3.52 \(\int \frac{A+B x^2}{b x^2-c x^4} \, dx\)

Optimal. Leaf size=41 \[ \frac{(A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{3/2} \sqrt{c}}-\frac{A}{b x} \]

[Out]

-(A/(b*x)) + ((b*B + A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b]])/(b^(3/2)*Sqrt[c])

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Rubi [A]  time = 0.0344379, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1593, 453, 208} \[ \frac{(A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{3/2} \sqrt{c}}-\frac{A}{b x} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(b*x^2 - c*x^4),x]

[Out]

-(A/(b*x)) + ((b*B + A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b]])/(b^(3/2)*Sqrt[c])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{A+B x^2}{b x^2-c x^4} \, dx &=\int \frac{A+B x^2}{x^2 \left (b-c x^2\right )} \, dx\\ &=-\frac{A}{b x}+\frac{(b B+A c) \int \frac{1}{b-c x^2} \, dx}{b}\\ &=-\frac{A}{b x}+\frac{(b B+A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{3/2} \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.023628, size = 41, normalized size = 1. \[ \frac{(A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{b^{3/2} \sqrt{c}}-\frac{A}{b x} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(b*x^2 - c*x^4),x]

[Out]

-(A/(b*x)) + ((b*B + A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b]])/(b^(3/2)*Sqrt[c])

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Maple [A]  time = 0.005, size = 39, normalized size = 1. \begin{align*} -{\frac{-Ac-Bb}{b}{\it Artanh} \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}-{\frac{A}{bx}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(-c*x^4+b*x^2),x)

[Out]

-(-A*c-B*b)/b/(b*c)^(1/2)*arctanh(x*c/(b*c)^(1/2))-A/b/x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(-c*x^4+b*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.747638, size = 230, normalized size = 5.61 \begin{align*} \left [\frac{{\left (B b + A c\right )} \sqrt{b c} x \log \left (\frac{c x^{2} + 2 \, \sqrt{b c} x + b}{c x^{2} - b}\right ) - 2 \, A b c}{2 \, b^{2} c x}, -\frac{{\left (B b + A c\right )} \sqrt{-b c} x \arctan \left (\frac{\sqrt{-b c} x}{b}\right ) + A b c}{b^{2} c x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(-c*x^4+b*x^2),x, algorithm="fricas")

[Out]

[1/2*((B*b + A*c)*sqrt(b*c)*x*log((c*x^2 + 2*sqrt(b*c)*x + b)/(c*x^2 - b)) - 2*A*b*c)/(b^2*c*x), -((B*b + A*c)
*sqrt(-b*c)*x*arctan(sqrt(-b*c)*x/b) + A*b*c)/(b^2*c*x)]

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Sympy [B]  time = 0.467934, size = 75, normalized size = 1.83 \begin{align*} - \frac{A}{b x} - \frac{\sqrt{\frac{1}{b^{3} c}} \left (A c + B b\right ) \log{\left (- b^{2} \sqrt{\frac{1}{b^{3} c}} + x \right )}}{2} + \frac{\sqrt{\frac{1}{b^{3} c}} \left (A c + B b\right ) \log{\left (b^{2} \sqrt{\frac{1}{b^{3} c}} + x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(-c*x**4+b*x**2),x)

[Out]

-A/(b*x) - sqrt(1/(b**3*c))*(A*c + B*b)*log(-b**2*sqrt(1/(b**3*c)) + x)/2 + sqrt(1/(b**3*c))*(A*c + B*b)*log(b
**2*sqrt(1/(b**3*c)) + x)/2

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Giac [A]  time = 1.25921, size = 51, normalized size = 1.24 \begin{align*} -\frac{{\left (B b + A c\right )} \arctan \left (\frac{c x}{\sqrt{-b c}}\right )}{\sqrt{-b c} b} - \frac{A}{b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(-c*x^4+b*x^2),x, algorithm="giac")

[Out]

-(B*b + A*c)*arctan(c*x/sqrt(-b*c))/(sqrt(-b*c)*b) - A/(b*x)