\(\int \frac {8+30 x^3}{(-10 x-40 x^3+75 x^4) \log (\frac {-2-8 x^2+15 x^3}{2 x^2})} \, dx\) [1999]

Optimal result

Integrand size = 46, antiderivative size = 20 \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {1}{5} \log \left (\log ^2\left (-4-\frac {1}{x^2}+\frac {15 x}{2}\right )\right ) \]

[Out]

1/5*ln(ln(15/2*x-1/x^2-4)^2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1608, 6816} \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {2}{5} \log \left (\log \left (-\frac {-15 x^3+8 x^2+2}{2 x^2}\right )\right ) \]

[In]

Int[(8 + 30*x^3)/((-10*x - 40*x^3 + 75*x^4)*Log[(-2 - 8*x^2 + 15*x^3)/(2*x^2)]),x]

[Out]

(2*Log[Log[-1/2*(2 + 8*x^2 - 15*x^3)/x^2]])/5

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - 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 {8+30 x^3}{x \left (-10-40 x^2+75 x^3\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx \\ & = \frac {2}{5} \log \left (\log \left (-\frac {2+8 x^2-15 x^3}{2 x^2}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {2}{5} \log \left (\log \left (-4-\frac {1}{x^2}+\frac {15 x}{2}\right )\right ) \]

[In]

Integrate[(8 + 30*x^3)/((-10*x - 40*x^3 + 75*x^4)*Log[(-2 - 8*x^2 + 15*x^3)/(2*x^2)]),x]

[Out]

(2*Log[Log[-4 - x^(-2) + (15*x)/2]])/5

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
norman	\(\frac {2 \ln \left (\ln \left (\frac {15 x^{3}-8 x^{2}-2}{2 x^{2}}\right )\right )}{5}\)	\(22\)
risch	\(\frac {2 \ln \left (\ln \left (\frac {15 x^{3}-8 x^{2}-2}{2 x^{2}}\right )\right )}{5}\)	\(22\)
parallelrisch	\(\frac {2 \ln \left (\ln \left (\frac {15 x^{3}-8 x^{2}-2}{2 x^{2}}\right )\right )}{5}\)	\(22\)
default	\(\frac {2 \ln \left (\ln \left (2\right )-\ln \left (\frac {15 x^{3}-8 x^{2}-2}{x^{2}}\right )\right )}{5}\)	\(26\)

[In]

int((30*x^3+8)/(75*x^4-40*x^3-10*x)/ln(1/2*(15*x^3-8*x^2-2)/x^2),x,method=_RETURNVERBOSE)

[Out]

2/5*ln(ln(1/2*(15*x^3-8*x^2-2)/x^2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {2}{5} \, \log \left (\log \left (\frac {15 \, x^{3} - 8 \, x^{2} - 2}{2 \, x^{2}}\right )\right ) \]

[In]

integrate((30*x^3+8)/(75*x^4-40*x^3-10*x)/log(1/2*(15*x^3-8*x^2-2)/x^2),x, algorithm="fricas")

[Out]

2/5*log(log(1/2*(15*x^3 - 8*x^2 - 2)/x^2))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {2 \log {\left (\log {\left (\frac {\frac {15 x^{3}}{2} - 4 x^{2} - 1}{x^{2}} \right )} \right )}}{5} \]

[In]

integrate((30*x**3+8)/(75*x**4-40*x**3-10*x)/ln(1/2*(15*x**3-8*x**2-2)/x**2),x)

[Out]

2*log(log((15*x**3/2 - 4*x**2 - 1)/x**2))/5

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {2}{5} \, \log \left (-\log \left (2\right ) + \log \left (15 \, x^{3} - 8 \, x^{2} - 2\right ) - 2 \, \log \left (x\right )\right ) \]

[In]

integrate((30*x^3+8)/(75*x^4-40*x^3-10*x)/log(1/2*(15*x^3-8*x^2-2)/x^2),x, algorithm="maxima")

[Out]

2/5*log(-log(2) + log(15*x^3 - 8*x^2 - 2) - 2*log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {2}{5} \, \log \left (\log \left (\frac {15 \, x^{3} - 8 \, x^{2} - 2}{2 \, x^{2}}\right )\right ) \]

[In]

integrate((30*x^3+8)/(75*x^4-40*x^3-10*x)/log(1/2*(15*x^3-8*x^2-2)/x^2),x, algorithm="giac")

[Out]

2/5*log(log(1/2*(15*x^3 - 8*x^2 - 2)/x^2))

Mupad [B] (verification not implemented)

Time = 9.57 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {8+30 x^3}{\left (-10 x-40 x^3+75 x^4\right ) \log \left (\frac {-2-8 x^2+15 x^3}{2 x^2}\right )} \, dx=\frac {2\,\ln \left (\ln \left (-\frac {-15\,x^3+8\,x^2+2}{2\,x^2}\right )\right )}{5} \]

[In]

int(-(30*x^3 + 8)/(log(-(4*x^2 - (15*x^3)/2 + 1)/x^2)*(10*x + 40*x^3 - 75*x^4)),x)

[Out]

(2*log(log(-(8*x^2 - 15*x^3 + 2)/(2*x^2))))/5