3.32 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^9} \, dx\)

Optimal. Leaf size=71 \[ b^2 \log (x) (3 A c+b B)-\frac{A b^3}{2 x^2}+\frac{1}{4} c^2 x^4 (A c+3 b B)+\frac{3}{2} b c x^2 (A c+b B)+\frac{1}{6} B c^3 x^6 \]

[Out]

-(A*b^3)/(2*x^2) + (3*b*c*(b*B + A*c)*x^2)/2 + (c^2*(3*b*B + A*c)*x^4)/4 + (B*c^3*x^6)/6 + b^2*(b*B + 3*A*c)*L
og[x]

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Rubi [A]  time = 0.0789122, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1584, 446, 76} \[ b^2 \log (x) (3 A c+b B)-\frac{A b^3}{2 x^2}+\frac{1}{4} c^2 x^4 (A c+3 b B)+\frac{3}{2} b c x^2 (A c+b B)+\frac{1}{6} B c^3 x^6 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^9,x]

[Out]

-(A*b^3)/(2*x^2) + (3*b*c*(b*B + A*c)*x^2)/2 + (c^2*(3*b*B + A*c)*x^4)/4 + (B*c^3*x^6)/6 + b^2*(b*B + 3*A*c)*L
og[x]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^3} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) (b+c x)^3}{x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (3 b c (b B+A c)+\frac{A b^3}{x^2}+\frac{b^2 (b B+3 A c)}{x}+c^2 (3 b B+A c) x+B c^3 x^2\right ) \, dx,x,x^2\right )\\ &=-\frac{A b^3}{2 x^2}+\frac{3}{2} b c (b B+A c) x^2+\frac{1}{4} c^2 (3 b B+A c) x^4+\frac{1}{6} B c^3 x^6+b^2 (b B+3 A c) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0273874, size = 73, normalized size = 1.03 \[ \log (x) \left (3 A b^2 c+b^3 B\right )-\frac{A b^3}{2 x^2}+\frac{1}{4} c^2 x^4 (A c+3 b B)+\frac{3}{2} b c x^2 (A c+b B)+\frac{1}{6} B c^3 x^6 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^9,x]

[Out]

-(A*b^3)/(2*x^2) + (3*b*c*(b*B + A*c)*x^2)/2 + (c^2*(3*b*B + A*c)*x^4)/4 + (B*c^3*x^6)/6 + (b^3*B + 3*A*b^2*c)
*Log[x]

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Maple [A]  time = 0.006, size = 75, normalized size = 1.1 \begin{align*}{\frac{B{c}^{3}{x}^{6}}{6}}+{\frac{A{x}^{4}{c}^{3}}{4}}+{\frac{3\,B{x}^{4}b{c}^{2}}{4}}+{\frac{3\,A{x}^{2}b{c}^{2}}{2}}+{\frac{3\,B{x}^{2}{b}^{2}c}{2}}+3\,A\ln \left ( x \right ){b}^{2}c+B\ln \left ( x \right ){b}^{3}-{\frac{A{b}^{3}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x)

[Out]

1/6*B*c^3*x^6+1/4*A*x^4*c^3+3/4*B*x^4*b*c^2+3/2*A*x^2*b*c^2+3/2*B*x^2*b^2*c+3*A*ln(x)*b^2*c+B*ln(x)*b^3-1/2*A*
b^3/x^2

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Maxima [A]  time = 1.18279, size = 100, normalized size = 1.41 \begin{align*} \frac{1}{6} \, B c^{3} x^{6} + \frac{1}{4} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + \frac{3}{2} \,{\left (B b^{2} c + A b c^{2}\right )} x^{2} - \frac{A b^{3}}{2 \, x^{2}} + \frac{1}{2} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x, algorithm="maxima")

[Out]

1/6*B*c^3*x^6 + 1/4*(3*B*b*c^2 + A*c^3)*x^4 + 3/2*(B*b^2*c + A*b*c^2)*x^2 - 1/2*A*b^3/x^2 + 1/2*(B*b^3 + 3*A*b
^2*c)*log(x^2)

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Fricas [A]  time = 0.497962, size = 171, normalized size = 2.41 \begin{align*} \frac{2 \, B c^{3} x^{8} + 3 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 18 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} - 6 \, A b^{3} + 12 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \log \left (x\right )}{12 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x, algorithm="fricas")

[Out]

1/12*(2*B*c^3*x^8 + 3*(3*B*b*c^2 + A*c^3)*x^6 + 18*(B*b^2*c + A*b*c^2)*x^4 - 6*A*b^3 + 12*(B*b^3 + 3*A*b^2*c)*
x^2*log(x))/x^2

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Sympy [A]  time = 0.409838, size = 78, normalized size = 1.1 \begin{align*} - \frac{A b^{3}}{2 x^{2}} + \frac{B c^{3} x^{6}}{6} + b^{2} \left (3 A c + B b\right ) \log{\left (x \right )} + x^{4} \left (\frac{A c^{3}}{4} + \frac{3 B b c^{2}}{4}\right ) + x^{2} \left (\frac{3 A b c^{2}}{2} + \frac{3 B b^{2} c}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**9,x)

[Out]

-A*b**3/(2*x**2) + B*c**3*x**6/6 + b**2*(3*A*c + B*b)*log(x) + x**4*(A*c**3/4 + 3*B*b*c**2/4) + x**2*(3*A*b*c*
*2/2 + 3*B*b**2*c/2)

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Giac [A]  time = 1.23608, size = 131, normalized size = 1.85 \begin{align*} \frac{1}{6} \, B c^{3} x^{6} + \frac{3}{4} \, B b c^{2} x^{4} + \frac{1}{4} \, A c^{3} x^{4} + \frac{3}{2} \, B b^{2} c x^{2} + \frac{3}{2} \, A b c^{2} x^{2} + \frac{1}{2} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} \log \left (x^{2}\right ) - \frac{B b^{3} x^{2} + 3 \, A b^{2} c x^{2} + A b^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x, algorithm="giac")

[Out]

1/6*B*c^3*x^6 + 3/4*B*b*c^2*x^4 + 1/4*A*c^3*x^4 + 3/2*B*b^2*c*x^2 + 3/2*A*b*c^2*x^2 + 1/2*(B*b^3 + 3*A*b^2*c)*
log(x^2) - 1/2*(B*b^3*x^2 + 3*A*b^2*c*x^2 + A*b^3)/x^2