3.8.69 \(\int \frac {\sqrt {a+c x^4}}{x^9} \, dx\) [769]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 = {272, 43, 44, 65, 214} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^4]/x^9,x]

[Out]

-1/8*Sqrt[a + c*x^4]/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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

Rule 272

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]]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+c x^4}}{x^9} \, dx &=\frac {1}{4} \text {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 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.09, size = 62, normalized size = 0.87 \begin {gather*} \frac {\left (-2 a-c x^4\right ) \sqrt {a+c x^4}}{16 a x^8}+\frac {c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{16 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^4]/x^9,x]

[Out]

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

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Maple [A]
time = 0.15, size = 85, normalized size = 1.20

method	result	size
risch	\(-\frac {\sqrt {x^{4} c +a}\, \left (x^{4} c +2 a \right )}{16 x^{8} a}+\frac {c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}\)	\(59\)
default	\(-\frac {\left (x^{4} c +a \right )^{\frac {3}{2}}}{8 a \,x^{8}}+\frac {c \left (x^{4} c +a \right )^{\frac {3}{2}}}{16 a^{2} x^{4}}+\frac {c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}-\frac {c^{2} \sqrt {x^{4} c +a}}{16 a^{2}}\)	\(85\)
elliptic	\(-\frac {\left (x^{4} c +a \right )^{\frac {3}{2}}}{8 a \,x^{8}}+\frac {c \left (x^{4} c +a \right )^{\frac {3}{2}}}{16 a^{2} x^{4}}+\frac {c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}-\frac {c^{2} \sqrt {x^{4} c +a}}{16 a^{2}}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 100, normalized size = 1.41 \begin {gather*} -\frac {c^{2} \log \left (\frac {\sqrt {c x^{4} + a} - \sqrt {a}}{\sqrt {c x^{4} + a} + \sqrt {a}}\right )}{32 \, a^{\frac {3}{2}}} - \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}} c^{2} + \sqrt {c x^{4} + a} a c^{2}}{16 \, {\left ({\left (c x^{4} + a\right )}^{2} a - 2 \, {\left (c x^{4} + a\right )} a^{2} + a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*c^2*log((sqrt(c*x^4 + a) - sqrt(a))/(sqrt(c*x^4 + a) + sqrt(a)))/a^(3/2) - 1/16*((c*x^4 + a)^(3/2)*c^2 +
sqrt(c*x^4 + a)*a*c^2)/((c*x^4 + a)^2*a - 2*(c*x^4 + a)*a^2 + a^3)

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Fricas [A]
time = 0.38, size = 133, normalized size = 1.87 \begin {gather*} \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 {gather*}

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 = 2.14, size = 95, normalized size = 1.34 \begin {gather*} - \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 {gather*}

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 = 0.52, size = 72, normalized size = 1.01 \begin {gather*} -\frac {\frac {c^{3} \arctan \left (\frac {\sqrt {c x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}} c^{3} + \sqrt {c x^{4} + a} a c^{3}}{a c^{2} x^{8}}}{16 \, c} \end {gather*}

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

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Mupad [B]
time = 1.44, size = 54, normalized size = 0.76 \begin {gather*} \frac {c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^4+a}}{\sqrt {a}}\right )}{16\,a^{3/2}}-\frac {\sqrt {c\,x^4+a}}{16\,x^8}-\frac {{\left (c\,x^4+a\right )}^{3/2}}{16\,a\,x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(1/2)/x^9,x)

[Out]

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

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