3.496 \(\int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx\)

Optimal. Leaf size=117 \[ -\frac {2 a c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 x^2 \left (c-\frac {c}{a x}\right )^{3/2}}+\frac {8 a^3 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{105 \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {8 a^3 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 \sqrt {c-\frac {c}{a x}}} \]

[Out]

8/105*a^3*c^2*(1-1/a^2/x^2)^(3/2)/(c-c/a/x)^(3/2)-2/7*a*c^2*(1-1/a^2/x^2)^(3/2)/(c-c/a/x)^(3/2)/x^2-8/35*a^3*c
*(1-1/a^2/x^2)^(3/2)/(c-c/a/x)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6178, 871, 795, 649} \[ \frac {8 a^3 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{105 \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {2 a c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 x^2 \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {8 a^3 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcCoth[a*x]*Sqrt[c - c/(a*x)])/x^4,x]

[Out]

(8*a^3*c^2*(1 - 1/(a^2*x^2))^(3/2))/(105*(c - c/(a*x))^(3/2)) - (8*a^3*c*(1 - 1/(a^2*x^2))^(3/2))/(35*Sqrt[c -
c/(a*x)]) - (2*a*c^2*(1 - 1/(a^2*x^2))^(3/2))/(7*(c - c/(a*x))^(3/2)*x^2)

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 871

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d +
e*x)^(m - 1)*(f + g*x)^n*(a + c*x^2)^(p + 1))/(c*(m - n - 1)), x] - Dist[(n*(e*f + d*g))/(e*(m - n - 1)), Int[
(d + e*x)^m*(f + g*x)^(n - 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0]
&& EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (IntegerQ[2*p
] || IntegerQ[n])

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx &=-\left (c \operatorname {Subst}\left (\int \frac {x^2 \sqrt {1-\frac {x^2}{a^2}}}{\sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )\right )\\ &=-\frac {2 a c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 \left (c-\frac {c}{a x}\right )^{3/2} x^2}+\frac {1}{7} (4 a c) \operatorname {Subst}\left (\int \frac {x \sqrt {1-\frac {x^2}{a^2}}}{\sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {8 a^3 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 \sqrt {c-\frac {c}{a x}}}-\frac {2 a c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 \left (c-\frac {c}{a x}\right )^{3/2} x^2}+\frac {1}{35} \left (4 a^2 c\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{\sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {8 a^3 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{105 \left (c-\frac {c}{a x}\right )^{3/2}}-\frac {8 a^3 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 \sqrt {c-\frac {c}{a x}}}-\frac {2 a c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 \left (c-\frac {c}{a x}\right )^{3/2} x^2}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 66, normalized size = 0.56 \[ -\frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \left (8 a^3 x^3-4 a^2 x^2+3 a x+15\right ) \sqrt {c-\frac {c}{a x}}}{105 x^2 (a x-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcCoth[a*x]*Sqrt[c - c/(a*x)])/x^4,x]

[Out]

(-2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(15 + 3*a*x - 4*a^2*x^2 + 8*a^3*x^3))/(105*x^2*(-1 + a*x))

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fricas [A]  time = 0.53, size = 77, normalized size = 0.66 \[ -\frac {2 \, {\left (8 \, a^{4} x^{4} + 4 \, a^{3} x^{3} - a^{2} x^{2} + 18 \, a x + 15\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{105 \, {\left (a x^{4} - x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(8*a^4*x^4 + 4*a^3*x^3 - a^2*x^2 + 18*a*x + 15)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(a*x^
4 - x^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, choosing root of [1,0,%%%{-2,[2,1,2]%%%}+%%%{-2,[2,0,2]%%%}+%%%{2,[1,1,1]%%%}+%%%{2,[0,0,0]%%%},
0,%%%{1,[4,2,4]%%%}+%%%{-2,[4,1,4]%%%}+%%%{1,[4,0,4]%%%}+%%%{-2,[3,2,3]%%%}+%%%{2,[3,1,3]%%%}+%%%{1,[2,2,2]%%%
}+%%%{2,[2,1,2]%%%}+%%%{-2,[2,0,2]%%%}+%%%{-2,[1,1,1]%%%}+%%%{1,[0,0,0]%%%}] at parameters values [86,-97,-82]
Warning, choosing root of [1,0,%%%{-2,[2,1,2]%%%}+%%%{-2,[2,0,2]%%%}+%%%{-2,[1,1,1]%%%}+%%%{2,[0,0,0]%%%},0,%%
%{1,[4,2,4]%%%}+%%%{-2,[4,1,4]%%%}+%%%{1,[4,0,4]%%%}+%%%{2,[3,2,3]%%%}+%%%{-2,[3,1,3]%%%}+%%%{1,[2,2,2]%%%}+%%
%{2,[2,1,2]%%%}+%%%{-2,[2,0,2]%%%}+%%%{2,[1,1,1]%%%}+%%%{1,[0,0,0]%%%}] at parameters values [7,-27,26]Warning
, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(a
*t_nostep+1)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 55, normalized size = 0.47 \[ -\frac {2 \left (a x +1\right ) \left (8 a^{2} x^{2}-12 a x +15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{105 x^{3} \sqrt {\frac {a x -1}{a x +1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(a*x+1)*(8*a^2*x^2-12*a*x+15)*(c*(a*x-1)/a/x)^(1/2)/x^3/((a*x-1)/(a*x+1))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a x}}}{x^{4} \sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c - c/(a*x))/(x^4*sqrt((a*x - 1)/(a*x + 1))), x)

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mupad [B]  time = 1.41, size = 100, normalized size = 0.85 \[ -\frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (8\,a^3\,x^3+12\,a^2\,x^2+11\,a\,x+29\right )\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3}-\frac {88\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*((a*x - 1)/(a*x + 1))^(1/2)*(11*a*x + 12*a^2*x^2 + 8*a^3*x^3 + 29)*((c*(a*x - 1))/(a*x))^(1/2))/(105*x^3)
- (88*((a*x - 1)/(a*x + 1))^(1/2)*((c*(a*x - 1))/(a*x))^(1/2))/(105*x^3*(a*x - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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