3.82.64 \(\int \frac {-15-5 x^2-10 x^3+(3 x+4 x^2-x^3-x^4) \log ^2(\frac {-3-4 x+x^2+x^3}{x})}{(-3 x-4 x^2+x^3+x^4) \log ^2(\frac {-3-4 x+x^2+x^3}{x})} \, dx\)

Optimal. Leaf size=23 \[ \frac {13}{4}-x+\frac {5}{\log \left (-4-\frac {3}{x}+x+x^2\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 2, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6688, 6686} \begin {gather*} \frac {5}{\log \left (x^2+x-\frac {3}{x}-4\right )}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 - 5*x^2 - 10*x^3 + (3*x + 4*x^2 - x^3 - x^4)*Log[(-3 - 4*x + x^2 + x^3)/x]^2)/((-3*x - 4*x^2 + x^3 +
x^4)*Log[(-3 - 4*x + x^2 + x^3)/x]^2),x]

[Out]

-x + 5/Log[-4 - 3/x + x + x^2]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.87 \begin {gather*} -x+\frac {5}{\log \left (-4-\frac {3}{x}+x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 - 5*x^2 - 10*x^3 + (3*x + 4*x^2 - x^3 - x^4)*Log[(-3 - 4*x + x^2 + x^3)/x]^2)/((-3*x - 4*x^2 +
x^3 + x^4)*Log[(-3 - 4*x + x^2 + x^3)/x]^2),x]

[Out]

-x + 5/Log[-4 - 3/x + x + x^2]

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fricas [A]  time = 0.50, size = 40, normalized size = 1.74 \begin {gather*} -\frac {x \log \left (\frac {x^{3} + x^{2} - 4 \, x - 3}{x}\right ) - 5}{\log \left (\frac {x^{3} + x^{2} - 4 \, x - 3}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4-x^3+4*x^2+3*x)*log((x^3+x^2-4*x-3)/x)^2-10*x^3-5*x^2-15)/(x^4+x^3-4*x^2-3*x)/log((x^3+x^2-4*x
-3)/x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.41, size = 24, normalized size = 1.04 \begin {gather*} -x + \frac {5}{\log \left (\frac {x^{3} + x^{2} - 4 \, x - 3}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4-x^3+4*x^2+3*x)*log((x^3+x^2-4*x-3)/x)^2-10*x^3-5*x^2-15)/(x^4+x^3-4*x^2-3*x)/log((x^3+x^2-4*x
-3)/x)^2,x, algorithm="giac")

[Out]

-x + 5/log((x^3 + x^2 - 4*x - 3)/x)

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maple [A]  time = 0.05, size = 25, normalized size = 1.09

method	result	size
default	\(-x +\frac {5}{\ln \left (\frac {x^{3}+x^{2}-4 x -3}{x}\right )}\)	\(25\)
risch	\(-x +\frac {5}{\ln \left (\frac {x^{3}+x^{2}-4 x -3}{x}\right )}\)	\(25\)
norman	\(\frac {5-x \ln \left (\frac {x^{3}+x^{2}-4 x -3}{x}\right )}{\ln \left (\frac {x^{3}+x^{2}-4 x -3}{x}\right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^4-x^3+4*x^2+3*x)*ln((x^3+x^2-4*x-3)/x)^2-10*x^3-5*x^2-15)/(x^4+x^3-4*x^2-3*x)/ln((x^3+x^2-4*x-3)/x)^2
,x,method=_RETURNVERBOSE)

[Out]

-x+5/ln((x^3+x^2-4*x-3)/x)

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maxima [A]  time = 0.40, size = 42, normalized size = 1.83 \begin {gather*} -\frac {x \log \left (x^{3} + x^{2} - 4 \, x - 3\right ) - x \log \relax (x) - 5}{\log \left (x^{3} + x^{2} - 4 \, x - 3\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4-x^3+4*x^2+3*x)*log((x^3+x^2-4*x-3)/x)^2-10*x^3-5*x^2-15)/(x^4+x^3-4*x^2-3*x)/log((x^3+x^2-4*x
-3)/x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.55, size = 29, normalized size = 1.26 \begin {gather*} \frac {5}{\ln \left (-\frac {-x^3-x^2+4\,x+3}{x}\right )}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2 + 10*x^3 - log(-(4*x - x^2 - x^3 + 3)/x)^2*(3*x + 4*x^2 - x^3 - x^4) + 15)/(log(-(4*x - x^2 - x^3 +
3)/x)^2*(3*x + 4*x^2 - x^3 - x^4)),x)

[Out]

5/log(-(4*x - x^2 - x^3 + 3)/x) - x

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sympy [A]  time = 0.18, size = 17, normalized size = 0.74 \begin {gather*} - x + \frac {5}{\log {\left (\frac {x^{3} + x^{2} - 4 x - 3}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**4-x**3+4*x**2+3*x)*ln((x**3+x**2-4*x-3)/x)**2-10*x**3-5*x**2-15)/(x**4+x**3-4*x**2-3*x)/ln((x*
*3+x**2-4*x-3)/x)**2,x)

[Out]

-x + 5/log((x**3 + x**2 - 4*x - 3)/x)

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