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

Optimal. Leaf size=124 \[ a^2 x (a C+3 A b)-\frac{a^3 A}{x}+\frac{3}{2} a^2 b B x^2+a^3 B \log (x)+\frac{1}{5} b^2 x^5 (3 a C+A b)+a b x^3 (a C+A b)+\frac{3}{4} a b^2 B x^4+\frac{D \left (a+b x^2\right )^4}{8 b}+\frac{1}{6} b^3 B x^6+\frac{1}{7} b^3 C x^7 \]

[Out]

-((a^3*A)/x) + a^2*(3*A*b + a*C)*x + (3*a^2*b*B*x^2)/2 + a*b*(A*b + a*C)*x^3 + (3*a*b^2*B*x^4)/4 + (b^2*(A*b +
 3*a*C)*x^5)/5 + (b^3*B*x^6)/6 + (b^3*C*x^7)/7 + (D*(a + b*x^2)^4)/(8*b) + a^3*B*Log[x]

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Rubi [A]  time = 0.10932, antiderivative size = 124, 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} \[ a^2 x (a C+3 A b)-\frac{a^3 A}{x}+\frac{3}{2} a^2 b B x^2+a^3 B \log (x)+\frac{1}{5} b^2 x^5 (3 a C+A b)+a b x^3 (a C+A b)+\frac{3}{4} a b^2 B x^4+\frac{D \left (a+b x^2\right )^4}{8 b}+\frac{1}{6} b^3 B x^6+\frac{1}{7} b^3 C x^7 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*A)/x) + a^2*(3*A*b + a*C)*x + (3*a^2*b*B*x^2)/2 + a*b*(A*b + a*C)*x^3 + (3*a*b^2*B*x^4)/4 + (b^2*(A*b +
 3*a*C)*x^5)/5 + (b^3*B*x^6)/6 + (b^3*C*x^7)/7 + (D*(a + b*x^2)^4)/(8*b) + a^3*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 )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx &=\frac{D \left (a+b x^2\right )^4}{8 b}+\int \frac{\left (a+b x^2\right )^3 \left (A+B x+C x^2\right )}{x^2} \, dx\\ &=\frac{D \left (a+b x^2\right )^4}{8 b}+\int \left (a^2 (3 A b+a C)+\frac{a^3 A}{x^2}+\frac{a^3 B}{x}+3 a^2 b B x+3 a b (A b+a C) x^2+3 a b^2 B x^3+b^2 (A b+3 a C) x^4+b^3 B x^5+b^3 C x^6\right ) \, dx\\ &=-\frac{a^3 A}{x}+a^2 (3 A b+a C) x+\frac{3}{2} a^2 b B x^2+a b (A b+a C) x^3+\frac{3}{4} a b^2 B x^4+\frac{1}{5} b^2 (A b+3 a C) x^5+\frac{1}{6} b^3 B x^6+\frac{1}{7} b^3 C x^7+\frac{D \left (a+b x^2\right )^4}{8 b}+a^3 B \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0800174, size = 123, normalized size = 0.99 \[ \frac{1}{4} a^2 b x (12 A+x (6 B+x (4 C+3 D x)))+a^3 \left (-\frac{A}{x}+C x+\frac{D x^2}{2}\right )+a^3 B \log (x)+\frac{1}{20} a b^2 x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+\frac{1}{840} b^3 x^5 (168 A+5 x (28 B+3 x (8 C+7 D x))) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(-(A/x) + C*x + (D*x^2)/2) + (a^2*b*x*(12*A + x*(6*B + x*(4*C + 3*D*x))))/4 + (a*b^2*x^3*(20*A + x*(15*B +
 2*x*(6*C + 5*D*x))))/20 + (b^3*x^5*(168*A + 5*x*(28*B + 3*x*(8*C + 7*D*x))))/840 + a^3*B*Log[x]

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Maple [A]  time = 0.006, size = 145, normalized size = 1.2 \begin{align*}{\frac{D{b}^{3}{x}^{8}}{8}}+{\frac{{b}^{3}C{x}^{7}}{7}}+{\frac{B{x}^{6}{b}^{3}}{6}}+{\frac{D{x}^{6}a{b}^{2}}{2}}+{\frac{A{x}^{5}{b}^{3}}{5}}+{\frac{3\,C{x}^{5}a{b}^{2}}{5}}+{\frac{3\,B{x}^{4}a{b}^{2}}{4}}+{\frac{3\,D{x}^{4}{a}^{2}b}{4}}+A{x}^{3}a{b}^{2}+C{x}^{3}{a}^{2}b+{\frac{3\,B{x}^{2}{a}^{2}b}{2}}+{\frac{D{x}^{2}{a}^{3}}{2}}+3\,A{a}^{2}bx+{a}^{3}Cx+{a}^{3}B\ln \left ( x \right ) -{\frac{A{a}^{3}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*D*b^3*x^8+1/7*b^3*C*x^7+1/6*B*x^6*b^3+1/2*D*x^6*a*b^2+1/5*A*x^5*b^3+3/5*C*x^5*a*b^2+3/4*B*x^4*a*b^2+3/4*D*
x^4*a^2*b+A*x^3*a*b^2+C*x^3*a^2*b+3/2*B*x^2*a^2*b+1/2*D*x^2*a^3+3*A*a^2*b*x+a^3*C*x+a^3*B*ln(x)-a^3*A/x

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Maxima [A]  time = 1.0072, size = 188, normalized size = 1.52 \begin{align*} \frac{1}{8} \, D b^{3} x^{8} + \frac{1}{7} \, C b^{3} x^{7} + \frac{1}{6} \,{\left (3 \, D a b^{2} + B b^{3}\right )} x^{6} + \frac{1}{5} \,{\left (3 \, C a b^{2} + A b^{3}\right )} x^{5} + \frac{3}{4} \,{\left (D a^{2} b + B a b^{2}\right )} x^{4} + B a^{3} \log \left (x\right ) +{\left (C a^{2} b + A a b^{2}\right )} x^{3} - \frac{A a^{3}}{x} + \frac{1}{2} \,{\left (D a^{3} + 3 \, B a^{2} b\right )} x^{2} +{\left (C a^{3} + 3 \, A a^{2} b\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*D*b^3*x^8 + 1/7*C*b^3*x^7 + 1/6*(3*D*a*b^2 + B*b^3)*x^6 + 1/5*(3*C*a*b^2 + A*b^3)*x^5 + 3/4*(D*a^2*b + B*a
*b^2)*x^4 + B*a^3*log(x) + (C*a^2*b + A*a*b^2)*x^3 - A*a^3/x + 1/2*(D*a^3 + 3*B*a^2*b)*x^2 + (C*a^3 + 3*A*a^2*
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)^3*(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 = 0.456226, size = 150, normalized size = 1.21 \begin{align*} - \frac{A a^{3}}{x} + B a^{3} \log{\left (x \right )} + \frac{C b^{3} x^{7}}{7} + \frac{D b^{3} x^{8}}{8} + x^{6} \left (\frac{B b^{3}}{6} + \frac{D a b^{2}}{2}\right ) + x^{5} \left (\frac{A b^{3}}{5} + \frac{3 C a b^{2}}{5}\right ) + x^{4} \left (\frac{3 B a b^{2}}{4} + \frac{3 D a^{2} b}{4}\right ) + x^{3} \left (A a b^{2} + C a^{2} b\right ) + x^{2} \left (\frac{3 B a^{2} b}{2} + \frac{D a^{3}}{2}\right ) + x \left (3 A a^{2} b + C a^{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a**3/x + B*a**3*log(x) + C*b**3*x**7/7 + D*b**3*x**8/8 + x**6*(B*b**3/6 + D*a*b**2/2) + x**5*(A*b**3/5 + 3*
C*a*b**2/5) + x**4*(3*B*a*b**2/4 + 3*D*a**2*b/4) + x**3*(A*a*b**2 + C*a**2*b) + x**2*(3*B*a**2*b/2 + D*a**3/2)
 + x*(3*A*a**2*b + C*a**3)

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Giac [A]  time = 1.15394, size = 196, normalized size = 1.58 \begin{align*} \frac{1}{8} \, D b^{3} x^{8} + \frac{1}{7} \, C b^{3} x^{7} + \frac{1}{2} \, D a b^{2} x^{6} + \frac{1}{6} \, B b^{3} x^{6} + \frac{3}{5} \, C a b^{2} x^{5} + \frac{1}{5} \, A b^{3} x^{5} + \frac{3}{4} \, D a^{2} b x^{4} + \frac{3}{4} \, B a b^{2} x^{4} + C a^{2} b x^{3} + A a b^{2} x^{3} + \frac{1}{2} \, D a^{3} x^{2} + \frac{3}{2} \, B a^{2} b x^{2} + C a^{3} x + 3 \, A a^{2} b x + B a^{3} \log \left ({\left | x \right |}\right ) - \frac{A a^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*D*b^3*x^8 + 1/7*C*b^3*x^7 + 1/2*D*a*b^2*x^6 + 1/6*B*b^3*x^6 + 3/5*C*a*b^2*x^5 + 1/5*A*b^3*x^5 + 3/4*D*a^2*
b*x^4 + 3/4*B*a*b^2*x^4 + C*a^2*b*x^3 + A*a*b^2*x^3 + 1/2*D*a^3*x^2 + 3/2*B*a^2*b*x^2 + C*a^3*x + 3*A*a^2*b*x
+ B*a^3*log(abs(x)) - A*a^3/x