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

Optimal. Leaf size=176 \[ -\frac{x^3 (3 A b-5 a C)}{6 a b^2}+\frac{x (3 A b-5 a C)}{2 b^3}-\frac{\sqrt{a} (3 A b-5 a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}-\frac{x^4 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac{x^2 (2 b B-3 a D)}{2 b^3}-\frac{a (2 b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}+\frac{D x^4}{4 b^2} \]

[Out]

((3*A*b - 5*a*C)*x)/(2*b^3) + ((2*b*B - 3*a*D)*x^2)/(2*b^3) - ((3*A*b - 5*a*C)*x^3)/(6*a*b^2) + (D*x^4)/(4*b^2
) - (x^4*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(2*a*b*(a + b*x^2)) - (Sqrt[a]*(3*A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/
Sqrt[a]])/(2*b^(7/2)) - (a*(2*b*B - 3*a*D)*Log[a + b*x^2])/(2*b^4)

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Rubi [A]  time = 0.267927, antiderivative size = 176, 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, 1802, 635, 205, 260} \[ -\frac{x^3 (3 A b-5 a C)}{6 a b^2}+\frac{x (3 A b-5 a C)}{2 b^3}-\frac{\sqrt{a} (3 A b-5 a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}-\frac{x^4 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac{x^2 (2 b B-3 a D)}{2 b^3}-\frac{a (2 b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}+\frac{D x^4}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A*b - 5*a*C)*x)/(2*b^3) + ((2*b*B - 3*a*D)*x^2)/(2*b^3) - ((3*A*b - 5*a*C)*x^3)/(6*a*b^2) + (D*x^4)/(4*b^2
) - (x^4*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(2*a*b*(a + b*x^2)) - (Sqrt[a]*(3*A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/
Sqrt[a]])/(2*b^(7/2)) - (a*(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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, 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^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^2} \, dx &=-\frac{x^4 \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{x^3 \left (-4 a \left (B-\frac{a D}{b}\right )+(3 A b-5 a C) x-2 a D x^2\right )}{a+b x^2} \, dx}{2 a b}\\ &=-\frac{x^4 \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{a (3 A b-5 a C)}{b^2}-\frac{2 a (2 b B-3 a D) x}{b^2}+\frac{(3 A b-5 a C) x^2}{b}-\frac{2 a D x^3}{b}+\frac{a^2 (3 A b-5 a C)+2 a^2 (2 b B-3 a D) x}{b^2 \left (a+b x^2\right )}\right ) \, dx}{2 a b}\\ &=\frac{(3 A b-5 a C) x}{2 b^3}+\frac{(2 b B-3 a D) x^2}{2 b^3}-\frac{(3 A b-5 a C) x^3}{6 a b^2}+\frac{D x^4}{4 b^2}-\frac{x^4 \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^2 (3 A b-5 a C)+2 a^2 (2 b B-3 a D) x}{a+b x^2} \, dx}{2 a b^3}\\ &=\frac{(3 A b-5 a C) x}{2 b^3}+\frac{(2 b B-3 a D) x^2}{2 b^3}-\frac{(3 A b-5 a C) x^3}{6 a b^2}+\frac{D x^4}{4 b^2}-\frac{x^4 \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{(a (3 A b-5 a C)) \int \frac{1}{a+b x^2} \, dx}{2 b^3}-\frac{(a (2 b B-3 a D)) \int \frac{x}{a+b x^2} \, dx}{b^3}\\ &=\frac{(3 A b-5 a C) x}{2 b^3}+\frac{(2 b B-3 a D) x^2}{2 b^3}-\frac{(3 A b-5 a C) x^3}{6 a b^2}+\frac{D x^4}{4 b^2}-\frac{x^4 \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{\sqrt{a} (3 A b-5 a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}-\frac{a (2 b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}\\ \end{align*}

Mathematica [A]  time = 0.125929, size = 139, normalized size = 0.79 \[ \frac{\frac{6 a \left (a^2 D-a b (B+C x)+A b^2 x\right )}{a+b x^2}+12 b x (A b-2 a C)+6 \sqrt{a} \sqrt{b} (5 a C-3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )+6 b x^2 (b B-2 a D)+6 a (3 a D-2 b B) \log \left (a+b x^2\right )+4 b^2 C x^3+3 b^2 D x^4}{12 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*b*(A*b - 2*a*C)*x + 6*b*(b*B - 2*a*D)*x^2 + 4*b^2*C*x^3 + 3*b^2*D*x^4 + (6*a*(a^2*D + A*b^2*x - a*b*(B + C
*x)))/(a + b*x^2) + 6*Sqrt[a]*Sqrt[b]*(-3*A*b + 5*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]] + 6*a*(-2*b*B + 3*a*D)*Log[
a + b*x^2])/(12*b^4)

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Maple [A]  time = 0.01, size = 201, normalized size = 1.1 \begin{align*}{\frac{D{x}^{4}}{4\,{b}^{2}}}+{\frac{C{x}^{3}}{3\,{b}^{2}}}+{\frac{B{x}^{2}}{2\,{b}^{2}}}-{\frac{D{x}^{2}a}{{b}^{3}}}+{\frac{Ax}{{b}^{2}}}-2\,{\frac{aCx}{{b}^{3}}}+{\frac{aAx}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{{a}^{2}Cx}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}-{\frac{{a}^{2}B}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{{a}^{3}D}{2\,{b}^{4} \left ( b{x}^{2}+a \right ) }}-{\frac{\ln \left ( b{x}^{2}+a \right ) Ba}{{b}^{3}}}+{\frac{3\,{a}^{2}\ln \left ( b{x}^{2}+a \right ) D}{2\,{b}^{4}}}-{\frac{3\,Aa}{2\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{a}^{2}C}{2\,{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)^2,x)

[Out]

1/4*D*x^4/b^2+1/3/b^2*C*x^3+1/2*B*x^2/b^2-1/b^3*D*x^2*a+1/b^2*A*x-2/b^3*a*C*x+1/2*a/b^2/(b*x^2+a)*A*x-1/2*a^2/
b^3/(b*x^2+a)*C*x-1/2/b^3*a^2/(b*x^2+a)*B+1/2*a^3/b^4/(b*x^2+a)*D-1/b^3*ln(b*x^2+a)*B*a+3/2*a^2/b^4*ln(b*x^2+a
)*D-3/2*a/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*A+5/2*a^2/b^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*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(x^4*(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^4*(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 = 3.68118, size = 333, normalized size = 1.89 \begin{align*} \frac{C x^{3}}{3 b^{2}} + \frac{D x^{4}}{4 b^{2}} + \left (\frac{a \left (- 2 B b + 3 D a\right )}{2 b^{4}} - \frac{\sqrt{- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right ) \log{\left (x + \frac{4 B a b - 6 D a^{2} + 4 b^{4} \left (\frac{a \left (- 2 B b + 3 D a\right )}{2 b^{4}} - \frac{\sqrt{- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right )}{- 3 A b^{2} + 5 C a b} \right )} + \left (\frac{a \left (- 2 B b + 3 D a\right )}{2 b^{4}} + \frac{\sqrt{- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right ) \log{\left (x + \frac{4 B a b - 6 D a^{2} + 4 b^{4} \left (\frac{a \left (- 2 B b + 3 D a\right )}{2 b^{4}} + \frac{\sqrt{- a b^{9}} \left (- 3 A b + 5 C a\right )}{4 b^{8}}\right )}{- 3 A b^{2} + 5 C a b} \right )} - \frac{B a^{2} b - D a^{3} + x \left (- A a b^{2} + C a^{2} b\right )}{2 a b^{4} + 2 b^{5} x^{2}} - \frac{x^{2} \left (- B b + 2 D a\right )}{2 b^{3}} - \frac{x \left (- A b + 2 C a\right )}{b^{3}} \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)**2,x)

[Out]

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

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Giac [A]  time = 1.18027, size = 215, normalized size = 1.22 \begin{align*} \frac{{\left (5 \, C a^{2} - 3 \, A a b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{3}} + \frac{{\left (3 \, D a^{2} - 2 \, B a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac{D a^{3} - B a^{2} b -{\left (C a^{2} b - A a b^{2}\right )} x}{2 \,{\left (b x^{2} + a\right )} b^{4}} + \frac{3 \, D b^{6} x^{4} + 4 \, C b^{6} x^{3} - 12 \, D a b^{5} x^{2} + 6 \, B b^{6} x^{2} - 24 \, C a b^{5} x + 12 \, A b^{6} x}{12 \, b^{8}} \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)^2,x, algorithm="giac")

[Out]

1/2*(5*C*a^2 - 3*A*a*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/2*(3*D*a^2 - 2*B*a*b)*log(b*x^2 + a)/b^4 + 1
/2*(D*a^3 - B*a^2*b - (C*a^2*b - A*a*b^2)*x)/((b*x^2 + a)*b^4) + 1/12*(3*D*b^6*x^4 + 4*C*b^6*x^3 - 12*D*a*b^5*
x^2 + 6*B*b^6*x^2 - 24*C*a*b^5*x + 12*A*b^6*x)/b^8