\(\int \frac {-4 x^3+3 x^5}{(-1+x^2)^5} \, dx\) [465]

Optimal result

Integrand size = 19, antiderivative size = 40 \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=\frac {1}{8 \left (1-x^2\right )^4}+\frac {1}{3 \left (1-x^2\right )^3}-\frac {3}{4 \left (1-x^2\right )^2} \]

[Out]

1/8/(-x^2+1)^4+1/3/(-x^2+1)^3-3/4/(-x^2+1)^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1607, 457, 78} \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=-\frac {3}{4 \left (1-x^2\right )^2}+\frac {1}{3 \left (1-x^2\right )^3}+\frac {1}{8 \left (1-x^2\right )^4} \]

[In]

Int[(-4*x^3 + 3*x^5)/(-1 + x^2)^5,x]

[Out]

1/(8*(1 - x^2)^4) + 1/(3*(1 - x^2)^3) - 3/(4*(1 - x^2)^2)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3 \left (-4+3 x^2\right )}{\left (-1+x^2\right )^5} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x (-4+3 x)}{(-1+x)^5} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {1}{(-1+x)^5}+\frac {2}{(-1+x)^4}+\frac {3}{(-1+x)^3}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{8 \left (1-x^2\right )^4}+\frac {1}{3 \left (1-x^2\right )^3}-\frac {3}{4 \left (1-x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.58 \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=\frac {-7+28 x^2-18 x^4}{24 \left (-1+x^2\right )^4} \]

[In]

Integrate[(-4*x^3 + 3*x^5)/(-1 + x^2)^5,x]

[Out]

(-7 + 28*x^2 - 18*x^4)/(24*(-1 + x^2)^4)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.52

method	result	size
norman	\(\frac {-\frac {3}{4} x^{4}+\frac {7}{6} x^{2}-\frac {7}{24}}{\left (x^{2}-1\right )^{4}}\)	\(21\)
risch	\(\frac {-\frac {3}{4} x^{4}+\frac {7}{6} x^{2}-\frac {7}{24}}{\left (x^{2}-1\right )^{4}}\)	\(21\)
gosper	\(-\frac {18 x^{4}-28 x^{2}+7}{24 \left (x^{2}-1\right )^{4}}\)	\(22\)
parallelrisch	\(\frac {-18 x^{4}+28 x^{2}-7}{24 \left (x^{2}-1\right )^{4}}\)	\(22\)
meijerg	\(-\frac {x^{6} \left (-x^{2}+4\right )}{8 \left (-x^{2}+1\right )^{4}}+\frac {x^{4} \left (x^{4}-4 x^{2}+6\right )}{6 \left (-x^{2}+1\right )^{4}}\)	\(47\)
default	\(\frac {1}{128 \left (-1+x \right )^{4}}-\frac {11}{192 \left (-1+x \right )^{3}}-\frac {27}{256 \left (-1+x \right )^{2}}+\frac {27}{256 \left (-1+x \right )}+\frac {1}{128 \left (1+x \right )^{4}}+\frac {11}{192 \left (1+x \right )^{3}}-\frac {27}{256 \left (1+x \right )^{2}}-\frac {27}{256 \left (1+x \right )}\)	\(58\)

[In]

int((3*x^5-4*x^3)/(x^2-1)^5,x,method=_RETURNVERBOSE)

[Out]

(-3/4*x^4+7/6*x^2-7/24)/(x^2-1)^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=-\frac {18 \, x^{4} - 28 \, x^{2} + 7}{24 \, {\left (x^{8} - 4 \, x^{6} + 6 \, x^{4} - 4 \, x^{2} + 1\right )}} \]

[In]

integrate((3*x^5-4*x^3)/(x^2-1)^5,x, algorithm="fricas")

[Out]

-1/24*(18*x^4 - 28*x^2 + 7)/(x^8 - 4*x^6 + 6*x^4 - 4*x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=\frac {- 18 x^{4} + 28 x^{2} - 7}{24 x^{8} - 96 x^{6} + 144 x^{4} - 96 x^{2} + 24} \]

[In]

integrate((3*x**5-4*x**3)/(x**2-1)**5,x)

[Out]

(-18*x**4 + 28*x**2 - 7)/(24*x**8 - 96*x**6 + 144*x**4 - 96*x**2 + 24)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=-\frac {18 \, x^{4} - 28 \, x^{2} + 7}{24 \, {\left (x^{8} - 4 \, x^{6} + 6 \, x^{4} - 4 \, x^{2} + 1\right )}} \]

[In]

integrate((3*x^5-4*x^3)/(x^2-1)^5,x, algorithm="maxima")

[Out]

-1/24*(18*x^4 - 28*x^2 + 7)/(x^8 - 4*x^6 + 6*x^4 - 4*x^2 + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.52 \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=-\frac {18 \, x^{4} - 28 \, x^{2} + 7}{24 \, {\left (x^{2} - 1\right )}^{4}} \]

[In]

integrate((3*x^5-4*x^3)/(x^2-1)^5,x, algorithm="giac")

[Out]

-1/24*(18*x^4 - 28*x^2 + 7)/(x^2 - 1)^4

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int \frac {-4 x^3+3 x^5}{\left (-1+x^2\right )^5} \, dx=-\frac {\frac {3\,x^4}{4}-\frac {7\,x^2}{6}+\frac {7}{24}}{x^8-4\,x^6+6\,x^4-4\,x^2+1} \]

[In]

int(-(4*x^3 - 3*x^5)/(x^2 - 1)^5,x)

[Out]

-((3*x^4)/4 - (7*x^2)/6 + 7/24)/(6*x^4 - 4*x^2 - 4*x^6 + x^8 + 1)