3.3.70 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^2}{x^{15}} \, dx\)

Optimal. Leaf size=56 \[ -\frac {a^4}{14 x^{14}}-\frac {a^3 b}{3 x^{12}}-\frac {3 a^2 b^2}{5 x^{10}}-\frac {a b^3}{2 x^8}-\frac {b^4}{6 x^6} \]

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Rubi [A]  time = 0.04, antiderivative size = 56, 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}{5 x^{10}}-\frac {a^3 b}{3 x^{12}}-\frac {a^4}{14 x^{14}}-\frac {a b^3}{2 x^8}-\frac {b^4}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^15,x]

[Out]

-a^4/(14*x^14) - (a^3*b)/(3*x^12) - (3*a^2*b^2)/(5*x^10) - (a*b^3)/(2*x^8) - b^4/(6*x^6)

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^{15}} \, dx &=\frac {\int \frac {\left (a b+b^2 x^2\right )^4}{x^{15}} \, dx}{b^4}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a b+b^2 x\right )^4}{x^8} \, dx,x,x^2\right )}{2 b^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^4 b^4}{x^8}+\frac {4 a^3 b^5}{x^7}+\frac {6 a^2 b^6}{x^6}+\frac {4 a b^7}{x^5}+\frac {b^8}{x^4}\right ) \, dx,x,x^2\right )}{2 b^4}\\ &=-\frac {a^4}{14 x^{14}}-\frac {a^3 b}{3 x^{12}}-\frac {3 a^2 b^2}{5 x^{10}}-\frac {a b^3}{2 x^8}-\frac {b^4}{6 x^6}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 56, normalized size = 1.00 \begin {gather*} -\frac {a^4}{14 x^{14}}-\frac {a^3 b}{3 x^{12}}-\frac {3 a^2 b^2}{5 x^{10}}-\frac {a b^3}{2 x^8}-\frac {b^4}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^15,x]

[Out]

-1/14*a^4/x^14 - (a^3*b)/(3*x^12) - (3*a^2*b^2)/(5*x^10) - (a*b^3)/(2*x^8) - b^4/(6*x^6)

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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^{15}} \, 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^15,x]

[Out]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^15, x]

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fricas [A]  time = 0.59, size = 48, normalized size = 0.86 \begin {gather*} -\frac {35 \, b^{4} x^{8} + 105 \, a b^{3} x^{6} + 126 \, a^{2} b^{2} x^{4} + 70 \, a^{3} b x^{2} + 15 \, a^{4}}{210 \, x^{14}} \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^15,x, algorithm="fricas")

[Out]

-1/210*(35*b^4*x^8 + 105*a*b^3*x^6 + 126*a^2*b^2*x^4 + 70*a^3*b*x^2 + 15*a^4)/x^14

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giac [A]  time = 0.15, size = 48, normalized size = 0.86 \begin {gather*} -\frac {35 \, b^{4} x^{8} + 105 \, a b^{3} x^{6} + 126 \, a^{2} b^{2} x^{4} + 70 \, a^{3} b x^{2} + 15 \, a^{4}}{210 \, x^{14}} \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^15,x, algorithm="giac")

[Out]

-1/210*(35*b^4*x^8 + 105*a*b^3*x^6 + 126*a^2*b^2*x^4 + 70*a^3*b*x^2 + 15*a^4)/x^14

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maple [A]  time = 0.01, size = 47, normalized size = 0.84 \begin {gather*} -\frac {b^{4}}{6 x^{6}}-\frac {a \,b^{3}}{2 x^{8}}-\frac {3 a^{2} b^{2}}{5 x^{10}}-\frac {a^{3} b}{3 x^{12}}-\frac {a^{4}}{14 x^{14}} \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^15,x)

[Out]

-1/14*a^4/x^14-1/3*a^3*b/x^12-3/5*a^2*b^2/x^10-1/2*a*b^3/x^8-1/6*b^4/x^6

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maxima [A]  time = 1.43, size = 48, normalized size = 0.86 \begin {gather*} -\frac {35 \, b^{4} x^{8} + 105 \, a b^{3} x^{6} + 126 \, a^{2} b^{2} x^{4} + 70 \, a^{3} b x^{2} + 15 \, a^{4}}{210 \, x^{14}} \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^15,x, algorithm="maxima")

[Out]

-1/210*(35*b^4*x^8 + 105*a*b^3*x^6 + 126*a^2*b^2*x^4 + 70*a^3*b*x^2 + 15*a^4)/x^14

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mupad [B]  time = 4.33, size = 48, normalized size = 0.86 \begin {gather*} -\frac {\frac {a^4}{14}+\frac {a^3\,b\,x^2}{3}+\frac {3\,a^2\,b^2\,x^4}{5}+\frac {a\,b^3\,x^6}{2}+\frac {b^4\,x^8}{6}}{x^{14}} \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^15,x)

[Out]

-(a^4/14 + (b^4*x^8)/6 + (a^3*b*x^2)/3 + (a*b^3*x^6)/2 + (3*a^2*b^2*x^4)/5)/x^14

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sympy [A]  time = 0.45, size = 51, normalized size = 0.91 \begin {gather*} \frac {- 15 a^{4} - 70 a^{3} b x^{2} - 126 a^{2} b^{2} x^{4} - 105 a b^{3} x^{6} - 35 b^{4} x^{8}}{210 x^{14}} \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**15,x)

[Out]

(-15*a**4 - 70*a**3*b*x**2 - 126*a**2*b**2*x**4 - 105*a*b**3*x**6 - 35*b**4*x**8)/(210*x**14)

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