3.102 \(\int \frac{x^4 (A+B x+C x^2+D x^3)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=185 \[ \frac{x^3 (4 x (b B-2 a D)-5 a C+A b)}{8 a b^2 \left (a+b x^2\right )}-\frac{3 x (A b-5 a C)}{8 a b^3}+\frac{3 (A b-5 a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{7/2}}-\frac{x^4 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}-\frac{x^2 (b B-3 a D)}{2 a b^3}+\frac{(b B-3 a D) \log \left (a+b x^2\right )}{2 b^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.338183, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {1804, 801, 635, 205, 260} \[ \frac{x^3 (4 x (b B-2 a D)-5 a C+A b)}{8 a b^2 \left (a+b x^2\right )}-\frac{3 x (A b-5 a C)}{8 a b^3}+\frac{3 (A b-5 a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{7/2}}-\frac{x^4 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}-\frac{x^2 (b B-3 a D)}{2 a b^3}+\frac{(b B-3 a D) \log \left (a+b x^2\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^3,x]

[Out]

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

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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

Mathematica [A]  time = 0.116181, size = 139, normalized size = 0.75 \[ \frac{\frac{2 a \left (a^2 D-a b (B+C x)+A b^2 x\right )}{\left (a+b x^2\right )^2}+\frac{-12 a^2 D+8 a b B+9 a b C x-5 A b^2 x}{a+b x^2}+\frac{3 \sqrt{b} (A b-5 a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a}}+4 (b B-3 a D) \log \left (a+b x^2\right )+8 b C x+4 b D x^2}{8 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^3,x]

[Out]

(8*b*C*x + 4*b*D*x^2 + (8*a*b*B - 12*a^2*D - 5*A*b^2*x + 9*a*b*C*x)/(a + b*x^2) + (2*a*(a^2*D + A*b^2*x - a*b*
(B + C*x)))/(a + b*x^2)^2 + (3*Sqrt[b]*(A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[a] + 4*(b*B - 3*a*D)*Lo
g[a + b*x^2])/(8*b^4)

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 235, normalized size = 1.3 \begin{align*}{\frac{D{x}^{2}}{2\,{b}^{3}}}+{\frac{Cx}{{b}^{3}}}-{\frac{5\,A{x}^{3}}{8\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{9\,C{x}^{3}a}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{Ba{x}^{2}}{{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{3\,D{x}^{2}{a}^{2}}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{3\,aAx}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{7\,{a}^{2}Cx}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{3\,{a}^{2}B}{4\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{5\,{a}^{3}D}{4\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{\ln \left ( b{x}^{2}+a \right ) B}{2\,{b}^{3}}}-{\frac{3\,\ln \left ( b{x}^{2}+a \right ) aD}{2\,{b}^{4}}}+{\frac{3\,A}{8\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{15\,aC}{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^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x)

[Out]

1/2/b^3*D*x^2+1/b^3*C*x-5/8/b/(b*x^2+a)^2*A*x^3+9/8/b^2/(b*x^2+a)^2*C*x^3*a+1/b^2/(b*x^2+a)^2*B*x^2*a-3/2/b^3/
(b*x^2+a)^2*D*x^2*a^2-3/8/b^2/(b*x^2+a)^2*A*a*x+7/8/b^3/(b*x^2+a)^2*a^2*C*x+3/4/b^3/(b*x^2+a)^2*a^2*B-5/4/b^4/
(b*x^2+a)^2*a^3*D+1/2/b^3*ln(b*x^2+a)*B-3/2/b^4*ln(b*x^2+a)*a*D+3/8/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*A-
15/8/b^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*a*C

________________________________________________________________________________________

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^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [B]  time = 23.3071, size = 357, normalized size = 1.93 \begin{align*} \frac{C x}{b^{3}} + \frac{D x^{2}}{2 b^{3}} + \left (- \frac{- B b + 3 D a}{2 b^{4}} - \frac{3 \sqrt{- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right ) \log{\left (x + \frac{8 B a b - 24 D a^{2} - 16 a b^{4} \left (- \frac{- B b + 3 D a}{2 b^{4}} - \frac{3 \sqrt{- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right )}{- 3 A b^{2} + 15 C a b} \right )} + \left (- \frac{- B b + 3 D a}{2 b^{4}} + \frac{3 \sqrt{- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right ) \log{\left (x + \frac{8 B a b - 24 D a^{2} - 16 a b^{4} \left (- \frac{- B b + 3 D a}{2 b^{4}} + \frac{3 \sqrt{- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right )}{- 3 A b^{2} + 15 C a b} \right )} + \frac{6 B a^{2} b - 10 D a^{3} + x^{3} \left (- 5 A b^{3} + 9 C a b^{2}\right ) + x^{2} \left (8 B a b^{2} - 12 D a^{2} b\right ) + x \left (- 3 A a b^{2} + 7 C a^{2} b\right )}{8 a^{2} b^{4} + 16 a b^{5} x^{2} + 8 b^{6} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**3,x)

[Out]

C*x/b**3 + D*x**2/(2*b**3) + (-(-B*b + 3*D*a)/(2*b**4) - 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8))*log(x + (
8*B*a*b - 24*D*a**2 - 16*a*b**4*(-(-B*b + 3*D*a)/(2*b**4) - 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8)))/(-3*A
*b**2 + 15*C*a*b)) + (-(-B*b + 3*D*a)/(2*b**4) + 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8))*log(x + (8*B*a*b
- 24*D*a**2 - 16*a*b**4*(-(-B*b + 3*D*a)/(2*b**4) + 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8)))/(-3*A*b**2 +
15*C*a*b)) + (6*B*a**2*b - 10*D*a**3 + x**3*(-5*A*b**3 + 9*C*a*b**2) + x**2*(8*B*a*b**2 - 12*D*a**2*b) + x*(-3
*A*a*b**2 + 7*C*a**2*b))/(8*a**2*b**4 + 16*a*b**5*x**2 + 8*b**6*x**4)

________________________________________________________________________________________

Giac [A]  time = 1.19736, size = 212, normalized size = 1.15 \begin{align*} -\frac{3 \,{\left (5 \, C a - A b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} b^{3}} - \frac{{\left (3 \, D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac{D b^{3} x^{2} + 2 \, C b^{3} x}{2 \, b^{6}} - \frac{10 \, D a^{3} - 6 \, B a^{2} b -{\left (9 \, C a b^{2} - 5 \, A b^{3}\right )} x^{3} + 4 \,{\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2} -{\left (7 \, C a^{2} b - 3 \, A a b^{2}\right )} x}{8 \,{\left (b x^{2} + a\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-3/8*(5*C*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*(3*D*a - B*b)*log(b*x^2 + a)/b^4 + 1/2*(D*b^3*x
^2 + 2*C*b^3*x)/b^6 - 1/8*(10*D*a^3 - 6*B*a^2*b - (9*C*a*b^2 - 5*A*b^3)*x^3 + 4*(3*D*a^2*b - 2*B*a*b^2)*x^2 -
(7*C*a^2*b - 3*A*a*b^2)*x)/((b*x^2 + a)^2*b^4)