3.1.31 \(\int \frac {(A+B x^2) (b x^2+c x^4)^3}{x^8} \, dx\)

Optimal. Leaf size=65 \[ -\frac {A b^3}{x}+b^2 x (3 A c+b B)+\frac {1}{5} c^2 x^5 (A c+3 b B)+b c x^3 (A c+b B)+\frac {1}{7} B c^3 x^7 \]

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Rubi [A]  time = 0.05, antiderivative size = 65, 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*} b^2 x (3 A c+b B)-\frac {A b^3}{x}+\frac {1}{5} c^2 x^5 (A c+3 b B)+b c x^3 (A c+b B)+\frac {1}{7} B c^3 x^7 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^8,x]

[Out]

-((A*b^3)/x) + b^2*(b*B + 3*A*c)*x + b*c*(b*B + A*c)*x^3 + (c^2*(3*b*B + A*c)*x^5)/5 + (B*c^3*x^7)/7

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^8} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^2} \, dx\\ &=\int \left (b^2 (b B+3 A c)+\frac {A b^3}{x^2}+3 b c (b B+A c) x^2+c^2 (3 b B+A c) x^4+B c^3 x^6\right ) \, dx\\ &=-\frac {A b^3}{x}+b^2 (b B+3 A c) x+b c (b B+A c) x^3+\frac {1}{5} c^2 (3 b B+A c) x^5+\frac {1}{7} B c^3 x^7\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 65, normalized size = 1.00 \begin {gather*} -\frac {A b^3}{x}+b^2 x (3 A c+b B)+\frac {1}{5} c^2 x^5 (A c+3 b B)+b c x^3 (A c+b B)+\frac {1}{7} B c^3 x^7 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^8,x]

[Out]

-((A*b^3)/x) + b^2*(b*B + 3*A*c)*x + b*c*(b*B + A*c)*x^3 + (c^2*(3*b*B + A*c)*x^5)/5 + (B*c^3*x^7)/7

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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^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^8,x]

[Out]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^8, x]

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fricas [A]  time = 0.39, size = 75, normalized size = 1.15 \begin {gather*} \frac {5 \, B c^{3} x^{8} + 7 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 35 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} - 35 \, A b^{3} + 35 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{35 \, x} \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^8,x, algorithm="fricas")

[Out]

1/35*(5*B*c^3*x^8 + 7*(3*B*b*c^2 + A*c^3)*x^6 + 35*(B*b^2*c + A*b*c^2)*x^4 - 35*A*b^3 + 35*(B*b^3 + 3*A*b^2*c)
*x^2)/x

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giac [A]  time = 0.16, size = 70, normalized size = 1.08 \begin {gather*} \frac {1}{7} \, B c^{3} x^{7} + \frac {3}{5} \, B b c^{2} x^{5} + \frac {1}{5} \, A c^{3} x^{5} + B b^{2} c x^{3} + A b c^{2} x^{3} + B b^{3} x + 3 \, A b^{2} c x - \frac {A b^{3}}{x} \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^8,x, algorithm="giac")

[Out]

1/7*B*c^3*x^7 + 3/5*B*b*c^2*x^5 + 1/5*A*c^3*x^5 + B*b^2*c*x^3 + A*b*c^2*x^3 + B*b^3*x + 3*A*b^2*c*x - A*b^3/x

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maple [A]  time = 0.05, size = 71, normalized size = 1.09 \begin {gather*} \frac {B \,c^{3} x^{7}}{7}+\frac {A \,c^{3} x^{5}}{5}+\frac {3 B b \,c^{2} x^{5}}{5}+A b \,c^{2} x^{3}+B \,b^{2} c \,x^{3}+3 A \,b^{2} c x +B \,b^{3} x -\frac {A \,b^{3}}{x} \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^8,x)

[Out]

1/7*B*c^3*x^7+1/5*A*x^5*c^3+3/5*B*x^5*b*c^2+A*x^3*b*c^2+B*x^3*b^2*c+3*A*c*b^2*x+B*b^3*x-A*b^3/x

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maxima [A]  time = 1.34, size = 69, normalized size = 1.06 \begin {gather*} \frac {1}{7} \, B c^{3} x^{7} + \frac {1}{5} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{5} + {\left (B b^{2} c + A b c^{2}\right )} x^{3} - \frac {A b^{3}}{x} + {\left (B b^{3} + 3 \, A b^{2} c\right )} x \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^8,x, algorithm="maxima")

[Out]

1/7*B*c^3*x^7 + 1/5*(3*B*b*c^2 + A*c^3)*x^5 + (B*b^2*c + A*b*c^2)*x^3 - A*b^3/x + (B*b^3 + 3*A*b^2*c)*x

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mupad [B]  time = 0.03, size = 65, normalized size = 1.00 \begin {gather*} x\,\left (B\,b^3+3\,A\,c\,b^2\right )+x^5\,\left (\frac {A\,c^3}{5}+\frac {3\,B\,b\,c^2}{5}\right )-\frac {A\,b^3}{x}+\frac {B\,c^3\,x^7}{7}+b\,c\,x^3\,\left (A\,c+B\,b\right ) \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^8,x)

[Out]

x*(B*b^3 + 3*A*b^2*c) + x^5*((A*c^3)/5 + (3*B*b*c^2)/5) - (A*b^3)/x + (B*c^3*x^7)/7 + b*c*x^3*(A*c + B*b)

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sympy [A]  time = 0.17, size = 68, normalized size = 1.05 \begin {gather*} - \frac {A b^{3}}{x} + \frac {B c^{3} x^{7}}{7} + x^{5} \left (\frac {A c^{3}}{5} + \frac {3 B b c^{2}}{5}\right ) + x^{3} \left (A b c^{2} + B b^{2} c\right ) + x \left (3 A b^{2} c + B b^{3}\right ) \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**8,x)

[Out]

-A*b**3/x + B*c**3*x**7/7 + x**5*(A*c**3/5 + 3*B*b*c**2/5) + x**3*(A*b*c**2 + B*b**2*c) + x*(3*A*b**2*c + B*b*
*3)

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