3.225 \(\int \frac{\sqrt{b x^2+c x^4}}{x^3} \, dx\)

Optimal. Leaf size=52 \[ \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )-\frac{\sqrt{b x^2+c x^4}}{x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0770799, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2018, 662, 620, 206} \[ \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )-\frac{\sqrt{b x^2+c x^4}}{x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^3,x]

[Out]

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

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(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]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 620

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 206

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

Rubi steps

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

Mathematica [A]  time = 0.093011, size = 60, normalized size = 1.15 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\frac{\sqrt{c} x \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{\frac{c x^2}{b}+1}}-1\right )}{x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^3,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-1 + (Sqrt[c]*x*ArcSinh[(Sqrt[c]*x)/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (c*x^2)/b])))/x^2

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 86, normalized size = 1.7 \begin{align*} -{\frac{1}{b{x}^{2}}\sqrt{c{x}^{4}+b{x}^{2}} \left ( -\sqrt{c{x}^{2}+b}{c}^{{\frac{3}{2}}}{x}^{2}+ \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}\sqrt{c}-\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) xbc \right ){\frac{1}{\sqrt{c{x}^{2}+b}}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(1/2)/x^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.5805, size = 261, normalized size = 5.02 \begin{align*} \left [\frac{\sqrt{c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \, \sqrt{c x^{4} + b x^{2}}}{2 \, x^{2}}, -\frac{\sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) + \sqrt{c x^{4} + b x^{2}}}{x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} \left (b + c x^{2}\right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**3,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.25569, size = 82, normalized size = 1.58 \begin{align*} -\frac{1}{2} \, \sqrt{c} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \, b \sqrt{c} \mathrm{sgn}\left (x\right )}{{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="giac")

[Out]

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