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

Optimal. Leaf size=93 \[ \frac{(a C+A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2}}-\frac{a \left (B-\frac{a D}{b}\right )-x (A b-a C)}{2 a b \left (a+b x^2\right )}+\frac{D \log \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0654198, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1814, 635, 205, 260} \[ \frac{(a C+A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2}}-\frac{a \left (B-\frac{a D}{b}\right )-x (A b-a C)}{2 a b \left (a+b x^2\right )}+\frac{D \log \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(B - (a*D)/b) - (A*b - a*C)*x)/(2*a*b*(a + b*x^2)) + ((A*b + a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(3/2)*
b^(3/2)) + (D*Log[a + b*x^2])/(2*b^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 635

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

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]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.0812819, size = 83, normalized size = 0.89 \[ \frac{\frac{a^2 D-a b (B+C x)+A b^2 x}{a \left (a+b x^2\right )}+\frac{\sqrt{b} (a C+A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2}}+D \log \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 97, normalized size = 1. \begin{align*}{\frac{1}{b{x}^{2}+a} \left ({\frac{ \left ( Ab-aC \right ) x}{2\,ab}}-{\frac{Bb-aD}{2\,{b}^{2}}} \right ) }+{\frac{D\ln \left ( b{x}^{2}+a \right ) }{2\,{b}^{2}}}+{\frac{A}{2\,a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{C}{2\,b}\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)^2,x)

[Out]

(1/2*(A*b-C*a)/a/b*x-1/2*(B*b-D*a)/b^2)/(b*x^2+a)+1/2*D*ln(b*x^2+a)/b^2+1/2/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/
2))*A+1/2/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)^2,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)^2,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [B]  time = 2.15369, size = 233, normalized size = 2.51 \begin{align*} \left (\frac{D}{2 b^{2}} - \frac{\sqrt{- a^{3} b^{5}} \left (A b + C a\right )}{4 a^{3} b^{4}}\right ) \log{\left (x + \frac{- 2 D a^{2} + 4 a^{2} b^{2} \left (\frac{D}{2 b^{2}} - \frac{\sqrt{- a^{3} b^{5}} \left (A b + C a\right )}{4 a^{3} b^{4}}\right )}{A b^{2} + C a b} \right )} + \left (\frac{D}{2 b^{2}} + \frac{\sqrt{- a^{3} b^{5}} \left (A b + C a\right )}{4 a^{3} b^{4}}\right ) \log{\left (x + \frac{- 2 D a^{2} + 4 a^{2} b^{2} \left (\frac{D}{2 b^{2}} + \frac{\sqrt{- a^{3} b^{5}} \left (A b + C a\right )}{4 a^{3} b^{4}}\right )}{A b^{2} + C a b} \right )} - \frac{B a b - D a^{2} + x \left (- A b^{2} + C a b\right )}{2 a^{2} b^{2} + 2 a b^{3} x^{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)**2,x)

[Out]

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

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Giac [A]  time = 1.17426, size = 119, normalized size = 1.28 \begin{align*} \frac{D \log \left (b x^{2} + a\right )}{2 \, b^{2}} + \frac{{\left (C a + A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a b} - \frac{{\left (C a - A b\right )} x - \frac{D a^{2} - B a b}{b}}{2 \,{\left (b x^{2} + a\right )} a b} \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)^2,x, algorithm="giac")

[Out]

1/2*D*log(b*x^2 + a)/b^2 + 1/2*(C*a + A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b) - 1/2*((C*a - A*b)*x - (D*a^2
- B*a*b)/b)/((b*x^2 + a)*a*b)