3.108 \(\int \frac{a+b x^2}{1-x^2} \, dx\)

Optimal. Leaf size=11 \[ (a+b) \tanh ^{-1}(x)-b x \]

[Out]

-(b*x) + (a + b)*ArcTanh[x]

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Rubi [A]  time = 0.0066391, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {388, 206} \[ (a+b) \tanh ^{-1}(x)-b x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)/(1 - x^2),x]

[Out]

-(b*x) + (a + b)*ArcTanh[x]

Rule 388

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

Rule 206

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

Rubi steps

\begin{align*} \int \frac{a+b x^2}{1-x^2} \, dx &=-b x-(-a-b) \int \frac{1}{1-x^2} \, dx\\ &=-b x+(a+b) \tanh ^{-1}(x)\\ \end{align*}

Mathematica [B]  time = 0.0085583, size = 28, normalized size = 2.55 \[ \frac{1}{2} (-(a+b) \log (1-x)+(a+b) \log (x+1)-2 b x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)/(1 - x^2),x]

[Out]

(-2*b*x - (a + b)*Log[1 - x] + (a + b)*Log[1 + x])/2

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Maple [B]  time = 0.003, size = 34, normalized size = 3.1 \begin{align*} -bx-{\frac{\ln \left ( -1+x \right ) a}{2}}-{\frac{\ln \left ( -1+x \right ) b}{2}}+{\frac{\ln \left ( 1+x \right ) a}{2}}+{\frac{\ln \left ( 1+x \right ) b}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)/(-x^2+1),x)

[Out]

-b*x-1/2*ln(-1+x)*a-1/2*ln(-1+x)*b+1/2*ln(1+x)*a+1/2*ln(1+x)*b

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Maxima [B]  time = 0.996524, size = 31, normalized size = 2.82 \begin{align*} -b x + \frac{1}{2} \,{\left (a + b\right )} \log \left (x + 1\right ) - \frac{1}{2} \,{\left (a + b\right )} \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/(-x^2+1),x, algorithm="maxima")

[Out]

-b*x + 1/2*(a + b)*log(x + 1) - 1/2*(a + b)*log(x - 1)

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Fricas [B]  time = 1.27101, size = 76, normalized size = 6.91 \begin{align*} -b x + \frac{1}{2} \,{\left (a + b\right )} \log \left (x + 1\right ) - \frac{1}{2} \,{\left (a + b\right )} \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/(-x^2+1),x, algorithm="fricas")

[Out]

-b*x + 1/2*(a + b)*log(x + 1) - 1/2*(a + b)*log(x - 1)

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Sympy [B]  time = 0.300267, size = 22, normalized size = 2. \begin{align*} - b x - \frac{\left (a + b\right ) \log{\left (x - 1 \right )}}{2} + \frac{\left (a + b\right ) \log{\left (x + 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)/(-x**2+1),x)

[Out]

-b*x - (a + b)*log(x - 1)/2 + (a + b)*log(x + 1)/2

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Giac [B]  time = 1.14409, size = 34, normalized size = 3.09 \begin{align*} -b x + \frac{1}{2} \,{\left (a + b\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{2} \,{\left (a + b\right )} \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/(-x^2+1),x, algorithm="giac")

[Out]

-b*x + 1/2*(a + b)*log(abs(x + 1)) - 1/2*(a + b)*log(abs(x - 1))