\(\int \frac {x^7}{(2-5 x^2)^3} \, dx\) [185]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=-\frac {x^2}{250}+\frac {2}{625 \left (2-5 x^2\right )^2}-\frac {6}{625 \left (2-5 x^2\right )}-\frac {3}{625} \log \left (2-5 x^2\right ) \]

[Out]

-1/250*x^2+2/625/(-5*x^2+2)^2-6/625/(-5*x^2+2)-3/625*ln(-5*x^2+2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=-\frac {x^2}{250}-\frac {6}{625 \left (2-5 x^2\right )}+\frac {2}{625 \left (2-5 x^2\right )^2}-\frac {3}{625} \log \left (2-5 x^2\right ) \]

[In]

Int[x^7/(2 - 5*x^2)^3,x]

[Out]

-1/250*x^2 + 2/(625*(2 - 5*x^2)^2) - 6/(625*(2 - 5*x^2)) - (3*Log[2 - 5*x^2])/625

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^3}{(2-5 x)^3} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {1}{125}-\frac {8}{125 (-2+5 x)^3}-\frac {12}{125 (-2+5 x)^2}-\frac {6}{125 (-2+5 x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {x^2}{250}+\frac {2}{625 \left (2-5 x^2\right )^2}-\frac {6}{625 \left (2-5 x^2\right )}-\frac {3}{625} \log \left (2-5 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=-\frac {12-150 x^4+125 x^6+6 \left (2-5 x^2\right )^2 \log \left (-2+5 x^2\right )}{1250 \left (2-5 x^2\right )^2} \]

[In]

Integrate[x^7/(2 - 5*x^2)^3,x]

[Out]

-1/1250*(12 - 150*x^4 + 125*x^6 + 6*(2 - 5*x^2)^2*Log[-2 + 5*x^2])/(2 - 5*x^2)^2

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74

method	result	size
norman	\(\frac {\frac {12}{125} x^{2}-\frac {1}{10} x^{6}-\frac {18}{625}}{\left (5 x^{2}-2\right )^{2}}-\frac {3 \ln \left (5 x^{2}-2\right )}{625}\)	\(34\)
risch	\(-\frac {x^{2}}{250}+\frac {\frac {6 x^{2}}{125}-\frac {2}{125}}{\left (5 x^{2}-2\right )^{2}}-\frac {3 \ln \left (5 x^{2}-2\right )}{625}\)	\(35\)
meijerg	\(-\frac {x^{2} \left (25 x^{4}-45 x^{2}+12\right )}{1000 \left (1-\frac {5 x^{2}}{2}\right )^{2}}-\frac {3 \ln \left (1-\frac {5 x^{2}}{2}\right )}{625}\)	\(38\)
default	\(-\frac {x^{2}}{250}+\frac {6}{625 \left (5 x^{2}-2\right )}-\frac {3 \ln \left (5 x^{2}-2\right )}{625}+\frac {2}{625 \left (5 x^{2}-2\right )^{2}}\)	\(39\)
parallelrisch	\(-\frac {250 x^{6}+300 \ln \left (x^{2}-\frac {2}{5}\right ) x^{4}-450 x^{4}-240 \ln \left (x^{2}-\frac {2}{5}\right ) x^{2}+120 x^{2}+48 \ln \left (x^{2}-\frac {2}{5}\right )}{2500 \left (5 x^{2}-2\right )^{2}}\)	\(58\)

[In]

int(x^7/(-5*x^2+2)^3,x,method=_RETURNVERBOSE)

[Out]

(12/125*x^2-1/10*x^6-18/625)/(5*x^2-2)^2-3/625*ln(5*x^2-2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20 \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=-\frac {125 \, x^{6} - 100 \, x^{4} - 40 \, x^{2} + 6 \, {\left (25 \, x^{4} - 20 \, x^{2} + 4\right )} \log \left (5 \, x^{2} - 2\right ) + 20}{1250 \, {\left (25 \, x^{4} - 20 \, x^{2} + 4\right )}} \]

[In]

integrate(x^7/(-5*x^2+2)^3,x, algorithm="fricas")

[Out]

-1/1250*(125*x^6 - 100*x^4 - 40*x^2 + 6*(25*x^4 - 20*x^2 + 4)*log(5*x^2 - 2) + 20)/(25*x^4 - 20*x^2 + 4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=- \frac {x^{2}}{250} - \frac {2 - 6 x^{2}}{3125 x^{4} - 2500 x^{2} + 500} - \frac {3 \log {\left (5 x^{2} - 2 \right )}}{625} \]

[In]

integrate(x**7/(-5*x**2+2)**3,x)

[Out]

-x**2/250 - (2 - 6*x**2)/(3125*x**4 - 2500*x**2 + 500) - 3*log(5*x**2 - 2)/625

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=-\frac {1}{250} \, x^{2} + \frac {2 \, {\left (3 \, x^{2} - 1\right )}}{125 \, {\left (25 \, x^{4} - 20 \, x^{2} + 4\right )}} - \frac {3}{625} \, \log \left (5 \, x^{2} - 2\right ) \]

[In]

integrate(x^7/(-5*x^2+2)^3,x, algorithm="maxima")

[Out]

-1/250*x^2 + 2/125*(3*x^2 - 1)/(25*x^4 - 20*x^2 + 4) - 3/625*log(5*x^2 - 2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=-\frac {1}{250} \, x^{2} + \frac {225 \, x^{4} - 120 \, x^{2} + 16}{1250 \, {\left (5 \, x^{2} - 2\right )}^{2}} - \frac {3}{625} \, \log \left ({\left | 5 \, x^{2} - 2 \right |}\right ) \]

[In]

integrate(x^7/(-5*x^2+2)^3,x, algorithm="giac")

[Out]

-1/250*x^2 + 1/1250*(225*x^4 - 120*x^2 + 16)/(5*x^2 - 2)^2 - 3/625*log(abs(5*x^2 - 2))

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {x^7}{\left (2-5 x^2\right )^3} \, dx=\frac {\frac {6\,x^2}{3125}-\frac {2}{3125}}{x^4-\frac {4\,x^2}{5}+\frac {4}{25}}-\frac {3\,\ln \left (x^2-\frac {2}{5}\right )}{625}-\frac {x^2}{250} \]

[In]

int(-x^7/(5*x^2 - 2)^3,x)

[Out]

((6*x^2)/3125 - 2/3125)/(x^4 - (4*x^2)/5 + 4/25) - (3*log(x^2 - 2/5))/625 - x^2/250