\(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx\) [547]

Optimal result

Integrand size = 26, antiderivative size = 72 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx=-\frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 a x^6}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{12 a^2 x^6} \]

[Out]

1/12*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/a^2/x^6-1/4*(b*x^2+a)*((b*x^2+a)^2)^(1/2)/a/x^6

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1124} \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{12 a^2 x^6}-\frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 a x^6} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^7,x]

[Out]

-1/4*((a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a*x^6) + (a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(12*a^2*x^6)

Rule 1124

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^2
+ c*x^4)^(p + 1)/(4*a*d*(p + 1)*(2*p + 1))), x] - Simp[(d*x)^(m + 1)*(2*a + b*x^2)*((a + b*x^2 + c*x^4)^p/(4*a
*d*(2*p + 1))), x] /; FreeQ[{a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[m + 4*p + 5,
0] && NeQ[p, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 a x^6}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{12 a^2 x^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.54 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (2 a+3 b x^2\right )}{12 x^6 \left (a+b x^2\right )} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^7,x]

[Out]

-1/12*(Sqrt[(a + b*x^2)^2]*(2*a + 3*b*x^2))/(x^6*(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 = 0.33

method	result	size
pseudoelliptic	\(-\frac {\left (3 b \,x^{2}+2 a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{12 x^{6}}\)	\(24\)
risch	\(\frac {\left (-\frac {b \,x^{2}}{4}-\frac {a}{6}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x^{6} \left (b \,x^{2}+a \right )}\)	\(35\)
gosper	\(-\frac {\left (3 b \,x^{2}+2 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{12 x^{6} \left (b \,x^{2}+a \right )}\)	\(36\)
default	\(-\frac {\left (3 b \,x^{2}+2 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{12 x^{6} \left (b \,x^{2}+a \right )}\)	\(36\)

[In]

int(((b*x^2+a)^2)^(1/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

-1/12*(3*b*x^2+2*a)*csgn(b*x^2+a)/x^6

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.21 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx=-\frac {3 \, b x^{2} + 2 \, a}{12 \, x^{6}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^7,x, algorithm="fricas")

[Out]

-1/12*(3*b*x^2 + 2*a)/x^6

Sympy [F]

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

[In]

integrate(((b*x**2+a)**2)**(1/2)/x**7,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.21 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx=-\frac {3 \, b x^{2} + 2 \, a}{12 \, x^{6}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^7,x, algorithm="maxima")

[Out]

-1/12*(3*b*x^2 + 2*a)/x^6

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.43 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx=-\frac {3 \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a \mathrm {sgn}\left (b x^{2} + a\right )}{12 \, x^{6}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^7,x, algorithm="giac")

[Out]

-1/12*(3*b*x^2*sgn(b*x^2 + a) + 2*a*sgn(b*x^2 + a))/x^6

Mupad [B] (verification not implemented)

Time = 13.40 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^7} \, dx=-\frac {\left (3\,b\,x^2+2\,a\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{12\,x^6\,\left (b\,x^2+a\right )} \]

[In]

int(((a + b*x^2)^2)^(1/2)/x^7,x)

[Out]

-((2*a + 3*b*x^2)*((a + b*x^2)^2)^(1/2))/(12*x^6*(a + b*x^2))