3.769 \(\int \frac{\sqrt{a+c x^4}}{x^9} \, dx\)

Optimal. Leaf size=71 \[ \frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^4}}{\sqrt{a}}\right )}{16 a^{3/2}}-\frac{c \sqrt{a+c x^4}}{16 a x^4}-\frac{\sqrt{a+c x^4}}{8 x^8} \]

[Out]

-Sqrt[a + c*x^4]/(8*x^8) - (c*Sqrt[a + c*x^4])/(16*a*x^4) + (c^2*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/(16*a^(3/2)
)

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Rubi [A]  time = 0.0392335, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 51, 63, 208} \[ \frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^4}}{\sqrt{a}}\right )}{16 a^{3/2}}-\frac{c \sqrt{a+c x^4}}{16 a x^4}-\frac{\sqrt{a+c x^4}}{8 x^8} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^4]/x^9,x]

[Out]

-Sqrt[a + c*x^4]/(8*x^8) - (c*Sqrt[a + c*x^4])/(16*a*x^4) + (c^2*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/(16*a^(3/2)
)

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+c x^4}}{x^9} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{\sqrt{a+c x}}{x^3} \, dx,x,x^4\right )\\ &=-\frac{\sqrt{a+c x^4}}{8 x^8}+\frac{1}{16} c \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+c x}} \, dx,x,x^4\right )\\ &=-\frac{\sqrt{a+c x^4}}{8 x^8}-\frac{c \sqrt{a+c x^4}}{16 a x^4}-\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^4\right )}{32 a}\\ &=-\frac{\sqrt{a+c x^4}}{8 x^8}-\frac{c \sqrt{a+c x^4}}{16 a x^4}-\frac{c \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^4}\right )}{16 a}\\ &=-\frac{\sqrt{a+c x^4}}{8 x^8}-\frac{c \sqrt{a+c x^4}}{16 a x^4}+\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^4}}{\sqrt{a}}\right )}{16 a^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0084714, size = 39, normalized size = 0.55 \[ -\frac{c^2 \left (a+c x^4\right )^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{c x^4}{a}+1\right )}{6 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^4]/x^9,x]

[Out]

-(c^2*(a + c*x^4)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (c*x^4)/a])/(6*a^3)

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Maple [A]  time = 0.011, size = 85, normalized size = 1.2 \begin{align*} -{\frac{1}{8\,a{x}^{8}} \left ( c{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{c}{16\,{a}^{2}{x}^{4}} \left ( c{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{c}^{2}}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{4}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{{c}^{2}}{16\,{a}^{2}}\sqrt{c{x}^{4}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(1/2)/x^9,x)

[Out]

-1/8/a/x^8*(c*x^4+a)^(3/2)+1/16*c/a^2/x^4*(c*x^4+a)^(3/2)+1/16*c^2/a^(3/2)*ln((2*a+2*a^(1/2)*(c*x^4+a)^(1/2))/
x^2)-1/16*c^2/a^2*(c*x^4+a)^(1/2)

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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+a)^(1/2)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52014, size = 316, normalized size = 4.45 \begin{align*} \left [\frac{\sqrt{a} c^{2} x^{8} \log \left (\frac{c x^{4} + 2 \, \sqrt{c x^{4} + a} \sqrt{a} + 2 \, a}{x^{4}}\right ) - 2 \,{\left (a c x^{4} + 2 \, a^{2}\right )} \sqrt{c x^{4} + a}}{32 \, a^{2} x^{8}}, -\frac{\sqrt{-a} c^{2} x^{8} \arctan \left (\frac{\sqrt{c x^{4} + a} \sqrt{-a}}{a}\right ) +{\left (a c x^{4} + 2 \, a^{2}\right )} \sqrt{c x^{4} + a}}{16 \, a^{2} x^{8}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x^9,x, algorithm="fricas")

[Out]

[1/32*(sqrt(a)*c^2*x^8*log((c*x^4 + 2*sqrt(c*x^4 + a)*sqrt(a) + 2*a)/x^4) - 2*(a*c*x^4 + 2*a^2)*sqrt(c*x^4 + a
))/(a^2*x^8), -1/16*(sqrt(-a)*c^2*x^8*arctan(sqrt(c*x^4 + a)*sqrt(-a)/a) + (a*c*x^4 + 2*a^2)*sqrt(c*x^4 + a))/
(a^2*x^8)]

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Sympy [A]  time = 5.3359, size = 95, normalized size = 1.34 \begin{align*} - \frac{a}{8 \sqrt{c} x^{10} \sqrt{\frac{a}{c x^{4}} + 1}} - \frac{3 \sqrt{c}}{16 x^{6} \sqrt{\frac{a}{c x^{4}} + 1}} - \frac{c^{\frac{3}{2}}}{16 a x^{2} \sqrt{\frac{a}{c x^{4}} + 1}} + \frac{c^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x^{2}} \right )}}{16 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(1/2)/x**9,x)

[Out]

-a/(8*sqrt(c)*x**10*sqrt(a/(c*x**4) + 1)) - 3*sqrt(c)/(16*x**6*sqrt(a/(c*x**4) + 1)) - c**(3/2)/(16*a*x**2*sqr
t(a/(c*x**4) + 1)) + c**2*asinh(sqrt(a)/(sqrt(c)*x**2))/(16*a**(3/2))

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Giac [A]  time = 1.39045, size = 84, normalized size = 1.18 \begin{align*} -\frac{1}{16} \, c^{2}{\left (\frac{\arctan \left (\frac{\sqrt{c x^{4} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{{\left (c x^{4} + a\right )}^{\frac{3}{2}} + \sqrt{c x^{4} + a} a}{a c^{2} x^{8}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x^9,x, algorithm="giac")

[Out]

-1/16*c^2*(arctan(sqrt(c*x^4 + a)/sqrt(-a))/(sqrt(-a)*a) + ((c*x^4 + a)^(3/2) + sqrt(c*x^4 + a)*a)/(a*c^2*x^8)
)