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

Optimal. Leaf size=116 \[ -\frac{4 a^2 D-b x (a C+3 A b)}{8 a^2 b^2 \left (a+b x^2\right )}+\frac{(a C+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}}-\frac{a \left (B-\frac{a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0682323, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {1814, 639, 205} \[ -\frac{4 a^2 D-b x (a C+3 A b)}{8 a^2 b^2 \left (a+b x^2\right )}+\frac{(a C+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}}-\frac{a \left (B-\frac{a D}{b}\right )-x (A b-a C)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

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

Mathematica [A]  time = 0.0990765, size = 104, normalized size = 0.9 \[ \frac{\frac{\sqrt{a} \left (-a^2 b (2 B+x (C+4 D x))-2 a^3 D+a b^2 x \left (5 A+C x^2\right )+3 A b^3 x^3\right )}{\left (a+b x^2\right )^2}+\sqrt{b} (a C+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 111, normalized size = 1. \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( 3\,Ab+aC \right ){x}^{3}}{8\,{a}^{2}}}-{\frac{D{x}^{2}}{2\,b}}+{\frac{ \left ( 5\,Ab-aC \right ) x}{8\,ab}}-{\frac{Bb+aD}{4\,{b}^{2}}} \right ) }+{\frac{3\,A}{8\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{C}{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((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x)

[Out]

(1/8*(3*A*b+C*a)/a^2*x^3-1/2*D*x^2/b+1/8*(5*A*b-C*a)/a/b*x-1/4*(B*b+D*a)/b^2)/(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))*C

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [A]  time = 7.9541, size = 184, normalized size = 1.59 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} b^{3}}} \left (3 A b + C 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 + C a\right ) \log{\left (a^{3} b \sqrt{- \frac{1}{a^{5} b^{3}}} + x \right )}}{16} + \frac{- 2 B a^{2} b - 2 D a^{3} - 4 D a^{2} b x^{2} + x^{3} \left (3 A b^{3} + C a b^{2}\right ) + x \left (5 A a b^{2} - C a^{2} b\right )}{8 a^{4} b^{2} + 16 a^{3} b^{3} x^{2} + 8 a^{2} b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21808, size = 143, normalized size = 1.23 \begin{align*} \frac{{\left (C a + 3 \, A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{2} b} + \frac{C a b^{2} x^{3} + 3 \, A b^{3} x^{3} - 4 \, D a^{2} b x^{2} - C a^{2} b x + 5 \, A a b^{2} x - 2 \, D a^{3} - 2 \, B a^{2} b}{8 \,{\left (b x^{2} + a\right )}^{2} a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(C*a + 3*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/8*(C*a*b^2*x^3 + 3*A*b^3*x^3 - 4*D*a^2*b*x^2 - C
*a^2*b*x + 5*A*a*b^2*x - 2*D*a^3 - 2*B*a^2*b)/((b*x^2 + a)^2*a^2*b^2)