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

Optimal. Leaf size=154 \[ -\frac{x^2 (A b-2 a C)}{2 a b^2}+\frac{(A b-2 a C) \log \left (a+b x^2\right )}{2 b^3}-\frac{x^3 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac{x (3 b B-5 a D)}{2 b^3}-\frac{\sqrt{a} (3 b B-5 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}+\frac{D x^3}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.241322, antiderivative size = 154, 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} \[ -\frac{x^2 (A b-2 a C)}{2 a b^2}+\frac{(A b-2 a C) \log \left (a+b x^2\right )}{2 b^3}-\frac{x^3 \left (a \left (B-\frac{a D}{b}\right )-x (A b-a C)\right )}{2 a b \left (a+b x^2\right )}+\frac{x (3 b B-5 a D)}{2 b^3}-\frac{\sqrt{a} (3 b B-5 a D) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}+\frac{D x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0770374, size = 128, normalized size = 0.83 \[ \frac{a (-a (C+D x)+A b+b B x)}{2 b^3 \left (a+b x^2\right )}+\frac{(A b-2 a C) \log \left (a+b x^2\right )}{2 b^3}+\frac{x (b B-2 a D)}{b^3}+\frac{\sqrt{a} (5 a D-3 b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}+\frac{C x^2}{2 b^2}+\frac{D x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 177, normalized size = 1.2 \begin{align*}{\frac{D{x}^{3}}{3\,{b}^{2}}}+{\frac{C{x}^{2}}{2\,{b}^{2}}}+{\frac{Bx}{{b}^{2}}}-2\,{\frac{aDx}{{b}^{3}}}+{\frac{Bax}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{{a}^{2}Dx}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{aA}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{{a}^{2}C}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{\ln \left ( b{x}^{2}+a \right ) A}{2\,{b}^{2}}}-{\frac{\ln \left ( b{x}^{2}+a \right ) aC}{{b}^{3}}}-{\frac{3\,Ba}{2\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{a}^{2}D}{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^3*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^2,x)

[Out]

1/3*D*x^3/b^2+1/2/b^2*C*x^2+1/b^2*B*x-2/b^3*a*D*x+1/2/b^2/(b*x^2+a)*B*x*a-1/2/b^3/(b*x^2+a)*a^2*D*x+1/2/b^2*a/
(b*x^2+a)*A-1/2/b^3/(b*x^2+a)*a^2*C+1/2/b^2*ln(b*x^2+a)*A-1/b^3*ln(b*x^2+a)*a*C-3/2/b^2/(a*b)^(1/2)*arctan(b*x
/(a*b)^(1/2))*B*a+5/2/b^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*a^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)^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^3*(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 = 3.46395, size = 287, normalized size = 1.86 \begin{align*} \frac{C x^{2}}{2 b^{2}} + \frac{D x^{3}}{3 b^{2}} + \left (- \frac{- A b + 2 C a}{2 b^{3}} - \frac{\sqrt{- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right ) \log{\left (x + \frac{- 2 A b + 4 C a + 4 b^{3} \left (- \frac{- A b + 2 C a}{2 b^{3}} - \frac{\sqrt{- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right )}{- 3 B b + 5 D a} \right )} + \left (- \frac{- A b + 2 C a}{2 b^{3}} + \frac{\sqrt{- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right ) \log{\left (x + \frac{- 2 A b + 4 C a + 4 b^{3} \left (- \frac{- A b + 2 C a}{2 b^{3}} + \frac{\sqrt{- a b^{7}} \left (- 3 B b + 5 D a\right )}{4 b^{7}}\right )}{- 3 B b + 5 D a} \right )} - \frac{- A a b + C a^{2} + x \left (- B a b + D a^{2}\right )}{2 a b^{3} + 2 b^{4} x^{2}} - \frac{x \left (- B b + 2 D a\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)**2,x)

[Out]

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

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Giac [A]  time = 1.20425, size = 177, normalized size = 1.15 \begin{align*} -\frac{{\left (2 \, C a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac{{\left (5 \, D a^{2} - 3 \, B a b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{3}} - \frac{C a^{2} - A a b +{\left (D a^{2} - B a b\right )} x}{2 \,{\left (b x^{2} + a\right )} b^{3}} + \frac{2 \, D b^{4} x^{3} + 3 \, C b^{4} x^{2} - 12 \, D a b^{3} x + 6 \, B b^{4} x}{6 \, b^{6}} \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)^2,x, algorithm="giac")

[Out]

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