Integrand size = 24, antiderivative size = 238 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{3 a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
1/4*c^2*(c-c/a^2/x^2)^(1/2)/a^5/(1-1/a^2/x^2)^(1/2)/x^4-1/3*c^2*(c-c/a^2/x ^2)^(1/2)/a^4/(1-1/a^2/x^2)^(1/2)/x^3-c^2*(c-c/a^2/x^2)^(1/2)/a^3/(1-1/a^2 /x^2)^(1/2)/x^2+2*c^2*(c-c/a^2/x^2)^(1/2)/a^2/(1-1/a^2/x^2)^(1/2)/x+c^2*(c -c/a^2/x^2)^(1/2)*x/(1-1/a^2/x^2)^(1/2)-c^2*(c-c/a^2/x^2)^(1/2)*ln(x)/a/(1 -1/a^2/x^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.32 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} \left (\frac {1}{4 a^5 x^4}-\frac {1}{3 a^4 x^3}-\frac {1}{a^3 x^2}+\frac {2}{a^2 x}+x-\frac {\log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \] Input:
Integrate[(c - c/(a^2*x^2))^(5/2)/E^ArcCoth[a*x],x]
Output:
((c - c/(a^2*x^2))^(5/2)*(1/(4*a^5*x^4) - 1/(3*a^4*x^3) - 1/(a^3*x^2) + 2/ (a^2*x) + x - Log[x]/a))/(1 - 1/(a^2*x^2))^(5/2)
Time = 0.68 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.34, 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, 99, 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.
\(\displaystyle \int \left (c-\frac {c}{a^2 x^2}\right )^{5/2} e^{-\coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \int e^{-\coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{5/2}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \int -\frac {(1-a x)^3 (a x+1)^2}{x^5}dx}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^3 (a x+1)^2}{x^5}dx}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \int \left (-a^5+\frac {a^4}{x}+\frac {2 a^3}{x^2}-\frac {2 a^2}{x^3}-\frac {a}{x^4}+\frac {1}{x^5}\right )dx}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \left (a^5 (-x)+a^4 \log (x)-\frac {2 a^3}{x}+\frac {a^2}{x^2}+\frac {a}{3 x^3}-\frac {1}{4 x^4}\right )}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(c - c/(a^2*x^2))^(5/2)/E^ArcCoth[a*x],x]
Output:
-((c^2*Sqrt[c - c/(a^2*x^2)]*(-1/4*1/x^4 + a/(3*x^3) + a^2/x^2 - (2*a^3)/x - a^5*x + a^4*Log[x]))/(a^5*Sqrt[1 - 1/(a^2*x^2)]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.40
method | result | size |
default | \(-\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \sqrt {\frac {a x -1}{a x +1}}\, x \left (-12 a^{5} x^{5}+12 \ln \left (x \right ) x^{4} a^{4}-24 a^{3} x^{3}+12 a^{2} x^{2}+4 a x -3\right )}{12 \left (a x -1\right ) \left (a^{2} x^{2}-1\right )^{2}}\) | \(96\) |
Input:
int((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/12*(c*(a^2*x^2-1)/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2)*x*(-12*a^5*x^5 +12*ln(x)*x^4*a^4-24*a^3*x^3+12*a^2*x^2+4*a*x-3)/(a*x-1)/(a^2*x^2-1)^2
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.31 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {{\left (12 \, a^{5} c^{2} x^{5} - 12 \, a^{4} c^{2} x^{4} \log \left (x\right ) + 24 \, a^{3} c^{2} x^{3} - 12 \, a^{2} c^{2} x^{2} - 4 \, a c^{2} x + 3 \, c^{2}\right )} \sqrt {a^{2} c}}{12 \, a^{6} x^{4}} \] Input:
integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="fricas ")
Output:
1/12*(12*a^5*c^2*x^5 - 12*a^4*c^2*x^4*log(x) + 24*a^3*c^2*x^3 - 12*a^2*c^2 *x^2 - 4*a*c^2*x + 3*c^2)*sqrt(a^2*c)/(a^6*x^4)
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(5/2)*((a*x-1)/(a*x+1))**(1/2),x)
Output:
Timed out
\[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="maxima ")
Output:
integrate((c - c/(a^2*x^2))^(5/2)*sqrt((a*x - 1)/(a*x + 1)), x)
Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.50 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {1}{12} \, {\left (\frac {12 \, c^{2} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a} - \frac {12 \, c^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{2}} + \frac {24 \, a^{3} c^{2} x^{3} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 12 \, a^{2} c^{2} x^{2} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 4 \, a c^{2} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 3 \, c^{2} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{6} x^{4}}\right )} \sqrt {c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="giac")
Output:
1/12*(12*c^2*x*sgn(a*x + 1)*sgn(x)/a - 12*c^2*log(abs(x))*sgn(a*x + 1)*sgn (x)/a^2 + (24*a^3*c^2*x^3*sgn(a*x + 1)*sgn(x) - 12*a^2*c^2*x^2*sgn(a*x + 1 )*sgn(x) - 4*a*c^2*x*sgn(a*x + 1)*sgn(x) + 3*c^2*sgn(a*x + 1)*sgn(x))/(a^6 *x^4))*sqrt(c)*abs(a)
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int {\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}} \,d x \] Input:
int((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(1/2),x)
Output:
int((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(1/2), x)
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.26 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} \left (-12 \,\mathrm {log}\left (a x \right ) a^{4} x^{4}+12 a^{5} x^{5}-36 a^{4} x^{4}+24 a^{3} x^{3}-12 a^{2} x^{2}-4 a x +3\right )}{12 a^{5} x^{4}} \] Input:
int((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x)
Output:
(sqrt(c)*c**2*( - 12*log(a*x)*a**4*x**4 + 12*a**5*x**5 - 36*a**4*x**4 + 24 *a**3*x**3 - 12*a**2*x**2 - 4*a*x + 3))/(12*a**5*x**4)