3.1.39 \(\int \frac {(a+b x^2)^3}{x^{11}} \, dx\)

Optimal. Leaf size=40 \[ \frac {b \left (a+b x^2\right )^4}{40 a^2 x^8}-\frac {\left (a+b x^2\right )^4}{10 a x^{10}} \]

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {266, 45, 37} \begin {gather*} \frac {b \left (a+b x^2\right )^4}{40 a^2 x^8}-\frac {\left (a+b x^2\right )^4}{10 a x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3/x^11,x]

[Out]

-(a + b*x^2)^4/(10*a*x^10) + (b*(a + b*x^2)^4)/(40*a^2*x^8)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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+b x^2\right )^3}{x^{11}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^3}{x^6} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^4}{10 a x^{10}}-\frac {b \operatorname {Subst}\left (\int \frac {(a+b x)^3}{x^5} \, dx,x,x^2\right )}{10 a}\\ &=-\frac {\left (a+b x^2\right )^4}{10 a x^{10}}+\frac {b \left (a+b x^2\right )^4}{40 a^2 x^8}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 43, normalized size = 1.08 \begin {gather*} -\frac {a^3}{10 x^{10}}-\frac {3 a^2 b}{8 x^8}-\frac {a b^2}{2 x^6}-\frac {b^3}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^3/x^11,x]

[Out]

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

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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 )^3}{x^{11}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x^2)^3/x^11,x]

[Out]

IntegrateAlgebraic[(a + b*x^2)^3/x^11, x]

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fricas [A]  time = 0.99, size = 37, normalized size = 0.92 \begin {gather*} -\frac {10 \, b^{3} x^{6} + 20 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 4 \, a^{3}}{40 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^11,x, algorithm="fricas")

[Out]

-1/40*(10*b^3*x^6 + 20*a*b^2*x^4 + 15*a^2*b*x^2 + 4*a^3)/x^10

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giac [A]  time = 1.02, size = 37, normalized size = 0.92 \begin {gather*} -\frac {10 \, b^{3} x^{6} + 20 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 4 \, a^{3}}{40 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^11,x, algorithm="giac")

[Out]

-1/40*(10*b^3*x^6 + 20*a*b^2*x^4 + 15*a^2*b*x^2 + 4*a^3)/x^10

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maple [A]  time = 0.00, size = 36, normalized size = 0.90 \begin {gather*} -\frac {b^{3}}{4 x^{4}}-\frac {a \,b^{2}}{2 x^{6}}-\frac {3 a^{2} b}{8 x^{8}}-\frac {a^{3}}{10 x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^3/x^11,x)

[Out]

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

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maxima [A]  time = 1.37, size = 37, normalized size = 0.92 \begin {gather*} -\frac {10 \, b^{3} x^{6} + 20 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 4 \, a^{3}}{40 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^11,x, algorithm="maxima")

[Out]

-1/40*(10*b^3*x^6 + 20*a*b^2*x^4 + 15*a^2*b*x^2 + 4*a^3)/x^10

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mupad [B]  time = 0.06, size = 37, normalized size = 0.92 \begin {gather*} -\frac {\frac {a^3}{10}+\frac {3\,a^2\,b\,x^2}{8}+\frac {a\,b^2\,x^4}{2}+\frac {b^3\,x^6}{4}}{x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^3/x^11,x)

[Out]

-(a^3/10 + (b^3*x^6)/4 + (3*a^2*b*x^2)/8 + (a*b^2*x^4)/2)/x^10

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sympy [A]  time = 0.29, size = 39, normalized size = 0.98 \begin {gather*} \frac {- 4 a^{3} - 15 a^{2} b x^{2} - 20 a b^{2} x^{4} - 10 b^{3} x^{6}}{40 x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**3/x**11,x)

[Out]

(-4*a**3 - 15*a**2*b*x**2 - 20*a*b**2*x**4 - 10*b**3*x**6)/(40*x**10)

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