\(\int \frac {(a+c x^4)^{3/2}}{x^9} \, dx\) [788]

Optimal result

Integrand size = 15, antiderivative size = 68 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {3 c \sqrt {a+c x^4}}{16 x^4}-\frac {\left (a+c x^4\right )^{3/2}}{8 x^8}-\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 43, 65, 214} \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {3 c \sqrt {a+c x^4}}{16 x^4}-\frac {\left (a+c x^4\right )^{3/2}}{8 x^8} \]

[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 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 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*} \text {integral}& = \frac {1}{4} \text {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) \text {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 ) \text {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) \text {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*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=\frac {\left (-2 a-5 c x^4\right ) \sqrt {a+c x^4}}{16 x^8}-\frac {3 c^2 \text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}} \]

[In]

Integrate[(a + c*x^4)^(3/2)/x^9,x]

[Out]

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

Maple [A] (verified)

Time = 4.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {\sqrt {x^{4} c +a}\, \left (5 x^{4} c +2 a \right )}{16 x^{8}}-\frac {3 c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{16 \sqrt {a}}\)	\(57\)
default	\(-\frac {a \sqrt {x^{4} c +a}}{8 x^{8}}-\frac {5 c \sqrt {x^{4} c +a}}{16 x^{4}}-\frac {3 c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{16 \sqrt {a}}\)	\(63\)
elliptic	\(-\frac {a \sqrt {x^{4} c +a}}{8 x^{8}}-\frac {5 c \sqrt {x^{4} c +a}}{16 x^{4}}-\frac {3 c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{16 \sqrt {a}}\)	\(63\)
pseudoelliptic	\(\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{4} c +a}}{\sqrt {a}}\right ) c^{2} x^{8}-5 c \,x^{4} \sqrt {x^{4} c +a}\, \sqrt {a}-2 a^{\frac {3}{2}} \sqrt {x^{4} c +a}}{16 x^{8} \sqrt {a}}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.03 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=\left [\frac {3 \, \sqrt {a} c^{2} x^{8} \log \left (\frac {c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, {\left (5 \, a c x^{4} + 2 \, a^{2}\right )} \sqrt {c x^{4} + a}}{32 \, a x^{8}}, \frac {3 \, \sqrt {-a} c^{2} x^{8} \arctan \left (\frac {\sqrt {c x^{4} + a} \sqrt {-a}}{a}\right ) - {\left (5 \, a c x^{4} + 2 \, a^{2}\right )} \sqrt {c x^{4} + a}}{16 \, a x^{8}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=- \frac {a \sqrt {c} \sqrt {\frac {a}{c x^{4}} + 1}}{8 x^{6}} - \frac {5 c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{4}} + 1}}{16 x^{2}} - \frac {3 c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x^{2}} \right )}}{16 \sqrt {a}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=\frac {3 \, c^{2} \log \left (\frac {\sqrt {c x^{4} + a} - \sqrt {a}}{\sqrt {c x^{4} + a} + \sqrt {a}}\right )}{32 \, \sqrt {a}} - \frac {5 \, {\left (c x^{4} + a\right )}^{\frac {3}{2}} c^{2} - 3 \, \sqrt {c x^{4} + a} a c^{2}}{16 \, {\left ({\left (c x^{4} + a\right )}^{2} - 2 \, {\left (c x^{4} + a\right )} a + a^{2}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=\frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (c x^{4} + a\right )}^{\frac {3}{2}} c^{3} - 3 \, \sqrt {c x^{4} + a} a c^{3}}{c^{2} x^{8}}}{16 \, c} \]

[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

Mupad [B] (verification not implemented)

Time = 6.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^9} \, dx=\frac {3\,a\,\sqrt {c\,x^4+a}}{16\,x^8}-\frac {3\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^4+a}}{\sqrt {a}}\right )}{16\,\sqrt {a}}-\frac {5\,{\left (c\,x^4+a\right )}^{3/2}}{16\,x^8} \]

[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)