\(\int e^{3 \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{3/2} \, dx\) [849]

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

Optimal result

Integrand size = 24, antiderivative size = 148 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=-\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}-\frac {3 c \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 c \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6332, 6328, 45} \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {c x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 c \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Int[E^(3*ArcCoth[a*x])*(c - c/(a^2*x^2))^(3/2),x]

[Out]

-1/2*(c*Sqrt[c - c/(a^2*x^2)])/(a^3*Sqrt[1 - 1/(a^2*x^2)]*x^2) - (3*c*Sqrt[c - c/(a^2*x^2)])/(a^2*Sqrt[1 - 1/(
a^2*x^2)]*x) + (c*Sqrt[c - c/(a^2*x^2)]*x)/Sqrt[1 - 1/(a^2*x^2)] + (3*c*Sqrt[c - c/(a^2*x^2)]*Log[x])/(a*Sqrt[
1 - 1/(a^2*x^2)])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6328

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

Rule 6332

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.39 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} \left (-\frac {1}{2 a^3 x^2}-\frac {3}{a^2 x}+x+\frac {3 \log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \]

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - c/(a^2*x^2))^(3/2),x]

[Out]

((c - c/(a^2*x^2))^(3/2)*(-1/2*1/(a^3*x^2) - 3/(a^2*x) + x + (3*Log[x])/a))/(1 - 1/(a^2*x^2))^(3/2)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.47

method result size
default \(\frac {\left (2 a^{3} x^{3}+6 a^{2} \ln \left (x \right ) x^{2}-6 a x -1\right ) {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {3}{2}} x}{2 \left (a x +1\right )^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) \(69\)

[In]

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

[Out]

1/2*(2*a^3*x^3+6*a^2*ln(x)*x^2-6*a*x-1)*(c*(a^2*x^2-1)/a^2/x^2)^(3/2)*x/(a*x+1)^3/((a*x-1)/(a*x+1))^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.30 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {{\left (2 \, a^{3} c x^{3} + 6 \, a^{2} c x^{2} \log \left (x\right ) - 6 \, a c x - c\right )} \sqrt {a^{2} c}}{2 \, a^{4} x^{2}} \]

[In]

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

[Out]

1/2*(2*a^3*c*x^3 + 6*a^2*c*x^2*log(x) - 6*a*c*x - c)*sqrt(a^2*c)/(a^4*x^2)

Sympy [F(-1)]

Timed out. \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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