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

Optimal. Leaf size=36 \[ \frac{1}{2} (d+e) \tanh ^{-1}(x)-\frac{(3 d+5 e) \tanh ^{-1}\left (\sqrt{\frac{3}{5}} x\right )}{2 \sqrt{15}} \]

[Out]

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

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Rubi [A]  time = 0.0399891, antiderivative size = 36, normalized size of antiderivative = 1., 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} \[ \frac{1}{2} (d+e) \tanh ^{-1}(x)-\frac{(3 d+5 e) \tanh ^{-1}\left (\sqrt{\frac{3}{5}} x\right )}{2 \sqrt{15}} \]

Antiderivative was successfully verified.

[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 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]

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])

Rubi steps

\begin{align*} \int \frac{d+e x^2}{5-8 x^2+3 x^4} \, dx &=-\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]  time = 0.0399562, size = 72, normalized size = 2. \[ \frac{1}{60} \left (\sqrt{15} (3 d+5 e) \log \left (\sqrt{15}-3 x\right )-15 (d+e) \log (1-x)+15 (d+e) \log (x+1)-\sqrt{15} (3 d+5 e) \log \left (3 x+\sqrt{15}\right )\right ) \]

Antiderivative was successfully verified.

[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

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Maple [B]  time = 0.049, size = 56, normalized size = 1.6 \begin{align*}{\frac{\ln \left ( 1+x \right ) d}{4}}+{\frac{\ln \left ( 1+x \right ) e}{4}}-{\frac{\ln \left ( -1+x \right ) d}{4}}-{\frac{\ln \left ( -1+x \right ) e}{4}}-{\frac{\sqrt{15}d}{10}{\it Artanh} \left ({\frac{x\sqrt{15}}{5}} \right ) }-{\frac{\sqrt{15}e}{6}{\it Artanh} \left ({\frac{x\sqrt{15}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*ln(1+x)*d+1/4*ln(1+x)*e-1/4*ln(-1+x)*d-1/4*ln(-1+x)*e-1/10*15^(1/2)*arctanh(1/5*x*15^(1/2))*d-1/6*15^(1/2)
*arctanh(1/5*x*15^(1/2))*e

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Maxima [A]  time = 1.48087, size = 69, normalized size = 1.92 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 1.467, size = 163, normalized size = 4.53 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [B]  time = 1.04993, size = 474, normalized size = 13.17 \begin{align*} \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} \left (3 d + 5 e\right ) \log{\left (x + \frac{- \frac{17 \sqrt{15} d^{3} \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} \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} \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} \left (3 d + 5 e\right ) \log{\left (x + \frac{\frac{17 \sqrt{15} d^{3} \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} \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} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [B]  time = 1.10593, size = 81, normalized size = 2.25 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))