Integrand size = 24, antiderivative size = 324 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{6 a^7 \sqrt {1-\frac {1}{a^2 x^2}} x^6}-\frac {3 c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{5 a^6 \sqrt {1-\frac {1}{a^2 x^2}} x^5}+\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {5 c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{3 a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}-\frac {5 c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}-\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c^3 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
1/6*c^3*(c-c/a^2/x^2)^(1/2)/a^7/(1-1/a^2/x^2)^(1/2)/x^6-3/5*c^3*(c-c/a^2/x ^2)^(1/2)/a^6/(1-1/a^2/x^2)^(1/2)/x^5+1/4*c^3*(c-c/a^2/x^2)^(1/2)/a^5/(1-1 /a^2/x^2)^(1/2)/x^4+5/3*c^3*(c-c/a^2/x^2)^(1/2)/a^4/(1-1/a^2/x^2)^(1/2)/x^ 3-5/2*c^3*(c-c/a^2/x^2)^(1/2)/a^3/(1-1/a^2/x^2)^(1/2)/x^2-c^3*(c-c/a^2/x^2 )^(1/2)/a^2/(1-1/a^2/x^2)^(1/2)/x+c^3*(c-c/a^2/x^2)^(1/2)*x/(1-1/a^2/x^2)^ (1/2)-3*c^3*(c-c/a^2/x^2)^(1/2)*ln(x)/a/(1-1/a^2/x^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.30 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{7/2} \left (\frac {1}{6 a^7 x^6}-\frac {3}{5 a^6 x^5}+\frac {1}{4 a^5 x^4}+\frac {5}{3 a^4 x^3}-\frac {5}{2 a^3 x^2}-\frac {1}{a^2 x}+x-\frac {3 \log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2}} \] Input:
Integrate[(c - c/(a^2*x^2))^(7/2)/E^(3*ArcCoth[a*x]),x]
Output:
((c - c/(a^2*x^2))^(7/2)*(1/(6*a^7*x^6) - 3/(5*a^6*x^5) + 1/(4*a^5*x^4) + 5/(3*a^4*x^3) - 5/(2*a^3*x^2) - 1/(a^2*x) + x - (3*Log[x])/a))/(1 - 1/(a^2 *x^2))^(7/2)
Time = 0.71 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.32, 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.
\(\displaystyle \int \left (c-\frac {c}{a^2 x^2}\right )^{7/2} e^{-3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{7/2}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \int -\frac {(1-a x)^5 (a x+1)^2}{x^7}dx}{a^7 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^5 (a x+1)^2}{x^7}dx}{a^7 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \int \left (-a^7+\frac {3 a^6}{x}-\frac {a^5}{x^2}-\frac {5 a^4}{x^3}+\frac {5 a^3}{x^4}+\frac {a^2}{x^5}-\frac {3 a}{x^6}+\frac {1}{x^7}\right )dx}{a^7 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \left (a^7 (-x)+3 a^6 \log (x)+\frac {a^5}{x}+\frac {5 a^4}{2 x^2}-\frac {5 a^3}{3 x^3}-\frac {a^2}{4 x^4}+\frac {3 a}{5 x^5}-\frac {1}{6 x^6}\right )}{a^7 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(c - c/(a^2*x^2))^(7/2)/E^(3*ArcCoth[a*x]),x]
Output:
-((c^3*Sqrt[c - c/(a^2*x^2)]*(-1/6*1/x^6 + (3*a)/(5*x^5) - a^2/(4*x^4) - ( 5*a^3)/(3*x^3) + (5*a^4)/(2*x^2) + a^5/x - a^7*x + 3*a^6*Log[x]))/(a^7*Sqr t[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 = 112, normalized size of antiderivative = 0.35
method | result | size |
default | \(-\frac {\left (-60 a^{7} x^{7}+180 a^{6} \ln \left (x \right ) x^{6}+60 a^{5} x^{5}+150 a^{4} x^{4}-100 a^{3} x^{3}-15 a^{2} x^{2}+36 a x -10\right ) x {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{60 \left (a x -1\right )^{3} \left (a^{2} x^{2}-1\right )^{2}}\) | \(112\) |
Input:
int((c-c/a^2/x^2)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/60*(-60*a^7*x^7+180*a^6*ln(x)*x^6+60*a^5*x^5+150*a^4*x^4-100*a^3*x^3-15 *a^2*x^2+36*a*x-10)*x*(c*(a^2*x^2-1)/a^2/x^2)^(7/2)*((a*x-1)/(a*x+1))^(3/2 )/(a*x-1)^3/(a^2*x^2-1)^2
Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.30 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {{\left (60 \, a^{7} c^{3} x^{7} - 180 \, a^{6} c^{3} x^{6} \log \left (x\right ) - 60 \, a^{5} c^{3} x^{5} - 150 \, a^{4} c^{3} x^{4} + 100 \, a^{3} c^{3} x^{3} + 15 \, a^{2} c^{3} x^{2} - 36 \, a c^{3} x + 10 \, c^{3}\right )} \sqrt {a^{2} c}}{60 \, a^{8} x^{6}} \] Input:
integrate((c-c/a^2/x^2)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas ")
Output:
1/60*(60*a^7*c^3*x^7 - 180*a^6*c^3*x^6*log(x) - 60*a^5*c^3*x^5 - 150*a^4*c ^3*x^4 + 100*a^3*c^3*x^3 + 15*a^2*c^3*x^2 - 36*a*c^3*x + 10*c^3)*sqrt(a^2* c)/(a^8*x^6)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(7/2)*((a*x-1)/(a*x+1))**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima ")
Output:
integrate((c - c/(a^2*x^2))^(7/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.48 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {1}{60} \, {\left (\frac {60 \, c^{3} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a} - \frac {180 \, c^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{2}} - \frac {60 \, a^{5} c^{3} x^{5} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 150 \, a^{4} c^{3} x^{4} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 100 \, a^{3} c^{3} x^{3} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 15 \, a^{2} c^{3} x^{2} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 36 \, a c^{3} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 10 \, c^{3} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{8} x^{6}}\right )} \sqrt {c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")
Output:
1/60*(60*c^3*x*sgn(a*x + 1)*sgn(x)/a - 180*c^3*log(abs(x))*sgn(a*x + 1)*sg n(x)/a^2 - (60*a^5*c^3*x^5*sgn(a*x + 1)*sgn(x) + 150*a^4*c^3*x^4*sgn(a*x + 1)*sgn(x) - 100*a^3*c^3*x^3*sgn(a*x + 1)*sgn(x) - 15*a^2*c^3*x^2*sgn(a*x + 1)*sgn(x) + 36*a*c^3*x*sgn(a*x + 1)*sgn(x) - 10*c^3*sgn(a*x + 1)*sgn(x)) /(a^8*x^6))*sqrt(c)*abs(a)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\int {\left (c-\frac {c}{a^2\,x^2}\right )}^{7/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:
int((c - c/(a^2*x^2))^(7/2)*((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int((c - c/(a^2*x^2))^(7/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.24 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {\sqrt {c}\, c^{3} \left (-180 \,\mathrm {log}\left (a x \right ) a^{6} x^{6}+60 a^{7} x^{7}-200 a^{6} x^{6}-60 a^{5} x^{5}-150 a^{4} x^{4}+100 a^{3} x^{3}+15 a^{2} x^{2}-36 a x +10\right )}{60 a^{7} x^{6}} \] Input:
int((c-c/a^2/x^2)^(7/2)*((a*x-1)/(a*x+1))^(3/2),x)
Output:
(sqrt(c)*c**3*( - 180*log(a*x)*a**6*x**6 + 60*a**7*x**7 - 200*a**6*x**6 - 60*a**5*x**5 - 150*a**4*x**4 + 100*a**3*x**3 + 15*a**2*x**2 - 36*a*x + 10) )/(60*a**7*x**6)