\(\int e^{-3 \coth ^{-1}(a x)} (c-\frac {c}{a x}) \, dx\) [432]

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

Optimal result

Integrand size = 20, antiderivative size = 75 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {c \csc ^{-1}(a x)}{a}-\frac {4 c \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \]

[Out]

c*arccsc(a*x)/a-4*c*arctanh((1-1/a^2/x^2)^(1/2))/a+8*c*(a-1/x)/a^2/(1-1/a^2/x^2)^(1/2)+c*x*(1-1/a^2/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6312, 1819, 1821, 858, 222, 272, 65, 214} \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=-\frac {4 c \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}+\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c x \sqrt {1-\frac {1}{a^2 x^2}}+\frac {c \csc ^{-1}(a x)}{a} \]

[In]

Int[(c - c/(a*x))/E^(3*ArcCoth[a*x]),x]

[Out]

(8*c*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + c*Sqrt[1 - 1/(a^2*x^2)]*x + (c*ArcCsc[a*x])/a - (4*c*ArcTanh[
Sqrt[1 - 1/(a^2*x^2)]])/a

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 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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 272

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

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
 a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
 x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1821

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
 R = PolynomialRemainder[Pq, c*x, x]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 6312

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^4}{x^2 \left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c^3} \\ & = \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\text {Subst}\left (\int \frac {-c^4+\frac {4 c^4 x}{a}+\frac {c^4 x^2}{a^2}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^3} \\ & = \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {\text {Subst}\left (\int \frac {-\frac {4 c^4}{a}-\frac {c^4 x}{a^2}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^3} \\ & = \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {c \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^2}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a} \\ & = \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {c \csc ^{-1}(a x)}{a}+\frac {(2 c) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{a} \\ & = \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {c \csc ^{-1}(a x)}{a}-(4 a c) \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right ) \\ & = \frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {c \csc ^{-1}(a x)}{a}-\frac {4 c \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.54 (sec) , antiderivative size = 234, normalized size of antiderivative = 3.12 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {5 a^2 c x^2 \left ((1+a x) \left (\sqrt {1+\frac {1}{a x}} \left (2-3 a x+a^2 x^2\right )+6 a \sqrt {1-\frac {1}{a x}} x \arcsin \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )-2 a \sqrt {1-\frac {1}{a x}} x \arcsin \left (\frac {1}{a x}\right )\right )-4 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {1+\frac {1}{a x}} x^2 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )+\sqrt {2} c (-1+a x)^3 (1+a x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},\frac {1}{2} \left (1-\frac {1}{a x}\right )\right )}{5 a^4 \sqrt {1-\frac {1}{a x}} x^3 (1+a x)} \]

[In]

Integrate[(c - c/(a*x))/E^(3*ArcCoth[a*x]),x]

[Out]

(5*a^2*c*x^2*((1 + a*x)*(Sqrt[1 + 1/(a*x)]*(2 - 3*a*x + a^2*x^2) + 6*a*Sqrt[1 - 1/(a*x)]*x*ArcSin[Sqrt[1 - 1/(
a*x)]/Sqrt[2]] - 2*a*Sqrt[1 - 1/(a*x)]*x*ArcSin[1/(a*x)]) - 4*a^2*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[1 + 1/(a*x)]*x^2*
ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]) + Sqrt[2]*c*(-1 + a*x)^3*(1 + a*x)*Hypergeometric2F1[3/2, 5/2, 7/2, (1 - 1/(a*
x))/2])/(5*a^4*Sqrt[1 - 1/(a*x)]*x^3*(1 + a*x))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(370\) vs. \(2(69)=138\).

Time = 0.14 (sec) , antiderivative size = 371, normalized size of antiderivative = 4.95

method result size
default \(\frac {\left (\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{2} x^{2}+a^{2} x^{2} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )-4 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+4 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} x^{2}+2 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x +2 a x \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )-8 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -4 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+8 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}+\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {a^{2}}-4 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )+4 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\right ) c \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x -1\right )}\) \(371\)

[In]

int((c-c/a/x)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.23 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=-\frac {2 \, c \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + 4 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 4 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (a c x + 9 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a} \]

[In]

integrate((c-c/a/x)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

-(2*c*arctan(sqrt((a*x - 1)/(a*x + 1))) + 4*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 4*c*log(sqrt((a*x - 1)/(a*x
 + 1)) - 1) - (a*c*x + 9*c)*sqrt((a*x - 1)/(a*x + 1)))/a

Sympy [F]

\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \left (\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x^{2} + x}\, dx + \int \left (- \frac {2 a \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\right )\, dx + \int \frac {a^{2} x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\, dx\right )}{a} \]

[In]

integrate((c-c/a/x)*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.80 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=-2 \, a {\left (\frac {c \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} + \frac {c \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} + \frac {2 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {2 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {4 \, c \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2}}\right )} \]

[In]

integrate((c-c/a/x)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\int { {\left (c - \frac {c}{a x}\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((c-c/a/x)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.43 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {2\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}-\frac {2\,c\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {8\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a}+\frac {c\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{a} \]

[In]

int((c - c/(a*x))*((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

(2*c*((a*x - 1)/(a*x + 1))^(1/2))/(a - (a*(a*x - 1))/(a*x + 1)) - (2*c*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a +
(c*atan(((a*x - 1)/(a*x + 1))^(1/2)*1i)*8i)/a + (8*c*((a*x - 1)/(a*x + 1))^(1/2))/a