\(\int \frac {x}{\sqrt {b x^2+c x^4}} \, dx\) [263]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2038, 634, 212} \[ \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}} \]

[In]

Int[x/Sqrt[b*x^2 + c*x^4],x]

[Out]

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

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 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 2038

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(a*x^Simplify[j/n]
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && IntegerQ[Simplify[j
/n]] && EqQ[Simplify[m - n + 1], 0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b+c x^2}}\right )}{\sqrt {c} \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

Integrate[x/Sqrt[b*x^2 + c*x^4],x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35

method	result	size
pseudoelliptic	\(\frac {-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )}{2 \sqrt {c}}\)	\(42\)
default	\(\frac {x \sqrt {c \,x^{2}+b}\, \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right )}{\sqrt {c \,x^{4}+b \,x^{2}}\, \sqrt {c}}\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.39 \[ \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx=\left [\frac {\log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right )}{c}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx=\frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, \sqrt {c}} \]

[In]

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

[Out]

1/2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx=\frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{2 \, \sqrt {c}} - \frac {\log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/2*log(abs(b))*sgn(x)/sqrt(c) - log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(sqrt(c)*sgn(x))

Mupad [B] (verification not implemented)

Time = 13.85 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {x}{\sqrt {b x^2+c x^4}} \, dx=\frac {\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,\sqrt {c}} \]

[In]

int(x/(b*x^2 + c*x^4)^(1/2),x)

[Out]

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