\(\int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx\) [451]

Optimal result

Integrand size = 24, antiderivative size = 70 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx=-\frac {5}{a \sqrt {c-\frac {c}{a x}}}+\frac {x}{\sqrt {c-\frac {c}{a x}}}+\frac {5 \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6302, 6268, 25, 528, 382, 79, 53, 65, 214} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a \sqrt {c}}+\frac {x}{\sqrt {c-\frac {c}{a x}}}-\frac {5}{a \sqrt {c-\frac {c}{a x}}} \]

[In]

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

[Out]

-5/(a*Sqrt[c - c/(a*x)]) + x/Sqrt[c - c/(a*x)] + (5*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(a*Sqrt[c])

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 53

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

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx=\frac {a x-5 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\frac {1}{a x}\right )}{a \sqrt {c-\frac {c}{a x}}} \]

[In]

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

[Out]

(a*x - 5*Hypergeometric2F1[-1/2, 1, 1/2, 1 - 1/(a*x)])/(a*Sqrt[c - c/(a*x)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(149\) vs. \(2(60)=120\).

Time = 0.54 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.14

method	result	size
risch	\(\frac {a x -1}{a \sqrt {\frac {c \left (a x -1\right )}{a x}}}+\frac {\left (\frac {5 \ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-a c x}\right )}{2 a \sqrt {a^{2} c}}-\frac {4 \sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+\left (x -\frac {1}{a}\right ) a c}}{a^{3} c \left (x -\frac {1}{a}\right )}\right ) \sqrt {c \left (a x -1\right ) a x}}{\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x}\)	\(150\)
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (10 a^{\frac {5}{2}} \sqrt {\left (a x -1\right ) x}\, x^{2}+5 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{2} x^{2}-8 a^{\frac {3}{2}} \left (\left (a x -1\right ) x \right )^{\frac {3}{2}}-20 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}} x -10 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a x +10 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+5 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right )\right )}{2 \sqrt {\left (a x -1\right ) x}\, c \left (a x -1\right )^{2} \sqrt {a}}\)	\(194\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.51 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx=\left [\frac {5 \, {\left (a x - 1\right )} \sqrt {c} \log \left (-2 \, a c x - 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right ) + 2 \, {\left (a^{2} x^{2} - 5 \, a x\right )} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} c x - a c\right )}}, -\frac {5 \, {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right ) - {\left (a^{2} x^{2} - 5 \, a x\right )} \sqrt {\frac {a c x - c}{a x}}}{a^{2} c x - a c}\right ] \]

[In]

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

[Out]

[1/2*(5*(a*x - 1)*sqrt(c)*log(-2*a*c*x - 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) + 2*(a^2*x^2 - 5*a*x)*sqrt
((a*c*x - c)/(a*x)))/(a^2*c*x - a*c), -(5*(a*x - 1)*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - (a^2
*x^2 - 5*a*x)*sqrt((a*c*x - c)/(a*x)))/(a^2*c*x - a*c)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (60) = 120\).

Time = 0.35 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.43 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx=\frac {5 \, \log \left (c^{2} {\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{6 \, a \sqrt {c}} - \frac {5 \, \log \left ({\left | 2 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{3} \sqrt {c} {\left | a \right |} - 5 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{2} a c + 4 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} c^{\frac {3}{2}} {\left | a \right |} - a c^{2} \right |}\right ) \mathrm {sgn}\left (x\right )}{6 \, a \sqrt {c}} + \frac {\sqrt {a^{2} c x^{2} - a c x} {\left | a \right |} \mathrm {sgn}\left (x\right )}{a^{2} c} \]

[In]

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

[Out]

5/6*log(c^2*abs(a))*sgn(x)/(a*sqrt(c)) - 5/6*log(abs(2*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^3*sqrt(c)*abs
(a) - 5*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^2*a*c + 4*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*c^(3/2)*
abs(a) - a*c^2))*sgn(x)/(a*sqrt(c)) + sqrt(a^2*c*x^2 - a*c*x)*abs(a)*sgn(x)/(a^2*c)

Mupad [F(-1)]

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

[In]

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

[Out]

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