3.91.91 \(\int \frac {20 x-15 x^2}{(-4-10 x^2+5 x^3) \log (4+10 x^2-5 x^3)} \, dx\)

Optimal. Leaf size=26 \[ 3-\log \left (\frac {5}{2}\right )-\log \left (\log \left (2 \left (2-\frac {5}{2} (-2+x) x^2\right )\right )\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 16, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1593, 6684} \begin {gather*} -\log \left (\log \left (-5 x^3+10 x^2+4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20*x - 15*x^2)/((-4 - 10*x^2 + 5*x^3)*Log[4 + 10*x^2 - 5*x^3]),x]

[Out]

-Log[Log[4 + 10*x^2 - 5*x^3]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(20-15 x) x}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx\\ &=-\log \left (\log \left (4+10 x^2-5 x^3\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 16, normalized size = 0.62 \begin {gather*} -\log \left (\log \left (4+10 x^2-5 x^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*x - 15*x^2)/((-4 - 10*x^2 + 5*x^3)*Log[4 + 10*x^2 - 5*x^3]),x]

[Out]

-Log[Log[4 + 10*x^2 - 5*x^3]]

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fricas [A]  time = 1.30, size = 16, normalized size = 0.62 \begin {gather*} -\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^2+20*x)/(5*x^3-10*x^2-4)/log(-5*x^3+10*x^2+4),x, algorithm="fricas")

[Out]

-log(log(-5*x^3 + 10*x^2 + 4))

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giac [A]  time = 0.19, size = 16, normalized size = 0.62 \begin {gather*} -\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^2+20*x)/(5*x^3-10*x^2-4)/log(-5*x^3+10*x^2+4),x, algorithm="giac")

[Out]

-log(log(-5*x^3 + 10*x^2 + 4))

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maple [A]  time = 0.03, size = 17, normalized size = 0.65




method result size



norman \(-\ln \left (\ln \left (-5 x^{3}+10 x^{2}+4\right )\right )\) \(17\)
risch \(-\ln \left (\ln \left (-5 x^{3}+10 x^{2}+4\right )\right )\) \(17\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-15*x^2+20*x)/(5*x^3-10*x^2-4)/ln(-5*x^3+10*x^2+4),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(-5*x^3+10*x^2+4))

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maxima [A]  time = 0.46, size = 16, normalized size = 0.62 \begin {gather*} -\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^2+20*x)/(5*x^3-10*x^2-4)/log(-5*x^3+10*x^2+4),x, algorithm="maxima")

[Out]

-log(log(-5*x^3 + 10*x^2 + 4))

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mupad [B]  time = 7.31, size = 16, normalized size = 0.62 \begin {gather*} -\ln \left (\ln \left (-5\,x^3+10\,x^2+4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x - 15*x^2)/(log(10*x^2 - 5*x^3 + 4)*(10*x^2 - 5*x^3 + 4)),x)

[Out]

-log(log(10*x^2 - 5*x^3 + 4))

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sympy [A]  time = 0.21, size = 15, normalized size = 0.58 \begin {gather*} - \log {\left (\log {\left (- 5 x^{3} + 10 x^{2} + 4 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x**2+20*x)/(5*x**3-10*x**2-4)/ln(-5*x**3+10*x**2+4),x)

[Out]

-log(log(-5*x**3 + 10*x**2 + 4))

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