\(\int \frac {x^3}{(2-5 x^2)^7} \, dx\) [184]

Optimal result

Integrand size = 13, antiderivative size = 27 \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=\frac {1}{150 \left (2-5 x^2\right )^6}-\frac {1}{250 \left (2-5 x^2\right )^5} \]

[Out]

1/150/(-5*x^2+2)^6-1/250/(-5*x^2+2)^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, 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} \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=\frac {1}{150 \left (2-5 x^2\right )^6}-\frac {1}{250 \left (2-5 x^2\right )^5} \]

[In]

Int[x^3/(2 - 5*x^2)^7,x]

[Out]

1/(150*(2 - 5*x^2)^6) - 1/(250*(2 - 5*x^2)^5)

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}{(2-5 x)^7} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {2}{5 (-2+5 x)^7}-\frac {1}{5 (-2+5 x)^6}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{150 \left (2-5 x^2\right )^6}-\frac {1}{250 \left (2-5 x^2\right )^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=\frac {-1+15 x^2}{750 \left (2-5 x^2\right )^6} \]

[In]

Integrate[x^3/(2 - 5*x^2)^7,x]

[Out]

(-1 + 15*x^2)/(750*(2 - 5*x^2)^6)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67

method	result	size
norman	\(\frac {\frac {x^{2}}{50}-\frac {1}{750}}{\left (5 x^{2}-2\right )^{6}}\)	\(18\)
gosper	\(\frac {15 x^{2}-1}{750 \left (5 x^{2}-2\right )^{6}}\)	\(19\)
risch	\(\frac {\frac {x^{2}}{50}-\frac {1}{750}}{\left (5 x^{2}-2\right )^{6}}\)	\(19\)
default	\(\frac {1}{150 \left (5 x^{2}-2\right )^{6}}+\frac {1}{250 \left (5 x^{2}-2\right )^{5}}\)	\(24\)
meijerg	\(\frac {x^{4} \left (\frac {625}{16} x^{8}-\frac {375}{4} x^{6}+\frac {375}{4} x^{4}-50 x^{2}+15\right )}{7680 \left (1-\frac {5 x^{2}}{2}\right )^{6}}\)	\(37\)
parallelrisch	\(\frac {125 x^{12}-300 x^{10}+300 x^{8}-160 x^{6}+48 x^{4}}{384 \left (5 x^{2}-2\right )^{6}}\)	\(38\)

[In]

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

[Out]

(1/50*x^2-1/750)/(5*x^2-2)^6

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=\frac {15 \, x^{2} - 1}{750 \, {\left (15625 \, x^{12} - 37500 \, x^{10} + 37500 \, x^{8} - 20000 \, x^{6} + 6000 \, x^{4} - 960 \, x^{2} + 64\right )}} \]

[In]

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

[Out]

1/750*(15*x^2 - 1)/(15625*x^12 - 37500*x^10 + 37500*x^8 - 20000*x^6 + 6000*x^4 - 960*x^2 + 64)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=- \frac {1 - 15 x^{2}}{11718750 x^{12} - 28125000 x^{10} + 28125000 x^{8} - 15000000 x^{6} + 4500000 x^{4} - 720000 x^{2} + 48000} \]

[In]

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

[Out]

-(1 - 15*x**2)/(11718750*x**12 - 28125000*x**10 + 28125000*x**8 - 15000000*x**6 + 4500000*x**4 - 720000*x**2 +
48000)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=\frac {15 \, x^{2} - 1}{750 \, {\left (15625 \, x^{12} - 37500 \, x^{10} + 37500 \, x^{8} - 20000 \, x^{6} + 6000 \, x^{4} - 960 \, x^{2} + 64\right )}} \]

[In]

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

[Out]

1/750*(15*x^2 - 1)/(15625*x^12 - 37500*x^10 + 37500*x^8 - 20000*x^6 + 6000*x^4 - 960*x^2 + 64)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=\frac {15 \, x^{2} - 1}{750 \, {\left (5 \, x^{2} - 2\right )}^{6}} \]

[In]

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

[Out]

1/750*(15*x^2 - 1)/(5*x^2 - 2)^6

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {x^3}{\left (2-5 x^2\right )^7} \, dx=\frac {15\,x^2-1}{750\,{\left (5\,x^2-2\right )}^6} \]

[In]

int(-x^3/(5*x^2 - 2)^7,x)

[Out]

(15*x^2 - 1)/(750*(5*x^2 - 2)^6)