Integrand size = 22, antiderivative size = 359 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c^3 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{24 a c^3 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)^3}-\frac {11 \sqrt {1-\frac {1}{a^2 x^2}}}{32 a c^3 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)^2}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a c^3 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{32 a c^3 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}-\frac {5 \sqrt {1-\frac {1}{a^2 x^2}}}{16 a c^3 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)}+\frac {51 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{32 a c^3 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {19 \sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{32 a c^3 \sqrt {c-\frac {c}{a^2 x^2}}} \] Output:
(1-1/a^2/x^2)^(1/2)*x/c^3/(c-c/a^2/x^2)^(1/2)+1/24*(1-1/a^2/x^2)^(1/2)/a/c ^3/(c-c/a^2/x^2)^(1/2)/(-a*x+1)^3-11/32*(1-1/a^2/x^2)^(1/2)/a/c^3/(c-c/a^2 /x^2)^(1/2)/(-a*x+1)^2+3/2*(1-1/a^2/x^2)^(1/2)/a/c^3/(c-c/a^2/x^2)^(1/2)/( -a*x+1)+1/32*(1-1/a^2/x^2)^(1/2)/a/c^3/(c-c/a^2/x^2)^(1/2)/(a*x+1)^2-5/16* (1-1/a^2/x^2)^(1/2)/a/c^3/(c-c/a^2/x^2)^(1/2)/(a*x+1)+51/32*(1-1/a^2/x^2)^ (1/2)*ln(-a*x+1)/a/c^3/(c-c/a^2/x^2)^(1/2)-19/32*(1-1/a^2/x^2)^(1/2)*ln(a* x+1)/a/c^3/(c-c/a^2/x^2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.33 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{7/2} \left (96 x-\frac {4}{a (-1+a x)^3}-\frac {33}{a (-1+a x)^2}+\frac {3}{a (1+a x)^2}+\frac {144}{a-a^2 x}-\frac {30}{a+a^2 x}+\frac {153 \log (1-a x)}{a}-\frac {57 \log (1+a x)}{a}\right )}{96 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \] Input:
Integrate[E^ArcCoth[a*x]/(c - c/(a^2*x^2))^(7/2),x]
Output:
((1 - 1/(a^2*x^2))^(7/2)*(96*x - 4/(a*(-1 + a*x)^3) - 33/(a*(-1 + a*x)^2) + 3/(a*(1 + a*x)^2) + 144/(a - a^2*x) - 30/(a + a^2*x) + (153*Log[1 - a*x] )/a - (57*Log[1 + a*x])/a))/(96*(c - c/(a^2*x^2))^(7/2))
Time = 0.80 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6751, 6747, 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.
| \(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 6751 |
| \(\displaystyle \frac {\sqrt {1-\frac {1}{a^2 x^2}} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2}}dx}{c^3 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
| \(\Big \downarrow \) 6747 |
| \(\displaystyle \frac {a^7 \sqrt {1-\frac {1}{a^2 x^2}} \int \frac {x^7}{(1-a x)^4 (a x+1)^3}dx}{c^3 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a^7 \sqrt {1-\frac {1}{a^2 x^2}} \int \left (-\frac {19}{32 a^7 (a x+1)}+\frac {5}{16 a^7 (a x+1)^2}-\frac {1}{16 a^7 (a x+1)^3}+\frac {1}{a^7}+\frac {51}{32 a^7 (a x-1)}+\frac {3}{2 a^7 (a x-1)^2}+\frac {11}{16 a^7 (a x-1)^3}+\frac {1}{8 a^7 (a x-1)^4}\right )dx}{c^3 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^7 \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {3}{2 a^8 (1-a x)}-\frac {5}{16 a^8 (a x+1)}-\frac {11}{32 a^8 (1-a x)^2}+\frac {1}{32 a^8 (a x+1)^2}+\frac {1}{24 a^8 (1-a x)^3}+\frac {51 \log (1-a x)}{32 a^8}-\frac {19 \log (a x+1)}{32 a^8}+\frac {x}{a^7}\right )}{c^3 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
Input:
Int[E^ArcCoth[a*x]/(c - c/(a^2*x^2))^(7/2),x]
Output:
(a^7*Sqrt[1 - 1/(a^2*x^2)]*(x/a^7 + 1/(24*a^8*(1 - a*x)^3) - 11/(32*a^8*(1 - a*x)^2) + 3/(2*a^8*(1 - a*x)) + 1/(32*a^8*(1 + a*x)^2) - 5/(16*a^8*(1 + a*x)) + (51*Log[1 - a*x])/(32*a^8) - (19*Log[1 + a*x])/(32*a^8)))/(c^3*Sq rt[c - c/(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 = 0.09 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.69
| method | result | size |
| default | \(-\frac {\left (a x -1\right ) \left (a x +1\right ) \left (-96 x^{6} a^{6}+57 \ln \left (a x +1\right ) x^{5} a^{5}-153 \ln \left (a x -1\right ) x^{5} a^{5}+96 a^{5} x^{5}-57 \ln \left (a x +1\right ) x^{4} a^{4}+153 \ln \left (a x -1\right ) x^{4} a^{4}+366 a^{4} x^{4}-114 \ln \left (a x +1\right ) x^{3} a^{3}+306 a^{3} \ln \left (a x -1\right ) x^{3}-222 a^{3} x^{3}+114 \ln \left (a x +1\right ) x^{2} a^{2}-306 a^{2} \ln \left (a x -1\right ) x^{2}-338 a^{2} x^{2}+57 \ln \left (a x +1\right ) x a -153 a \ln \left (a x -1\right ) x +122 a x -57 \ln \left (a x +1\right )+153 \ln \left (a x -1\right )+88\right )}{96 \sqrt {\frac {a x -1}{a x +1}}\, a^{8} x^{7} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}}}\) | \(247\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/96/((a*x-1)/(a*x+1))^(1/2)*(a*x-1)*(a*x+1)*(-96*x^6*a^6+57*ln(a*x+1)*x^ 5*a^5-153*ln(a*x-1)*x^5*a^5+96*a^5*x^5-57*ln(a*x+1)*x^4*a^4+153*ln(a*x-1)* x^4*a^4+366*a^4*x^4-114*ln(a*x+1)*x^3*a^3+306*a^3*ln(a*x-1)*x^3-222*a^3*x^ 3+114*ln(a*x+1)*x^2*a^2-306*a^2*ln(a*x-1)*x^2-338*a^2*x^2+57*ln(a*x+1)*x*a -153*a*ln(a*x-1)*x+122*a*x-57*ln(a*x+1)+153*ln(a*x-1)+88)/a^8/x^7/(c*(a^2* x^2-1)/a^2/x^2)^(7/2)
Time = 0.10 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.57 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {{\left (96 \, a^{6} x^{6} - 96 \, a^{5} x^{5} - 366 \, a^{4} x^{4} + 222 \, a^{3} x^{3} + 338 \, a^{2} x^{2} - 122 \, a x - 57 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \log \left (a x + 1\right ) + 153 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \log \left (a x - 1\right ) - 88\right )} \sqrt {a^{2} c}}{96 \, {\left (a^{7} c^{4} x^{5} - a^{6} c^{4} x^{4} - 2 \, a^{5} c^{4} x^{3} + 2 \, a^{4} c^{4} x^{2} + a^{3} c^{4} x - a^{2} c^{4}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(7/2),x, algorithm="fric as")
Output:
1/96*(96*a^6*x^6 - 96*a^5*x^5 - 366*a^4*x^4 + 222*a^3*x^3 + 338*a^2*x^2 - 122*a*x - 57*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*log(a*x + 1) + 153*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*log(a*x - 1) - 88)*sqrt(a^2*c)/(a^7*c^4*x^5 - a^6*c^4*x^4 - 2*a^5*c^4*x^3 + 2*a^4* c^4*x^2 + a^3*c^4*x - a^2*c^4)
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)/(c-c/a**2/x**2)**(7/2),x)
Output:
Timed out
\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int { \frac {1}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(7/2),x, algorithm="maxi ma")
Output:
integrate(1/((c - c/(a^2*x^2))^(7/2)*sqrt((a*x - 1)/(a*x + 1))), x)
Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(7/2),x, algorithm="giac ")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int \frac {1}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{7/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(1/((c - c/(a^2*x^2))^(7/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(1/((c - c/(a^2*x^2))^(7/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {\sqrt {c}\, \left (153 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}-153 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-306 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+306 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+153 \,\mathrm {log}\left (a x -1\right ) a x -153 \,\mathrm {log}\left (a x -1\right )-57 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}+57 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+114 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-114 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-57 \,\mathrm {log}\left (a x +1\right ) a x +57 \,\mathrm {log}\left (a x +1\right )+96 a^{6} x^{6}-462 a^{5} x^{5}+954 a^{3} x^{3}-394 a^{2} x^{2}-488 a x +278\right )}{96 a \,c^{4} \left (a^{5} x^{5}-a^{4} x^{4}-2 a^{3} x^{3}+2 a^{2} x^{2}+a x -1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(7/2),x)
Output:
(sqrt(c)*(153*log(a*x - 1)*a**5*x**5 - 153*log(a*x - 1)*a**4*x**4 - 306*lo g(a*x - 1)*a**3*x**3 + 306*log(a*x - 1)*a**2*x**2 + 153*log(a*x - 1)*a*x - 153*log(a*x - 1) - 57*log(a*x + 1)*a**5*x**5 + 57*log(a*x + 1)*a**4*x**4 + 114*log(a*x + 1)*a**3*x**3 - 114*log(a*x + 1)*a**2*x**2 - 57*log(a*x + 1 )*a*x + 57*log(a*x + 1) + 96*a**6*x**6 - 462*a**5*x**5 + 954*a**3*x**3 - 3 94*a**2*x**2 - 488*a*x + 278))/(96*a*c**4*(a**5*x**5 - a**4*x**4 - 2*a**3* x**3 + 2*a**2*x**2 + a*x - 1))