3.3.64 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^2}{x^9} \, dx\)

Optimal. Leaf size=50 \[ -\frac {a^4}{8 x^8}-\frac {2 a^3 b}{3 x^6}-\frac {3 a^2 b^2}{2 x^4}-\frac {2 a b^3}{x^2}+b^4 \log (x) \]

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 266, 43} \begin {gather*} -\frac {3 a^2 b^2}{2 x^4}-\frac {2 a^3 b}{3 x^6}-\frac {a^4}{8 x^8}-\frac {2 a b^3}{x^2}+b^4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^9,x]

[Out]

-a^4/(8*x^8) - (2*a^3*b)/(3*x^6) - (3*a^2*b^2)/(2*x^4) - (2*a*b^3)/x^2 + b^4*Log[x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^9} \, dx &=\frac {\int \frac {\left (a b+b^2 x^2\right )^4}{x^9} \, dx}{b^4}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a b+b^2 x\right )^4}{x^5} \, dx,x,x^2\right )}{2 b^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^4 b^4}{x^5}+\frac {4 a^3 b^5}{x^4}+\frac {6 a^2 b^6}{x^3}+\frac {4 a b^7}{x^2}+\frac {b^8}{x}\right ) \, dx,x,x^2\right )}{2 b^4}\\ &=-\frac {a^4}{8 x^8}-\frac {2 a^3 b}{3 x^6}-\frac {3 a^2 b^2}{2 x^4}-\frac {2 a b^3}{x^2}+b^4 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 50, normalized size = 1.00 \begin {gather*} -\frac {a^4}{8 x^8}-\frac {2 a^3 b}{3 x^6}-\frac {3 a^2 b^2}{2 x^4}-\frac {2 a b^3}{x^2}+b^4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^9,x]

[Out]

-1/8*a^4/x^8 - (2*a^3*b)/(3*x^6) - (3*a^2*b^2)/(2*x^4) - (2*a*b^3)/x^2 + b^4*Log[x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^9} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^9,x]

[Out]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^9, x]

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fricas [A]  time = 0.82, size = 50, normalized size = 1.00 \begin {gather*} \frac {24 \, b^{4} x^{8} \log \relax (x) - 48 \, a b^{3} x^{6} - 36 \, a^{2} b^{2} x^{4} - 16 \, a^{3} b x^{2} - 3 \, a^{4}}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x^9,x, algorithm="fricas")

[Out]

1/24*(24*b^4*x^8*log(x) - 48*a*b^3*x^6 - 36*a^2*b^2*x^4 - 16*a^3*b*x^2 - 3*a^4)/x^8

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giac [A]  time = 0.15, size = 58, normalized size = 1.16 \begin {gather*} \frac {1}{2} \, b^{4} \log \left (x^{2}\right ) - \frac {25 \, b^{4} x^{8} + 48 \, a b^{3} x^{6} + 36 \, a^{2} b^{2} x^{4} + 16 \, a^{3} b x^{2} + 3 \, a^{4}}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x^9,x, algorithm="giac")

[Out]

1/2*b^4*log(x^2) - 1/24*(25*b^4*x^8 + 48*a*b^3*x^6 + 36*a^2*b^2*x^4 + 16*a^3*b*x^2 + 3*a^4)/x^8

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maple [A]  time = 0.00, size = 45, normalized size = 0.90 \begin {gather*} b^{4} \ln \relax (x )-\frac {2 a \,b^{3}}{x^{2}}-\frac {3 a^{2} b^{2}}{2 x^{4}}-\frac {2 a^{3} b}{3 x^{6}}-\frac {a^{4}}{8 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^2/x^9,x)

[Out]

-1/8*a^4/x^8-2/3*a^3*b/x^6-3/2*a^2*b^2/x^4-2*a*b^3/x^2+b^4*ln(x)

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maxima [A]  time = 1.38, size = 50, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, b^{4} \log \left (x^{2}\right ) - \frac {48 \, a b^{3} x^{6} + 36 \, a^{2} b^{2} x^{4} + 16 \, a^{3} b x^{2} + 3 \, a^{4}}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x^9,x, algorithm="maxima")

[Out]

1/2*b^4*log(x^2) - 1/24*(48*a*b^3*x^6 + 36*a^2*b^2*x^4 + 16*a^3*b*x^2 + 3*a^4)/x^8

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mupad [B]  time = 0.05, size = 47, normalized size = 0.94 \begin {gather*} b^4\,\ln \relax (x)-\frac {\frac {a^4}{8}+\frac {2\,a^3\,b\,x^2}{3}+\frac {3\,a^2\,b^2\,x^4}{2}+2\,a\,b^3\,x^6}{x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^2/x^9,x)

[Out]

b^4*log(x) - (a^4/8 + (2*a^3*b*x^2)/3 + 2*a*b^3*x^6 + (3*a^2*b^2*x^4)/2)/x^8

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sympy [A]  time = 0.37, size = 49, normalized size = 0.98 \begin {gather*} b^{4} \log {\relax (x )} + \frac {- 3 a^{4} - 16 a^{3} b x^{2} - 36 a^{2} b^{2} x^{4} - 48 a b^{3} x^{6}}{24 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**2/x**9,x)

[Out]

b**4*log(x) + (-3*a**4 - 16*a**3*b*x**2 - 36*a**2*b**2*x**4 - 48*a*b**3*x**6)/(24*x**8)

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