\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^9} \, dx\) [567]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=-\frac {\left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 a x^8} \]

[Out]

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

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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=-\frac {\left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 a x^8} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^9,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (a^3+4 a^2 b x^2+6 a b^2 x^4+4 b^3 x^6\right )}{8 x^8 \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^9,x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(-\frac {\left (2 b \,x^{2}+a \right ) \left (2 b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{8 x^{8}}\)	\(41\)
gosper	\(-\frac {\left (4 b^{3} x^{6}+6 b^{2} x^{4} a +4 a^{2} b \,x^{2}+a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{8 x^{8} \left (b \,x^{2}+a \right )^{3}}\)	\(56\)
default	\(-\frac {\left (4 b^{3} x^{6}+6 b^{2} x^{4} a +4 a^{2} b \,x^{2}+a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{8 x^{8} \left (b \,x^{2}+a \right )^{3}}\)	\(56\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{2} b^{3} x^{6}-\frac {3}{4} b^{2} x^{4} a -\frac {1}{2} a^{2} b \,x^{2}-\frac {1}{8} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{8}}\)	\(57\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^9,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)/x^8

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=-\frac {4 \, b^{3} x^{6} + 6 \, a b^{2} x^{4} + 4 \, a^{2} b x^{2} + a^{3}}{8 \, x^{8}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^9,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)/x^8

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{9}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**9,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**9, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=-\frac {b^{3}}{2 \, x^{2}} - \frac {3 \, a b^{2}}{4 \, x^{4}} - \frac {a^{2} b}{2 \, x^{6}} - \frac {a^{3}}{8 \, x^{8}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^9,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (28) = 56\).

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.66 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=-\frac {4 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{8 \, x^{8}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^9,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 3.68 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^9} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^2\,\left (b\,x^2+a\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^4\,\left (b\,x^2+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^6\,\left (b\,x^2+a\right )} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^9,x)

[Out]

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