\(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx\) [544]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.01 (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} \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx=\frac {b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {a \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

[In]

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

[Out]

(b*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*(a + b*x^2)) + (a*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} \, 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}+b^2 x\right ) \, dx}{a b+b^2 x^2} \\ & = \frac {b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {a \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. \(454\) vs. \(2(75)=150\).

Time = 0.97 (sec) , antiderivative size = 454, normalized size of antiderivative = 6.05 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx=\frac {-2 a \sqrt {a^2} b x^2-2 \sqrt {a^2} b^2 x^4+2 a b x^2 \sqrt {\left (a+b x^2\right )^2}-2 a \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right ) \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-2 \left (\left (a^2\right )^{3/2}+a \sqrt {a^2} b x^2-a^2 \sqrt {\left (a+b x^2\right )^2}\right ) \log \left (x^2\right )+\left (a^2\right )^{3/2} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+a \sqrt {a^2} b x^2 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-a^2 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+\left (a^2\right )^{3/2} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+a \sqrt {a^2} b x^2 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-a^2 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )}{4 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

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

[Out]

(-2*a*Sqrt[a^2]*b*x^2 - 2*Sqrt[a^2]*b^2*x^4 + 2*a*b*x^2*Sqrt[(a + b*x^2)^2] - 2*a*(a^2 + a*b*x^2 - Sqrt[a^2]*S
qrt[(a + b*x^2)^2])*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 2*((a^2)^(3/2) + a*Sqrt[a^2]*b*x^2 -
a^2*Sqrt[(a + b*x^2)^2])*Log[x^2] + (a^2)^(3/2)*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + a*Sqrt[a^2]*b*x
^2*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] - a^2*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*
x^2)^2]] + (a^2)^(3/2)*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]] + a*Sqrt[a^2]*b*x^2*Log[Sqrt[a^2] + b*x^2
- Sqrt[(a + b*x^2)^2]] - a^2*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]])/(4*(a^2 + a*b*x
^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))

Maple [C] (warning: unable to verify)

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

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.35

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (b \,x^{2}+a +a \ln \left (b \,x^{2}\right )\right )}{2}\)	\(26\)
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (b \,x^{2}+2 a \ln \left (x \right )\right )}{2 b \,x^{2}+2 a}\)	\(34\)
risch	\(\frac {b \,x^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 b \,x^{2}+2 a}+\frac {a \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.15 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx=\frac {1}{2} \, b x^{2} + a \log \left (x\right ) \]

[In]

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

[Out]

1/2*b*x^2 + a*log(x)

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt((a + b*x**2)**2)/x, x)

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

1/2*b*x^2 + 1/2*a*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.40 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx=\frac {1}{2} \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{2} \, a \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.81 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.45 \[ \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2}-\frac {\ln \left (a\,b+\frac {a^2}{x^2}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^2}\right )\,\sqrt {a^2}}{2}+\frac {a\,b\,\ln \left (a\,b+\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {b^2}+b^2\,x^2\right )}{2\,\sqrt {b^2}} \]

[In]

int(((a + b*x^2)^2)^(1/2)/x,x)

[Out]

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