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

Optimal. Leaf size=15 \[ \frac {x}{\log \left (4-x+16 x^2\right )} \]

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Rubi [F]  time = 0.89, 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 {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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 15, normalized size = 1.00 \begin {gather*} \frac {x}{\log \left (4-x+16 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.75, size = 15, normalized size = 1.00 \begin {gather*} \frac {x}{\log \left (16 \, x^{2} - x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.40, size = 15, normalized size = 1.00 \begin {gather*} \frac {x}{\log \left (16 \, x^{2} - x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.44, size = 16, normalized size = 1.07

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.50, size = 15, normalized size = 1.00 \begin {gather*} \frac {x}{\log \left (16 \, x^{2} - x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.19, size = 15, normalized size = 1.00 \begin {gather*} \frac {x}{\ln \left (16\,x^2-x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.13, size = 10, normalized size = 0.67 \begin {gather*} \frac {x}{\log {\left (16 x^{2} - x + 4 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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