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

Optimal. Leaf size=90 \[ -\frac{a^2 A}{x}+a^2 B \log (x)+\frac{1}{3} b x^3 (2 a C+A b)+a x (a C+2 A b)+a b B x^2+\frac{D \left (a+b x^2\right )^3}{6 b}+\frac{1}{4} b^2 B x^4+\frac{1}{5} b^2 C x^5 \]

[Out]

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

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Rubi [A]  time = 0.080234, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {1583, 1628} \[ -\frac{a^2 A}{x}+a^2 B \log (x)+\frac{1}{3} b x^3 (2 a C+A b)+a x (a C+2 A b)+a b B x^2+\frac{D \left (a+b x^2\right )^3}{6 b}+\frac{1}{4} b^2 B x^4+\frac{1}{5} b^2 C x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1583

Int[(Px_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(Coeff[Px, x, n - m - 1]*(a + b*x^n)^(p
+ 1))/(b*n*(p + 1)), x] + Int[(Px - Coeff[Px, x, n - m - 1]*x^(n - m - 1))*x^m*(a + b*x^n)^p, x] /; FreeQ[{a,
b, m, n}, x] && PolyQ[Px, x] && IGtQ[p, 1] && IGtQ[n - m, 0] && NeQ[Coeff[Px, x, n - m - 1], 0]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.0503056, size = 88, normalized size = 0.98 \[ a^2 \left (-\frac{A}{x}+C x+\frac{D x^2}{2}\right )+a^2 B \log (x)+\frac{1}{6} a b x (12 A+x (6 B+x (4 C+3 D x)))+\frac{1}{60} b^2 x^3 (20 A+x (15 B+2 x (6 C+5 D x))) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 98, normalized size = 1.1 \begin{align*}{\frac{D{b}^{2}{x}^{6}}{6}}+{\frac{{b}^{2}C{x}^{5}}{5}}+{\frac{{b}^{2}B{x}^{4}}{4}}+{\frac{D{x}^{4}ab}{2}}+{\frac{A{x}^{3}{b}^{2}}{3}}+{\frac{2\,C{x}^{3}ab}{3}}+B{x}^{2}ab+{\frac{D{x}^{2}{a}^{2}}{2}}+2\,Aabx+{a}^{2}Cx+{a}^{2}B\ln \left ( x \right ) -{\frac{A{a}^{2}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.994366, size = 130, normalized size = 1.44 \begin{align*} \frac{1}{6} \, D b^{2} x^{6} + \frac{1}{5} \, C b^{2} x^{5} + \frac{1}{4} \,{\left (2 \, D a b + B b^{2}\right )} x^{4} + \frac{1}{3} \,{\left (2 \, C a b + A b^{2}\right )} x^{3} + B a^{2} \log \left (x\right ) + \frac{1}{2} \,{\left (D a^{2} + 2 \, B a b\right )} x^{2} - \frac{A a^{2}}{x} +{\left (C a^{2} + 2 \, A a b\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: UnboundLocalError

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Sympy [A]  time = 0.37863, size = 99, normalized size = 1.1 \begin{align*} - \frac{A a^{2}}{x} + B a^{2} \log{\left (x \right )} + \frac{C b^{2} x^{5}}{5} + \frac{D b^{2} x^{6}}{6} + x^{4} \left (\frac{B b^{2}}{4} + \frac{D a b}{2}\right ) + x^{3} \left (\frac{A b^{2}}{3} + \frac{2 C a b}{3}\right ) + x^{2} \left (B a b + \frac{D a^{2}}{2}\right ) + x \left (2 A a b + C a^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22434, size = 132, normalized size = 1.47 \begin{align*} \frac{1}{6} \, D b^{2} x^{6} + \frac{1}{5} \, C b^{2} x^{5} + \frac{1}{2} \, D a b x^{4} + \frac{1}{4} \, B b^{2} x^{4} + \frac{2}{3} \, C a b x^{3} + \frac{1}{3} \, A b^{2} x^{3} + \frac{1}{2} \, D a^{2} x^{2} + B a b x^{2} + C a^{2} x + 2 \, A a b x + B a^{2} \log \left ({\left | x \right |}\right ) - \frac{A a^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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