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

Optimal. Leaf size=27 \[ \frac{1}{150 \left (2-5 x^2\right )^6}-\frac{1}{250 \left (2-5 x^2\right )^5} \]

[Out]

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

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Rubi [A]  time = 0.0191322, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac{1}{150 \left (2-5 x^2\right )^6}-\frac{1}{250 \left (2-5 x^2\right )^5} \]

Antiderivative was successfully verified.

[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 266

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]]

Rule 43

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])

Rubi steps

\begin{align*} \int \frac{x^3}{\left (2-5 x^2\right )^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(2-5 x)^7} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{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]  time = 0.0074893, size = 20, normalized size = 0.74 \[ \frac{15 x^2-1}{750 \left (2-5 x^2\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(2 - 5*x^2)^7,x]

[Out]

(-1 + 15*x^2)/(750*(2 - 5*x^2)^6)

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Maple [A]  time = 0.007, size = 24, normalized size = 0.9 \begin{align*}{\frac{1}{150\, \left ( 5\,{x}^{2}-2 \right ) ^{6}}}+{\frac{1}{250\, \left ( 5\,{x}^{2}-2 \right ) ^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.934018, size = 58, normalized size = 2.15 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.74723, size = 130, normalized size = 4.81 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.18098, size = 37, normalized size = 1.37 \begin{align*} \frac{15 x^{2} - 1}{11718750 x^{12} - 28125000 x^{10} + 28125000 x^{8} - 15000000 x^{6} + 4500000 x^{4} - 720000 x^{2} + 48000} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.05182, size = 24, normalized size = 0.89 \begin{align*} \frac{15 \, x^{2} - 1}{750 \,{\left (5 \, x^{2} - 2\right )}^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/750*(15*x^2 - 1)/(5*x^2 - 2)^6