3.788 \(\int \frac {(a+c x^4)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=68 \[ -\frac {3 c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {3 c \sqrt {a+c x^4}}{16 x^4}-\frac {\left (a+c x^4\right )^{3/2}}{8 x^8} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 47, 63, 208} \[ -\frac {3 c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {3 c \sqrt {a+c x^4}}{16 x^4}-\frac {\left (a+c x^4\right )^{3/2}}{8 x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x^9,x]

[Out]

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

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 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]

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]]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 76, normalized size = 1.12 \[ -\frac {2 a^2+3 c^2 x^8 \sqrt {\frac {c x^4}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {c x^4}{a}+1}\right )+7 a c x^4+5 c^2 x^8}{16 x^8 \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x^9,x]

[Out]

-1/16*(2*a^2 + 7*a*c*x^4 + 5*c^2*x^8 + 3*c^2*x^8*Sqrt[1 + (c*x^4)/a]*ArcTanh[Sqrt[1 + (c*x^4)/a]])/(x^8*Sqrt[a
+ c*x^4])

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fricas [A]  time = 0.76, size = 138, normalized size = 2.03 \[ \left [\frac {3 \, \sqrt {a} c^{2} x^{8} \log \left (\frac {c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, {\left (5 \, a c x^{4} + 2 \, a^{2}\right )} \sqrt {c x^{4} + a}}{32 \, a x^{8}}, \frac {3 \, \sqrt {-a} c^{2} x^{8} \arctan \left (\frac {\sqrt {c x^{4} + a} \sqrt {-a}}{a}\right ) - {\left (5 \, a c x^{4} + 2 \, a^{2}\right )} \sqrt {c x^{4} + a}}{16 \, a x^{8}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^9,x, algorithm="fricas")

[Out]

[1/32*(3*sqrt(a)*c^2*x^8*log((c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(a) + 2*a)/x^4) - 2*(5*a*c*x^4 + 2*a^2)*sqrt(c*x^4
+ a))/(a*x^8), 1/16*(3*sqrt(-a)*c^2*x^8*arctan(sqrt(c*x^4 + a)*sqrt(-a)/a) - (5*a*c*x^4 + 2*a^2)*sqrt(c*x^4 +
a))/(a*x^8)]

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giac [A]  time = 0.16, size = 70, normalized size = 1.03 \[ \frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (c x^{4} + a\right )}^{\frac {3}{2}} c^{3} - 3 \, \sqrt {c x^{4} + a} a c^{3}}{c^{2} x^{8}}}{16 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^9,x, algorithm="giac")

[Out]

1/16*(3*c^3*arctan(sqrt(c*x^4 + a)/sqrt(-a))/sqrt(-a) - (5*(c*x^4 + a)^(3/2)*c^3 - 3*sqrt(c*x^4 + a)*a*c^3)/(c
^2*x^8))/c

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maple [A]  time = 0.02, size = 63, normalized size = 0.93 \[ -\frac {3 c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {5 \sqrt {c \,x^{4}+a}\, c}{16 x^{4}}-\frac {\sqrt {c \,x^{4}+a}\, a}{8 x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(3/2)/x^9,x)

[Out]

-5/16*c*(c*x^4+a)^(1/2)/x^4-3/16*c^2/a^(1/2)*ln((2*a+2*(c*x^4+a)^(1/2)*a^(1/2))/x^2)-1/8*a/x^8*(c*x^4+a)^(1/2)

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maxima [A]  time = 2.94, size = 98, normalized size = 1.44 \[ \frac {3 \, c^{2} \log \left (\frac {\sqrt {c x^{4} + a} - \sqrt {a}}{\sqrt {c x^{4} + a} + \sqrt {a}}\right )}{32 \, \sqrt {a}} - \frac {5 \, {\left (c x^{4} + a\right )}^{\frac {3}{2}} c^{2} - 3 \, \sqrt {c x^{4} + a} a c^{2}}{16 \, {\left ({\left (c x^{4} + a\right )}^{2} - 2 \, {\left (c x^{4} + a\right )} a + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^9,x, algorithm="maxima")

[Out]

3/32*c^2*log((sqrt(c*x^4 + a) - sqrt(a))/(sqrt(c*x^4 + a) + sqrt(a)))/sqrt(a) - 1/16*(5*(c*x^4 + a)^(3/2)*c^2
- 3*sqrt(c*x^4 + a)*a*c^2)/((c*x^4 + a)^2 - 2*(c*x^4 + a)*a + a^2)

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mupad [B]  time = 1.49, size = 52, normalized size = 0.76 \[ \frac {3\,a\,\sqrt {c\,x^4+a}}{16\,x^8}-\frac {3\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^4+a}}{\sqrt {a}}\right )}{16\,\sqrt {a}}-\frac {5\,{\left (c\,x^4+a\right )}^{3/2}}{16\,x^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(3/2)/x^9,x)

[Out]

(3*a*(a + c*x^4)^(1/2))/(16*x^8) - (3*c^2*atanh((a + c*x^4)^(1/2)/a^(1/2)))/(16*a^(1/2)) - (5*(a + c*x^4)^(3/2
))/(16*x^8)

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sympy [A]  time = 4.98, size = 75, normalized size = 1.10 \[ - \frac {a \sqrt {c} \sqrt {\frac {a}{c x^{4}} + 1}}{8 x^{6}} - \frac {5 c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{4}} + 1}}{16 x^{2}} - \frac {3 c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x^{2}} \right )}}{16 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(3/2)/x**9,x)

[Out]

-a*sqrt(c)*sqrt(a/(c*x**4) + 1)/(8*x**6) - 5*c**(3/2)*sqrt(a/(c*x**4) + 1)/(16*x**2) - 3*c**2*asinh(sqrt(a)/(s
qrt(c)*x**2))/(16*sqrt(a))

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