Integrand size = 24, antiderivative size = 148 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}-\frac {3 c \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
1/2*c*(c-c/a^2/x^2)^(1/2)/a^3/(1-1/a^2/x^2)^(1/2)/x^2-3*c*(c-c/a^2/x^2)^(1 /2)/a^2/(1-1/a^2/x^2)^(1/2)/x+c*(c-c/a^2/x^2)^(1/2)*x/(1-1/a^2/x^2)^(1/2)- 3*c*(c-c/a^2/x^2)^(1/2)*ln(x)/a/(1-1/a^2/x^2)^(1/2)
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.39 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} \left (\frac {1}{2 a^3 x^2}-\frac {3}{a^2 x}+x-\frac {3 \log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \] Input:
Integrate[(c - c/(a^2*x^2))^(3/2)/E^(3*ArcCoth[a*x]),x]
Output:
((c - c/(a^2*x^2))^(3/2)*(1/(2*a^3*x^2) - 3/(a^2*x) + x - (3*Log[x])/a))/( 1 - 1/(a^2*x^2))^(3/2)
Time = 0.63 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.42, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6751, 6747, 25, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (c-\frac {c}{a^2 x^2}\right )^{3/2} e^{-3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {c \sqrt {c-\frac {c}{a^2 x^2}} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{3/2}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {c \sqrt {c-\frac {c}{a^2 x^2}} \int -\frac {(1-a x)^3}{x^3}dx}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {c \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^3}{x^3}dx}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {c \sqrt {c-\frac {c}{a^2 x^2}} \int \left (-a^3+\frac {3 a^2}{x}-\frac {3 a}{x^2}+\frac {1}{x^3}\right )dx}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c \sqrt {c-\frac {c}{a^2 x^2}} \left (a^3 (-x)+3 a^2 \log (x)+\frac {3 a}{x}-\frac {1}{2 x^2}\right )}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(c - c/(a^2*x^2))^(3/2)/E^(3*ArcCoth[a*x]),x]
Output:
-((c*Sqrt[c - c/(a^2*x^2)]*(-1/2*1/x^2 + (3*a)/x - a^3*x + 3*a^2*Log[x]))/ (a^3*Sqrt[1 - 1/(a^2*x^2)]))
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[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] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[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])
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.47
method | result | size |
default | \(-\frac {\left (-2 a^{3} x^{3}+6 a^{2} \ln \left (x \right ) x^{2}+6 a x -1\right ) x {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {3}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{2 \left (a x -1\right )^{3}}\) | \(69\) |
Input:
int((c-c/a^2/x^2)^(3/2)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2*(-2*a^3*x^3+6*a^2*ln(x)*x^2+6*a*x-1)*x*(c*(a^2*x^2-1)/a^2/x^2)^(3/2)* ((a*x-1)/(a*x+1))^(3/2)/(a*x-1)^3
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.28 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {{\left (2 \, a^{3} c x^{3} - 6 \, a^{2} c x^{2} \log \left (x\right ) - 6 \, a c x + c\right )} \sqrt {a^{2} c}}{2 \, a^{4} x^{2}} \] Input:
integrate((c-c/a^2/x^2)^(3/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas ")
Output:
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)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(3/2)*((a*x-1)/(a*x+1))**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {3}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(3/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima ")
Output:
integrate((c - c/(a^2*x^2))^(3/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.49 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {1}{2} \, {\left (\frac {2 \, c x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a} - \frac {6 \, c \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{2}} - \frac {6 \, a c x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - c \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{4} x^{2}}\right )} \sqrt {c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(3/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")
Output:
1/2*(2*c*x*sgn(a*x + 1)*sgn(x)/a - 6*c*log(abs(x))*sgn(a*x + 1)*sgn(x)/a^2 - (6*a*c*x*sgn(a*x + 1)*sgn(x) - c*sgn(a*x + 1)*sgn(x))/(a^4*x^2))*sqrt(c )*abs(a)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\int {\left (c-\frac {c}{a^2\,x^2}\right )}^{3/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:
int((c - c/(a^2*x^2))^(3/2)*((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int((c - c/(a^2*x^2))^(3/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.25 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {\sqrt {c}\, c \left (-6 \,\mathrm {log}\left (a x \right ) a^{2} x^{2}+2 a^{3} x^{3}-6 a x +1\right )}{2 a^{3} x^{2}} \] Input:
int((c-c/a^2/x^2)^(3/2)*((a*x-1)/(a*x+1))^(3/2),x)
Output:
(sqrt(c)*c*( - 6*log(a*x)*a**2*x**2 + 2*a**3*x**3 - 6*a*x + 1))/(2*a**3*x* *2)