3.539 \(\int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\)

Optimal. Leaf size=150 \[ -\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {7 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{5 c}-\frac {56}{15} a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}-\frac {224 a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}{15 \sqrt {c-\frac {c}{a x}}} \]

[Out]

-a^2*(c-c/a/x)^(7/2)/c^3/(1-1/a^2/x^2)^(1/2)-7/5*a^2*(c-c/a/x)^(3/2)*(1-1/a^2/x^2)^(1/2)/c-224/15*a^2*c*(1-1/a
^2/x^2)^(1/2)/(c-c/a/x)^(1/2)-56/15*a^2*(1-1/a^2/x^2)^(1/2)*(c-c/a/x)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6178, 789, 657, 649} \[ -\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {7 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{5 c}-\frac {56}{15} a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}-\frac {224 a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}{15 \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c/(a*x)]/(E^(3*ArcCoth[a*x])*x^3),x]

[Out]

(-224*a^2*c*Sqrt[1 - 1/(a^2*x^2)])/(15*Sqrt[c - c/(a*x)]) - (56*a^2*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)])/1
5 - (7*a^2*Sqrt[1 - 1/(a^2*x^2)]*(c - c/(a*x))^(3/2))/(5*c) - (a^2*(c - c/(a*x))^(7/2))/(c^3*Sqrt[1 - 1/(a^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 657

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*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 789

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

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^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x \left (c-\frac {c x}{a}\right )^{7/2}}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=-\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {(7 a) \operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{5/2}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 c^2}\\ &=-\frac {7 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{5 c}-\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {(28 a) \operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{3/2}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{5 c}\\ &=-\frac {56}{15} a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}-\frac {7 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{5 c}-\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {1}{15} (112 a) \operatorname {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {224 a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}{15 \sqrt {c-\frac {c}{a x}}}-\frac {56}{15} a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}-\frac {7 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{5 c}-\frac {a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 70, normalized size = 0.47 \[ -\frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \left (158 a^3 x^3+79 a^2 x^2-16 a x+3\right ) \sqrt {c-\frac {c}{a x}}}{15 x \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - c/(a*x)]/(E^(3*ArcCoth[a*x])*x^3),x]

[Out]

(-2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(3 - 16*a*x + 79*a^2*x^2 + 158*a^3*x^3))/(15*x*(-1 + a^2*x^2))

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fricas [A]  time = 0.50, size = 69, normalized size = 0.46 \[ -\frac {2 \, {\left (158 \, a^{3} x^{3} + 79 \, a^{2} x^{2} - 16 \, a x + 3\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{15 \, {\left (a x^{3} - x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^3,x, algorithm="fricas")

[Out]

-2/15*(158*a^3*x^3 + 79*a^2*x^2 - 16*a*x + 3)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(a*x^3 - x^2)

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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((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^3,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),abs(a*x+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 = 70, normalized size = 0.47 \[ -\frac {2 \left (a x +1\right ) \left (158 x^{3} a^{3}+79 a^{2} x^{2}-16 a x +3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{15 x^{2} \left (a x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^3,x)

[Out]

-2/15*(a*x+1)*(158*a^3*x^3+79*a^2*x^2-16*a*x+3)*(c*(a*x-1)/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^2/(a*x-1)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a x}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^3,x, algorithm="maxima")

[Out]

integrate(sqrt(c - c/(a*x))*((a*x - 1)/(a*x + 1))^(3/2)/x^3, x)

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mupad [B]  time = 1.44, size = 62, normalized size = 0.41 \[ -\frac {2\,\sqrt {c-\frac {c}{a\,x}}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (158\,a^3\,x^3+79\,a^2\,x^2-16\,a\,x+3\right )}{15\,x^2\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - c/(a*x))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x^3,x)

[Out]

-(2*(c - c/(a*x))^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(79*a^2*x^2 - 16*a*x + 158*a^3*x^3 + 3))/(15*x^2*(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((c-c/a/x)**(1/2)*((a*x-1)/(a*x+1))**(3/2)/x**3,x)

[Out]

Timed out

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