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

Optimal. Leaf size=101 \[ -\frac{x \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac{(b B-3 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 \sqrt{a} b^{5/2}}+\frac{C \log \left (a+b x^2\right )}{2 b^2}+\frac{D x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.117818, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1804, 1810, 635, 205, 260} \[ -\frac{x \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac{(b B-3 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 \sqrt{a} b^{5/2}}+\frac{C \log \left (a+b x^2\right )}{2 b^2}+\frac{D x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1804

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

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

Mathematica [A]  time = 0.0485397, size = 92, normalized size = 0.91 \[ \frac{a C+a D x-A b-b B x}{2 b^2 \left (a+b x^2\right )}-\frac{(3 a D-b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 \sqrt{a} b^{5/2}}+\frac{C \log \left (a+b x^2\right )}{2 b^2}+\frac{D x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 127, normalized size = 1.3 \begin{align*}{\frac{Dx}{{b}^{2}}}-{\frac{Bx}{2\,b \left ( b{x}^{2}+a \right ) }}+{\frac{aDx}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{A}{2\,b \left ( b{x}^{2}+a \right ) }}+{\frac{aC}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{C\ln \left ( b{x}^{2}+a \right ) }{2\,{b}^{2}}}+{\frac{B}{2\,b}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,aD}{2\,{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*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^2,x)

[Out]

D*x/b^2-1/2/b/(b*x^2+a)*B*x+1/2/b^2/(b*x^2+a)*a*D*x-1/2/b/(b*x^2+a)*A+1/2/b^2/(b*x^2+a)*a*C+1/2*C*ln(b*x^2+a)/
b^2+1/2/b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*B-3/2/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*a*D

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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*(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(x*(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 = 2.65598, size = 212, normalized size = 2.1 \begin{align*} \frac{D x}{b^{2}} + \left (\frac{C}{2 b^{2}} - \frac{\sqrt{- a b^{5}} \left (- B b + 3 D a\right )}{4 a b^{5}}\right ) \log{\left (x + \frac{2 C a - 4 a b^{2} \left (\frac{C}{2 b^{2}} - \frac{\sqrt{- a b^{5}} \left (- B b + 3 D a\right )}{4 a b^{5}}\right )}{- B b + 3 D a} \right )} + \left (\frac{C}{2 b^{2}} + \frac{\sqrt{- a b^{5}} \left (- B b + 3 D a\right )}{4 a b^{5}}\right ) \log{\left (x + \frac{2 C a - 4 a b^{2} \left (\frac{C}{2 b^{2}} + \frac{\sqrt{- a b^{5}} \left (- B b + 3 D a\right )}{4 a b^{5}}\right )}{- B b + 3 D a} \right )} + \frac{- A b + C a + x \left (- B b + D a\right )}{2 a b^{2} + 2 b^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14644, size = 109, normalized size = 1.08 \begin{align*} \frac{D x}{b^{2}} + \frac{C \log \left (b x^{2} + a\right )}{2 \, b^{2}} - \frac{{\left (3 \, D a - B b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{2}} + \frac{C a - A b +{\left (D a - B b\right )} x}{2 \,{\left (b x^{2} + a\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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