\(\int \frac {20 x-15 x^2}{(-4-10 x^2+5 x^3) \log (4+10 x^2-5 x^3)} \, dx\) [9091]

Optimal result

Integrand size = 39, antiderivative size = 26 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=3-\log \left (\frac {5}{2}\right )-\log \left (\log \left (2 \left (2-\frac {5}{2} (-2+x) x^2\right )\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1607, 6816} \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 x^3+10 x^2+4\right )\right ) \]

[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 1607

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 6816

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (4+10 x^2-5 x^3\right )\right ) \]

[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]]

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \]

[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))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=- \log {\left (\log {\left (- 5 x^{3} + 10 x^{2} + 4 \right )} \right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 14.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\ln \left (\ln \left (-5\,x^3+10\,x^2+4\right )\right ) \]

[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))