3.469 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^3}{x^{17}} \, dx\)

Optimal. Leaf size=40 \[ \frac{b \left (a+b x^2\right )^7}{112 a^2 x^{14}}-\frac{\left (a+b x^2\right )^7}{16 a x^{16}} \]

[Out]

-(a + b*x^2)^7/(16*a*x^16) + (b*(a + b*x^2)^7)/(112*a^2*x^14)

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Rubi [A]  time = 0.0251543, 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} \[ \frac{b \left (a+b x^2\right )^7}{112 a^2 x^{14}}-\frac{\left (a+b x^2\right )^7}{16 a x^{16}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^17,x]

[Out]

-(a + b*x^2)^7/(16*a*x^16) + (b*(a + b*x^2)^7)/(112*a^2*x^14)

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 )^3}{x^{17}} \, dx &=\frac{\int \frac{\left (a b+b^2 x^2\right )^6}{x^{17}} \, dx}{b^6}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^6}{x^9} \, dx,x,x^2\right )}{2 b^6}\\ &=-\frac{\left (a+b x^2\right )^7}{16 a x^{16}}-\frac{\operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^6}{x^8} \, dx,x,x^2\right )}{16 a b^5}\\ &=-\frac{\left (a+b x^2\right )^7}{16 a x^{16}}+\frac{b \left (a+b x^2\right )^7}{112 a^2 x^{14}}\\ \end{align*}

Mathematica [A]  time = 0.0046634, size = 78, normalized size = 1.95 \[ -\frac{5 a^4 b^2}{4 x^{12}}-\frac{2 a^3 b^3}{x^{10}}-\frac{15 a^2 b^4}{8 x^8}-\frac{3 a^5 b}{7 x^{14}}-\frac{a^6}{16 x^{16}}-\frac{a b^5}{x^6}-\frac{b^6}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^17,x]

[Out]

-a^6/(16*x^16) - (3*a^5*b)/(7*x^14) - (5*a^4*b^2)/(4*x^12) - (2*a^3*b^3)/x^10 - (15*a^2*b^4)/(8*x^8) - (a*b^5)
/x^6 - b^6/(4*x^4)

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Maple [A]  time = 0.053, size = 69, normalized size = 1.7 \begin{align*} -{\frac{{a}^{6}}{16\,{x}^{16}}}-{\frac{{b}^{6}}{4\,{x}^{4}}}-{\frac{15\,{a}^{2}{b}^{4}}{8\,{x}^{8}}}-{\frac{3\,{a}^{5}b}{7\,{x}^{14}}}-{\frac{a{b}^{5}}{{x}^{6}}}-2\,{\frac{{a}^{3}{b}^{3}}{{x}^{10}}}-{\frac{5\,{a}^{4}{b}^{2}}{4\,{x}^{12}}} \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)^3/x^17,x)

[Out]

-1/16*a^6/x^16-1/4*b^6/x^4-15/8*a^2*b^4/x^8-3/7*a^5*b/x^14-a*b^5/x^6-2*a^3*b^3/x^10-5/4*a^4*b^2/x^12

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Maxima [A]  time = 0.975522, size = 95, normalized size = 2.38 \begin{align*} -\frac{28 \, b^{6} x^{12} + 112 \, a b^{5} x^{10} + 210 \, a^{2} b^{4} x^{8} + 224 \, a^{3} b^{3} x^{6} + 140 \, a^{4} b^{2} x^{4} + 48 \, a^{5} b x^{2} + 7 \, a^{6}}{112 \, x^{16}} \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)^3/x^17,x, algorithm="maxima")

[Out]

-1/112*(28*b^6*x^12 + 112*a*b^5*x^10 + 210*a^2*b^4*x^8 + 224*a^3*b^3*x^6 + 140*a^4*b^2*x^4 + 48*a^5*b*x^2 + 7*
a^6)/x^16

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Fricas [A]  time = 1.72615, size = 163, normalized size = 4.08 \begin{align*} -\frac{28 \, b^{6} x^{12} + 112 \, a b^{5} x^{10} + 210 \, a^{2} b^{4} x^{8} + 224 \, a^{3} b^{3} x^{6} + 140 \, a^{4} b^{2} x^{4} + 48 \, a^{5} b x^{2} + 7 \, a^{6}}{112 \, x^{16}} \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)^3/x^17,x, algorithm="fricas")

[Out]

-1/112*(28*b^6*x^12 + 112*a*b^5*x^10 + 210*a^2*b^4*x^8 + 224*a^3*b^3*x^6 + 140*a^4*b^2*x^4 + 48*a^5*b*x^2 + 7*
a^6)/x^16

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Sympy [B]  time = 0.81678, size = 75, normalized size = 1.88 \begin{align*} - \frac{7 a^{6} + 48 a^{5} b x^{2} + 140 a^{4} b^{2} x^{4} + 224 a^{3} b^{3} x^{6} + 210 a^{2} b^{4} x^{8} + 112 a b^{5} x^{10} + 28 b^{6} x^{12}}{112 x^{16}} \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)**3/x**17,x)

[Out]

-(7*a**6 + 48*a**5*b*x**2 + 140*a**4*b**2*x**4 + 224*a**3*b**3*x**6 + 210*a**2*b**4*x**8 + 112*a*b**5*x**10 +
28*b**6*x**12)/(112*x**16)

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Giac [A]  time = 1.12667, size = 95, normalized size = 2.38 \begin{align*} -\frac{28 \, b^{6} x^{12} + 112 \, a b^{5} x^{10} + 210 \, a^{2} b^{4} x^{8} + 224 \, a^{3} b^{3} x^{6} + 140 \, a^{4} b^{2} x^{4} + 48 \, a^{5} b x^{2} + 7 \, a^{6}}{112 \, x^{16}} \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)^3/x^17,x, algorithm="giac")

[Out]

-1/112*(28*b^6*x^12 + 112*a*b^5*x^10 + 210*a^2*b^4*x^8 + 224*a^3*b^3*x^6 + 140*a^4*b^2*x^4 + 48*a^5*b*x^2 + 7*
a^6)/x^16