-8 \ln \relax (x )-i \pi \mathrm {csgn}\left (i \left (x -\frac {3}{4}\right )\right )^{2} \mathrm {csgn}\left (i \left (x -\frac {3}{4}\right )^{2}\right )-i \pi \mathrm {csgn}\left (\frac {i \left (x -\frac {3}{4}\right )^{2}}{x^{4}}\right )^{3}+i \pi \mathrm {csgn}\left (i x^{4}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-i \pi \mathrm {csgn}\left (i \left (x -\frac {3}{4}\right )^{2}\right )^{3}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{x^{4}}\right ) \mathrm {csgn}\left (i \left (x -\frac {3}{4}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {3}{4}\right )^{2}}{x^{4}}\right )+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+4 \ln \left (x -\frac {3}{4}\right )+i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )+2 i \pi \,\mathrm {csgn}\left (i \left (x -\frac {3}{4}\right )\right ) \mathrm {csgn}\left (i \left (x -\frac {3}{4}\right )^{2}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (x -\frac {3}{4}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {3}{4}\right )^{2}}{x^{4}}\right )^{2}-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \,\mathrm {csgn}\left (\frac {i}{x^{4}}\right ) \mathrm {csgn}\left (\frac {i \left (x -\frac {3}{4}\right )^{2}}{x^{4}}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}}{2 \ln \left (-5 \ln \relax (x )+10-x \right )}\)	\(390\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((20*x^2-15*x)*ln(x)+4*x^3-43*x^2+30*x)*ln(-5*ln(x)+10-x)^2+((-40*x+60)*ln(x)-8*x^2+92*x-120)*ln(-5*ln(x)
+10-x)+(-4*x^2-17*x+15)*ln((16*x^2-24*x+9)/x^4))/((20*x^2-15*x)*ln(x)+4*x^3-43*x^2+30*x)/ln(-5*ln(x)+10-x)^2,x
,method=_RETURNVERBOSE)

[Out]

x+1/2*(8*ln(2)-8*ln(x)-I*Pi*csgn(I*(x-3/4))^2*csgn(I*(x-3/4)^2)-I*Pi*csgn(I/x^4*(x-3/4)^2)^3+I*Pi*csgn(I*x^4)^
3+I*Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-I*Pi*csgn(I*(x-3/4)^2)^3-I*Pi*csgn(I*x)*csgn(I*x^4)^2-I*Pi*csgn(I/x^4
)*csgn(I*(x-3/4)^2)*csgn(I/x^4*(x-3/4)^2)+I*Pi*csgn(I*x^2)*csgn(I*x)^2+I*Pi*csgn(I*x^2)^3-I*Pi*csgn(I*x^2)*csg
n(I*x^3)^2+4*ln(x-3/4)+I*Pi*csgn(I*x^3)^3+I*Pi*csgn(I*x)*csgn(I*x^3)*csgn(I*x^4)+2*I*Pi*csgn(I*(x-3/4))*csgn(I
*(x-3/4)^2)^2+I*Pi*csgn(I*(x-3/4)^2)*csgn(I/x^4*(x-3/4)^2)^2-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+I*Pi*csgn(I/x^4)*c
sgn(I/x^4*(x-3/4)^2)^2-I*Pi*csgn(I*x^3)*csgn(I*x^4)^2-I*Pi*csgn(I*x)*csgn(I*x^3)^2)/ln(-5*ln(x)+10-x)

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maxima [A]  time = 0.42, size = 38, normalized size = 1.19 \begin {gather*} \frac {x \log \left (-x - 5 \, \log \relax (x) + 10\right ) + 2 \, \log \left (4 \, x - 3\right ) - 4 \, \log \relax (x)}{\log \left (-x - 5 \, \log \relax (x) + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2-15*x)*log(x)+4*x^3-43*x^2+30*x)*log(-5*log(x)+10-x)^2+((-40*x+60)*log(x)-8*x^2+92*x-120)*l
og(-5*log(x)+10-x)+(-4*x^2-17*x+15)*log((16*x^2-24*x+9)/x^4))/((20*x^2-15*x)*log(x)+4*x^3-43*x^2+30*x)/log(-5*
log(x)+10-x)^2,x, algorithm="maxima")

[Out]

(x*log(-x - 5*log(x) + 10) + 2*log(4*x - 3) - 4*log(x))/log(-x - 5*log(x) + 10)

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mupad [B]  time = 5.05, size = 30, normalized size = 0.94 \begin {gather*} x+\frac {\ln \left (\frac {16\,x^2-24\,x+9}{x^4}\right )}{\ln \left (10-5\,\ln \relax (x)-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(10 - 5*log(x) - x)*(log(x)*(40*x - 60) - 92*x + 8*x^2 + 120) + log((16*x^2 - 24*x + 9)/x^4)*(17*x +
4*x^2 - 15) - log(10 - 5*log(x) - x)^2*(30*x - log(x)*(15*x - 20*x^2) - 43*x^2 + 4*x^3))/(log(10 - 5*log(x) -
x)^2*(30*x - log(x)*(15*x - 20*x^2) - 43*x^2 + 4*x^3)),x)

[Out]

x + log((16*x^2 - 24*x + 9)/x^4)/log(10 - 5*log(x) - x)

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sympy [A]  time = 0.61, size = 26, normalized size = 0.81 \begin {gather*} x + \frac {\log {\left (\frac {16 x^{2} - 24 x + 9}{x^{4}} \right )}}{\log {\left (- x - 5 \log {\relax (x )} + 10 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x**2-15*x)*ln(x)+4*x**3-43*x**2+30*x)*ln(-5*ln(x)+10-x)**2+((-40*x+60)*ln(x)-8*x**2+92*x-120)*
ln(-5*ln(x)+10-x)+(-4*x**2-17*x+15)*ln((16*x**2-24*x+9)/x**4))/((20*x**2-15*x)*ln(x)+4*x**3-43*x**2+30*x)/ln(-
5*ln(x)+10-x)**2,x)

[Out]

x + log((16*x**2 - 24*x + 9)/x**4)/log(-x - 5*log(x) + 10)

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