\(\int \frac {x-32 x^2+(4-x+16 x^2) \log (4-x+16 x^2)}{(4-x+16 x^2) \log ^2(4-x+16 x^2)} \, dx\) [482]

Optimal result

Integrand size = 55, antiderivative size = 15 \[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\frac {x}{\log \left (4-x+16 x^2\right )} \]

[Out]

x/ln(16*x^2-x+4)

Rubi [F]

\[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx \]

[In]

Int[(x - 32*x^2 + (4 - x + 16*x^2)*Log[4 - x + 16*x^2])/((4 - x + 16*x^2)*Log[4 - x + 16*x^2]^2),x]

[Out]

-2*Defer[Int][Log[4 - x + 16*x^2]^(-2), x] + ((256*I)*Defer[Int][1/((1 + I*Sqrt[255] - 32*x)*Log[4 - x + 16*x^
2]^2), x])/Sqrt[255] - ((255 - I*Sqrt[255])*Defer[Int][1/((-1 - I*Sqrt[255] + 32*x)*Log[4 - x + 16*x^2]^2), x]
)/255 + ((256*I)*Defer[Int][1/((-1 + I*Sqrt[255] + 32*x)*Log[4 - x + 16*x^2]^2), x])/Sqrt[255] - ((255 + I*Sqr
t[255])*Defer[Int][1/((-1 + I*Sqrt[255] + 32*x)*Log[4 - x + 16*x^2]^2), x])/255 + Defer[Int][Log[4 - x + 16*x^
2]^(-1), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(1-32 x) x}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )}+\frac {1}{\log \left (4-x+16 x^2\right )}\right ) \, dx \\ & = \int \frac {(1-32 x) x}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx+\int \frac {1}{\log \left (4-x+16 x^2\right )} \, dx \\ & = \int \left (-\frac {2}{\log ^2\left (4-x+16 x^2\right )}+\frac {8-x}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )}\right ) \, dx+\int \frac {1}{\log \left (4-x+16 x^2\right )} \, dx \\ & = -\left (2 \int \frac {1}{\log ^2\left (4-x+16 x^2\right )} \, dx\right )+\int \frac {8-x}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx+\int \frac {1}{\log \left (4-x+16 x^2\right )} \, dx \\ & = -\left (2 \int \frac {1}{\log ^2\left (4-x+16 x^2\right )} \, dx\right )+\int \left (\frac {8}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )}-\frac {x}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )}\right ) \, dx+\int \frac {1}{\log \left (4-x+16 x^2\right )} \, dx \\ & = -\left (2 \int \frac {1}{\log ^2\left (4-x+16 x^2\right )} \, dx\right )+8 \int \frac {1}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx-\int \frac {x}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx+\int \frac {1}{\log \left (4-x+16 x^2\right )} \, dx \\ & = -\left (2 \int \frac {1}{\log ^2\left (4-x+16 x^2\right )} \, dx\right )+8 \int \left (\frac {32 i}{\sqrt {255} \left (1+i \sqrt {255}-32 x\right ) \log ^2\left (4-x+16 x^2\right )}+\frac {32 i}{\sqrt {255} \left (-1+i \sqrt {255}+32 x\right ) \log ^2\left (4-x+16 x^2\right )}\right ) \, dx-\int \left (\frac {1-\frac {i}{\sqrt {255}}}{\left (-1-i \sqrt {255}+32 x\right ) \log ^2\left (4-x+16 x^2\right )}+\frac {1+\frac {i}{\sqrt {255}}}{\left (-1+i \sqrt {255}+32 x\right ) \log ^2\left (4-x+16 x^2\right )}\right ) \, dx+\int \frac {1}{\log \left (4-x+16 x^2\right )} \, dx \\ & = -\left (2 \int \frac {1}{\log ^2\left (4-x+16 x^2\right )} \, dx\right )+\frac {(256 i) \int \frac {1}{\left (1+i \sqrt {255}-32 x\right ) \log ^2\left (4-x+16 x^2\right )} \, dx}{\sqrt {255}}+\frac {(256 i) \int \frac {1}{\left (-1+i \sqrt {255}+32 x\right ) \log ^2\left (4-x+16 x^2\right )} \, dx}{\sqrt {255}}-\frac {1}{255} \left (255-i \sqrt {255}\right ) \int \frac {1}{\left (-1-i \sqrt {255}+32 x\right ) \log ^2\left (4-x+16 x^2\right )} \, dx-\frac {1}{255} \left (255+i \sqrt {255}\right ) \int \frac {1}{\left (-1+i \sqrt {255}+32 x\right ) \log ^2\left (4-x+16 x^2\right )} \, dx+\int \frac {1}{\log \left (4-x+16 x^2\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\frac {x}{\log \left (4-x+16 x^2\right )} \]

[In]

Integrate[(x - 32*x^2 + (4 - x + 16*x^2)*Log[4 - x + 16*x^2])/((4 - x + 16*x^2)*Log[4 - x + 16*x^2]^2),x]

[Out]

x/Log[4 - x + 16*x^2]

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
norman	\(\frac {x}{\ln \left (16 x^{2}-x +4\right )}\)	\(16\)
risch	\(\frac {x}{\ln \left (16 x^{2}-x +4\right )}\)	\(16\)
parallelrisch	\(\frac {x}{\ln \left (16 x^{2}-x +4\right )}\)	\(16\)

[In]

int(((16*x^2-x+4)*ln(16*x^2-x+4)-32*x^2+x)/(16*x^2-x+4)/ln(16*x^2-x+4)^2,x,method=_RETURNVERBOSE)

[Out]

x/ln(16*x^2-x+4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\frac {x}{\log \left (16 \, x^{2} - x + 4\right )} \]

[In]

integrate(((16*x^2-x+4)*log(16*x^2-x+4)-32*x^2+x)/(16*x^2-x+4)/log(16*x^2-x+4)^2,x, algorithm="fricas")

[Out]

x/log(16*x^2 - x + 4)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\frac {x}{\log {\left (16 x^{2} - x + 4 \right )}} \]

[In]

integrate(((16*x**2-x+4)*ln(16*x**2-x+4)-32*x**2+x)/(16*x**2-x+4)/ln(16*x**2-x+4)**2,x)

[Out]

x/log(16*x**2 - x + 4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\frac {x}{\log \left (16 \, x^{2} - x + 4\right )} \]

[In]

integrate(((16*x^2-x+4)*log(16*x^2-x+4)-32*x^2+x)/(16*x^2-x+4)/log(16*x^2-x+4)^2,x, algorithm="maxima")

[Out]

x/log(16*x^2 - x + 4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\frac {x}{\log \left (16 \, x^{2} - x + 4\right )} \]

[In]

integrate(((16*x^2-x+4)*log(16*x^2-x+4)-32*x^2+x)/(16*x^2-x+4)/log(16*x^2-x+4)^2,x, algorithm="giac")

[Out]

x/log(16*x^2 - x + 4)

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {x-32 x^2+\left (4-x+16 x^2\right ) \log \left (4-x+16 x^2\right )}{\left (4-x+16 x^2\right ) \log ^2\left (4-x+16 x^2\right )} \, dx=\frac {x}{\ln \left (16\,x^2-x+4\right )} \]

[In]

int((x + log(16*x^2 - x + 4)*(16*x^2 - x + 4) - 32*x^2)/(log(16*x^2 - x + 4)^2*(16*x^2 - x + 4)),x)

[Out]

x/log(16*x^2 - x + 4)