3.1.35 \(\int \frac {(A+B x^2) (b x^2+c x^4)^3}{x^{12}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {A b^3}{5 x^5}-\frac {b^2 (3 A c+b B)}{3 x^3}+c^2 x (A c+3 b B)-\frac {3 b c (A c+b B)}{x}+\frac {1}{3} B c^3 x^3 \]

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Rubi [A]  time = 0.05, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1584, 448} \begin {gather*} -\frac {b^2 (3 A c+b B)}{3 x^3}-\frac {A b^3}{5 x^5}+c^2 x (A c+3 b B)-\frac {3 b c (A c+b B)}{x}+\frac {1}{3} B c^3 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^12,x]

[Out]

-(A*b^3)/(5*x^5) - (b^2*(b*B + 3*A*c))/(3*x^3) - (3*b*c*(b*B + A*c))/x + c^2*(3*b*B + A*c)*x + (B*c^3*x^3)/3

Rule 448

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

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^{12}} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^6} \, dx\\ &=\int \left (c^2 (3 b B+A c)+\frac {A b^3}{x^6}+\frac {b^2 (b B+3 A c)}{x^4}+\frac {3 b c (b B+A c)}{x^2}+B c^3 x^2\right ) \, dx\\ &=-\frac {A b^3}{5 x^5}-\frac {b^2 (b B+3 A c)}{3 x^3}-\frac {3 b c (b B+A c)}{x}+c^2 (3 b B+A c) x+\frac {1}{3} B c^3 x^3\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 68, normalized size = 1.00 \begin {gather*} -\frac {A b^3}{5 x^5}-\frac {b^2 (3 A c+b B)}{3 x^3}+c^2 x (A c+3 b B)-\frac {3 b c (A c+b B)}{x}+\frac {1}{3} B c^3 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^12,x]

[Out]

-1/5*(A*b^3)/x^5 - (b^2*(b*B + 3*A*c))/(3*x^3) - (3*b*c*(b*B + A*c))/x + c^2*(3*b*B + A*c)*x + (B*c^3*x^3)/3

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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^{12}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^12,x]

[Out]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^12, x]

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fricas [A]  time = 0.38, size = 75, normalized size = 1.10 \begin {gather*} \frac {5 \, B c^{3} x^{8} + 15 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} - 45 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} - 3 \, A b^{3} - 5 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{15 \, x^{5}} \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^12,x, algorithm="fricas")

[Out]

1/15*(5*B*c^3*x^8 + 15*(3*B*b*c^2 + A*c^3)*x^6 - 45*(B*b^2*c + A*b*c^2)*x^4 - 3*A*b^3 - 5*(B*b^3 + 3*A*b^2*c)*
x^2)/x^5

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giac [A]  time = 0.16, size = 75, normalized size = 1.10 \begin {gather*} \frac {1}{3} \, B c^{3} x^{3} + 3 \, B b c^{2} x + A c^{3} x - \frac {45 \, B b^{2} c x^{4} + 45 \, A b c^{2} x^{4} + 5 \, B b^{3} x^{2} + 15 \, A b^{2} c x^{2} + 3 \, A b^{3}}{15 \, x^{5}} \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^12,x, algorithm="giac")

[Out]

1/3*B*c^3*x^3 + 3*B*b*c^2*x + A*c^3*x - 1/15*(45*B*b^2*c*x^4 + 45*A*b*c^2*x^4 + 5*B*b^3*x^2 + 15*A*b^2*c*x^2 +
3*A*b^3)/x^5

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maple [A]  time = 0.05, size = 64, normalized size = 0.94 \begin {gather*} \frac {B \,c^{3} x^{3}}{3}+A \,c^{3} x +3 B b \,c^{2} x -\frac {3 \left (A c +b B \right ) b c}{x}-\frac {A \,b^{3}}{5 x^{5}}-\frac {\left (3 A c +b B \right ) b^{2}}{3 x^{3}} \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^12,x)

[Out]

1/3*B*c^3*x^3+A*c^3*x+3*B*b*c^2*x-1/5*A*b^3/x^5-1/3*b^2*(3*A*c+B*b)/x^3-3*b*c*(A*c+B*b)/x

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maxima [A]  time = 1.33, size = 73, normalized size = 1.07 \begin {gather*} \frac {1}{3} \, B c^{3} x^{3} + {\left (3 \, B b c^{2} + A c^{3}\right )} x - \frac {45 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 3 \, A b^{3} + 5 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{15 \, x^{5}} \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^12,x, algorithm="maxima")

[Out]

1/3*B*c^3*x^3 + (3*B*b*c^2 + A*c^3)*x - 1/15*(45*(B*b^2*c + A*b*c^2)*x^4 + 3*A*b^3 + 5*(B*b^3 + 3*A*b^2*c)*x^2
)/x^5

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mupad [B]  time = 0.06, size = 73, normalized size = 1.07 \begin {gather*} x\,\left (A\,c^3+3\,B\,b\,c^2\right )-\frac {x^4\,\left (3\,B\,b^2\,c+3\,A\,b\,c^2\right )+\frac {A\,b^3}{5}+x^2\,\left (\frac {B\,b^3}{3}+A\,c\,b^2\right )}{x^5}+\frac {B\,c^3\,x^3}{3} \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^12,x)

[Out]

x*(A*c^3 + 3*B*b*c^2) - (x^4*(3*A*b*c^2 + 3*B*b^2*c) + (A*b^3)/5 + x^2*((B*b^3)/3 + A*b^2*c))/x^5 + (B*c^3*x^3
)/3

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sympy [A]  time = 0.68, size = 78, normalized size = 1.15 \begin {gather*} \frac {B c^{3} x^{3}}{3} + x \left (A c^{3} + 3 B b c^{2}\right ) + \frac {- 3 A b^{3} + x^{4} \left (- 45 A b c^{2} - 45 B b^{2} c\right ) + x^{2} \left (- 15 A b^{2} c - 5 B b^{3}\right )}{15 x^{5}} \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**12,x)

[Out]

B*c**3*x**3/3 + x*(A*c**3 + 3*B*b*c**2) + (-3*A*b**3 + x**4*(-45*A*b*c**2 - 45*B*b**2*c) + x**2*(-15*A*b**2*c
- 5*B*b**3))/(15*x**5)

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