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

Optimal. Leaf size=151 \[ \frac{a^{3/2} (A b-a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{7/2}}+\frac{a^2 (b B-a D) \log \left (a+b x^2\right )}{2 b^4}+\frac{x^3 (A b-a C)}{3 b^2}-\frac{a x (A b-a C)}{b^3}+\frac{x^4 (b B-a D)}{4 b^2}-\frac{a x^2 (b B-a D)}{2 b^3}+\frac{C x^5}{5 b}+\frac{D x^6}{6 b} \]

[Out]

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

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Rubi [A]  time = 0.144934, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1802, 635, 205, 260} \[ \frac{a^{3/2} (A b-a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{7/2}}+\frac{a^2 (b B-a D) \log \left (a+b x^2\right )}{2 b^4}+\frac{x^3 (A b-a C)}{3 b^2}-\frac{a x (A b-a C)}{b^3}+\frac{x^4 (b B-a D)}{4 b^2}-\frac{a x^2 (b B-a D)}{2 b^3}+\frac{C x^5}{5 b}+\frac{D x^6}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 )}{a+b x^2} \, dx &=\int \left (-\frac{a (A b-a C)}{b^3}-\frac{a (b B-a D) x}{b^3}+\frac{(A b-a C) x^2}{b^2}+\frac{(b B-a D) x^3}{b^2}+\frac{C x^4}{b}+\frac{D x^5}{b}+\frac{a^2 (A b-a C)+a^2 (b B-a D) x}{b^3 \left (a+b x^2\right )}\right ) \, dx\\ &=-\frac{a (A b-a C) x}{b^3}-\frac{a (b B-a D) x^2}{2 b^3}+\frac{(A b-a C) x^3}{3 b^2}+\frac{(b B-a D) x^4}{4 b^2}+\frac{C x^5}{5 b}+\frac{D x^6}{6 b}+\frac{\int \frac{a^2 (A b-a C)+a^2 (b B-a D) x}{a+b x^2} \, dx}{b^3}\\ &=-\frac{a (A b-a C) x}{b^3}-\frac{a (b B-a D) x^2}{2 b^3}+\frac{(A b-a C) x^3}{3 b^2}+\frac{(b B-a D) x^4}{4 b^2}+\frac{C x^5}{5 b}+\frac{D x^6}{6 b}+\frac{\left (a^2 (A b-a C)\right ) \int \frac{1}{a+b x^2} \, dx}{b^3}+\frac{\left (a^2 (b B-a D)\right ) \int \frac{x}{a+b x^2} \, dx}{b^3}\\ &=-\frac{a (A b-a C) x}{b^3}-\frac{a (b B-a D) x^2}{2 b^3}+\frac{(A b-a C) x^3}{3 b^2}+\frac{(b B-a D) x^4}{4 b^2}+\frac{C x^5}{5 b}+\frac{D x^6}{6 b}+\frac{a^{3/2} (A b-a C) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{7/2}}+\frac{a^2 (b B-a D) \log \left (a+b x^2\right )}{2 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0730102, size = 130, normalized size = 0.86 \[ \frac{b x \left (30 a^2 (2 C+D x)-5 a b (12 A+x (6 B+x (4 C+3 D x)))+b^2 x^2 (20 A+x (15 B+2 x (6 C+5 D x)))\right )-60 a^{3/2} \sqrt{b} (a C-A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )-30 a^2 (a D-b B) \log \left (a+b x^2\right )}{60 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*(30*a^2*(2*C + D*x) - 5*a*b*(12*A + x*(6*B + x*(4*C + 3*D*x))) + b^2*x^2*(20*A + x*(15*B + 2*x*(6*C + 5*D
*x)))) - 60*a^(3/2)*Sqrt[b]*(-(A*b) + a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]] - 30*a^2*(-(b*B) + a*D)*Log[a + b*x^2])
/(60*b^4)

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Maple [A]  time = 0.005, size = 176, normalized size = 1.2 \begin{align*}{\frac{D{x}^{6}}{6\,b}}+{\frac{C{x}^{5}}{5\,b}}+{\frac{B{x}^{4}}{4\,b}}-{\frac{D{x}^{4}a}{4\,{b}^{2}}}+{\frac{A{x}^{3}}{3\,b}}-{\frac{C{x}^{3}a}{3\,{b}^{2}}}-{\frac{Ba{x}^{2}}{2\,{b}^{2}}}+{\frac{D{x}^{2}{a}^{2}}{2\,{b}^{3}}}-{\frac{aAx}{{b}^{2}}}+{\frac{{a}^{2}Cx}{{b}^{3}}}+{\frac{{a}^{2}\ln \left ( b{x}^{2}+a \right ) B}{2\,{b}^{3}}}-{\frac{{a}^{3}\ln \left ( b{x}^{2}+a \right ) D}{2\,{b}^{4}}}+{\frac{A{a}^{2}}{{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{a}^{3}C}{{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),x)

[Out]

1/6*D*x^6/b+1/5*C*x^5/b+1/4/b*B*x^4-1/4/b^2*D*x^4*a+1/3/b*A*x^3-1/3/b^2*C*x^3*a-1/2/b^2*B*x^2*a+1/2/b^3*D*x^2*
a^2-1/b^2*A*a*x+1/b^3*a^2*C*x+1/2*a^2/b^3*ln(b*x^2+a)*B-1/2*a^3/b^4*ln(b*x^2+a)*D+a^2/b^2/(a*b)^(1/2)*arctan(b
*x/(a*b)^(1/2))*A-a^3/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),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),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [B]  time = 1.11458, size = 308, normalized size = 2.04 \begin{align*} \frac{C x^{5}}{5 b} + \frac{D x^{6}}{6 b} + \left (- \frac{a^{2} \left (- B b + D a\right )}{2 b^{4}} - \frac{\sqrt{- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right ) \log{\left (x + \frac{B a^{2} b - D a^{3} - 2 b^{4} \left (- \frac{a^{2} \left (- B b + D a\right )}{2 b^{4}} - \frac{\sqrt{- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right )}{- A a b^{2} + C a^{2} b} \right )} + \left (- \frac{a^{2} \left (- B b + D a\right )}{2 b^{4}} + \frac{\sqrt{- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right ) \log{\left (x + \frac{B a^{2} b - D a^{3} - 2 b^{4} \left (- \frac{a^{2} \left (- B b + D a\right )}{2 b^{4}} + \frac{\sqrt{- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right )}{- A a b^{2} + C a^{2} b} \right )} - \frac{x^{4} \left (- B b + D a\right )}{4 b^{2}} - \frac{x^{3} \left (- A b + C a\right )}{3 b^{2}} + \frac{x^{2} \left (- B a b + D a^{2}\right )}{2 b^{3}} + \frac{x \left (- A a b + C a^{2}\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),x)

[Out]

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

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Giac [A]  time = 1.19921, size = 217, normalized size = 1.44 \begin{align*} -\frac{{\left (C a^{3} - A a^{2} b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} - \frac{{\left (D a^{3} - B a^{2} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac{10 \, D b^{5} x^{6} + 12 \, C b^{5} x^{5} - 15 \, D a b^{4} x^{4} + 15 \, B b^{5} x^{4} - 20 \, C a b^{4} x^{3} + 20 \, A b^{5} x^{3} + 30 \, D a^{2} b^{3} x^{2} - 30 \, B a b^{4} x^{2} + 60 \, C a^{2} b^{3} x - 60 \, A a b^{4} x}{60 \, b^{6}} \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),x, algorithm="giac")

[Out]

-(C*a^3 - A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*(D*a^3 - B*a^2*b)*log(b*x^2 + a)/b^4 + 1/60*(10
*D*b^5*x^6 + 12*C*b^5*x^5 - 15*D*a*b^4*x^4 + 15*B*b^5*x^4 - 20*C*a*b^4*x^3 + 20*A*b^5*x^3 + 30*D*a^2*b^3*x^2 -
30*B*a*b^4*x^2 + 60*C*a^2*b^3*x - 60*A*a*b^4*x)/b^6