3.1.22 \(\int \frac {(A+B x^2) (b x^2+c x^4)^2}{x^{10}} \, dx\)

Optimal. Leaf size=48 \[ -\frac {A b^2}{5 x^5}-\frac {b (2 A c+b B)}{3 x^3}-\frac {c (A c+2 b B)}{x}+B c^2 x \]

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Rubi [A]  time = 0.04, antiderivative size = 48, 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 {A b^2}{5 x^5}-\frac {b (2 A c+b B)}{3 x^3}-\frac {c (A c+2 b B)}{x}+B c^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^10,x]

[Out]

-(A*b^2)/(5*x^5) - (b*(b*B + 2*A*c))/(3*x^3) - (c*(2*b*B + A*c))/x + B*c^2*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 )^2}{x^{10}} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^2}{x^6} \, dx\\ &=\int \left (B c^2+\frac {A b^2}{x^6}+\frac {b (b B+2 A c)}{x^4}+\frac {c (2 b B+A c)}{x^2}\right ) \, dx\\ &=-\frac {A b^2}{5 x^5}-\frac {b (b B+2 A c)}{3 x^3}-\frac {c (2 b B+A c)}{x}+B c^2 x\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 1.00 \begin {gather*} -\frac {A b^2}{5 x^5}-\frac {b (2 A c+b B)}{3 x^3}-\frac {c (A c+2 b B)}{x}+B c^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^10,x]

[Out]

-1/5*(A*b^2)/x^5 - (b*(b*B + 2*A*c))/(3*x^3) - (c*(2*b*B + A*c))/x + B*c^2*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 )^2}{x^{10}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^10,x]

[Out]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^10, x]

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fricas [A]  time = 0.37, size = 53, normalized size = 1.10 \begin {gather*} \frac {15 \, B c^{2} x^{6} - 15 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} - 3 \, A b^{2} - 5 \, {\left (B b^{2} + 2 \, A b 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)^2/x^10,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 53, normalized size = 1.10 \begin {gather*} B c^{2} x - \frac {30 \, B b c x^{4} + 15 \, A c^{2} x^{4} + 5 \, B b^{2} x^{2} + 10 \, A b c x^{2} + 3 \, A b^{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)^2/x^10,x, algorithm="giac")

[Out]

B*c^2*x - 1/15*(30*B*b*c*x^4 + 15*A*c^2*x^4 + 5*B*b^2*x^2 + 10*A*b*c*x^2 + 3*A*b^2)/x^5

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maple [A]  time = 0.06, size = 45, normalized size = 0.94 \begin {gather*} B \,c^{2} x -\frac {\left (A c +2 b B \right ) c}{x}-\frac {A \,b^{2}}{5 x^{5}}-\frac {\left (2 A c +b B \right ) b}{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)^2/x^10,x)

[Out]

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

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maxima [A]  time = 1.39, size = 51, normalized size = 1.06 \begin {gather*} B c^{2} x - \frac {15 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 3 \, A b^{2} + 5 \, {\left (B b^{2} + 2 \, A b 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)^2/x^10,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 50, normalized size = 1.04 \begin {gather*} B\,c^2\,x-\frac {x^2\,\left (\frac {B\,b^2}{3}+\frac {2\,A\,c\,b}{3}\right )+x^4\,\left (A\,c^2+2\,B\,b\,c\right )+\frac {A\,b^2}{5}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^10,x)

[Out]

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

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sympy [A]  time = 0.61, size = 54, normalized size = 1.12 \begin {gather*} B c^{2} x + \frac {- 3 A b^{2} + x^{4} \left (- 15 A c^{2} - 30 B b c\right ) + x^{2} \left (- 10 A b c - 5 B b^{2}\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)**2/x**10,x)

[Out]

B*c**2*x + (-3*A*b**2 + x**4*(-15*A*c**2 - 30*B*b*c) + x**2*(-10*A*b*c - 5*B*b**2))/(15*x**5)

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