\(\int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx\) [96]

Optimal result

Integrand size = 22, antiderivative size = 36 \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {1}{2} (d+e) \text {arctanh}(x)-\frac {(3 d+5 e) \text {arctanh}\left (\sqrt {\frac {3}{5}} x\right )}{2 \sqrt {15}} \]

[Out]

1/2*(d+e)*arctanh(x)-1/30*(3*d+5*e)*arctanh(1/5*x*15^(1/2))*15^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, 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} \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {1}{2} \text {arctanh}(x) (d+e)-\frac {\text {arctanh}\left (\sqrt {\frac {3}{5}} x\right ) (3 d+5 e)}{2 \sqrt {15}} \]

[In]

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

[Out]

((d + e)*ArcTanh[x])/2 - ((3*d + 5*e)*ArcTanh[Sqrt[3/5]*x])/(2*Sqrt[15])

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 {1}{2} (3 (d+e)) \int \frac {1}{-3+3 x^2} \, dx\right )+\frac {1}{2} (3 d+5 e) \int \frac {1}{-5+3 x^2} \, dx \\ & = \frac {1}{2} (d+e) \tanh ^{-1}(x)-\frac {(3 d+5 e) \tanh ^{-1}\left (\sqrt {\frac {3}{5}} x\right )}{2 \sqrt {15}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.00 \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {1}{60} \left (\sqrt {15} (3 d+5 e) \log \left (\sqrt {15}-3 x\right )-15 (d+e) \log (1-x)+15 (d+e) \log (1+x)-\sqrt {15} (3 d+5 e) \log \left (\sqrt {15}+3 x\right )\right ) \]

[In]

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

[Out]

(Sqrt[15]*(3*d + 5*e)*Log[Sqrt[15] - 3*x] - 15*(d + e)*Log[1 - x] + 15*(d + e)*Log[1 + x] - Sqrt[15]*(3*d + 5*
e)*Log[Sqrt[15] + 3*x])/60

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25

method	result	size
default	\(\left (\frac {d}{4}+\frac {e}{4}\right ) \ln \left (x +1\right )-\frac {\left (\frac {3 d}{2}+\frac {5 e}{2}\right ) \operatorname {arctanh}\left (\frac {x \sqrt {15}}{5}\right ) \sqrt {15}}{15}+\left (-\frac {d}{4}-\frac {e}{4}\right ) \ln \left (x -1\right )\)	\(45\)
risch	\(\frac {\sqrt {15}\, \ln \left (-x \sqrt {15}+5\right ) d}{20}+\frac {\sqrt {15}\, \ln \left (-x \sqrt {15}+5\right ) e}{12}-\frac {\sqrt {15}\, \ln \left (x \sqrt {15}+5\right ) d}{20}-\frac {\sqrt {15}\, \ln \left (x \sqrt {15}+5\right ) e}{12}+\frac {\ln \left (x +1\right ) d}{4}+\frac {\ln \left (x +1\right ) e}{4}-\frac {\ln \left (1-x \right ) d}{4}-\frac {\ln \left (1-x \right ) e}{4}\)	\(92\)

[In]

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

[Out]

(1/4*d+1/4*e)*ln(x+1)-1/15*(3/2*d+5/2*e)*arctanh(1/5*x*15^(1/2))*15^(1/2)+(-1/4*d-1/4*e)*ln(x-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (27) = 54\).

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.53 \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {1}{60} \, \sqrt {15} {\left (3 \, d + 5 \, e\right )} \log \left (\frac {3 \, x^{2} - 2 \, \sqrt {15} x + 5}{3 \, x^{2} - 5}\right ) + \frac {1}{4} \, {\left (d + e\right )} \log \left (x + 1\right ) - \frac {1}{4} \, {\left (d + e\right )} \log \left (x - 1\right ) \]

[In]

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

[Out]

1/60*sqrt(15)*(3*d + 5*e)*log((3*x^2 - 2*sqrt(15)*x + 5)/(3*x^2 - 5)) + 1/4*(d + e)*log(x + 1) - 1/4*(d + e)*l
og(x - 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (31) = 62\).

Time = 0.79 (sec) , antiderivative size = 474, normalized size of antiderivative = 13.17 \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {\left (d + e\right ) \log {\left (x + \frac {- 51 d^{3} \left (d + e\right ) - 180 d^{2} e \left (d + e\right ) - 225 d e^{2} \left (d + e\right ) + 60 d \left (d + e\right )^{3} - 100 e^{3} \left (d + e\right ) + 75 e \left (d + e\right )^{3}}{9 d^{4} + 24 d^{3} e - 40 d e^{3} - 25 e^{4}} \right )}}{4} - \frac {\left (d + e\right ) \log {\left (x + \frac {51 d^{3} \left (d + e\right ) + 180 d^{2} e \left (d + e\right ) + 225 d e^{2} \left (d + e\right ) - 60 d \left (d + e\right )^{3} + 100 e^{3} \left (d + e\right ) - 75 e \left (d + e\right )^{3}}{9 d^{4} + 24 d^{3} e - 40 d e^{3} - 25 e^{4}} \right )}}{4} + \frac {\sqrt {15} \cdot \left (3 d + 5 e\right ) \log {\left (x + \frac {- \frac {17 \sqrt {15} d^{3} \cdot \left (3 d + 5 e\right )}{5} - 12 \sqrt {15} d^{2} e \left (3 d + 5 e\right ) - 15 \sqrt {15} d e^{2} \cdot \left (3 d + 5 e\right ) + \frac {4 \sqrt {15} d \left (3 d + 5 e\right )^{3}}{15} - \frac {20 \sqrt {15} e^{3} \cdot \left (3 d + 5 e\right )}{3} + \frac {\sqrt {15} e \left (3 d + 5 e\right )^{3}}{3}}{9 d^{4} + 24 d^{3} e - 40 d e^{3} - 25 e^{4}} \right )}}{60} - \frac {\sqrt {15} \cdot \left (3 d + 5 e\right ) \log {\left (x + \frac {\frac {17 \sqrt {15} d^{3} \cdot \left (3 d + 5 e\right )}{5} + 12 \sqrt {15} d^{2} e \left (3 d + 5 e\right ) + 15 \sqrt {15} d e^{2} \cdot \left (3 d + 5 e\right ) - \frac {4 \sqrt {15} d \left (3 d + 5 e\right )^{3}}{15} + \frac {20 \sqrt {15} e^{3} \cdot \left (3 d + 5 e\right )}{3} - \frac {\sqrt {15} e \left (3 d + 5 e\right )^{3}}{3}}{9 d^{4} + 24 d^{3} e - 40 d e^{3} - 25 e^{4}} \right )}}{60} \]

[In]

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

[Out]

(d + e)*log(x + (-51*d**3*(d + e) - 180*d**2*e*(d + e) - 225*d*e**2*(d + e) + 60*d*(d + e)**3 - 100*e**3*(d +
e) + 75*e*(d + e)**3)/(9*d**4 + 24*d**3*e - 40*d*e**3 - 25*e**4))/4 - (d + e)*log(x + (51*d**3*(d + e) + 180*d
**2*e*(d + e) + 225*d*e**2*(d + e) - 60*d*(d + e)**3 + 100*e**3*(d + e) - 75*e*(d + e)**3)/(9*d**4 + 24*d**3*e
- 40*d*e**3 - 25*e**4))/4 + sqrt(15)*(3*d + 5*e)*log(x + (-17*sqrt(15)*d**3*(3*d + 5*e)/5 - 12*sqrt(15)*d**2*
e*(3*d + 5*e) - 15*sqrt(15)*d*e**2*(3*d + 5*e) + 4*sqrt(15)*d*(3*d + 5*e)**3/15 - 20*sqrt(15)*e**3*(3*d + 5*e)
/3 + sqrt(15)*e*(3*d + 5*e)**3/3)/(9*d**4 + 24*d**3*e - 40*d*e**3 - 25*e**4))/60 - sqrt(15)*(3*d + 5*e)*log(x
+ (17*sqrt(15)*d**3*(3*d + 5*e)/5 + 12*sqrt(15)*d**2*e*(3*d + 5*e) + 15*sqrt(15)*d*e**2*(3*d + 5*e) - 4*sqrt(1
5)*d*(3*d + 5*e)**3/15 + 20*sqrt(15)*e**3*(3*d + 5*e)/3 - sqrt(15)*e*(3*d + 5*e)**3/3)/(9*d**4 + 24*d**3*e - 4
0*d*e**3 - 25*e**4))/60

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.42 \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {1}{60} \, \sqrt {15} {\left (3 \, d + 5 \, e\right )} \log \left (\frac {3 \, x - \sqrt {15}}{3 \, x + \sqrt {15}}\right ) + \frac {1}{4} \, {\left (d + e\right )} \log \left (x + 1\right ) - \frac {1}{4} \, {\left (d + e\right )} \log \left (x - 1\right ) \]

[In]

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

[Out]

1/60*sqrt(15)*(3*d + 5*e)*log((3*x - sqrt(15))/(3*x + sqrt(15))) + 1/4*(d + e)*log(x + 1) - 1/4*(d + e)*log(x
- 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (27) = 54\).

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.58 \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {1}{60} \, \sqrt {15} {\left (3 \, d + 5 \, e\right )} \log \left (\frac {{\left | 6 \, x - 2 \, \sqrt {15} \right |}}{{\left | 6 \, x + 2 \, \sqrt {15} \right |}}\right ) + \frac {1}{4} \, {\left (d + e\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{4} \, {\left (d + e\right )} \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

1/60*sqrt(15)*(3*d + 5*e)*log(abs(6*x - 2*sqrt(15))/abs(6*x + 2*sqrt(15))) + 1/4*(d + e)*log(abs(x + 1)) - 1/4
*(d + e)*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 13.41 (sec) , antiderivative size = 290, normalized size of antiderivative = 8.06 \[ \int \frac {d+e x^2}{5-8 x^2+3 x^4} \, dx=\frac {\sqrt {15}\,\mathrm {atanh}\left (\frac {54\,\sqrt {15}\,d^3\,x}{25\,\left (-\frac {54\,d^3}{5}-18\,d^2\,e+18\,d\,e^2+30\,e^3\right )}-\frac {6\,\sqrt {15}\,e^3\,x}{-\frac {54\,d^3}{5}-18\,d^2\,e+18\,d\,e^2+30\,e^3}-\frac {18\,\sqrt {15}\,d\,e^2\,x}{5\,\left (-\frac {54\,d^3}{5}-18\,d^2\,e+18\,d\,e^2+30\,e^3\right )}+\frac {18\,\sqrt {15}\,d^2\,e\,x}{5\,\left (-\frac {54\,d^3}{5}-18\,d^2\,e+18\,d\,e^2+30\,e^3\right )}\right )\,\left (3\,d+5\,e\right )}{30}-\mathrm {atanh}\left (\frac {18\,d^3\,x}{-18\,d^3-18\,d^2\,e+30\,d\,e^2+30\,e^3}-\frac {30\,e^3\,x}{-18\,d^3-18\,d^2\,e+30\,d\,e^2+30\,e^3}-\frac {30\,d\,e^2\,x}{-18\,d^3-18\,d^2\,e+30\,d\,e^2+30\,e^3}+\frac {18\,d^2\,e\,x}{-18\,d^3-18\,d^2\,e+30\,d\,e^2+30\,e^3}\right )\,\left (\frac {d}{2}+\frac {e}{2}\right ) \]

[In]

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

[Out]

(15^(1/2)*atanh((54*15^(1/2)*d^3*x)/(25*(18*d*e^2 - 18*d^2*e - (54*d^3)/5 + 30*e^3)) - (6*15^(1/2)*e^3*x)/(18*
d*e^2 - 18*d^2*e - (54*d^3)/5 + 30*e^3) - (18*15^(1/2)*d*e^2*x)/(5*(18*d*e^2 - 18*d^2*e - (54*d^3)/5 + 30*e^3)
) + (18*15^(1/2)*d^2*e*x)/(5*(18*d*e^2 - 18*d^2*e - (54*d^3)/5 + 30*e^3)))*(3*d + 5*e))/30 - atanh((18*d^3*x)/
(30*d*e^2 - 18*d^2*e - 18*d^3 + 30*e^3) - (30*e^3*x)/(30*d*e^2 - 18*d^2*e - 18*d^3 + 30*e^3) - (30*d*e^2*x)/(3
0*d*e^2 - 18*d^2*e - 18*d^3 + 30*e^3) + (18*d^2*e*x)/(30*d*e^2 - 18*d^2*e - 18*d^3 + 30*e^3))*(d/2 + e/2)