\(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx\) [548]

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx=-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx=-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^9,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 {\sqrt {a^2+2 a b x+b^2 x^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 {a b+b^2 x}{x^5} \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (\frac {a b}{x^5}+\frac {b^2}{x^4}\right ) \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )} \\ & = -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (3 a+4 b x^2\right )}{24 x^8 \left (a+b x^2\right )} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^9,x]

[Out]

-1/24*(Sqrt[(a + b*x^2)^2]*(3*a + 4*b*x^2))/(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 = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.30

method	result	size
pseudoelliptic	\(-\frac {\left (4 b \,x^{2}+3 a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{24 x^{8}}\)	\(24\)
risch	\(\frac {\left (-\frac {b \,x^{2}}{6}-\frac {a}{8}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x^{8} \left (b \,x^{2}+a \right )}\)	\(35\)
gosper	\(-\frac {\left (4 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{24 x^{8} \left (b \,x^{2}+a \right )}\)	\(36\)
default	\(-\frac {\left (4 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{24 x^{8} \left (b \,x^{2}+a \right )}\)	\(36\)

[In]

int(((b*x^2+a)^2)^(1/2)/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/24*(4*b*x^2+3*a)*csgn(b*x^2+a)/x^8

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx=-\frac {4 \, b x^{2} + 3 \, a}{24 \, x^{8}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^9,x, algorithm="fricas")

[Out]

-1/24*(4*b*x^2 + 3*a)/x^8

Sympy [F]

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

[In]

integrate(((b*x**2+a)**2)**(1/2)/x**9,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx=-\frac {4 \, b x^{2} + 3 \, a}{24 \, x^{8}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^9,x, algorithm="maxima")

[Out]

-1/24*(4*b*x^2 + 3*a)/x^8

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx=-\frac {4 \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a \mathrm {sgn}\left (b x^{2} + a\right )}{24 \, x^{8}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^9,x, algorithm="giac")

[Out]

-1/24*(4*b*x^2*sgn(b*x^2 + a) + 3*a*sgn(b*x^2 + a))/x^8

Mupad [B] (verification not implemented)

Time = 13.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx=-\frac {\left (4\,b\,x^2+3\,a\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{24\,x^8\,\left (b\,x^2+a\right )} \]

[In]

int(((a + b*x^2)^2)^(1/2)/x^9,x)

[Out]

-((3*a + 4*b*x^2)*((a + b*x^2)^2)^(1/2))/(24*x^8*(a + b*x^2))