3.647 \(\int \frac{x^7}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=41 \[ \frac{x^8}{8 a \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

x^8/(8*a*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.0404306, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1111, 646, 37} \[ \frac{x^8}{8 a \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

x^8/(8*a*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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

Mathematica [A]  time = 0.0170194, size = 61, normalized size = 1.49 \[ \frac{-4 a^2 b x^2-a^3-6 a b^2 x^4-4 b^3 x^6}{8 b^4 \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(-a^3 - 4*a^2*b*x^2 - 6*a*b^2*x^4 - 4*b^3*x^6)/(8*b^4*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.173, size = 54, normalized size = 1.3 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 4\,{b}^{3}{x}^{6}+6\,a{x}^{4}{b}^{2}+4\,{a}^{2}b{x}^{2}+{a}^{3} \right ) }{8\,{b}^{4}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

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

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Maxima [B]  time = 0.995088, size = 197, normalized size = 4.8 \begin{align*} -\frac{x^{4}}{2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{a^{2}}{3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{3}{2}} b^{4}} + \frac{a^{2}}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{3}} - \frac{a}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{2} b} - \frac{a^{3} b}{8 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{4}} + \frac{a^{3}}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{4} b^{3}} \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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.30553, size = 159, normalized size = 3.88 \begin{align*} -\frac{4 \, b^{3} x^{6} + 6 \, a b^{2} x^{4} + 4 \, a^{2} b x^{2} + a^{3}}{8 \,{\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}}\, dx \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)**(5/2),x)

[Out]

Integral(x**7/((a + b*x**2)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \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)^(5/2),x, algorithm="giac")

[Out]

sage0*x