\(\int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx\) [95]

Optimal result

Integrand size = 22, antiderivative size = 28 \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=\frac {5 \text {arctanh}(x)}{2}-\frac {7}{2} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} x\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1180, 213} \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=\frac {5 \text {arctanh}(x)}{2}-\frac {7}{2} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} x\right ) \]

[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 213

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

Rule 1180

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=\frac {1}{20} \left (7 \sqrt {15} \log \left (\sqrt {15}-3 x\right )-25 \log (1-x)+25 \log (1+x)-7 \sqrt {15} \log \left (\sqrt {15}+3 x\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {5 \ln \left (x +1\right )}{4}-\frac {7 \,\operatorname {arctanh}\left (\frac {x \sqrt {15}}{5}\right ) \sqrt {15}}{10}-\frac {5 \ln \left (x -1\right )}{4}\)	\(26\)
risch	\(-\frac {5 \ln \left (x -1\right )}{4}+\frac {7 \sqrt {15}\, \ln \left (3 x -\sqrt {15}\right )}{20}-\frac {7 \sqrt {15}\, \ln \left (3 x +\sqrt {15}\right )}{20}+\frac {5 \ln \left (x +1\right )}{4}\)	\(42\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (17) = 34\).

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=\frac {7}{20} \, \sqrt {5} \sqrt {3} \log \left (-\frac {2 \, \sqrt {5} \sqrt {3} x - 3 \, x^{2} - 5}{3 \, x^{2} - 5}\right ) + \frac {5}{4} \, \log \left (x + 1\right ) - \frac {5}{4} \, \log \left (x - 1\right ) \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=- \frac {5 \log {\left (x - 1 \right )}}{4} + \frac {5 \log {\left (x + 1 \right )}}{4} + \frac {7 \sqrt {15} \log {\left (x - \frac {\sqrt {15}}{3} \right )}}{20} - \frac {7 \sqrt {15} \log {\left (x + \frac {\sqrt {15}}{3} \right )}}{20} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (17) = 34\).

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=\frac {7}{20} \, \sqrt {15} \log \left (\frac {3 \, x - \sqrt {15}}{3 \, x + \sqrt {15}}\right ) + \frac {5}{4} \, \log \left (x + 1\right ) - \frac {5}{4} \, \log \left (x - 1\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (17) = 34\).

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=\frac {7}{20} \, \sqrt {15} \log \left (\frac {{\left | 6 \, x - 2 \, \sqrt {15} \right |}}{{\left | 6 \, x + 2 \, \sqrt {15} \right |}}\right ) + \frac {5}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {5}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 13.52 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {2+3 x^2}{5-8 x^2+3 x^4} \, dx=\frac {5\,\mathrm {atanh}\left (x\right )}{2}-\frac {7\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,x}{5}\right )}{10} \]

[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