3.103 \(\int \frac{A+B x^2}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=92 \[ \frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}}+\frac{x (a B+3 A b)}{8 a^2 b \left (a+b x^2\right )}+\frac{x (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

[Out]

((A*b - a*B)*x)/(4*a*b*(a + b*x^2)^2) + ((3*A*b + a*B)*x)/(8*a^2*b*(a + b*x^2)) + ((3*A*b + a*B)*ArcTan[(Sqrt[
b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0324289, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {385, 199, 205} \[ \frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}}+\frac{x (a B+3 A b)}{8 a^2 b \left (a+b x^2\right )}+\frac{x (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(a + b*x^2)^3,x]

[Out]

((A*b - a*B)*x)/(4*a*b*(a + b*x^2)^2) + ((3*A*b + a*B)*x)/(8*a^2*b*(a + b*x^2)) + ((3*A*b + a*B)*ArcTan[(Sqrt[
b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(3/2))

Rule 385

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rubi steps

\begin{align*} \int \frac{A+B x^2}{\left (a+b x^2\right )^3} \, dx &=\frac{(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac{(3 A b+a B) \int \frac{1}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac{(3 A b+a B) x}{8 a^2 b \left (a+b x^2\right )}+\frac{(3 A b+a B) \int \frac{1}{a+b x^2} \, dx}{8 a^2 b}\\ &=\frac{(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac{(3 A b+a B) x}{8 a^2 b \left (a+b x^2\right )}+\frac{(3 A b+a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0635813, size = 84, normalized size = 0.91 \[ \frac{x \left (a^2 (-B)+a b \left (5 A+B x^2\right )+3 A b^2 x^2\right )}{8 a^2 b \left (a+b x^2\right )^2}+\frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(a + b*x^2)^3,x]

[Out]

(x*(-(a^2*B) + 3*A*b^2*x^2 + a*b*(5*A + B*x^2)))/(8*a^2*b*(a + b*x^2)^2) + ((3*A*b + a*B)*ArcTan[(Sqrt[b]*x)/S
qrt[a]])/(8*a^(5/2)*b^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 90, normalized size = 1. \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( 3\,Ab+Ba \right ){x}^{3}}{8\,{a}^{2}}}+{\frac{ \left ( 5\,Ab-Ba \right ) x}{8\,ab}} \right ) }+{\frac{3\,A}{8\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{B}{8\,ab}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(b*x^2+a)^3,x)

[Out]

(1/8*(3*A*b+B*a)/a^2*x^3+1/8*(5*A*b-B*a)/a/b*x)/(b*x^2+a)^2+3/8/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*A+1/8/
a/b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*B

________________________________________________________________________________________

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)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.28903, size = 621, normalized size = 6.75 \begin{align*} \left [\frac{2 \,{\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} -{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{4} + B a^{3} + 3 \, A a^{2} b + 2 \,{\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right ) - 2 \,{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{16 \,{\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, \frac{{\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} +{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{4} + B a^{3} + 3 \, A a^{2} b + 2 \,{\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) -{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{8 \,{\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(2*(B*a^2*b^2 + 3*A*a*b^3)*x^3 - ((B*a*b^2 + 3*A*b^3)*x^4 + B*a^3 + 3*A*a^2*b + 2*(B*a^2*b + 3*A*a*b^2)*
x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*(B*a^3*b - 5*A*a^2*b^2)*x)/(a^3*b^4*x^4 + 2*
a^4*b^3*x^2 + a^5*b^2), 1/8*((B*a^2*b^2 + 3*A*a*b^3)*x^3 + ((B*a*b^2 + 3*A*b^3)*x^4 + B*a^3 + 3*A*a^2*b + 2*(B
*a^2*b + 3*A*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - (B*a^3*b - 5*A*a^2*b^2)*x)/(a^3*b^4*x^4 + 2*a^4*b^3
*x^2 + a^5*b^2)]

________________________________________________________________________________________

Sympy [A]  time = 0.725714, size = 150, normalized size = 1.63 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} b^{3}}} \left (3 A b + B a\right ) \log{\left (- a^{3} b \sqrt{- \frac{1}{a^{5} b^{3}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{5} b^{3}}} \left (3 A b + B a\right ) \log{\left (a^{3} b \sqrt{- \frac{1}{a^{5} b^{3}}} + x \right )}}{16} + \frac{x^{3} \left (3 A b^{2} + B a b\right ) + x \left (5 A a b - B a^{2}\right )}{8 a^{4} b + 16 a^{3} b^{2} x^{2} + 8 a^{2} b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(b*x**2+a)**3,x)

[Out]

-sqrt(-1/(a**5*b**3))*(3*A*b + B*a)*log(-a**3*b*sqrt(-1/(a**5*b**3)) + x)/16 + sqrt(-1/(a**5*b**3))*(3*A*b + B
*a)*log(a**3*b*sqrt(-1/(a**5*b**3)) + x)/16 + (x**3*(3*A*b**2 + B*a*b) + x*(5*A*a*b - B*a**2))/(8*a**4*b + 16*
a**3*b**2*x**2 + 8*a**2*b**3*x**4)

________________________________________________________________________________________

Giac [A]  time = 1.19365, size = 105, normalized size = 1.14 \begin{align*} \frac{{\left (B a + 3 \, A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{2} b} + \frac{B a b x^{3} + 3 \, A b^{2} x^{3} - B a^{2} x + 5 \, A a b x}{8 \,{\left (b x^{2} + a\right )}^{2} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/8*(B*a + 3*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/8*(B*a*b*x^3 + 3*A*b^2*x^3 - B*a^2*x + 5*A*a*b*x
)/((b*x^2 + a)^2*a^2*b)