### 3.531 $$\int \frac{e^{-2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^4} \, dx$$

Optimal. Leaf size=113 $-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}-4 a^3 \sqrt{c-\frac{c}{a x}}+4 \sqrt{2} a^3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )$

[Out]

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

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Rubi [A]  time = 0.416017, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {6167, 6133, 25, 514, 446, 88, 50, 63, 208} $-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}-4 a^3 \sqrt{c-\frac{c}{a x}}+4 \sqrt{2} a^3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[c - c/(a*x)]/(E^(2*ArcCoth[a*x])*x^4),x]

[Out]

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6133

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Int[(u*(c + d/x)^p*(1 + a*x)^(n/
2))/(1 - a*x)^(n/2), x] /; FreeQ[{a, c, d, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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]

Rubi steps

\begin{align*} \int \frac{e^{-2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^4} \, dx &=-\int \frac{e^{-2 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^4} \, dx\\ &=-\int \frac{\sqrt{c-\frac{c}{a x}} (1-a x)}{x^4 (1+a x)} \, dx\\ &=\frac{a \int \frac{\left (c-\frac{c}{a x}\right )^{3/2}}{x^3 (1+a x)} \, dx}{c}\\ &=\frac{a \int \frac{\left (c-\frac{c}{a x}\right )^{3/2}}{\left (a+\frac{1}{x}\right ) x^4} \, dx}{c}\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{x^2 \left (c-\frac{c x}{a}\right )^{3/2}}{a+x} \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{a \operatorname{Subst}\left (\int \left (\frac{a^2 \left (c-\frac{c x}{a}\right )^{3/2}}{a+x}-\frac{a \left (c-\frac{c x}{a}\right )^{5/2}}{c}\right ) \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}-\frac{a^3 \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{3/2}}{a+x} \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}-\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-\frac{c x}{a}}}{a+x} \, dx,x,\frac{1}{x}\right )\\ &=-4 a^3 \sqrt{c-\frac{c}{a x}}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}-\left (4 a^3 c\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )\\ &=-4 a^3 \sqrt{c-\frac{c}{a x}}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )\\ &=-4 a^3 \sqrt{c-\frac{c}{a x}}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}-\frac{2 a^3 \left (c-\frac{c}{a x}\right )^{7/2}}{7 c^3}+4 \sqrt{2} a^3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.134165, size = 87, normalized size = 0.77 $\frac{2 \left (-52 a^3 x^3+16 a^2 x^2-9 a x+3\right ) \sqrt{c-\frac{c}{a x}}}{21 x^3}+4 \sqrt{2} a^3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[c - c/(a*x)]/(E^(2*ArcCoth[a*x])*x^4),x]

[Out]

(2*Sqrt[c - c/(a*x)]*(3 - 9*a*x + 16*a^2*x^2 - 52*a^3*x^3))/(21*x^3) + 4*Sqrt[2]*a^3*Sqrt[c]*ArcTanh[Sqrt[c -
c/(a*x)]/(Sqrt[2]*Sqrt[c])]

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Maple [B]  time = 0.168, size = 302, normalized size = 2.7 \begin{align*}{\frac{1}{21\,{x}^{4}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -126\,\sqrt{a{x}^{2}-x}{a}^{9/2}\sqrt{{a}^{-1}}{x}^{5}+42\,{a}^{9/2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}{x}^{5}+84\, \left ( a{x}^{2}-x \right ) ^{3/2}{a}^{7/2}\sqrt{{a}^{-1}}{x}^{3}+63\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \sqrt{{a}^{-1}}{x}^{5}{a}^{4}-42\,{a}^{7/2}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}a-3\,ax+1}{ax+1}} \right ){x}^{5}-63\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \sqrt{{a}^{-1}}{x}^{5}{a}^{4}-20\,{a}^{5/2}\sqrt{{a}^{-1}} \left ( a{x}^{2}-x \right ) ^{3/2}{x}^{2}+12\,{a}^{3/2} \left ( a{x}^{2}-x \right ) ^{3/2}x\sqrt{{a}^{-1}}-6\, \left ( a{x}^{2}-x \right ) ^{3/2}\sqrt{a}\sqrt{{a}^{-1}} \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{a}^{-1}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(c*(a*x-1)/a/x)^(1/2)/x^4*(-126*(a*x^2-x)^(1/2)*a^(9/2)*(1/a)^(1/2)*x^5+42*a^(9/2)*(1/a)^(1/2)*((a*x-1)*x
)^(1/2)*x^5+84*(a*x^2-x)^(3/2)*a^(7/2)*(1/a)^(1/2)*x^3+63*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*
(1/a)^(1/2)*x^5*a^4-42*a^(7/2)*2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*x^5-63*
ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*x^5*a^4-20*a^(5/2)*(1/a)^(1/2)*(a*x^2-x)^(3/
2)*x^2+12*a^(3/2)*(a*x^2-x)^(3/2)*x*(1/a)^(1/2)-6*(a*x^2-x)^(3/2)*a^(1/2)*(1/a)^(1/2))/((a*x-1)*x)^(1/2)/a^(1/
2)/(1/a)^(1/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.7059, size = 466, normalized size = 4.12 \begin{align*} \left [\frac{2 \,{\left (21 \, \sqrt{2} a^{3} \sqrt{c} x^{3} \log \left (-\frac{2 \, \sqrt{2} a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + 3 \, a c x - c}{a x + 1}\right ) -{\left (52 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 9 \, a x - 3\right )} \sqrt{\frac{a c x - c}{a x}}\right )}}{21 \, x^{3}}, -\frac{2 \,{\left (42 \, \sqrt{2} a^{3} \sqrt{-c} x^{3} \arctan \left (\frac{\sqrt{2} \sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{2 \, c}\right ) +{\left (52 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 9 \, a x - 3\right )} \sqrt{\frac{a c x - c}{a x}}\right )}}{21 \, x^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/21*(21*sqrt(2)*a^3*sqrt(c)*x^3*log(-(2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + 3*a*c*x - c)/(a*x + 1)
) - (52*a^3*x^3 - 16*a^2*x^2 + 9*a*x - 3)*sqrt((a*c*x - c)/(a*x)))/x^3, -2/21*(42*sqrt(2)*a^3*sqrt(-c)*x^3*arc
tan(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) + (52*a^3*x^3 - 16*a^2*x^2 + 9*a*x - 3)*sqrt((a*c*x - c)/(
a*x)))/x^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (-1 + \frac{1}{a x}\right )} \left (a x - 1\right )}{x^{4} \left (a x + 1\right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.2953, size = 481, normalized size = 4.26 \begin{align*} -\frac{4 \, \sqrt{2} a^{4} c \arctan \left (\frac{\sqrt{2}{\left ({\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )} a + \sqrt{c}{\left | a \right |}\right )}}{2 \, a \sqrt{-c}}\right )}{\sqrt{-c}{\left | a \right |} \mathrm{sgn}\left (x\right )} - \frac{2 \,{\left (84 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{6} a^{7} c - 84 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{5} a^{6} c^{\frac{3}{2}}{\left | a \right |} + 112 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{4} a^{7} c^{2} - 105 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{3} a^{6} c^{\frac{5}{2}}{\left | a \right |} + 63 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{2} a^{7} c^{3} - 21 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )} a^{6} c^{\frac{7}{2}}{\left | a \right |} + 3 \, a^{7} c^{4}\right )}}{21 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}^{7} a^{3}{\left | a \right |} \mathrm{sgn}\left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*sqrt(2)*a^4*c*arctan(1/2*sqrt(2)*((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a + sqrt(c)*abs(a))/(a*sqrt(-c)
))/(sqrt(-c)*abs(a)*sgn(x)) - 2/21*(84*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^6*a^7*c - 84*(sqrt(a^2*c)*x -
sqrt(a^2*c*x^2 - a*c*x))^5*a^6*c^(3/2)*abs(a) + 112*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^4*a^7*c^2 - 105
*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^3*a^6*c^(5/2)*abs(a) + 63*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))
^2*a^7*c^3 - 21*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a^6*c^(7/2)*abs(a) + 3*a^7*c^4)/((sqrt(a^2*c)*x - sq
rt(a^2*c*x^2 - a*c*x))^7*a^3*abs(a)*sgn(x))