3.95 \(\int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx\)

Optimal. Leaf size=28 \[ \frac {5}{2} \tanh ^{-1}(x)-\frac {7}{2} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} x\right ) \]

[Out]

5/2*arctanh(x)-7/10*arctanh(1/5*x*15^(1/2))*15^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1166, 207} \[ \frac {5}{2} \tanh ^{-1}(x)-\frac {7}{2} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} x\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)/(5 - 8*x^2 + 3*x^4),x]

[Out]

(5*ArcTanh[x])/2 - (7*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*x])/2

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx &=-\left (\frac {15}{2} \int \frac {1}{-3+3 x^2} \, dx\right )+\frac {21}{2} \int \frac {1}{-5+3 x^2} \, dx\\ &=\frac {5}{2} \tanh ^{-1}(x)-\frac {7}{2} \sqrt {\frac {3}{5}} \tanh ^{-1}\left (\sqrt {\frac {3}{5}} x\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 1.89 \[ \frac {1}{20} \left (7 \sqrt {15} \log \left (\sqrt {15}-3 x\right )-25 \log (1-x)+25 \log (x+1)-7 \sqrt {15} \log \left (3 x+\sqrt {15}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x^2)/(5 - 8*x^2 + 3*x^4),x]

[Out]

(7*Sqrt[15]*Log[Sqrt[15] - 3*x] - 25*Log[1 - x] + 25*Log[1 + x] - 7*Sqrt[15]*Log[Sqrt[15] + 3*x])/20

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fricas [B]  time = 0.40, size = 49, normalized size = 1.75 \[ \frac {7}{20} \, \sqrt {5} \sqrt {3} \log \left (-\frac {2 \, \sqrt {5} \sqrt {3} x - 3 \, x^{2} - 5}{3 \, x^{2} - 5}\right ) + \frac {5}{4} \, \log \left (x + 1\right ) - \frac {5}{4} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/(3*x^4-8*x^2+5),x, algorithm="fricas")

[Out]

7/20*sqrt(5)*sqrt(3)*log(-(2*sqrt(5)*sqrt(3)*x - 3*x^2 - 5)/(3*x^2 - 5)) + 5/4*log(x + 1) - 5/4*log(x - 1)

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giac [B]  time = 0.17, size = 44, normalized size = 1.57 \[ \frac {7}{20} \, \sqrt {15} \log \left (\frac {{\left | 6 \, x - 2 \, \sqrt {15} \right |}}{{\left | 6 \, x + 2 \, \sqrt {15} \right |}}\right ) + \frac {5}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {5}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/(3*x^4-8*x^2+5),x, algorithm="giac")

[Out]

7/20*sqrt(15)*log(abs(6*x - 2*sqrt(15))/abs(6*x + 2*sqrt(15))) + 5/4*log(abs(x + 1)) - 5/4*log(abs(x - 1))

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maple [A]  time = 0.01, size = 26, normalized size = 0.93 \[ -\frac {7 \sqrt {15}\, \arctanh \left (\frac {\sqrt {15}\, x}{5}\right )}{10}+\frac {5 \ln \left (x +1\right )}{4}-\frac {5 \ln \left (x -1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2)/(3*x^4-8*x^2+5),x)

[Out]

-7/10*arctanh(1/5*x*15^(1/2))*15^(1/2)+5/4*ln(x+1)-5/4*ln(x-1)

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maxima [B]  time = 2.36, size = 38, normalized size = 1.36 \[ \frac {7}{20} \, \sqrt {15} \log \left (\frac {3 \, x - \sqrt {15}}{3 \, x + \sqrt {15}}\right ) + \frac {5}{4} \, \log \left (x + 1\right ) - \frac {5}{4} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)/(3*x^4-8*x^2+5),x, algorithm="maxima")

[Out]

7/20*sqrt(15)*log((3*x - sqrt(15))/(3*x + sqrt(15))) + 5/4*log(x + 1) - 5/4*log(x - 1)

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mupad [B]  time = 4.39, size = 17, normalized size = 0.61 \[ \frac {5\,\mathrm {atanh}\relax (x)}{2}-\frac {7\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,x}{5}\right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2 + 2)/(3*x^4 - 8*x^2 + 5),x)

[Out]

(5*atanh(x))/2 - (7*15^(1/2)*atanh((15^(1/2)*x)/5))/10

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sympy [B]  time = 0.61, size = 53, normalized size = 1.89 \[ - \frac {5 \log {\left (x - 1 \right )}}{4} + \frac {5 \log {\left (x + 1 \right )}}{4} + \frac {7 \sqrt {15} \log {\left (x - \frac {\sqrt {15}}{3} \right )}}{20} - \frac {7 \sqrt {15} \log {\left (x + \frac {\sqrt {15}}{3} \right )}}{20} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+2)/(3*x**4-8*x**2+5),x)

[Out]

-5*log(x - 1)/4 + 5*log(x + 1)/4 + 7*sqrt(15)*log(x - sqrt(15)/3)/20 - 7*sqrt(15)*log(x + sqrt(15)/3)/20

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