\(\int \frac {(-8 x+12 x^2+72 x^3+32 x^4) \log (x)+(64-216 x-148 x^2-44 x^3-16 x^4+(224 x+284 x^2+60 x^3+32 x^4) \log (x)) \log (8+x+x^2)}{(8 x+x^2+x^3) \log ^2(x)} \, dx\) [7508]

Optimal result

Integrand size = 90, antiderivative size = 21 \[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 (2+x) (-1+4 x) \log \left (8+x+x^2\right )}{\log (x)} \]

[Out]

4*(2+x)/ln(x)*ln(x^2+x+8)*(-1+4*x)

Rubi [F]

\[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx \]

[In]

Int[((-8*x + 12*x^2 + 72*x^3 + 32*x^4)*Log[x] + (64 - 216*x - 148*x^2 - 44*x^3 - 16*x^4 + (224*x + 284*x^2 + 6
0*x^3 + 32*x^4)*Log[x])*Log[8 + x + x^2])/((8*x + x^2 + x^3)*Log[x]^2),x]

[Out]

28*Log[8 + x + x^2]*LogIntegral[x] + 4*Defer[Int][((2 + x)*(1 + 2*x)*(-1 + 4*x))/((8 + x + x^2)*Log[x]), x] -
28*Defer[Int][Log[8 + x + x^2]/Log[x]^2, x] + 8*Defer[Int][Log[8 + x + x^2]/(x*Log[x]^2), x] - 16*Defer[Int][(
x*Log[8 + x + x^2])/Log[x]^2, x] + 32*Defer[Int][(x*Log[8 + x + x^2])/Log[x], x] - ((56*I)*Defer[Int][LogInteg
ral[x]/(-1 + I*Sqrt[31] - 2*x), x])/Sqrt[31] - (56*(31 + I*Sqrt[31])*Defer[Int][LogIntegral[x]/(1 - I*Sqrt[31]
+ 2*x), x])/31 - ((56*I)*Defer[Int][LogIntegral[x]/(1 + I*Sqrt[31] + 2*x), x])/Sqrt[31] - (56*(31 - I*Sqrt[31
])*Defer[Int][LogIntegral[x]/(1 + I*Sqrt[31] + 2*x), x])/31

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{x \left (8+x+x^2\right ) \log ^2(x)} \, dx \\ & = \int \left (\frac {4 (2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)}+\frac {4 \left (2-7 x-4 x^2+7 x \log (x)+8 x^2 \log (x)\right ) \log \left (8+x+x^2\right )}{x \log ^2(x)}\right ) \, dx \\ & = 4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+4 \int \frac {\left (2-7 x-4 x^2+7 x \log (x)+8 x^2 \log (x)\right ) \log \left (8+x+x^2\right )}{x \log ^2(x)} \, dx \\ & = 4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+4 \int \left (-\frac {7 \log \left (8+x+x^2\right )}{\log ^2(x)}+\frac {2 \log \left (8+x+x^2\right )}{x \log ^2(x)}-\frac {4 x \log \left (8+x+x^2\right )}{\log ^2(x)}+\frac {7 \log \left (8+x+x^2\right )}{\log (x)}+\frac {8 x \log \left (8+x+x^2\right )}{\log (x)}\right ) \, dx \\ & = 4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+8 \int \frac {\log \left (8+x+x^2\right )}{x \log ^2(x)} \, dx-16 \int \frac {x \log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {\log \left (8+x+x^2\right )}{\log ^2(x)} \, dx+28 \int \frac {\log \left (8+x+x^2\right )}{\log (x)} \, dx+32 \int \frac {x \log \left (8+x+x^2\right )}{\log (x)} \, dx \\ & = 28 \log \left (8+x+x^2\right ) \operatorname {LogIntegral}(x)+4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+8 \int \frac {\log \left (8+x+x^2\right )}{x \log ^2(x)} \, dx-16 \int \frac {x \log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {\log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {(1+2 x) \operatorname {LogIntegral}(x)}{8+x+x^2} \, dx+32 \int \frac {x \log \left (8+x+x^2\right )}{\log (x)} \, dx \\ & = 28 \log \left (8+x+x^2\right ) \operatorname {LogIntegral}(x)+4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+8 \int \frac {\log \left (8+x+x^2\right )}{x \log ^2(x)} \, dx-16 \int \frac {x \log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {\log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \left (\frac {\operatorname {LogIntegral}(x)}{8+x+x^2}+\frac {2 x \operatorname {LogIntegral}(x)}{8+x+x^2}\right ) \, dx+32 \int \frac {x \log \left (8+x+x^2\right )}{\log (x)} \, dx \\ & = 28 \log \left (8+x+x^2\right ) \operatorname {LogIntegral}(x)+4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+8 \int \frac {\log \left (8+x+x^2\right )}{x \log ^2(x)} \, dx-16 \int \frac {x \log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {\log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {\operatorname {LogIntegral}(x)}{8+x+x^2} \, dx+32 \int \frac {x \log \left (8+x+x^2\right )}{\log (x)} \, dx-56 \int \frac {x \operatorname {LogIntegral}(x)}{8+x+x^2} \, dx \\ & = 28 \log \left (8+x+x^2\right ) \operatorname {LogIntegral}(x)+4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+8 \int \frac {\log \left (8+x+x^2\right )}{x \log ^2(x)} \, dx-16 \int \frac {x \log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {\log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \left (\frac {2 i \operatorname {LogIntegral}(x)}{\sqrt {31} \left (-1+i \sqrt {31}-2 x\right )}+\frac {2 i \operatorname {LogIntegral}(x)}{\sqrt {31} \left (1+i \sqrt {31}+2 x\right )}\right ) \, dx+32 \int \frac {x \log \left (8+x+x^2\right )}{\log (x)} \, dx-56 \int \left (\frac {\left (1+\frac {i}{\sqrt {31}}\right ) \operatorname {LogIntegral}(x)}{1-i \sqrt {31}+2 x}+\frac {\left (1-\frac {i}{\sqrt {31}}\right ) \operatorname {LogIntegral}(x)}{1+i \sqrt {31}+2 x}\right ) \, dx \\ & = 28 \log \left (8+x+x^2\right ) \operatorname {LogIntegral}(x)+4 \int \frac {(2+x) (1+2 x) (-1+4 x)}{\left (8+x+x^2\right ) \log (x)} \, dx+8 \int \frac {\log \left (8+x+x^2\right )}{x \log ^2(x)} \, dx-16 \int \frac {x \log \left (8+x+x^2\right )}{\log ^2(x)} \, dx-28 \int \frac {\log \left (8+x+x^2\right )}{\log ^2(x)} \, dx+32 \int \frac {x \log \left (8+x+x^2\right )}{\log (x)} \, dx-\frac {(56 i) \int \frac {\operatorname {LogIntegral}(x)}{-1+i \sqrt {31}-2 x} \, dx}{\sqrt {31}}-\frac {(56 i) \int \frac {\operatorname {LogIntegral}(x)}{1+i \sqrt {31}+2 x} \, dx}{\sqrt {31}}-\frac {1}{31} \left (56 \left (31-i \sqrt {31}\right )\right ) \int \frac {\operatorname {LogIntegral}(x)}{1+i \sqrt {31}+2 x} \, dx-\frac {1}{31} \left (56 \left (31+i \sqrt {31}\right )\right ) \int \frac {\operatorname {LogIntegral}(x)}{1-i \sqrt {31}+2 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \left (-2+7 x+4 x^2\right ) \log \left (8+x+x^2\right )}{\log (x)} \]

[In]

Integrate[((-8*x + 12*x^2 + 72*x^3 + 32*x^4)*Log[x] + (64 - 216*x - 148*x^2 - 44*x^3 - 16*x^4 + (224*x + 284*x
^2 + 60*x^3 + 32*x^4)*Log[x])*Log[8 + x + x^2])/((8*x + x^2 + x^3)*Log[x]^2),x]

[Out]

(4*(-2 + 7*x + 4*x^2)*Log[8 + x + x^2])/Log[x]

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14

method	result	size
risch	\(\frac {4 \left (4 x^{2}+7 x -2\right ) \ln \left (x^{2}+x +8\right )}{\ln \left (x \right )}\)	\(24\)
parallelrisch	\(\frac {32 \ln \left (x^{2}+x +8\right ) x^{2}+56 \ln \left (x^{2}+x +8\right ) x -16 \ln \left (x^{2}+x +8\right )}{2 \ln \left (x \right )}\)	\(39\)

[In]

int((((32*x^4+60*x^3+284*x^2+224*x)*ln(x)-16*x^4-44*x^3-148*x^2-216*x+64)*ln(x^2+x+8)+(32*x^4+72*x^3+12*x^2-8*
x)*ln(x))/(x^3+x^2+8*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

4*(4*x^2+7*x-2)/ln(x)*ln(x^2+x+8)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \, {\left (4 \, x^{2} + 7 \, x - 2\right )} \log \left (x^{2} + x + 8\right )}{\log \left (x\right )} \]

[In]

integrate((((32*x^4+60*x^3+284*x^2+224*x)*log(x)-16*x^4-44*x^3-148*x^2-216*x+64)*log(x^2+x+8)+(32*x^4+72*x^3+1
2*x^2-8*x)*log(x))/(x^3+x^2+8*x)/log(x)^2,x, algorithm="fricas")

[Out]

4*(4*x^2 + 7*x - 2)*log(x^2 + x + 8)/log(x)

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((((32*x**4+60*x**3+284*x**2+224*x)*ln(x)-16*x**4-44*x**3-148*x**2-216*x+64)*ln(x**2+x+8)+(32*x**4+72
*x**3+12*x**2-8*x)*ln(x))/(x**3+x**2+8*x)/ln(x)**2,x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \, {\left (4 \, x^{2} + 7 \, x - 2\right )} \log \left (x^{2} + x + 8\right )}{\log \left (x\right )} \]

[In]

integrate((((32*x^4+60*x^3+284*x^2+224*x)*log(x)-16*x^4-44*x^3-148*x^2-216*x+64)*log(x^2+x+8)+(32*x^4+72*x^3+1
2*x^2-8*x)*log(x))/(x^3+x^2+8*x)/log(x)^2,x, algorithm="maxima")

[Out]

4*(4*x^2 + 7*x - 2)*log(x^2 + x + 8)/log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \, {\left (4 \, x^{2} + 7 \, x - 2\right )} \log \left (x^{2} + x + 8\right )}{\log \left (x\right )} \]

[In]

integrate((((32*x^4+60*x^3+284*x^2+224*x)*log(x)-16*x^4-44*x^3-148*x^2-216*x+64)*log(x^2+x+8)+(32*x^4+72*x^3+1
2*x^2-8*x)*log(x))/(x^3+x^2+8*x)/log(x)^2,x, algorithm="giac")

[Out]

4*(4*x^2 + 7*x - 2)*log(x^2 + x + 8)/log(x)

Mupad [B] (verification not implemented)

Time = 13.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-8 x+12 x^2+72 x^3+32 x^4\right ) \log (x)+\left (64-216 x-148 x^2-44 x^3-16 x^4+\left (224 x+284 x^2+60 x^3+32 x^4\right ) \log (x)\right ) \log \left (8+x+x^2\right )}{\left (8 x+x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4\,\ln \left (x^2+x+8\right )\,\left (4\,x^2+7\,x-2\right )}{\ln \left (x\right )} \]

[In]

int((log(x)*(12*x^2 - 8*x + 72*x^3 + 32*x^4) - log(x + x^2 + 8)*(216*x - log(x)*(224*x + 284*x^2 + 60*x^3 + 32
*x^4) + 148*x^2 + 44*x^3 + 16*x^4 - 64))/(log(x)^2*(8*x + x^2 + x^3)),x)

[Out]

(4*log(x + x^2 + 8)*(7*x + 4*x^2 - 2))/log(x)