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

Optimal. Leaf size=155 \[ -\frac{x^3 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}-\frac{x^2 (4 a C-x (3 b B-7 a D))}{8 a b^2 \left (a+b x^2\right )}-\frac{3 x (b B-5 a D)}{8 a b^3}+\frac{3 (b B-5 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{7/2}}+\frac{C \log \left (a+b x^2\right )}{2 b^3} \]

[Out]

(-3*(b*B - 5*a*D)*x)/(8*a*b^3) - (x^3*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(4*a*b*(a + b*x^2)^2) - (x^2*(4*a*C -
(3*b*B - 7*a*D)*x))/(8*a*b^2*(a + b*x^2)) + (3*(b*B - 5*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2))
+ (C*Log[a + b*x^2])/(2*b^3)

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Rubi [A]  time = 0.232444, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {1804, 774, 635, 205, 260} \[ -\frac{x^3 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}-\frac{x^2 (4 a C-x (3 b B-7 a D))}{8 a b^2 \left (a+b x^2\right )}-\frac{3 x (b B-5 a D)}{8 a b^3}+\frac{3 (b B-5 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{7/2}}+\frac{C \log \left (a+b x^2\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*B - 5*a*D)*x)/(8*a*b^3) - (x^3*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(4*a*b*(a + b*x^2)^2) - (x^2*(4*a*C -
(3*b*B - 7*a*D)*x))/(8*a*b^2*(a + b*x^2)) + (3*(b*B - 5*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2))
+ (C*Log[a + b*x^2])/(2*b^3)

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 774

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

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^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx &=-\frac{x^3 \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{x^2 \left (-3 a \left (B-\frac{a D}{b}\right )-4 a C x-4 a D x^2\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=-\frac{x^3 \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{x^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac{\int \frac{x \left (8 a^2 C-3 a (b B-5 a D) x\right )}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=-\frac{3 (b B-5 a D) x}{8 a b^3}-\frac{x^3 \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{x^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac{\int \frac{3 a^2 (b B-5 a D)+8 a^2 b C x}{a+b x^2} \, dx}{8 a^2 b^3}\\ &=-\frac{3 (b B-5 a D) x}{8 a b^3}-\frac{x^3 \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{x^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac{C \int \frac{x}{a+b x^2} \, dx}{b^2}+\frac{(3 (b B-5 a D)) \int \frac{1}{a+b x^2} \, dx}{8 b^3}\\ &=-\frac{3 (b B-5 a D) x}{8 a b^3}-\frac{x^3 \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{x^2 (4 a C-(3 b B-7 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac{3 (b B-5 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{7/2}}+\frac{C \log \left (a+b x^2\right )}{2 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0758234, size = 126, normalized size = 0.81 \[ \frac{a (-a (C+D x)+A b+b B x)}{4 b^3 \left (a+b x^2\right )^2}+\frac{8 a C+9 a D x-4 A b-5 b B x}{8 b^3 \left (a+b x^2\right )}+\frac{3 (b B-5 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{7/2}}+\frac{C \log \left (a+b x^2\right )}{2 b^3}+\frac{D x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(D*x)/b^3 + (-4*A*b + 8*a*C - 5*b*B*x + 9*a*D*x)/(8*b^3*(a + b*x^2)) + (a*(A*b + b*B*x - a*(C + D*x)))/(4*b^3*
(a + b*x^2)^2) + (3*(b*B - 5*a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2)) + (C*Log[a + b*x^2])/(2*b^3
)

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Maple [A]  time = 0.01, size = 206, normalized size = 1.3 \begin{align*}{\frac{Dx}{{b}^{3}}}-{\frac{5\,B{x}^{3}}{8\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{9\,D{x}^{3}a}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{A{x}^{2}}{2\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{C{x}^{2}a}{{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{3\,Bax}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{7\,{a}^{2}Dx}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{Aa}{4\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{3\,{a}^{2}C}{4\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{C\ln \left ( b{x}^{2}+a \right ) }{2\,{b}^{3}}}+{\frac{3\,B}{8\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{15\,aD}{8\,{b}^{3}}\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^3*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x)

[Out]

D/b^3*x-5/8/b/(b*x^2+a)^2*B*x^3+9/8/b^2/(b*x^2+a)^2*D*x^3*a-1/2/b/(b*x^2+a)^2*A*x^2+1/b^2/(b*x^2+a)^2*C*x^2*a-
3/8/b^2/(b*x^2+a)^2*B*x*a+7/8/b^3/(b*x^2+a)^2*a^2*D*x-1/4/b^2/(b*x^2+a)^2*A*a+3/4/b^3/(b*x^2+a)^2*a^2*C+1/2*C*
ln(b*x^2+a)/b^3+3/8/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*B-15/8/b^3/(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^3*(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^3*(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 [B]  time = 21.8294, size = 282, normalized size = 1.82 \begin{align*} \frac{D x}{b^{3}} + \left (\frac{C}{2 b^{3}} - \frac{3 \sqrt{- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right ) \log{\left (x + \frac{8 C a - 16 a b^{3} \left (\frac{C}{2 b^{3}} - \frac{3 \sqrt{- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right )}{- 3 B b + 15 D a} \right )} + \left (\frac{C}{2 b^{3}} + \frac{3 \sqrt{- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right ) \log{\left (x + \frac{8 C a - 16 a b^{3} \left (\frac{C}{2 b^{3}} + \frac{3 \sqrt{- a b^{7}} \left (- B b + 5 D a\right )}{16 a b^{7}}\right )}{- 3 B b + 15 D a} \right )} + \frac{- 2 A a b + 6 C a^{2} + x^{3} \left (- 5 B b^{2} + 9 D a b\right ) + x^{2} \left (- 4 A b^{2} + 8 C a b\right ) + x \left (- 3 B a b + 7 D a^{2}\right )}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

D*x/b**3 + (C/(2*b**3) - 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7))*log(x + (8*C*a - 16*a*b**3*(C/(2*b**3) -
3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7)))/(-3*B*b + 15*D*a)) + (C/(2*b**3) + 3*sqrt(-a*b**7)*(-B*b + 5*D*a)
/(16*a*b**7))*log(x + (8*C*a - 16*a*b**3*(C/(2*b**3) + 3*sqrt(-a*b**7)*(-B*b + 5*D*a)/(16*a*b**7)))/(-3*B*b +
15*D*a)) + (-2*A*a*b + 6*C*a**2 + x**3*(-5*B*b**2 + 9*D*a*b) + x**2*(-4*A*b**2 + 8*C*a*b) + x*(-3*B*a*b + 7*D*
a**2))/(8*a**2*b**3 + 16*a*b**4*x**2 + 8*b**5*x**4)

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Giac [A]  time = 1.13859, size = 165, normalized size = 1.06 \begin{align*} \frac{D x}{b^{3}} + \frac{C \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac{3 \,{\left (5 \, D a - B b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} b^{3}} + \frac{{\left (9 \, D a b - 5 \, B b^{2}\right )} x^{3} + 6 \, C a^{2} - 2 \, A a b + 4 \,{\left (2 \, C a b - A b^{2}\right )} x^{2} +{\left (7 \, D a^{2} - 3 \, B a b\right )} x}{8 \,{\left (b x^{2} + a\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

D*x/b^3 + 1/2*C*log(b*x^2 + a)/b^3 - 3/8*(5*D*a - B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/8*((9*D*a*b -
5*B*b^2)*x^3 + 6*C*a^2 - 2*A*a*b + 4*(2*C*a*b - A*b^2)*x^2 + (7*D*a^2 - 3*B*a*b)*x)/((b*x^2 + a)^2*b^3)