\(\int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx\) [271]

Optimal result

Integrand size = 19, antiderivative size = 59 \[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}+\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}} \]

[Out]

1/2*c*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(3/2)-1/2*(c*x^4+b*x^2)^(1/2)/b/x^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, 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 = {2050, 2033, 212} \[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3} \]

[In]

Int[1/(x^2*Sqrt[b*x^2 + c*x^4]),x]

[Out]

-1/2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))

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 2050

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {b x^2+c x^4}}{2 b x^3}-\frac {c \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{2 b} \\ & = -\frac {\sqrt {b x^2+c x^4}}{2 b x^3}+\frac {c \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{2 b} \\ & = -\frac {\sqrt {b x^2+c x^4}}{2 b x^3}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=\frac {-\sqrt {b} \left (b+c x^2\right )+c x^2 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{2 b^{3/2} x \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

Integrate[1/(x^2*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(-(Sqrt[b]*(b + c*x^2)) + c*x^2*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]])/(2*b^(3/2)*x*Sqrt[x^2*(b + c
*x^2)])

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24

method	result	size
default	\(-\frac {\sqrt {c \,x^{2}+b}\, \left (-c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b \,x^{2}+\sqrt {c \,x^{2}+b}\, b^{\frac {3}{2}}\right )}{2 x \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {5}{2}}}\)	\(73\)
risch	\(-\frac {c \,x^{2}+b}{2 b x \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x \sqrt {c \,x^{2}+b}}{2 b^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(82\)

[In]

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

[Out]

-1/2/x*(c*x^2+b)^(1/2)*(-c*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*b*x^2+(c*x^2+b)^(1/2)*b^(3/2))/(c*x^4+b*x^2)^(1
/2)/b^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.25 \[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=\left [\frac {\sqrt {b} c x^{3} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} b}{4 \, b^{2} x^{3}}, -\frac {\sqrt {-b} c x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}} b}{2 \, b^{2} x^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=\int \frac {1}{x^{2} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

[In]

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

[Out]

Integral(1/(x**2*sqrt(x**2*(b + c*x**2))), x)

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2}} x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=-\frac {\frac {c^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\sqrt {c x^{2} + b} c}{b x^{2}}}{2 \, c \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx=-\frac {\left (\frac {\sqrt {c}\,x^2\,\sqrt {c+\frac {b}{x^2}}}{2\,b}+\frac {c^{3/2}\,x^3\,\mathrm {asin}\left (\frac {\sqrt {b}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,1{}\mathrm {i}}{2\,b^{3/2}}\right )\,\sqrt {\frac {b}{c\,x^2}+1}}{x\,\sqrt {c\,x^4+b\,x^2}} \]

[In]

int(1/(x^2*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

-(((c^(1/2)*x^2*(c + b/x^2)^(1/2))/(2*b) + (c^(3/2)*x^3*asin((b^(1/2)*1i)/(c^(1/2)*x))*1i)/(2*b^(3/2)))*(b/(c*
x^2) + 1)^(1/2))/(x*(b*x^2 + c*x^4)^(1/2))