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

Optimal. Leaf size=28 \[ \frac {4}{3 \left (-5+5 x^2\right )}+\log \left (\frac {\left (x-x^2\right )^2}{x^2}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {28, 1814, 627, 31} \begin {gather*} 2 \log (1-x)-\frac {4}{15 \left (1-x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
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 1814

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.79 \begin {gather*} \frac {2}{15} \left (\frac {2}{-1+x^2}+15 \log (1-x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 1.01, size = 22, normalized size = 0.79 \begin {gather*} \frac {2 \, {\left (15 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) + 2\right )}}{15 \, {\left (x^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.23, size = 20, normalized size = 0.71 \begin {gather*} \frac {4}{15 \, {\left (x + 1\right )} {\left (x - 1\right )}} + 2 \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.03, size = 17, normalized size = 0.61

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

Verification of antiderivative is not currently implemented for this CAS.

[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(x-1)

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maxima [A]  time = 0.68, size = 16, normalized size = 0.57 \begin {gather*} \frac {4}{15 \, {\left (x^{2} - 1\right )}} + 2 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.05, size = 18, normalized size = 0.64 \begin {gather*} 2\,\ln \left (x-1\right )+\frac {4}{15\,\left (x^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.09, size = 14, normalized size = 0.50 \begin {gather*} 2 \log {\left (x - 1 \right )} + \frac {4}{15 x^{2} - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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