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

Optimal. Leaf size=89 \[ \frac{(3 a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{5/2}}+\frac{x (A b-5 a B)}{8 a b^2 \left (a+b x^2\right )}-\frac{x (A b-a B)}{4 b^2 \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0619753, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {455, 385, 205} \[ \frac{(3 a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{5/2}}+\frac{x (A b-5 a B)}{8 a b^2 \left (a+b x^2\right )}-\frac{x (A b-a B)}{4 b^2 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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

Mathematica [A]  time = 0.0824406, size = 83, normalized size = 0.93 \[ \frac{\frac{\sqrt{b} x \left (-3 a^2 B-a b \left (A+5 B x^2\right )+A b^2 x^2\right )}{a \left (a+b x^2\right )^2}+\frac{(3 a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2}}}{8 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 89, normalized size = 1. \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( Ab-5\,Ba \right ){x}^{3}}{8\,ab}}-{\frac{ \left ( Ab+3\,Ba \right ) x}{8\,{b}^{2}}} \right ) }+{\frac{A}{8\,ab}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,B}{8\,{b}^{2}}\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(x^2*(B*x^2+A)/(b*x^2+a)^3,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33385, size = 624, normalized size = 7.01 \begin{align*} \left [-\frac{2 \,{\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} +{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} + 3 \, B a^{3} + A a^{2} b + 2 \,{\left (3 \, B a^{2} b + 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 (3 \, B a^{3} b + A a^{2} b^{2}\right )} x}{16 \,{\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, -\frac{{\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} -{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} + 3 \, B a^{3} + A a^{2} b + 2 \,{\left (3 \, B a^{2} b + A a b^{2}\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) +{\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} x}{8 \,{\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.881778, size = 153, normalized size = 1.72 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{3} b^{5}}} \left (A b + 3 B a\right ) \log{\left (- a^{2} b^{2} \sqrt{- \frac{1}{a^{3} b^{5}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{3} b^{5}}} \left (A b + 3 B a\right ) \log{\left (a^{2} b^{2} \sqrt{- \frac{1}{a^{3} b^{5}}} + x \right )}}{16} - \frac{x^{3} \left (- A b^{2} + 5 B a b\right ) + x \left (A a b + 3 B a^{2}\right )}{8 a^{3} b^{2} + 16 a^{2} b^{3} x^{2} + 8 a b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.42951, size = 105, normalized size = 1.18 \begin{align*} \frac{{\left (3 \, B a + A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a b^{2}} - \frac{5 \, B a b x^{3} - A b^{2} x^{3} + 3 \, B a^{2} x + A a b x}{8 \,{\left (b x^{2} + a\right )}^{2} a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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