3.236 \(\int \frac{\sqrt{b x^2+c x^4}}{x^6} \, dx\)

Optimal. Leaf size=84 \[ \frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{3/2}}-\frac{c \sqrt{b x^2+c x^4}}{8 b x^3}-\frac{\sqrt{b x^2+c x^4}}{4 x^5} \]

[Out]

-Sqrt[b*x^2 + c*x^4]/(4*x^5) - (c*Sqrt[b*x^2 + c*x^4])/(8*b*x^3) + (c^2*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4
]])/(8*b^(3/2))

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Rubi [A]  time = 0.0980381, antiderivative size = 84, 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 = {2020, 2025, 2008, 206} \[ \frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{3/2}}-\frac{c \sqrt{b x^2+c x^4}}{8 b x^3}-\frac{\sqrt{b x^2+c x^4}}{4 x^5} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^6,x]

[Out]

-Sqrt[b*x^2 + c*x^4]/(4*x^5) - (c*Sqrt[b*x^2 + c*x^4])/(8*b*x^3) + (c^2*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4
]])/(8*b^(3/2))

Rule 2020

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 + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2025

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]

Rule 2008

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 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^6} \, dx &=-\frac{\sqrt{b x^2+c x^4}}{4 x^5}+\frac{1}{4} c \int \frac{1}{x^2 \sqrt{b x^2+c x^4}} \, dx\\ &=-\frac{\sqrt{b x^2+c x^4}}{4 x^5}-\frac{c \sqrt{b x^2+c x^4}}{8 b x^3}-\frac{c^2 \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx}{8 b}\\ &=-\frac{\sqrt{b x^2+c x^4}}{4 x^5}-\frac{c \sqrt{b x^2+c x^4}}{8 b x^3}+\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{b x^2+c x^4}}\right )}{8 b}\\ &=-\frac{\sqrt{b x^2+c x^4}}{4 x^5}-\frac{c \sqrt{b x^2+c x^4}}{8 b x^3}+\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0152151, size = 46, normalized size = 0.55 \[ -\frac{c^2 \left (x^2 \left (b+c x^2\right )\right )^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{c x^2}{b}+1\right )}{3 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^6,x]

[Out]

-(c^2*(x^2*(b + c*x^2))^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (c*x^2)/b])/(3*b^3*x^3)

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Maple [A]  time = 0.048, size = 106, normalized size = 1.3 \begin{align*}{\frac{1}{8\,{b}^{2}{x}^{5}}\sqrt{c{x}^{4}+b{x}^{2}} \left ( \sqrt{b}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{4}{c}^{2}-\sqrt{c{x}^{2}+b}{x}^{4}{c}^{2}+ \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{2}c-2\, \left ( c{x}^{2}+b \right ) ^{3/2}b \right ){\frac{1}{\sqrt{c{x}^{2}+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(1/2)/x^6,x)

[Out]

1/8*(c*x^4+b*x^2)^(1/2)*(b^(1/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*x^4*c^2-(c*x^2+b)^(1/2)*x^4*c^2+(c*x^2+b)
^(3/2)*x^2*c-2*(c*x^2+b)^(3/2)*b)/x^5/(c*x^2+b)^(1/2)/b^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{4} + b x^{2}}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^6,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^6, x)

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Fricas [A]  time = 1.62325, size = 356, normalized size = 4.24 \begin{align*} \left [\frac{\sqrt{b} c^{2} x^{5} \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}}{\left (b c x^{2} + 2 \, b^{2}\right )}}{16 \, b^{2} x^{5}}, -\frac{\sqrt{-b} c^{2} x^{5} \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}}{\left (b c x^{2} + 2 \, b^{2}\right )}}{8 \, b^{2} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^6,x, algorithm="fricas")

[Out]

[1/16*(sqrt(b)*c^2*x^5*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*c*
x^2 + 2*b^2))/(b^2*x^5), -1/8*(sqrt(-b)*c^2*x^5*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^
4 + b*x^2)*(b*c*x^2 + 2*b^2))/(b^2*x^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} \left (b + c x^{2}\right )}}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**6,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**6, x)

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Giac [A]  time = 1.30026, size = 86, normalized size = 1.02 \begin{align*} -\frac{1}{8} \, c^{2}{\left (\frac{\arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{{\left (c x^{2} + b\right )}^{\frac{3}{2}} + \sqrt{c x^{2} + b} b}{b c^{2} x^{4}}\right )} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/8*c^2*(arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b) + ((c*x^2 + b)^(3/2) + sqrt(c*x^2 + b)*b)/(b*c^2*x^4))
*sgn(x)