\(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx\) [555]

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1126, 14} \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^6,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {a b+b^2 x^2}{x^6} \, dx}{a b+b^2 x^2} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a b}{x^6}+\frac {b^2}{x^4}\right ) \, dx}{a b+b^2 x^2} \\ & = -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (3 a+5 b x^2\right )}{15 x^5 \left (a+b x^2\right )} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^6,x]

[Out]

-1/15*(Sqrt[(a + b*x^2)^2]*(3*a + 5*b*x^2))/(x^5*(a + b*x^2))

Maple [A] (verified)

Time = 1.93 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44

method	result	size
risch	\(\frac {\left (-\frac {b \,x^{2}}{3}-\frac {a}{5}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x^{5} \left (b \,x^{2}+a \right )}\)	\(35\)
gosper	\(-\frac {\left (5 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{15 \left (b \,x^{2}+a \right ) x^{5}}\)	\(36\)
default	\(-\frac {\left (5 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{15 \left (b \,x^{2}+a \right ) x^{5}}\)	\(36\)

[In]

int(((b*x^2+a)^2)^(1/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=-\frac {5 \, b x^{2} + 3 \, a}{15 \, x^{5}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^6,x, algorithm="fricas")

[Out]

-1/15*(5*b*x^2 + 3*a)/x^5

Sympy [F]

\[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=\int \frac {\sqrt {\left (a + b x^{2}\right )^{2}}}{x^{6}}\, dx \]

[In]

integrate(((b*x**2+a)**2)**(1/2)/x**6,x)

[Out]

Integral(sqrt((a + b*x**2)**2)/x**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=-\frac {5 \, b x^{2} + 3 \, a}{15 \, x^{5}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^6,x, algorithm="maxima")

[Out]

-1/15*(5*b*x^2 + 3*a)/x^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=-\frac {5 \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a \mathrm {sgn}\left (b x^{2} + a\right )}{15 \, x^{5}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/15*(5*b*x^2*sgn(b*x^2 + a) + 3*a*sgn(b*x^2 + a))/x^5

Mupad [B] (verification not implemented)

Time = 13.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^6} \, dx=-\frac {\left (5\,b\,x^2+3\,a\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{15\,x^5\,\left (b\,x^2+a\right )} \]

[In]

int(((a + b*x^2)^2)^(1/2)/x^6,x)

[Out]

-((3*a + 5*b*x^2)*((a + b*x^2)^2)^(1/2))/(15*x^5*(a + b*x^2))