3.38 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^{15}} \, dx\)

Optimal. Leaf size=63 \[ -\frac{A \left (b+c x^2\right )^4}{8 b x^8}-\frac{3 b^2 B c}{4 x^4}-\frac{b^3 B}{6 x^6}-\frac{3 b B c^2}{2 x^2}+B c^3 \log (x) \]

[Out]

-(b^3*B)/(6*x^6) - (3*b^2*B*c)/(4*x^4) - (3*b*B*c^2)/(2*x^2) - (A*(b + c*x^2)^4)/(8*b*x^8) + B*c^3*Log[x]

________________________________________________________________________________________

Rubi [A]  time = 0.0467981, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1584, 446, 78, 43} \[ -\frac{A \left (b+c x^2\right )^4}{8 b x^8}-\frac{3 b^2 B c}{4 x^4}-\frac{b^3 B}{6 x^6}-\frac{3 b B c^2}{2 x^2}+B c^3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^15,x]

[Out]

-(b^3*B)/(6*x^6) - (3*b^2*B*c)/(4*x^4) - (3*b*B*c^2)/(2*x^2) - (A*(b + c*x^2)^4)/(8*b*x^8) + B*c^3*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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^9} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) (b+c x)^3}{x^5} \, dx,x,x^2\right )\\ &=-\frac{A \left (b+c x^2\right )^4}{8 b x^8}+\frac{1}{2} B \operatorname{Subst}\left (\int \frac{(b+c x)^3}{x^4} \, dx,x,x^2\right )\\ &=-\frac{A \left (b+c x^2\right )^4}{8 b x^8}+\frac{1}{2} B \operatorname{Subst}\left (\int \left (\frac{b^3}{x^4}+\frac{3 b^2 c}{x^3}+\frac{3 b c^2}{x^2}+\frac{c^3}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac{b^3 B}{6 x^6}-\frac{3 b^2 B c}{4 x^4}-\frac{3 b B c^2}{2 x^2}-\frac{A \left (b+c x^2\right )^4}{8 b x^8}+B c^3 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0327315, size = 77, normalized size = 1.22 \[ B c^3 \log (x)-\frac{3 A \left (4 b^2 c x^2+b^3+6 b c^2 x^4+4 c^3 x^6\right )+2 b B x^2 \left (2 b^2+9 b c x^2+18 c^2 x^4\right )}{24 x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^15,x]

[Out]

-(2*b*B*x^2*(2*b^2 + 9*b*c*x^2 + 18*c^2*x^4) + 3*A*(b^3 + 4*b^2*c*x^2 + 6*b*c^2*x^4 + 4*c^3*x^6))/(24*x^8) + B
*c^3*Log[x]

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 76, normalized size = 1.2 \begin{align*} B{c}^{3}\ln \left ( x \right ) -{\frac{3\,Ab{c}^{2}}{4\,{x}^{4}}}-{\frac{3\,B{b}^{2}c}{4\,{x}^{4}}}-{\frac{A{c}^{3}}{2\,{x}^{2}}}-{\frac{3\,Bb{c}^{2}}{2\,{x}^{2}}}-{\frac{A{b}^{3}}{8\,{x}^{8}}}-{\frac{A{b}^{2}c}{2\,{x}^{6}}}-{\frac{B{b}^{3}}{6\,{x}^{6}}} \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^15,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.17294, size = 104, normalized size = 1.65 \begin{align*} \frac{1}{2} \, B c^{3} \log \left (x^{2}\right ) - \frac{12 \,{\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} + 3 \, A b^{3} + 4 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{24 \, x^{8}} \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^15,x, algorithm="maxima")

[Out]

1/2*B*c^3*log(x^2) - 1/24*(12*(3*B*b*c^2 + A*c^3)*x^6 + 18*(B*b^2*c + A*b*c^2)*x^4 + 3*A*b^3 + 4*(B*b^3 + 3*A*
b^2*c)*x^2)/x^8

________________________________________________________________________________________

Fricas [A]  time = 0.502951, size = 173, normalized size = 2.75 \begin{align*} \frac{24 \, B c^{3} x^{8} \log \left (x\right ) - 12 \,{\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} - 3 \, A b^{3} - 4 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{24 \, x^{8}} \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^15,x, algorithm="fricas")

[Out]

1/24*(24*B*c^3*x^8*log(x) - 12*(3*B*b*c^2 + A*c^3)*x^6 - 18*(B*b^2*c + A*b*c^2)*x^4 - 3*A*b^3 - 4*(B*b^3 + 3*A
*b^2*c)*x^2)/x^8

________________________________________________________________________________________

Sympy [A]  time = 2.66069, size = 76, normalized size = 1.21 \begin{align*} B c^{3} \log{\left (x \right )} - \frac{3 A b^{3} + x^{6} \left (12 A c^{3} + 36 B b c^{2}\right ) + x^{4} \left (18 A b c^{2} + 18 B b^{2} c\right ) + x^{2} \left (12 A b^{2} c + 4 B b^{3}\right )}{24 x^{8}} \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**15,x)

[Out]

B*c**3*log(x) - (3*A*b**3 + x**6*(12*A*c**3 + 36*B*b*c**2) + x**4*(18*A*b*c**2 + 18*B*b**2*c) + x**2*(12*A*b**
2*c + 4*B*b**3))/(24*x**8)

________________________________________________________________________________________

Giac [A]  time = 1.19666, size = 122, normalized size = 1.94 \begin{align*} \frac{1}{2} \, B c^{3} \log \left (x^{2}\right ) - \frac{25 \, B c^{3} x^{8} + 36 \, B b c^{2} x^{6} + 12 \, A c^{3} x^{6} + 18 \, B b^{2} c x^{4} + 18 \, A b c^{2} x^{4} + 4 \, B b^{3} x^{2} + 12 \, A b^{2} c x^{2} + 3 \, A b^{3}}{24 \, x^{8}} \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^15,x, algorithm="giac")

[Out]

1/2*B*c^3*log(x^2) - 1/24*(25*B*c^3*x^8 + 36*B*b*c^2*x^6 + 12*A*c^3*x^6 + 18*B*b^2*c*x^4 + 18*A*b*c^2*x^4 + 4*
B*b^3*x^2 + 12*A*b^2*c*x^2 + 3*A*b^3)/x^8