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

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

[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)

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Rubi [A]  time = 0.246082, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.16, 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.108857, 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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Maple [A]  time = 0.119, size = 55, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,ax+2 \right ) \left ( 8\,{a}^{2}{x}^{2}-12\,ax+15 \right ) }{105\,{x}^{3}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c - \frac{c}{a x}}}{x^{4} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

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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Fricas [A]  time = 1.58065, size = 162, normalized size = 1.38 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c - \frac{c}{a x}}}{x^{4} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

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]

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