3.516 \(\int \frac{x^7}{(a^2+2 a b x^2+b^2 x^4)^3} \, dx\)

Optimal. Leaf size=39 \[ \frac{x^8}{40 a^2 \left (a+b x^2\right )^4}+\frac{x^8}{10 a \left (a+b x^2\right )^5} \]

[Out]

x^8/(10*a*(a + b*x^2)^5) + x^8/(40*a^2*(a + b*x^2)^4)

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Rubi [A]  time = 0.0261123, antiderivative size = 39, 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} \[ \frac{x^8}{40 a^2 \left (a+b x^2\right )^4}+\frac{x^8}{10 a \left (a+b x^2\right )^5} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

x^8/(10*a*(a + b*x^2)^5) + x^8/(40*a^2*(a + b*x^2)^4)

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{x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx &=b^6 \int \frac{x^7}{\left (a b+b^2 x^2\right )^6} \, dx\\ &=\frac{1}{2} b^6 \operatorname{Subst}\left (\int \frac{x^3}{\left (a b+b^2 x\right )^6} \, dx,x,x^2\right )\\ &=\frac{x^8}{10 a \left (a+b x^2\right )^5}+\frac{b^5 \operatorname{Subst}\left (\int \frac{x^3}{\left (a b+b^2 x\right )^5} \, dx,x,x^2\right )}{10 a}\\ &=\frac{x^8}{10 a \left (a+b x^2\right )^5}+\frac{x^8}{40 a^2 \left (a+b x^2\right )^4}\\ \end{align*}

Mathematica [A]  time = 0.0137822, size = 46, normalized size = 1.18 \[ -\frac{5 a^2 b x^2+a^3+10 a b^2 x^4+10 b^3 x^6}{40 b^4 \left (a+b x^2\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

-(a^3 + 5*a^2*b*x^2 + 10*a*b^2*x^4 + 10*b^3*x^6)/(40*b^4*(a + b*x^2)^5)

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Maple [A]  time = 0.048, size = 65, normalized size = 1.7 \begin{align*}{\frac{a}{2\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{{a}^{3}}{10\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{5}}}-{\frac{3\,{a}^{2}}{8\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{4}}}-{\frac{1}{4\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b^2*x^4+2*a*b*x^2+a^2)^3,x)

[Out]

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

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Maxima [B]  time = 1.34569, size = 123, normalized size = 3.15 \begin{align*} -\frac{10 \, b^{3} x^{6} + 10 \, a b^{2} x^{4} + 5 \, a^{2} b x^{2} + a^{3}}{40 \,{\left (b^{9} x^{10} + 5 \, a b^{8} x^{8} + 10 \, a^{2} b^{7} x^{6} + 10 \, a^{3} b^{6} x^{4} + 5 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")

[Out]

-1/40*(10*b^3*x^6 + 10*a*b^2*x^4 + 5*a^2*b*x^2 + a^3)/(b^9*x^10 + 5*a*b^8*x^8 + 10*a^2*b^7*x^6 + 10*a^3*b^6*x^
4 + 5*a^4*b^5*x^2 + a^5*b^4)

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Fricas [B]  time = 1.69553, size = 189, normalized size = 4.85 \begin{align*} -\frac{10 \, b^{3} x^{6} + 10 \, a b^{2} x^{4} + 5 \, a^{2} b x^{2} + a^{3}}{40 \,{\left (b^{9} x^{10} + 5 \, a b^{8} x^{8} + 10 \, a^{2} b^{7} x^{6} + 10 \, a^{3} b^{6} x^{4} + 5 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")

[Out]

-1/40*(10*b^3*x^6 + 10*a*b^2*x^4 + 5*a^2*b*x^2 + a^3)/(b^9*x^10 + 5*a*b^8*x^8 + 10*a^2*b^7*x^6 + 10*a^3*b^6*x^
4 + 5*a^4*b^5*x^2 + a^5*b^4)

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Sympy [B]  time = 1.10337, size = 95, normalized size = 2.44 \begin{align*} - \frac{a^{3} + 5 a^{2} b x^{2} + 10 a b^{2} x^{4} + 10 b^{3} x^{6}}{40 a^{5} b^{4} + 200 a^{4} b^{5} x^{2} + 400 a^{3} b^{6} x^{4} + 400 a^{2} b^{7} x^{6} + 200 a b^{8} x^{8} + 40 b^{9} x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(b**2*x**4+2*a*b*x**2+a**2)**3,x)

[Out]

-(a**3 + 5*a**2*b*x**2 + 10*a*b**2*x**4 + 10*b**3*x**6)/(40*a**5*b**4 + 200*a**4*b**5*x**2 + 400*a**3*b**6*x**
4 + 400*a**2*b**7*x**6 + 200*a*b**8*x**8 + 40*b**9*x**10)

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Giac [A]  time = 1.13036, size = 59, normalized size = 1.51 \begin{align*} -\frac{10 \, b^{3} x^{6} + 10 \, a b^{2} x^{4} + 5 \, a^{2} b x^{2} + a^{3}}{40 \,{\left (b x^{2} + a\right )}^{5} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")

[Out]

-1/40*(10*b^3*x^6 + 10*a*b^2*x^4 + 5*a^2*b*x^2 + a^3)/((b*x^2 + a)^5*b^4)