3.3.64 \(\int \frac {3 x+4 x^2+x^3+(64 x+32 x^2) \log ^2(x) \log ^3(3+4 x+x^2)+(24+32 x+8 x^2) \log (x) \log ^4(3+4 x+x^2)}{3 x+4 x^2+x^3} \, dx\)

Optimal. Leaf size=18 \[ x+4 \log ^2(x) \log ^4((1+x) (3+x)) \]

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Rubi [F]  time = 14.53, 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 {3 x+4 x^2+x^3+\left (64 x+32 x^2\right ) \log ^2(x) \log ^3\left (3+4 x+x^2\right )+\left (24+32 x+8 x^2\right ) \log (x) \log ^4\left (3+4 x+x^2\right )}{3 x+4 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3*x + 4*x^2 + x^3 + (64*x + 32*x^2)*Log[x]^2*Log[3 + 4*x + x^2]^3 + (24 + 32*x + 8*x^2)*Log[x]*Log[3 + 4*
x + x^2]^4)/(3*x + 4*x^2 + x^3),x]

[Out]

x + 16*Defer[Int][(Log[x]^2*Log[3 + 4*x + x^2]^3)/(1 + x), x] + 16*Defer[Int][(Log[x]^2*Log[3 + 4*x + x^2]^3)/
(3 + x), x] + 8*Defer[Int][(Log[x]*Log[3 + 4*x + x^2]^4)/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x+4 x^2+x^3+\left (64 x+32 x^2\right ) \log ^2(x) \log ^3\left (3+4 x+x^2\right )+\left (24+32 x+8 x^2\right ) \log (x) \log ^4\left (3+4 x+x^2\right )}{x \left (3+4 x+x^2\right )} \, dx\\ &=\int \left (1+\frac {32 (2+x) \log ^2(x) \log ^3\left (3+4 x+x^2\right )}{3+4 x+x^2}+\frac {8 \log (x) \log ^4\left (3+4 x+x^2\right )}{x}\right ) \, dx\\ &=x+8 \int \frac {\log (x) \log ^4\left (3+4 x+x^2\right )}{x} \, dx+32 \int \frac {(2+x) \log ^2(x) \log ^3\left (3+4 x+x^2\right )}{3+4 x+x^2} \, dx\\ &=x+8 \int \frac {\log (x) \log ^4\left (3+4 x+x^2\right )}{x} \, dx+32 \int \left (\frac {\log ^2(x) \log ^3\left (3+4 x+x^2\right )}{2 (1+x)}+\frac {\log ^2(x) \log ^3\left (3+4 x+x^2\right )}{2 (3+x)}\right ) \, dx\\ &=x+8 \int \frac {\log (x) \log ^4\left (3+4 x+x^2\right )}{x} \, dx+16 \int \frac {\log ^2(x) \log ^3\left (3+4 x+x^2\right )}{1+x} \, dx+16 \int \frac {\log ^2(x) \log ^3\left (3+4 x+x^2\right )}{3+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.31, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 x+4 x^2+x^3+\left (64 x+32 x^2\right ) \log ^2(x) \log ^3\left (3+4 x+x^2\right )+\left (24+32 x+8 x^2\right ) \log (x) \log ^4\left (3+4 x+x^2\right )}{3 x+4 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(3*x + 4*x^2 + x^3 + (64*x + 32*x^2)*Log[x]^2*Log[3 + 4*x + x^2]^3 + (24 + 32*x + 8*x^2)*Log[x]*Log[
3 + 4*x + x^2]^4)/(3*x + 4*x^2 + x^3),x]

[Out]

Integrate[(3*x + 4*x^2 + x^3 + (64*x + 32*x^2)*Log[x]^2*Log[3 + 4*x + x^2]^3 + (24 + 32*x + 8*x^2)*Log[x]*Log[
3 + 4*x + x^2]^4)/(3*x + 4*x^2 + x^3), x]

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fricas [A]  time = 0.54, size = 19, normalized size = 1.06 \begin {gather*} 4 \, \log \left (x^{2} + 4 \, x + 3\right )^{4} \log \relax (x)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+32*x+24)*log(x)*log(x^2+4*x+3)^4+(32*x^2+64*x)*log(x)^2*log(x^2+4*x+3)^3+x^3+4*x^2+3*x)/(x^3
+4*x^2+3*x),x, algorithm="fricas")

[Out]

4*log(x^2 + 4*x + 3)^4*log(x)^2 + x

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giac [A]  time = 2.37, size = 19, normalized size = 1.06 \begin {gather*} 4 \, \log \left (x^{2} + 4 \, x + 3\right )^{4} \log \relax (x)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+32*x+24)*log(x)*log(x^2+4*x+3)^4+(32*x^2+64*x)*log(x)^2*log(x^2+4*x+3)^3+x^3+4*x^2+3*x)/(x^3
+4*x^2+3*x),x, algorithm="giac")

[Out]

4*log(x^2 + 4*x + 3)^4*log(x)^2 + x

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maple [A]  time = 0.08, size = 20, normalized size = 1.11




method result size



risch \(4 \ln \relax (x )^{2} \ln \left (x^{2}+4 x +3\right )^{4}+x\) \(20\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2+32*x+24)*ln(x)*ln(x^2+4*x+3)^4+(32*x^2+64*x)*ln(x)^2*ln(x^2+4*x+3)^3+x^3+4*x^2+3*x)/(x^3+4*x^2+3*x
),x,method=_RETURNVERBOSE)

[Out]

4*ln(x)^2*ln(x^2+4*x+3)^4+x

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maxima [B]  time = 0.85, size = 76, normalized size = 4.22 \begin {gather*} 4 \, \log \left (x + 3\right )^{4} \log \relax (x)^{2} + 16 \, \log \left (x + 3\right )^{3} \log \left (x + 1\right ) \log \relax (x)^{2} + 24 \, \log \left (x + 3\right )^{2} \log \left (x + 1\right )^{2} \log \relax (x)^{2} + 16 \, \log \left (x + 3\right ) \log \left (x + 1\right )^{3} \log \relax (x)^{2} + 4 \, \log \left (x + 1\right )^{4} \log \relax (x)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+32*x+24)*log(x)*log(x^2+4*x+3)^4+(32*x^2+64*x)*log(x)^2*log(x^2+4*x+3)^3+x^3+4*x^2+3*x)/(x^3
+4*x^2+3*x),x, algorithm="maxima")

[Out]

4*log(x + 3)^4*log(x)^2 + 16*log(x + 3)^3*log(x + 1)*log(x)^2 + 24*log(x + 3)^2*log(x + 1)^2*log(x)^2 + 16*log
(x + 3)*log(x + 1)^3*log(x)^2 + 4*log(x + 1)^4*log(x)^2 + x

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mupad [B]  time = 0.42, size = 19, normalized size = 1.06 \begin {gather*} 4\,{\ln \left (x^2+4\,x+3\right )}^4\,{\ln \relax (x)}^2+x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 4*x^2 + x^3 + log(4*x + x^2 + 3)^4*log(x)*(32*x + 8*x^2 + 24) + log(4*x + x^2 + 3)^3*log(x)^2*(64*x
 + 32*x^2))/(3*x + 4*x^2 + x^3),x)

[Out]

x + 4*log(4*x + x^2 + 3)^4*log(x)^2

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sympy [A]  time = 0.40, size = 19, normalized size = 1.06 \begin {gather*} x + 4 \log {\relax (x )}^{2} \log {\left (x^{2} + 4 x + 3 \right )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2+32*x+24)*ln(x)*ln(x**2+4*x+3)**4+(32*x**2+64*x)*ln(x)**2*ln(x**2+4*x+3)**3+x**3+4*x**2+3*x)
/(x**3+4*x**2+3*x),x)

[Out]

x + 4*log(x)**2*log(x**2 + 4*x + 3)**4

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