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

Optimal. Leaf size=31 \[ -\frac {A c+b B}{3 x^3}-\frac {A b}{5 x^5}-\frac {B c}{x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1584, 448} \[ -\frac {A c+b B}{3 x^3}-\frac {A b}{5 x^5}-\frac {B c}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.06 \[ \frac {-A c-b B}{3 x^3}-\frac {A b}{5 x^5}-\frac {B c}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 29, normalized size = 0.94 \[ -\frac {15 \, B c x^{4} + 5 \, {\left (B b + A c\right )} x^{2} + 3 \, A b}{15 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^8,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 31, normalized size = 1.00 \[ -\frac {15 \, B c x^{4} + 5 \, B b x^{2} + 5 \, A c x^{2} + 3 \, A b}{15 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^8,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 28, normalized size = 0.90 \[ -\frac {B c}{x}-\frac {A b}{5 x^{5}}-\frac {A c +b B}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)/x^8,x)

[Out]

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

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maxima [A]  time = 1.35, size = 29, normalized size = 0.94 \[ -\frac {15 \, B c x^{4} + 5 \, {\left (B b + A c\right )} x^{2} + 3 \, A b}{15 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^8,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 29, normalized size = 0.94 \[ -\frac {B\,c\,x^4+\left (\frac {A\,c}{3}+\frac {B\,b}{3}\right )\,x^2+\frac {A\,b}{5}}{x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4))/x^8,x)

[Out]

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

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sympy [A]  time = 0.39, size = 32, normalized size = 1.03 \[ \frac {- 3 A b - 15 B c x^{4} + x^{2} \left (- 5 A c - 5 B b\right )}{15 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)/x**8,x)

[Out]

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

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