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

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

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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) \end {gather*}

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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.06, size = 77, normalized size = 1.22 \begin {gather*} B c^3 \log (x)-\frac {3 A \left (b^3+4 b^2 c x^2+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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(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))/x^8 + B
*c^3*Log[x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 77, normalized size = 1.22 \begin {gather*} \frac {24 \, B c^{3} x^{8} \log \relax (x) - 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 {gather*}

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

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giac [A]  time = 0.17, size = 90, normalized size = 1.43 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.05, size = 76, normalized size = 1.21 \begin {gather*} B \,c^{3} \ln \relax (x )-\frac {A \,c^{3}}{2 x^{2}}-\frac {3 B b \,c^{2}}{2 x^{2}}-\frac {3 A b \,c^{2}}{4 x^{4}}-\frac {3 B \,b^{2} c}{4 x^{4}}-\frac {A \,b^{2} c}{2 x^{6}}-\frac {B \,b^{3}}{6 x^{6}}-\frac {A \,b^{3}}{8 x^{8}} \end {gather*}

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]

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

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maxima [A]  time = 1.32, size = 77, normalized size = 1.22 \begin {gather*} \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 {gather*}

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

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mupad [B]  time = 0.10, size = 75, normalized size = 1.19 \begin {gather*} B\,c^3\,\ln \relax (x)-\frac {x^4\,\left (\frac {3\,B\,b^2\,c}{4}+\frac {3\,A\,b\,c^2}{4}\right )+\frac {A\,b^3}{8}+x^2\,\left (\frac {B\,b^3}{6}+\frac {A\,c\,b^2}{2}\right )+x^6\,\left (\frac {A\,c^3}{2}+\frac {3\,B\,b\,c^2}{2}\right )}{x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.37, size = 82, normalized size = 1.30 \begin {gather*} B c^{3} \log {\relax (x )} + \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 {gather*}

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)

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