3.440 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^2}{x^{13}} \, dx\)

Optimal. Leaf size=40 \[ \frac{b \left (a+b x^2\right )^5}{60 a^2 x^{10}}-\frac{\left (a+b x^2\right )^5}{12 a x^{12}} \]

[Out]

-(a + b*x^2)^5/(12*a*x^12) + (b*(a + b*x^2)^5)/(60*a^2*x^10)

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Rubi [A]  time = 0.0245002, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {28, 266, 45, 37} \[ \frac{b \left (a+b x^2\right )^5}{60 a^2 x^{10}}-\frac{\left (a+b x^2\right )^5}{12 a x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^13,x]

[Out]

-(a + b*x^2)^5/(12*a*x^12) + (b*(a + b*x^2)^5)/(60*a^2*x^10)

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 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]]

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 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]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^{13}} \, dx &=\frac{\int \frac{\left (a b+b^2 x^2\right )^4}{x^{13}} \, dx}{b^4}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^4}{x^7} \, dx,x,x^2\right )}{2 b^4}\\ &=-\frac{\left (a+b x^2\right )^5}{12 a x^{12}}-\frac{\operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^4}{x^6} \, dx,x,x^2\right )}{12 a b^3}\\ &=-\frac{\left (a+b x^2\right )^5}{12 a x^{12}}+\frac{b \left (a+b x^2\right )^5}{60 a^2 x^{10}}\\ \end{align*}

Mathematica [A]  time = 0.0041385, size = 56, normalized size = 1.4 \[ -\frac{3 a^2 b^2}{4 x^8}-\frac{2 a^3 b}{5 x^{10}}-\frac{a^4}{12 x^{12}}-\frac{2 a b^3}{3 x^6}-\frac{b^4}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^13,x]

[Out]

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

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Maple [A]  time = 0.048, size = 47, normalized size = 1.2 \begin{align*} -{\frac{2\,{a}^{3}b}{5\,{x}^{10}}}-{\frac{{a}^{4}}{12\,{x}^{12}}}-{\frac{{b}^{4}}{4\,{x}^{4}}}-{\frac{3\,{b}^{2}{a}^{2}}{4\,{x}^{8}}}-{\frac{2\,a{b}^{3}}{3\,{x}^{6}}} \end{align*}

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^13,x)

[Out]

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

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Maxima [A]  time = 0.980356, size = 65, normalized size = 1.62 \begin{align*} -\frac{15 \, b^{4} x^{8} + 40 \, a b^{3} x^{6} + 45 \, a^{2} b^{2} x^{4} + 24 \, a^{3} b x^{2} + 5 \, a^{4}}{60 \, x^{12}} \end{align*}

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^13,x, algorithm="maxima")

[Out]

-1/60*(15*b^4*x^8 + 40*a*b^3*x^6 + 45*a^2*b^2*x^4 + 24*a^3*b*x^2 + 5*a^4)/x^12

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Fricas [A]  time = 1.69737, size = 108, normalized size = 2.7 \begin{align*} -\frac{15 \, b^{4} x^{8} + 40 \, a b^{3} x^{6} + 45 \, a^{2} b^{2} x^{4} + 24 \, a^{3} b x^{2} + 5 \, a^{4}}{60 \, x^{12}} \end{align*}

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^13,x, algorithm="fricas")

[Out]

-1/60*(15*b^4*x^8 + 40*a*b^3*x^6 + 45*a^2*b^2*x^4 + 24*a^3*b*x^2 + 5*a^4)/x^12

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Sympy [A]  time = 0.546981, size = 51, normalized size = 1.27 \begin{align*} - \frac{5 a^{4} + 24 a^{3} b x^{2} + 45 a^{2} b^{2} x^{4} + 40 a b^{3} x^{6} + 15 b^{4} x^{8}}{60 x^{12}} \end{align*}

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**13,x)

[Out]

-(5*a**4 + 24*a**3*b*x**2 + 45*a**2*b**2*x**4 + 40*a*b**3*x**6 + 15*b**4*x**8)/(60*x**12)

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Giac [A]  time = 1.14102, size = 65, normalized size = 1.62 \begin{align*} -\frac{15 \, b^{4} x^{8} + 40 \, a b^{3} x^{6} + 45 \, a^{2} b^{2} x^{4} + 24 \, a^{3} b x^{2} + 5 \, a^{4}}{60 \, x^{12}} \end{align*}

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^13,x, algorithm="giac")

[Out]

-1/60*(15*b^4*x^8 + 40*a*b^3*x^6 + 45*a^2*b^2*x^4 + 24*a^3*b*x^2 + 5*a^4)/x^12