Integrand size = 24, antiderivative size = 264 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^3}+\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}-\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \] Output:
(1-1/a^2/x^2)^(1/2)*x/c^2/(c-c/a^2/x^2)^(1/2)-1/6*(1-1/a^2/x^2)^(1/2)/a/c^ 2/(c-c/a^2/x^2)^(1/2)/(a*x+1)^3+9/8*(1-1/a^2/x^2)^(1/2)/a/c^2/(c-c/a^2/x^2 )^(1/2)/(a*x+1)^2-31/8*(1-1/a^2/x^2)^(1/2)/a/c^2/(c-c/a^2/x^2)^(1/2)/(a*x+ 1)+1/16*(1-1/a^2/x^2)^(1/2)*ln(-a*x+1)/a/c^2/(c-c/a^2/x^2)^(1/2)-49/16*(1- 1/a^2/x^2)^(1/2)*ln(a*x+1)/a/c^2/(c-c/a^2/x^2)^(1/2)
Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.36 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (48 x-\frac {8}{a (1+a x)^3}+\frac {54}{a (1+a x)^2}-\frac {186}{a+a^2 x}+\frac {3 \log (1-a x)}{a}-\frac {147 \log (1+a x)}{a}\right )}{48 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \] Input:
Integrate[1/(E^(3*ArcCoth[a*x])*(c - c/(a^2*x^2))^(5/2)),x]
Output:
((1 - 1/(a^2*x^2))^(5/2)*(48*x - 8/(a*(1 + a*x)^3) + 54/(a*(1 + a*x)^2) - 186/(a + a^2*x) + (3*Log[1 - a*x])/a - (147*Log[1 + a*x])/a))/(48*(c - c/( a^2*x^2))^(5/2))
Time = 0.75 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.43, 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.
\(\displaystyle \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {\sqrt {1-\frac {1}{a^2 x^2}} \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}}dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \int -\frac {x^5}{(1-a x) (a x+1)^4}dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \int \frac {x^5}{(1-a x) (a x+1)^4}dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \int \left (\frac {49}{16 a^5 (a x+1)}-\frac {31}{8 a^5 (a x+1)^2}+\frac {9}{4 a^5 (a x+1)^3}-\frac {1}{2 a^5 (a x+1)^4}-\frac {1}{a^5}-\frac {1}{16 a^5 (a x-1)}\right )dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {31}{8 a^6 (a x+1)}-\frac {9}{8 a^6 (a x+1)^2}+\frac {1}{6 a^6 (a x+1)^3}-\frac {\log (1-a x)}{16 a^6}+\frac {49 \log (a x+1)}{16 a^6}-\frac {x}{a^5}\right )}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
Input:
Int[1/(E^(3*ArcCoth[a*x])*(c - c/(a^2*x^2))^(5/2)),x]
Output:
-((a^5*Sqrt[1 - 1/(a^2*x^2)]*(-(x/a^5) + 1/(6*a^6*(1 + a*x)^3) - 9/(8*a^6* (1 + a*x)^2) + 31/(8*a^6*(1 + a*x)) - Log[1 - a*x]/(16*a^6) + (49*Log[1 + a*x])/(16*a^6)))/(c^2*Sqrt[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 = 0.08 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.66
method | result | size |
default | \(-\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x -1\right ) \left (a x +1\right ) \left (-48 a^{4} x^{4}+147 \ln \left (a x +1\right ) x^{3} a^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}-144 a^{3} x^{3}+441 \ln \left (a x +1\right ) x^{2} a^{2}-9 a^{2} \ln \left (a x -1\right ) x^{2}+42 a^{2} x^{2}+441 \ln \left (a x +1\right ) x a -9 a \ln \left (a x -1\right ) x +270 a x +147 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )+140\right )}{48 a^{6} x^{5} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\) | \(175\) |
Input:
int(((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/48*((a*x-1)/(a*x+1))^(3/2)*(a*x-1)*(a*x+1)*(-48*a^4*x^4+147*ln(a*x+1)*x ^3*a^3-3*a^3*ln(a*x-1)*x^3-144*a^3*x^3+441*ln(a*x+1)*x^2*a^2-9*a^2*ln(a*x- 1)*x^2+42*a^2*x^2+441*ln(a*x+1)*x*a-9*a*ln(a*x-1)*x+270*a*x+147*ln(a*x+1)- 3*ln(a*x-1)+140)/a^6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)
Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {{\left (48 \, a^{4} x^{4} + 144 \, a^{3} x^{3} - 42 \, a^{2} x^{2} - 270 \, a x - 147 \, {\left (a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1\right )} \log \left (a x - 1\right ) - 140\right )} \sqrt {a^{2} c}}{48 \, {\left (a^{5} c^{3} x^{3} + 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x + a^{2} c^{3}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="fricas ")
Output:
1/48*(48*a^4*x^4 + 144*a^3*x^3 - 42*a^2*x^2 - 270*a*x - 147*(a^3*x^3 + 3*a ^2*x^2 + 3*a*x + 1)*log(a*x + 1) + 3*(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)*log (a*x - 1) - 140)*sqrt(a^2*c)/(a^5*c^3*x^3 + 3*a^4*c^3*x^2 + 3*a^3*c^3*x + a^2*c^3)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(((a*x-1)/(a*x+1))**(3/2)/(c-c/a**2/x**2)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="maxima ")
Output:
integrate(((a*x - 1)/(a*x + 1))^(3/2)/(c - c/(a^2*x^2))^(5/2), x)
Exception generated. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),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. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}} \,d x \] Input:
int(((a*x - 1)/(a*x + 1))^(3/2)/(c - c/(a^2*x^2))^(5/2),x)
Output:
int(((a*x - 1)/(a*x + 1))^(3/2)/(c - c/(a^2*x^2))^(5/2), x)
Time = 0.14 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.56 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (3 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+9 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+9 \,\mathrm {log}\left (a x -1\right ) a x +3 \,\mathrm {log}\left (a x -1\right )-147 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-441 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-441 \,\mathrm {log}\left (a x +1\right ) a x -147 \,\mathrm {log}\left (a x +1\right )+48 a^{4} x^{4}+158 a^{3} x^{3}-228 a x -126\right )}{48 a \,c^{3} \left (a^{3} x^{3}+3 a^{2} x^{2}+3 a x +1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x)
Output:
(sqrt(c)*(3*log(a*x - 1)*a**3*x**3 + 9*log(a*x - 1)*a**2*x**2 + 9*log(a*x - 1)*a*x + 3*log(a*x - 1) - 147*log(a*x + 1)*a**3*x**3 - 441*log(a*x + 1)* a**2*x**2 - 441*log(a*x + 1)*a*x - 147*log(a*x + 1) + 48*a**4*x**4 + 158*a **3*x**3 - 228*a*x - 126))/(48*a*c**3*(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 1 ))