3.97.42 \(\int \frac {270 x^2-225 x^4}{-32+160 x^2-200 x^4+e (8-40 x^2+50 x^4)} \, dx\)

Optimal. Leaf size=25 \[ 4+\frac {9 x^2}{(-4+e) \left (\frac {4}{5 x}-2 x\right )} \]

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Rubi [A]  time = 0.12, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1593, 1992, 28, 449} \begin {gather*} -\frac {45 x^3}{2 (4-e) \left (2-5 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(270*x^2 - 225*x^4)/(-32 + 160*x^2 - 200*x^4 + E*(8 - 40*x^2 + 50*x^4)),x]

[Out]

(-45*x^3)/(2*(4 - E)*(2 - 5*x^2))

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 449

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

Rule 1593

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]

Rule 1992

Int[(u_)^(p_.)*((f_.)*(x_))^(m_.)*(z_)^(q_.), x_Symbol] :> Int[(f*x)^m*ExpandToSum[z, x]^q*ExpandToSum[u, x]^p
, x] /; FreeQ[{f, m, p, q}, x] && BinomialQ[z, x] && TrinomialQ[u, x] &&  !(BinomialMatchQ[z, x] && TrinomialM
atchQ[u, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2 \left (270-225 x^2\right )}{-32+160 x^2-200 x^4+e \left (8-40 x^2+50 x^4\right )} \, dx\\ &=\int \frac {x^2 \left (270-225 x^2\right )}{-8 (4-e)+40 (4-e) x^2-50 (4-e) x^4} \, dx\\ &=-\left ((50 (4-e)) \int \frac {x^2 \left (270-225 x^2\right )}{\left (20 (4-e)-50 (4-e) x^2\right )^2} \, dx\right )\\ &=-\frac {45 x^3}{2 (4-e) \left (2-5 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.84 \begin {gather*} -\frac {45 x^3}{2 (-4+e) \left (-2+5 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(270*x^2 - 225*x^4)/(-32 + 160*x^2 - 200*x^4 + E*(8 - 40*x^2 + 50*x^4)),x]

[Out]

(-45*x^3)/(2*(-4 + E)*(-2 + 5*x^2))

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fricas [A]  time = 0.48, size = 25, normalized size = 1.00 \begin {gather*} \frac {45 \, x^{3}}{2 \, {\left (20 \, x^{2} - {\left (5 \, x^{2} - 2\right )} e - 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-225*x^4+270*x^2)/((50*x^4-40*x^2+8)*exp(1)-200*x^4+160*x^2-32),x, algorithm="fricas")

[Out]

45/2*x^3/(20*x^2 - (5*x^2 - 2)*e - 8)

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giac [A]  time = 0.17, size = 28, normalized size = 1.12 \begin {gather*} -\frac {9 \, x}{2 \, {\left (e - 4\right )}} - \frac {9 \, x}{{\left (5 \, x^{2} - 2\right )} {\left (e - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-225*x^4+270*x^2)/((50*x^4-40*x^2+8)*exp(1)-200*x^4+160*x^2-32),x, algorithm="giac")

[Out]

-9/2*x/(e - 4) - 9*x/((5*x^2 - 2)*(e - 4))

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maple [A]  time = 0.06, size = 21, normalized size = 0.84

method	result	size
norman	\(-\frac {45 x^{3}}{2 \left ({\mathrm e}-4\right ) \left (5 x^{2}-2\right )}\)	\(21\)
default	\(\frac {-9 x -\frac {18 x}{5 \left (x^{2}-\frac {2}{5}\right )}}{2 \,{\mathrm e}-8}\)	\(25\)
gosper	\(-\frac {45 x^{3}}{2 \left (5 x^{2} {\mathrm e}-20 x^{2}-2 \,{\mathrm e}+8\right )}\)	\(26\)
risch	\(-\frac {9 x}{2 \,{\mathrm e}-8}+\frac {\left (-\frac {18 \,{\mathrm e}}{5}+\frac {72}{5}\right ) x}{\left (2 \,{\mathrm e}-8\right ) \left (x^{2} {\mathrm e}-4 x^{2}-\frac {2 \,{\mathrm e}}{5}+\frac {8}{5}\right )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-225*x^4+270*x^2)/((50*x^4-40*x^2+8)*exp(1)-200*x^4+160*x^2-32),x,method=_RETURNVERBOSE)

[Out]

-45/2/(exp(1)-4)*x^3/(5*x^2-2)

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maxima [A]  time = 0.35, size = 30, normalized size = 1.20 \begin {gather*} -\frac {9 \, x}{5 \, x^{2} {\left (e - 4\right )} - 2 \, e + 8} - \frac {9 \, x}{2 \, {\left (e - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-225*x^4+270*x^2)/((50*x^4-40*x^2+8)*exp(1)-200*x^4+160*x^2-32),x, algorithm="maxima")

[Out]

-9*x/(5*x^2*(e - 4) - 2*e + 8) - 9/2*x/(e - 4)

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mupad [B]  time = 5.54, size = 20, normalized size = 0.80 \begin {gather*} -\frac {45\,x^3}{2\,\left (5\,x^2-2\right )\,\left (\mathrm {e}-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((270*x^2 - 225*x^4)/(exp(1)*(50*x^4 - 40*x^2 + 8) + 160*x^2 - 200*x^4 - 32),x)

[Out]

-(45*x^3)/(2*(5*x^2 - 2)*(exp(1) - 4))

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sympy [A]  time = 0.31, size = 31, normalized size = 1.24 \begin {gather*} - \frac {9 x}{-8 + 2 e} - \frac {9 x}{x^{2} \left (-20 + 5 e\right ) - 2 e + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-225*x**4+270*x**2)/((50*x**4-40*x**2+8)*exp(1)-200*x**4+160*x**2-32),x)

[Out]

-9*x/(-8 + 2*E) - 9*x/(x**2*(-20 + 5*E) - 2*E + 8)

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