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

Optimal. Leaf size=130 \[ \frac{a^{3/2} (b B-a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{7/2}}+\frac{x^2 (A b-a C)}{2 b^2}-\frac{a (A b-a C) \log \left (a+b x^2\right )}{2 b^3}+\frac{x^3 (b B-a D)}{3 b^2}-\frac{a x (b B-a D)}{b^3}+\frac{C x^4}{4 b}+\frac{D x^5}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.123792, antiderivative size = 130, 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} \[ \frac{a^{3/2} (b B-a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{7/2}}+\frac{x^2 (A b-a C)}{2 b^2}-\frac{a (A b-a C) \log \left (a+b x^2\right )}{2 b^3}+\frac{x^3 (b B-a D)}{3 b^2}-\frac{a x (b B-a D)}{b^3}+\frac{C x^4}{4 b}+\frac{D x^5}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0903211, size = 114, normalized size = 0.88 \[ \frac{x \left (60 a^2 D-10 a b (6 B+x (3 C+2 D x))+b^2 x (30 A+x (20 B+3 x (5 C+4 D x)))\right )+30 a (a C-A b) \log \left (a+b x^2\right )}{60 b^3}-\frac{a^{3/2} (a D-b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*(-(b*B) + a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(7/2)) + (x*(60*a^2*D - 10*a*b*(6*B + x*(3*C + 2*D*x)
) + b^2*x*(30*A + x*(20*B + 3*x*(5*C + 4*D*x)))) + 30*a*(-(A*b) + a*C)*Log[a + b*x^2])/(60*b^3)

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Maple [A]  time = 0.006, size = 152, normalized size = 1.2 \begin{align*}{\frac{D{x}^{5}}{5\,b}}+{\frac{C{x}^{4}}{4\,b}}+{\frac{B{x}^{3}}{3\,b}}-{\frac{D{x}^{3}a}{3\,{b}^{2}}}+{\frac{A{x}^{2}}{2\,b}}-{\frac{C{x}^{2}a}{2\,{b}^{2}}}-{\frac{Bax}{{b}^{2}}}+{\frac{{a}^{2}Dx}{{b}^{3}}}-{\frac{a\ln \left ( b{x}^{2}+a \right ) A}{2\,{b}^{2}}}+{\frac{{a}^{2}\ln \left ( b{x}^{2}+a \right ) C}{2\,{b}^{3}}}+{\frac{{a}^{2}B}{{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{a}^{3}D}{{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),x)

[Out]

1/5*D*x^5/b+1/4*C*x^4/b+1/3/b*B*x^3-1/3/b^2*D*x^3*a+1/2/b*A*x^2-1/2/b^2*C*x^2*a-1/b^2*B*x*a+1/b^3*a^2*D*x-1/2*
a/b^2*ln(b*x^2+a)*A+1/2*a^2/b^3*ln(b*x^2+a)*C+a^2/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*B-a^3/b^3/(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^3*(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^3*(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 = 1.07699, size = 269, normalized size = 2.07 \begin{align*} \frac{C x^{4}}{4 b} + \frac{D x^{5}}{5 b} + \left (\frac{a \left (- A b + C a\right )}{2 b^{3}} - \frac{\sqrt{- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right ) \log{\left (x + \frac{- A a b + C a^{2} - 2 b^{3} \left (\frac{a \left (- A b + C a\right )}{2 b^{3}} - \frac{\sqrt{- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right )}{- B a b + D a^{2}} \right )} + \left (\frac{a \left (- A b + C a\right )}{2 b^{3}} + \frac{\sqrt{- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right ) \log{\left (x + \frac{- A a b + C a^{2} - 2 b^{3} \left (\frac{a \left (- A b + C a\right )}{2 b^{3}} + \frac{\sqrt{- a^{3} b^{7}} \left (- B b + D a\right )}{2 b^{7}}\right )}{- B a b + D a^{2}} \right )} - \frac{x^{3} \left (- B b + D a\right )}{3 b^{2}} - \frac{x^{2} \left (- A b + C a\right )}{2 b^{2}} + \frac{x \left (- B a b + D a^{2}\right )}{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),x)

[Out]

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

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Giac [A]  time = 1.18655, size = 185, normalized size = 1.42 \begin{align*} \frac{{\left (C a^{2} - A a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} - \frac{{\left (D a^{3} - B a^{2} b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} + \frac{12 \, D b^{4} x^{5} + 15 \, C b^{4} x^{4} - 20 \, D a b^{3} x^{3} + 20 \, B b^{4} x^{3} - 30 \, C a b^{3} x^{2} + 30 \, A b^{4} x^{2} + 60 \, D a^{2} b^{2} x - 60 \, B a b^{3} x}{60 \, b^{5}} \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),x, algorithm="giac")

[Out]

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