\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-\frac {c}{a x})^{5/2}} \, dx\) [477]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=-\frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}}+\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}+\frac {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a c^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6302, 6268, 25, 528, 382, 105, 157, 162, 65, 214} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a c^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}}+\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}-\frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}} \]

[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 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 65

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 105

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] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 157

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 162

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 382

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 528

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 6268

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 6302

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]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {e^{-2 \text {arctanh}(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 \text {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 {\text {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 {\text {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 {\text {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a c^2}-\frac {\text {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 {\text {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )}{c^3}+\frac {\text {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 {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a c^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.60 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\frac {a x-\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a-\frac {1}{x}}{2 a}\right )-\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\frac {1}{a x}\right )}{a c^2 \sqrt {c-\frac {c}{a x}}} \]

[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)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(98)=196\).

Time = 0.50 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.96

method result size
risch \(\frac {a x -1}{a \,c^{2} \sqrt {\frac {c \left (a x -1\right )}{a x}}}+\frac {\left (\frac {\ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-a c x}\right )}{2 a^{3} \sqrt {a^{2} c}}-\frac {\sqrt {2}\, \ln \left (\frac {4 c -3 \left (x +\frac {1}{a}\right ) a c +2 \sqrt {2}\, \sqrt {c}\, \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-3 \left (x +\frac {1}{a}\right ) a c +2 c}}{x +\frac {1}{a}}\right )}{4 a^{4} \sqrt {c}}-\frac {\sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+\left (x -\frac {1}{a}\right ) a c}}{a^{5} c \left (x -\frac {1}{a}\right )}\right ) a^{2} \sqrt {c \left (a x -1\right ) a x}}{c^{2} x \sqrt {\frac {c \left (a x -1\right )}{a x}}}\) \(227\)
default \(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (-8 \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}\, a^{\frac {7}{2}} x^{2}+4 \left (\left (a x -1\right ) x \right )^{\frac {3}{2}} \sqrt {\frac {1}{a}}\, a^{\frac {5}{2}}+\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) a^{\frac {5}{2}} x^{2}-2 \sqrt {\frac {1}{a}}\, \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{3} x^{2}+16 \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}\, a^{\frac {5}{2}} x -2 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) a^{\frac {3}{2}} x +4 \sqrt {\frac {1}{a}}\, \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{2} x -8 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}} \sqrt {\frac {1}{a}}+\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) \sqrt {a}-2 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \sqrt {\frac {1}{a}}\right )}{4 a^{\frac {3}{2}} \sqrt {\left (a x -1\right ) x}\, c^{3} \sqrt {\frac {1}{a}}\, \left (a x -1\right )^{2}}\) \(368\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.47 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {2 \, \sqrt {2} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + 3 \, a c x - c}{a x + 1}\right ) + 2 \, {\left (a x - 1\right )} \sqrt {c} \log \left (-2 \, a c x - 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right ) + 4 \, {\left (a^{2} x^{2} - 2 \, a x\right )} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}, -\frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) + 2 \, {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right ) - 2 \, {\left (a^{2} x^{2} - 2 \, a x\right )} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\int \frac {a x - 1}{\left (- c \left (-1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \]

[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)

Maxima [F]

\[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\int { \frac {a x - 1}{{\left (a x + 1\right )} {\left (c - \frac {c}{a x}\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\int \frac {a\,x-1}{{\left (c-\frac {c}{a\,x}\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \]

[In]

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

[Out]

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