3.256 \(\int \frac{(b x^2+c x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=81 \[ -\frac{3 c^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 \sqrt{b}}-\frac{3 c \sqrt{b x^2+c x^4}}{8 x^3}-\frac{\left (b x^2+c x^4\right )^{3/2}}{4 x^7} \]

[Out]

(-3*c*Sqrt[b*x^2 + c*x^4])/(8*x^3) - (b*x^2 + c*x^4)^(3/2)/(4*x^7) - (3*c^2*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c
*x^4]])/(8*Sqrt[b])

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Rubi [A]  time = 0.114552, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2020, 2008, 206} \[ -\frac{3 c^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 \sqrt{b}}-\frac{3 c \sqrt{b x^2+c x^4}}{8 x^3}-\frac{\left (b x^2+c x^4\right )^{3/2}}{4 x^7} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^8,x]

[Out]

(-3*c*Sqrt[b*x^2 + c*x^4])/(8*x^3) - (b*x^2 + c*x^4)^(3/2)/(4*x^7) - (3*c^2*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c
*x^4]])/(8*Sqrt[b])

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

Mathematica [A]  time = 0.0501643, size = 80, normalized size = 0.99 \[ -\frac{2 b^2+3 c^2 x^4 \sqrt{\frac{c x^2}{b}+1} \tanh ^{-1}\left (\sqrt{\frac{c x^2}{b}+1}\right )+7 b c x^2+5 c^2 x^4}{8 x^3 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^8,x]

[Out]

-(2*b^2 + 7*b*c*x^2 + 5*c^2*x^4 + 3*c^2*x^4*Sqrt[1 + (c*x^2)/b]*ArcTanh[Sqrt[1 + (c*x^2)/b]])/(8*x^3*Sqrt[x^2*
(b + c*x^2)])

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Maple [A]  time = 0.048, size = 125, normalized size = 1.5 \begin{align*} -{\frac{1}{8\,{b}^{2}{x}^{7}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( - \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{4}{c}^{2}+3\,{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{4}{c}^{2}+ \left ( c{x}^{2}+b \right ) ^{{\frac{5}{2}}}{x}^{2}c-3\,\sqrt{c{x}^{2}+b}{x}^{4}b{c}^{2}+2\, \left ( c{x}^{2}+b \right ) ^{5/2}b \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^8,x)

[Out]

-1/8*(c*x^4+b*x^2)^(3/2)*(-(c*x^2+b)^(3/2)*x^4*c^2+3*b^(3/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*x^4*c^2+(c*x^
2+b)^(5/2)*x^2*c-3*(c*x^2+b)^(1/2)*x^4*b*c^2+2*(c*x^2+b)^(5/2)*b)/x^7/(c*x^2+b)^(3/2)/b^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{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)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^8, x)

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Fricas [A]  time = 1.26905, size = 360, normalized size = 4.44 \begin{align*} \left [\frac{3 \, \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 (5 \, b c x^{2} + 2 \, b^{2}\right )}}{16 \, b x^{5}}, \frac{3 \, \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 (5 \, b c x^{2} + 2 \, b^{2}\right )}}{8 \, b x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}}{x^{8}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**8,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**8, x)

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Giac [A]  time = 1.23184, size = 85, normalized size = 1.05 \begin{align*} \frac{1}{8} \, c^{2}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{c x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{5 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} - 3 \, \sqrt{c x^{2} + 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)^(3/2)/x^8,x, algorithm="giac")

[Out]

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