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]

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

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Rubi [A]  time = 0.0218235, antiderivative size = 31, normalized size of antiderivative = 1., 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 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 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]

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*}

Mathematica [A]  time = 0.0108171, 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]

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

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Maple [A]  time = 0.004, size = 28, normalized size = 0.9 \begin{align*} -{\frac{Ac+Bb}{3\,{x}^{3}}}-{\frac{Ab}{5\,{x}^{5}}}-{\frac{Bc}{x}} \end{align*}

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]

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

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Maxima [A]  time = 1.12884, size = 39, normalized size = 1.26 \begin{align*} -\frac{15 \, B c x^{4} + 5 \,{\left (B b + A c\right )} x^{2} + 3 \, A b}{15 \, x^{5}} \end{align*}

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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Fricas [A]  time = 0.491082, size = 70, normalized size = 2.26 \begin{align*} -\frac{15 \, B c x^{4} + 5 \,{\left (B b + A c\right )} x^{2} + 3 \, A b}{15 \, x^{5}} \end{align*}

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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Sympy [A]  time = 0.504007, size = 32, normalized size = 1.03 \begin{align*} - \frac{3 A b + 15 B c x^{4} + x^{2} \left (5 A c + 5 B b\right )}{15 x^{5}} \end{align*}

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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Giac [A]  time = 1.14603, size = 42, normalized size = 1.35 \begin{align*} -\frac{15 \, B c x^{4} + 5 \, B b x^{2} + 5 \, A c x^{2} + 3 \, A b}{15 \, x^{5}} \end{align*}

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