3.1.9 \(\int \frac {(A+B x^2) (b x^2+c x^4)}{x^6} \, dx\)

Optimal. Leaf size=26 \[ -\frac {A c+b B}{x}-\frac {A b}{3 x^3}+B c x \]

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Rubi [A]  time = 0.02, antiderivative size = 26, 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} \begin {gather*} -\frac {A c+b B}{x}-\frac {A b}{3 x^3}+B c x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4))/x^6,x]

[Out]

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

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Mathematica [A]  time = 0.03, size = 27, normalized size = 1.04 \begin {gather*} \frac {-A c-b B}{x}-\frac {A b}{3 x^3}+B c x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4))/x^6,x]

[Out]

-1/3*(A*b)/x^3 + (-(b*B) - A*c)/x + B*c*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 )}{x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4))/x^6,x]

[Out]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4))/x^6, x]

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fricas [A]  time = 0.37, size = 29, normalized size = 1.12 \begin {gather*} \frac {3 \, B c x^{4} - 3 \, {\left (B b + A c\right )} x^{2} - A b}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^6,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 28, normalized size = 1.08 \begin {gather*} B c x - \frac {3 \, B b x^{2} + 3 \, A c x^{2} + A b}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^6,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 25, normalized size = 0.96 \begin {gather*} B c x -\frac {A b}{3 x^{3}}-\frac {A c +b B}{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)/x^6,x)

[Out]

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

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maxima [A]  time = 1.34, size = 26, normalized size = 1.00 \begin {gather*} B c x - \frac {3 \, {\left (B b + A c\right )} x^{2} + A b}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^6,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 26, normalized size = 1.00 \begin {gather*} B\,c\,x-\frac {\left (A\,c+B\,b\right )\,x^2+\frac {A\,b}{3}}{x^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))/x^6,x)

[Out]

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

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sympy [A]  time = 0.22, size = 27, normalized size = 1.04 \begin {gather*} B c x + \frac {- A b + x^{2} \left (- 3 A c - 3 B b\right )}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)/x**6,x)

[Out]

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

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