\(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx\) [545]

Optimal result

Integrand size = 26, antiderivative size = 75 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, 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} \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {b \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^3,x]

[Out]

-1/2*(a*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^2*(a + b*x^2)) + (b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[x])/(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^3} \, 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^3}+\frac {b^2}{x}\right ) \, dx}{a b+b^2 x^2} \\ & = -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(178\) vs. \(2(75)=150\).

Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.37 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {a \sqrt {a^2}-a \sqrt {\left (a+b x^2\right )^2}-2 a b x^2 \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-2 \sqrt {a^2} b x^2 \log \left (x^2\right )+\sqrt {a^2} b x^2 \log \left (a \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )+\sqrt {a^2} b x^2 \log \left (a \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{4 a x^2} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^3,x]

[Out]

(a*Sqrt[a^2] - a*Sqrt[(a + b*x^2)^2] - 2*a*b*x^2*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 2*Sqrt[a
^2]*b*x^2*Log[x^2] + Sqrt[a^2]*b*x^2*Log[a*(Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2])] + Sqrt[a^2]*b*x^2*Log[a*
(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/(4*a*x^2)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.37

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-\ln \left (b \,x^{2}\right ) x^{2} b +a \right )}{2 x^{2}}\)	\(28\)
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (2 \ln \left (x \right ) x^{2} b -a \right )}{2 x^{2} \left (b \,x^{2}+a \right )}\)	\(38\)
risch	\(-\frac {a \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 x^{2} \left (b \,x^{2}+a \right )}+\frac {b \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(52\)

[In]

int(((b*x^2+a)^2)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.23 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {2 \, b x^{2} \log \left (x\right ) - a}{2 \, x^{2}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^3,x, algorithm="fricas")

[Out]

1/2*(2*b*x^2*log(x) - a)/x^2

Sympy [F]

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

[In]

integrate(((b*x**2+a)**2)**(1/2)/x**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {1}{2} \, b \log \left (x^{2}\right ) - \frac {a}{2 \, x^{2}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^3,x, algorithm="maxima")

[Out]

1/2*b*log(x^2) - 1/2*a/x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {1}{2} \, b \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, x^{2}} \]

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/2*b*log(x^2)*sgn(b*x^2 + a) - 1/2*(b*x^2*sgn(b*x^2 + a) + a*sgn(b*x^2 + a))/x^2

Mupad [B] (verification not implemented)

Time = 13.74 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {\ln \left (a\,b+\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {b^2}+b^2\,x^2\right )\,\sqrt {b^2}}{2}-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^2}-\frac {a\,b\,\ln \left (a\,b+\frac {a^2}{x^2}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^2}\right )}{2\,\sqrt {a^2}} \]

[In]

int(((a + b*x^2)^2)^(1/2)/x^3,x)

[Out]

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