\(\int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx\) [234]

Optimal result

Integrand size = 19, antiderivative size = 50 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \]

[Out]

-arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))*b^(1/2)+(c*x^4+b*x^2)^(1/2)/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2046, 2033, 212} \[ \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \]

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^2,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {b x^2+c x^4}}{x}+b \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx \\ & = \frac {\sqrt {b x^2+c x^4}}{x}-b \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right ) \\ & = \frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {x \left (b+c x^2-\sqrt {b} \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{\sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^2,x]

[Out]

(x*(b + c*x^2 - Sqrt[b]*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/Sqrt[x^2*(b + c*x^2)]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26

method	result	size
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (\sqrt {c \,x^{2}+b}-\sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )\right )}{x \sqrt {c \,x^{2}+b}}\)	\(63\)

[In]

int((c*x^4+b*x^2)^(1/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

(c*x^4+b*x^2)^(1/2)/x/(c*x^2+b)^(1/2)*((c*x^2+b)^(1/2)-b^(1/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.34 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx=\left [\frac {\sqrt {b} x \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}}}{2 \, x}, \frac {\sqrt {-b} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}}}{x}\right ] \]

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^2,x, algorithm="fricas")

[Out]

[1/2*(sqrt(b)*x*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*sqrt(c*x^4 + b*x^2))/x, (sqrt(-b
)*x*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^4 + b*x^2))/x]

Sympy [F]

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

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**2,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**2, x)

Maxima [F]

\[ \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{2}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^2,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^2, x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {b \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + \sqrt {c x^{2} + b} \mathrm {sgn}\left (x\right ) - \frac {{\left (b \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + \sqrt {-b} \sqrt {b}\right )} \mathrm {sgn}\left (x\right )}{\sqrt {-b}} \]

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^2,x, algorithm="giac")

[Out]

b*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) + sqrt(c*x^2 + b)*sgn(x) - (b*arctan(sqrt(b)/sqrt(-b)) + sq
rt(-b)*sqrt(b))*sgn(x)/sqrt(-b)

Mupad [B] (verification not implemented)

Time = 13.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}}{x}+\frac {\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,\sqrt {c\,x^4+b\,x^2}\,1{}\mathrm {i}}{\sqrt {c}\,x^2\,\sqrt {\frac {b}{c\,x^2}+1}} \]

[In]

int((b*x^2 + c*x^4)^(1/2)/x^2,x)

[Out]

(b*x^2 + c*x^4)^(1/2)/x + (b^(1/2)*asin((b^(1/2)*1i)/(c^(1/2)*x))*(b*x^2 + c*x^4)^(1/2)*1i)/(c^(1/2)*x^2*(b/(c
*x^2) + 1)^(1/2))