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

Optimal. Leaf size=119 \[ \frac{(3 a D+b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{5/2}}-\frac{2 (a C+A b)-x (b B-5 a D)}{8 a b^2 \left (a+b x^2\right )}-\frac{x \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.114108, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1804, 1814, 12, 205} \[ \frac{(3 a D+b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{5/2}}-\frac{2 (a C+A b)-x (b B-5 a D)}{8 a b^2 \left (a+b x^2\right )}-\frac{x \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.111549, size = 99, normalized size = 0.83 \[ \frac{\frac{\sqrt{b} \left (-a^2 (2 C+3 D x)-a b \left (2 A+x \left (B+4 C x+5 D x^2\right )\right )+b^2 B x^3\right )}{a \left (a+b x^2\right )^2}+\frac{(3 a D+b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2}}}{8 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 110, normalized size = 0.9 \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( Bb-5\,aD \right ){x}^{3}}{8\,ab}}-{\frac{C{x}^{2}}{2\,b}}-{\frac{ \left ( Bb+3\,aD \right ) x}{8\,{b}^{2}}}-{\frac{Ab+aC}{4\,{b}^{2}}} \right ) }+{\frac{B}{8\,ab}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,D}{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*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x)

[Out]

(1/8*(B*b-5*D*a)/a/b*x^3-1/2*C*x^2/b-1/8*(B*b+3*D*a)/b^2*x-1/4*(A*b+C*a)/b^2)/(b*x^2+a)^2+1/8/a/b/(a*b)^(1/2)*
arctan(b*x/(a*b)^(1/2))*B+3/8/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*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)^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(x*(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 = 13.4165, size = 177, normalized size = 1.49 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{3} b^{5}}} \left (B b + 3 D 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 (B b + 3 D a\right ) \log{\left (a^{2} b^{2} \sqrt{- \frac{1}{a^{3} b^{5}}} + x \right )}}{16} - \frac{2 A a b + 2 C a^{2} + 4 C a b x^{2} + x^{3} \left (- B b^{2} + 5 D a b\right ) + x \left (B a b + 3 D 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*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**3,x)

[Out]

-sqrt(-1/(a**3*b**5))*(B*b + 3*D*a)*log(-a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/16 + sqrt(-1/(a**3*b**5))*(B*b +
3*D*a)*log(a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/16 - (2*A*a*b + 2*C*a**2 + 4*C*a*b*x**2 + x**3*(-B*b**2 + 5*D*a
*b) + x*(B*a*b + 3*D*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 = 1.19612, size = 131, normalized size = 1.1 \begin{align*} \frac{{\left (3 \, D a + B b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a b^{2}} - \frac{5 \, D a b x^{3} - B b^{2} x^{3} + 4 \, C a b x^{2} + 3 \, D a^{2} x + B a b x + 2 \, C a^{2} + 2 \, A a b}{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*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

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