### 3.477 $$\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-\frac{c}{a x})^{5/2}} \, dx$$

Optimal. Leaf size=116 $\frac{x}{c^2 \sqrt{c-\frac{c}{a x}}}-\frac{2}{a c^2 \sqrt{c-\frac{c}{a x}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a c^{5/2}}$

[Out]

-2/(a*c^2*Sqrt[c - c/(a*x)]) + x/(c^2*Sqrt[c - c/(a*x)]) + ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]]/(a*c^(5/2)) + Ar
cTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])]/(Sqrt[2]*a*c^(5/2))

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Rubi [A]  time = 0.23544, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.417, Rules used = {6167, 6133, 25, 514, 375, 103, 152, 156, 63, 208} $\frac{x}{c^2 \sqrt{c-\frac{c}{a x}}}-\frac{2}{a c^2 \sqrt{c-\frac{c}{a x}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a c^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^(5/2)),x]

[Out]

-2/(a*c^2*Sqrt[c - c/(a*x)]) + x/(c^2*Sqrt[c - c/(a*x)]) + ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]]/(a*c^(5/2)) + Ar
cTanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])]/(Sqrt[2]*a*c^(5/2))

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 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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)}}{\left (c-\frac{c}{a x}\right )^{5/2}} \, dx &=-\int \frac{e^{-2 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^{5/2}} \, dx\\ &=-\int \frac{1-a x}{\left (c-\frac{c}{a x}\right )^{5/2} (1+a x)} \, dx\\ &=\frac{a \int \frac{x}{\left (c-\frac{c}{a x}\right )^{3/2} (1+a x)} \, dx}{c}\\ &=\frac{a \int \frac{1}{\left (a+\frac{1}{x}\right ) \left (c-\frac{c}{a x}\right )^{3/2}} \, dx}{c}\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{1}{x^2 (a+x) \left (c-\frac{c x}{a}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{x}{c^2 \sqrt{c-\frac{c}{a x}}}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{c}{2}-\frac{3 c x}{2 a}}{x (a+x) \left (c-\frac{c x}{a}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{c^2}\\ &=-\frac{2}{a c^2 \sqrt{c-\frac{c}{a x}}}+\frac{x}{c^2 \sqrt{c-\frac{c}{a x}}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{c^2}{2}+\frac{c^2 x}{a}}{x (a+x) \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{c^4}\\ &=-\frac{2}{a c^2 \sqrt{c-\frac{c}{a x}}}+\frac{x}{c^2 \sqrt{c-\frac{c}{a x}}}-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{2 a c^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{2 a c^2}\\ &=-\frac{2}{a c^2 \sqrt{c-\frac{c}{a x}}}+\frac{x}{c^2 \sqrt{c-\frac{c}{a x}}}+\frac{\operatorname{Subst}\left (\int \frac{1}{a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )}{c^3}+\frac{\operatorname{Subst}\left (\int \frac{1}{2 a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )}{c^3}\\ &=-\frac{2}{a c^2 \sqrt{c-\frac{c}{a x}}}+\frac{x}{c^2 \sqrt{c-\frac{c}{a x}}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{5/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a c^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0547699, size = 70, normalized size = 0.6 $\frac{-\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a-\frac{1}{x}}{2 a}\right )-\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},1-\frac{1}{a x}\right )+a x}{a c^2 \sqrt{c-\frac{c}{a x}}}$

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^(5/2)),x]

[Out]

(a*x - Hypergeometric2F1[-1/2, 1, 1/2, (a - x^(-1))/(2*a)] - Hypergeometric2F1[-1/2, 1, 1/2, 1 - 1/(a*x)])/(a*
c^2*Sqrt[c - c/(a*x)])

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Maple [B]  time = 0.17, size = 368, normalized size = 3.2 \begin{align*} -{\frac{x}{4\,{c}^{3} \left ( ax-1 \right ) ^{2}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -8\,{a}^{7/2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}{x}^{2}+\ln \left ({\frac{1}{ax+1} \left ( 2\,\sqrt{2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}a-3\,ax+1 \right ) } \right ){a}^{{\frac{5}{2}}}\sqrt{2}{x}^{2}-2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \sqrt{{a}^{-1}}{x}^{2}{a}^{3}+4\,{a}^{5/2}\sqrt{{a}^{-1}} \left ( \left ( ax-1 \right ) x \right ) ^{3/2}+16\,{a}^{5/2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}x-2\,\ln \left ({\frac{2\,\sqrt{2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}a-3\,ax+1}{ax+1}} \right ){a}^{3/2}\sqrt{2}x+4\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \sqrt{{a}^{-1}}x{a}^{2}-8\,\sqrt{ \left ( ax-1 \right ) x}{a}^{3/2}\sqrt{{a}^{-1}}+\sqrt{2}\ln \left ({\frac{1}{ax+1} \left ( 2\,\sqrt{2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}a-3\,ax+1 \right ) } \right ) \sqrt{a}-2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) a\sqrt{{a}^{-1}} \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{a}^{-1}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(a*x-1)/(c-c/a/x)^(5/2),x)

[Out]

-1/4*(c*(a*x-1)/a/x)^(1/2)*x/a^(3/2)*(-8*a^(7/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*x^2+ln((2*2^(1/2)*(1/a)^(1/2)*(
(a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*a^(5/2)*2^(1/2)*x^2-2*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^2*a^3+4*a^(5/2)*(1/a)^(1/2)*((a*x-1)*x)^(3/2)+16*a^(5/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*x-2*ln
((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*a^(3/2)*2^(1/2)*x+4*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*a^2-8*((a*x-1)*x)^(1/2)*a^(3/2)*(1/a)^(1/2)+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))*a^(1/2)-2*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*
a*(1/a)^(1/2))/((a*x-1)*x)^(1/2)/c^3/(1/a)^(1/2)/(a*x-1)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x - 1}{{\left (a x + 1\right )}{\left (c - \frac{c}{a x}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x - 1)/((a*x + 1)*(c - c/(a*x))^(5/2)), x)

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Fricas [A]  time = 2.00711, size = 652, normalized size = 5.62 \begin{align*} \left [\frac{\sqrt{2}{\left (a x - 1\right )} \sqrt{c} \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 ) + 2 \,{\left (a x - 1\right )} \sqrt{c} \log \left (-2 \, a c x - 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right ) + 4 \,{\left (a^{2} x^{2} - 2 \, a x\right )} \sqrt{\frac{a c x - c}{a x}}}{4 \,{\left (a^{2} c^{3} x - a c^{3}\right )}}, -\frac{\sqrt{2}{\left (a x - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{2} \sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{2 \, c}\right ) + 2 \,{\left (a x - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right ) - 2 \,{\left (a^{2} x^{2} - 2 \, a x\right )} \sqrt{\frac{a c x - c}{a x}}}{2 \,{\left (a^{2} c^{3} x - a c^{3}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^(5/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(2)*(a*x - 1)*sqrt(c)*log(-(2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + 3*a*c*x - c)/(a*x + 1))
+ 2*(a*x - 1)*sqrt(c)*log(-2*a*c*x - 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) + 4*(a^2*x^2 - 2*a*x)*sqrt((a*
c*x - c)/(a*x)))/(a^2*c^3*x - a*c^3), -1/2*(sqrt(2)*(a*x - 1)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x
- c)/(a*x))/c) + 2*(a*x - 1)*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - 2*(a^2*x^2 - 2*a*x)*sqrt((
a*c*x - c)/(a*x)))/(a^2*c^3*x - a*c^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x - 1}{\left (- c \left (-1 + \frac{1}{a x}\right )\right )^{\frac{5}{2}} \left (a x + 1\right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)**(5/2),x)

[Out]

Integral((a*x - 1)/((-c*(-1 + 1/(a*x)))**(5/2)*(a*x + 1)), x)

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Giac [A]  time = 1.25974, size = 224, normalized size = 1.93 \begin{align*} -\frac{1}{2} \, a c{\left (\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a c x - c}{a x}}}{2 \, \sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{3}} + \frac{2 \, \arctan \left (\frac{\sqrt{\frac{a c x - c}{a x}}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{3}} + \frac{2 \,{\left (c - \frac{2 \,{\left (a c x - c\right )}}{a x}\right )}}{{\left (c \sqrt{\frac{a c x - c}{a x}} - \frac{{\left (a c x - c\right )} \sqrt{\frac{a c x - c}{a x}}}{a x}\right )} a^{2} c^{3}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x-1)/(a*x+1)/(c-c/a/x)^(5/2),x, algorithm="giac")

[Out]

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