3.237 \(\int \frac{\sqrt{b x^2+c x^4}}{x^8} \, dx\)

Optimal. Leaf size=112 \[ \frac{c^2 \sqrt{b x^2+c x^4}}{16 b^2 x^3}-\frac{c^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{16 b^{5/2}}-\frac{c \sqrt{b x^2+c x^4}}{24 b x^5}-\frac{\sqrt{b x^2+c x^4}}{6 x^7} \]

[Out]

-Sqrt[b*x^2 + c*x^4]/(6*x^7) - (c*Sqrt[b*x^2 + c*x^4])/(24*b*x^5) + (c^2*Sqrt[b*x^2 + c*x^4])/(16*b^2*x^3) - (
c^3*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(16*b^(5/2))

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Rubi [A]  time = 0.144887, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2008, 206} \[ \frac{c^2 \sqrt{b x^2+c x^4}}{16 b^2 x^3}-\frac{c^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{16 b^{5/2}}-\frac{c \sqrt{b x^2+c x^4}}{24 b x^5}-\frac{\sqrt{b x^2+c x^4}}{6 x^7} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^8,x]

[Out]

-Sqrt[b*x^2 + c*x^4]/(6*x^7) - (c*Sqrt[b*x^2 + c*x^4])/(24*b*x^5) + (c^2*Sqrt[b*x^2 + c*x^4])/(16*b^2*x^3) - (
c^3*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(16*b^(5/2))

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^8} \, dx &=-\frac{\sqrt{b x^2+c x^4}}{6 x^7}+\frac{1}{6} c \int \frac{1}{x^4 \sqrt{b x^2+c x^4}} \, dx\\ &=-\frac{\sqrt{b x^2+c x^4}}{6 x^7}-\frac{c \sqrt{b x^2+c x^4}}{24 b x^5}-\frac{c^2 \int \frac{1}{x^2 \sqrt{b x^2+c x^4}} \, dx}{8 b}\\ &=-\frac{\sqrt{b x^2+c x^4}}{6 x^7}-\frac{c \sqrt{b x^2+c x^4}}{24 b x^5}+\frac{c^2 \sqrt{b x^2+c x^4}}{16 b^2 x^3}+\frac{c^3 \int \frac{1}{\sqrt{b x^2+c x^4}} \, dx}{16 b^2}\\ &=-\frac{\sqrt{b x^2+c x^4}}{6 x^7}-\frac{c \sqrt{b x^2+c x^4}}{24 b x^5}+\frac{c^2 \sqrt{b x^2+c x^4}}{16 b^2 x^3}-\frac{c^3 \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{b x^2+c x^4}}\right )}{16 b^2}\\ &=-\frac{\sqrt{b x^2+c x^4}}{6 x^7}-\frac{c \sqrt{b x^2+c x^4}}{24 b x^5}+\frac{c^2 \sqrt{b x^2+c x^4}}{16 b^2 x^3}-\frac{c^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{16 b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0141793, size = 46, normalized size = 0.41 \[ \frac{c^3 \left (x^2 \left (b+c x^2\right )\right )^{3/2} \, _2F_1\left (\frac{3}{2},4;\frac{5}{2};\frac{c x^2}{b}+1\right )}{3 b^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^8,x]

[Out]

(c^3*(x^2*(b + c*x^2))^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, 1 + (c*x^2)/b])/(3*b^4*x^3)

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Maple [A]  time = 0.051, size = 128, normalized size = 1.1 \begin{align*} -{\frac{1}{48\,{x}^{7}{b}^{3}}\sqrt{c{x}^{4}+b{x}^{2}} \left ( 3\,\sqrt{b}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{6}{c}^{3}-3\,\sqrt{c{x}^{2}+b}{x}^{6}{c}^{3}+3\, \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{4}{c}^{2}-6\, \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{2}bc+8\, \left ( c{x}^{2}+b \right ) ^{3/2}{b}^{2} \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^8,x)

[Out]

-1/48*(c*x^4+b*x^2)^(1/2)*(3*b^(1/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*x^6*c^3-3*(c*x^2+b)^(1/2)*x^6*c^3+3*(
c*x^2+b)^(3/2)*x^4*c^2-6*(c*x^2+b)^(3/2)*x^2*b*c+8*(c*x^2+b)^(3/2)*b^2)/x^7/(c*x^2+b)^(1/2)/b^3

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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^{8}}\,{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^8,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^8, x)

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Fricas [A]  time = 1.67624, size = 410, normalized size = 3.66 \begin{align*} \left [\frac{3 \, \sqrt{b} c^{3} x^{7} \log \left (-\frac{c x^{3} + 2 \, b x - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{b}}{x^{3}}\right ) + 2 \,{\left (3 \, b c^{2} x^{4} - 2 \, b^{2} c x^{2} - 8 \, b^{3}\right )} \sqrt{c x^{4} + b x^{2}}}{96 \, b^{3} x^{7}}, \frac{3 \, \sqrt{-b} c^{3} x^{7} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-b}}{c x^{3} + b x}\right ) +{\left (3 \, b c^{2} x^{4} - 2 \, b^{2} c x^{2} - 8 \, b^{3}\right )} \sqrt{c x^{4} + b x^{2}}}{48 \, b^{3} x^{7}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^8,x, algorithm="fricas")

[Out]

[1/96*(3*sqrt(b)*c^3*x^7*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*b*c^2*x^4 - 2*b^2*c*
x^2 - 8*b^3)*sqrt(c*x^4 + b*x^2))/(b^3*x^7), 1/48*(3*sqrt(-b)*c^3*x^7*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x
^3 + b*x)) + (3*b*c^2*x^4 - 2*b^2*c*x^2 - 8*b^3)*sqrt(c*x^4 + b*x^2))/(b^3*x^7)]

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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^{8}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**8,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**8, x)

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Giac [A]  time = 1.34729, size = 111, normalized size = 0.99 \begin{align*} \frac{1}{48} \, c^{3}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} + \frac{3 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} - 8 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b - 3 \, \sqrt{c x^{2} + b} b^{2}}{b^{2} c^{3} x^{6}}\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^8,x, algorithm="giac")

[Out]

1/48*c^3*(3*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x^2 + b)^(5/2) - 8*(c*x^2 + b)^(3/2)*b - 3
*sqrt(c*x^2 + b)*b^2)/(b^2*c^3*x^6))*sgn(x)