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

Optimal. Leaf size=73 \[ \frac{(A b-a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}}+\frac{(b B-a D) \log \left (a+b x^2\right )}{2 b^2}+\frac{C x}{b}+\frac{D x^2}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0650568, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1810, 635, 205, 260} \[ \frac{(A b-a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}}+\frac{(b B-a D) \log \left (a+b x^2\right )}{2 b^2}+\frac{C x}{b}+\frac{D x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A]  time = 0.0404422, size = 68, normalized size = 0.93 \[ \frac{\frac{2 \sqrt{b} (A b-a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a}}+(b B-a D) \log \left (a+b x^2\right )+b x (2 C+D x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 83, normalized size = 1.1 \begin{align*}{\frac{D{x}^{2}}{2\,b}}+{\frac{Cx}{b}}+{\frac{\ln \left ( b{x}^{2}+a \right ) B}{2\,b}}-{\frac{\ln \left ( b{x}^{2}+a \right ) aD}{2\,{b}^{2}}}+{A\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{aC}{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),x)

[Out]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [B]  time = 0.959732, size = 219, normalized size = 3. \begin{align*} \frac{C x}{b} + \frac{D x^{2}}{2 b} + \left (- \frac{- B b + D a}{2 b^{2}} - \frac{\sqrt{- a b^{5}} \left (- A b + C a\right )}{2 a b^{4}}\right ) \log{\left (x + \frac{B a b - D a^{2} - 2 a b^{2} \left (- \frac{- B b + D a}{2 b^{2}} - \frac{\sqrt{- a b^{5}} \left (- A b + C a\right )}{2 a b^{4}}\right )}{- A b^{2} + C a b} \right )} + \left (- \frac{- B b + D a}{2 b^{2}} + \frac{\sqrt{- a b^{5}} \left (- A b + C a\right )}{2 a b^{4}}\right ) \log{\left (x + \frac{B a b - D a^{2} - 2 a b^{2} \left (- \frac{- B b + D a}{2 b^{2}} + \frac{\sqrt{- a b^{5}} \left (- A b + C a\right )}{2 a b^{4}}\right )}{- A b^{2} + C a b} \right )} \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),x)

[Out]

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

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Giac [A]  time = 1.19558, size = 89, normalized size = 1.22 \begin{align*} -\frac{{\left (C a - A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{\sqrt{a b} b} - \frac{{\left (D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{2}} + \frac{D b x^{2} + 2 \, C b x}{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),x, algorithm="giac")

[Out]

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