3.34 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^{11}} \, dx\)

Optimal. Leaf size=72 \[ -\frac{b^2 (3 A c+b B)}{2 x^2}-\frac{A b^3}{4 x^4}+\frac{1}{2} c^2 x^2 (A c+3 b B)+3 b c \log (x) (A c+b B)+\frac{1}{4} B c^3 x^4 \]

[Out]

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

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Rubi [A]  time = 0.0719043, antiderivative size = 72, 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} \[ -\frac{b^2 (3 A c+b B)}{2 x^2}-\frac{A b^3}{4 x^4}+\frac{1}{2} c^2 x^2 (A c+3 b B)+3 b c \log (x) (A c+b B)+\frac{1}{4} B c^3 x^4 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^11,x]

[Out]

-(A*b^3)/(4*x^4) - (b^2*(b*B + 3*A*c))/(2*x^2) + (c^2*(3*b*B + A*c)*x^2)/2 + (B*c^3*x^4)/4 + 3*b*c*(b*B + A*c)
*Log[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^{11}} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^5} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) (b+c x)^3}{x^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (c^2 (3 b B+A c)+\frac{A b^3}{x^3}+\frac{b^2 (b B+3 A c)}{x^2}+\frac{3 b c (b B+A c)}{x}+B c^3 x\right ) \, dx,x,x^2\right )\\ &=-\frac{A b^3}{4 x^4}-\frac{b^2 (b B+3 A c)}{2 x^2}+\frac{1}{2} c^2 (3 b B+A c) x^2+\frac{1}{4} B c^3 x^4+3 b c (b B+A c) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0304726, size = 73, normalized size = 1.01 \[ \frac{B x^2 \left (-2 b^3+6 b c^2 x^4+c^3 x^6\right )-A \left (6 b^2 c x^2+b^3-2 c^3 x^6\right )}{4 x^4}+3 b c \log (x) (A c+b B) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^11,x]

[Out]

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

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Maple [A]  time = 0.007, size = 76, normalized size = 1.1 \begin{align*}{\frac{B{c}^{3}{x}^{4}}{4}}+{\frac{A{x}^{2}{c}^{3}}{2}}+{\frac{3\,B{x}^{2}b{c}^{2}}{2}}+3\,A\ln \left ( x \right ) b{c}^{2}+3\,B\ln \left ( x \right ){b}^{2}c-{\frac{A{b}^{3}}{4\,{x}^{4}}}-{\frac{3\,A{b}^{2}c}{2\,{x}^{2}}}-{\frac{B{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^11,x)

[Out]

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

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Maxima [A]  time = 1.14411, size = 103, normalized size = 1.43 \begin{align*} \frac{1}{4} \, B c^{3} x^{4} + \frac{1}{2} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + \frac{3}{2} \,{\left (B b^{2} c + A b c^{2}\right )} \log \left (x^{2}\right ) - \frac{A b^{3} + 2 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{4 \, x^{4}} \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^11,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.4301, size = 163, normalized size = 2.26 \begin{align*} \frac{B c^{3} x^{8} + 2 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 12 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} \log \left (x\right ) - A b^{3} - 2 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{4 \, x^{4}} \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^11,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.72699, size = 73, normalized size = 1.01 \begin{align*} \frac{B c^{3} x^{4}}{4} + 3 b c \left (A c + B b\right ) \log{\left (x \right )} + x^{2} \left (\frac{A c^{3}}{2} + \frac{3 B b c^{2}}{2}\right ) - \frac{A b^{3} + x^{2} \left (6 A b^{2} c + 2 B b^{3}\right )}{4 x^{4}} \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**11,x)

[Out]

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

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Giac [A]  time = 1.2913, size = 132, normalized size = 1.83 \begin{align*} \frac{1}{4} \, B c^{3} x^{4} + \frac{3}{2} \, B b c^{2} x^{2} + \frac{1}{2} \, A c^{3} x^{2} + \frac{3}{2} \,{\left (B b^{2} c + A b c^{2}\right )} \log \left (x^{2}\right ) - \frac{9 \, B b^{2} c x^{4} + 9 \, A b c^{2} x^{4} + 2 \, B b^{3} x^{2} + 6 \, A b^{2} c x^{2} + A b^{3}}{4 \, x^{4}} \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^11,x, algorithm="giac")

[Out]

1/4*B*c^3*x^4 + 3/2*B*b*c^2*x^2 + 1/2*A*c^3*x^2 + 3/2*(B*b^2*c + A*b*c^2)*log(x^2) - 1/4*(9*B*b^2*c*x^4 + 9*A*
b*c^2*x^4 + 2*B*b^3*x^2 + 6*A*b^2*c*x^2 + A*b^3)/x^4