\(\int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx\) [1890]

Optimal result

Integrand size = 30, antiderivative size = 28 \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=\frac {4}{3 \left (-5+5 x^2\right )}+\log \left (\frac {\left (x-x^2\right )^2}{x^2}\right ) \]

[Out]

4/(15*x^2-15)+ln((-x^2+x)^2/x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {28, 1828, 641, 31} \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=2 \log (1-x)-\frac {4}{15 \left (1-x^2\right )} \]

[In]

Int[(-30 - 38*x + 30*x^2 + 30*x^3)/(15 - 30*x^2 + 15*x^4),x]

[Out]

-4/(15*(1 - x^2)) + 2*Log[1 - x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = 15 \int \frac {-30-38 x+30 x^2+30 x^3}{\left (-15+15 x^2\right )^2} \, dx \\ & = -\frac {4}{15 \left (1-x^2\right )}+\frac {1}{2} \int \frac {60+60 x}{-15+15 x^2} \, dx \\ & = -\frac {4}{15 \left (1-x^2\right )}+\frac {1}{2} \int \frac {1}{-\frac {1}{4}+\frac {x}{4}} \, dx \\ & = -\frac {4}{15 \left (1-x^2\right )}+2 \log (1-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=\frac {2}{15} \left (\frac {2}{-1+x^2}+15 \log (1-x)\right ) \]

[In]

Integrate[(-30 - 38*x + 30*x^2 + 30*x^3)/(15 - 30*x^2 + 15*x^4),x]

[Out]

(2*(2/(-1 + x^2) + 15*Log[1 - x]))/15

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61

method	result	size
norman	\(\frac {4}{15 \left (x^{2}-1\right )}+2 \ln \left (-1+x \right )\)	\(17\)
risch	\(\frac {4}{15 \left (x^{2}-1\right )}+2 \ln \left (-1+x \right )\)	\(17\)
default	\(\frac {2}{15 \left (-1+x \right )}+2 \ln \left (-1+x \right )-\frac {2}{15 \left (1+x \right )}\)	\(22\)
parallelrisch	\(\frac {30 \ln \left (-1+x \right ) x^{2}+4-30 \ln \left (-1+x \right )}{15 x^{2}-15}\)	\(27\)

[In]

int((30*x^3+30*x^2-38*x-30)/(15*x^4-30*x^2+15),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=\frac {2 \, {\left (15 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) + 2\right )}}{15 \, {\left (x^{2} - 1\right )}} \]

[In]

integrate((30*x^3+30*x^2-38*x-30)/(15*x^4-30*x^2+15),x, algorithm="fricas")

[Out]

2/15*(15*(x^2 - 1)*log(x - 1) + 2)/(x^2 - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50 \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=2 \log {\left (x - 1 \right )} + \frac {4}{15 x^{2} - 15} \]

[In]

integrate((30*x**3+30*x**2-38*x-30)/(15*x**4-30*x**2+15),x)

[Out]

2*log(x - 1) + 4/(15*x**2 - 15)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.57 \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=\frac {4}{15 \, {\left (x^{2} - 1\right )}} + 2 \, \log \left (x - 1\right ) \]

[In]

integrate((30*x^3+30*x^2-38*x-30)/(15*x^4-30*x^2+15),x, algorithm="maxima")

[Out]

4/15/(x^2 - 1) + 2*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=\frac {4}{15 \, {\left (x + 1\right )} {\left (x - 1\right )}} + 2 \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate((30*x^3+30*x^2-38*x-30)/(15*x^4-30*x^2+15),x, algorithm="giac")

[Out]

4/15/((x + 1)*(x - 1)) + 2*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {-30-38 x+30 x^2+30 x^3}{15-30 x^2+15 x^4} \, dx=2\,\ln \left (x-1\right )+\frac {4}{15\,\left (x^2-1\right )} \]

[In]

int(-(38*x - 30*x^2 - 30*x^3 + 30)/(15*x^4 - 30*x^2 + 15),x)

[Out]

2*log(x - 1) + 4/(15*(x^2 - 1))