\(\int \frac {x^7}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [647]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x^8}{8 a \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

1/8*x^8/a/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 37} \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x^8}{8 a \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[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 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 660

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 1125

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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {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 ) \text {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 [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(161\) vs. \(2(41)=82\).

Time = 0.50 (sec) , antiderivative size = 161, normalized size of antiderivative = 3.93 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {x^8 \left (a^5+a b^4 x^8-a^3 \sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}-a \sqrt {a^2} b^2 x^4 \sqrt {\left (a+b x^2\right )^2}+\sqrt {a^2} b x^2 \sqrt {\left (a+b x^2\right )^2} \left (a^2+b^2 x^4\right )\right )}{8 a^5 \left (a+b x^2\right )^3 \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )} \]

[In]

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

[Out]

-1/8*(x^8*(a^5 + a*b^4*x^8 - a^3*Sqrt[a^2]*Sqrt[(a + b*x^2)^2] - a*Sqrt[a^2]*b^2*x^4*Sqrt[(a + b*x^2)^2] + Sqr
t[a^2]*b*x^2*Sqrt[(a + b*x^2)^2]*(a^2 + b^2*x^4)))/(a^5*(a + b*x^2)^3*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(
a + b*x^2)^2])))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.22

method	result	size
pseudoelliptic	\(-\frac {\left (2 b \,x^{2}+a \right ) \left (2 b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{8 \left (b \,x^{2}+a \right )^{4} b^{4}}\)	\(50\)
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (4 b^{3} x^{6}+6 b^{2} x^{4} a +4 a^{2} b \,x^{2}+a^{3}\right )}{8 b^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(54\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (4 b^{3} x^{6}+6 b^{2} x^{4} a +4 a^{2} b \,x^{2}+a^{3}\right )}{8 b^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(54\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {x^{6}}{2 b}-\frac {3 a \,x^{4}}{4 b^{2}}-\frac {a^{2} x^{2}}{2 b^{3}}-\frac {a^{3}}{8 b^{4}}\right )}{\left (b \,x^{2}+a \right )^{5}}\)	\(59\)

[In]

int(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*(2*b*x^2+a)*(2*b^2*x^4+2*a*b*x^2+a^2)*csgn(b*x^2+a)/(b*x^2+a)^4/b^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (28) = 56\).

Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.95 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {4 \, b^{3} x^{6} + 6 \, a b^{2} x^{4} + 4 \, a^{2} b x^{2} + a^{3}}{8 \, {\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{7}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (28) = 56\).

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.95 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {4 \, b^{3} x^{6} + 6 \, a b^{2} x^{4} + 4 \, a^{2} b x^{2} + a^{3}}{8 \, {\left (b^{8} x^{8} + 4 \, a b^{7} x^{6} + 6 \, a^{2} b^{6} x^{4} + 4 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.32 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {4 \, b^{3} x^{6} + 6 \, a b^{2} x^{4} + 4 \, a^{2} b x^{2} + a^{3}}{8 \, {\left (b x^{2} + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 3.51 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,b^4\,{\left (b\,x^2+a\right )}^5}-\frac {a^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,b^4\,{\left (b\,x^2+a\right )}^4}-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,b^4\,{\left (b\,x^2+a\right )}^2}+\frac {3\,a\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,b^4\,{\left (b\,x^2+a\right )}^3} \]

[In]

int(x^7/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

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