3.62.53 \(\int \frac {75+25 x^2}{162-108 x^2+18 x^4} \, dx\)

Optimal. Leaf size=25 \[ 5 \left (2+2 \left (\frac {5 x}{36 \left (3-x^2\right )}-\log (4)\right )\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 14, normalized size of antiderivative = 0.56, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {28, 383} \begin {gather*} \frac {25 x}{18 \left (3-x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75 + 25*x^2)/(162 - 108*x^2 + 18*x^4),x]

[Out]

(25*x)/(18*(3 - 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 383

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=18 \int \frac {75+25 x^2}{\left (-54+18 x^2\right )^2} \, dx\\ &=\frac {25 x}{18 \left (3-x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.48 \begin {gather*} -\frac {25 x}{18 \left (-3+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75 + 25*x^2)/(162 - 108*x^2 + 18*x^4),x]

[Out]

(-25*x)/(18*(-3 + x^2))

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fricas [A]  time = 1.07, size = 10, normalized size = 0.40 \begin {gather*} -\frac {25 \, x}{18 \, {\left (x^{2} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x^2+75)/(18*x^4-108*x^2+162),x, algorithm="fricas")

[Out]

-25/18*x/(x^2 - 3)

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giac [A]  time = 0.14, size = 10, normalized size = 0.40 \begin {gather*} -\frac {25 \, x}{18 \, {\left (x^{2} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x^2+75)/(18*x^4-108*x^2+162),x, algorithm="giac")

[Out]

-25/18*x/(x^2 - 3)

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maple [A]  time = 0.04, size = 11, normalized size = 0.44

method	result	size
gosper	\(-\frac {25 x}{18 \left (x^{2}-3\right )}\)	\(11\)
default	\(-\frac {25 x}{18 \left (x^{2}-3\right )}\)	\(11\)
norman	\(-\frac {25 x}{18 \left (x^{2}-3\right )}\)	\(11\)
risch	\(-\frac {25 x}{18 \left (x^{2}-3\right )}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x^2+75)/(18*x^4-108*x^2+162),x,method=_RETURNVERBOSE)

[Out]

-25/18*x/(x^2-3)

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maxima [A]  time = 0.34, size = 10, normalized size = 0.40 \begin {gather*} -\frac {25 \, x}{18 \, {\left (x^{2} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x^2+75)/(18*x^4-108*x^2+162),x, algorithm="maxima")

[Out]

-25/18*x/(x^2 - 3)

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mupad [B]  time = 3.98, size = 12, normalized size = 0.48 \begin {gather*} -\frac {25\,x}{18\,\left (x^2-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x^2 + 75)/(18*x^4 - 108*x^2 + 162),x)

[Out]

-(25*x)/(18*(x^2 - 3))

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sympy [A]  time = 0.08, size = 10, normalized size = 0.40 \begin {gather*} - \frac {25 x}{18 x^{2} - 54} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x**2+75)/(18*x**4-108*x**2+162),x)

[Out]

-25*x/(18*x**2 - 54)

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