Integrand size = 24, antiderivative size = 267 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)^3}-\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)^2}+\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)}+\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \] 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)+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)-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)
Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.36 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (48 x-\frac {8}{a (-1+a x)^3}-\frac {54}{a (-1+a x)^2}+\frac {186}{a-a^2 x}+\frac {147 \log (1-a x)}{a}-\frac {3 \log (1+a x)}{a}\right )}{48 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \] Input:
Integrate[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) + (147*Log[1 - a*x])/a - (3*Log[1 + a*x])/a))/(48*(c - c /(a^2*x^2))^(5/2))
Time = 0.74 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.43, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, 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.
\(\displaystyle \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \int \frac {x^5}{(1-a x)^4 (a x+1)}dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \int \left (-\frac {1}{16 a^5 (a x+1)}+\frac {1}{a^5}+\frac {49}{16 a^5 (a x-1)}+\frac {31}{8 a^5 (a x-1)^2}+\frac {9}{4 a^5 (a x-1)^3}+\frac {1}{2 a^5 (a x-1)^4}\right )dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {31}{8 a^6 (1-a x)}-\frac {9}{8 a^6 (1-a x)^2}+\frac {1}{6 a^6 (1-a x)^3}+\frac {49 \log (1-a x)}{16 a^6}-\frac {\log (a x+1)}{16 a^6}+\frac {x}{a^5}\right )}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
Input:
Int[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)) + (49*Log[1 - a*x])/(16*a^6) - 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 = 0.09 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.66
method | result | size |
default | \(-\frac {\left (a x -1\right ) \left (a x +1\right ) \left (-48 a^{4} x^{4}+3 \ln \left (a x +1\right ) x^{3} a^{3}-147 a^{3} \ln \left (a x -1\right ) x^{3}+144 a^{3} x^{3}-9 \ln \left (a x +1\right ) x^{2} a^{2}+441 a^{2} \ln \left (a x -1\right ) x^{2}+42 a^{2} x^{2}+9 \ln \left (a x +1\right ) x a -441 a \ln \left (a x -1\right ) x -270 a x -3 \ln \left (a x +1\right )+147 \ln \left (a x -1\right )+140\right )}{48 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a^{6} x^{5} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\) | \(175\) |
Input:
int(1/((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+3*ln(a*x+1)*x^3 *a^3-147*a^3*ln(a*x-1)*x^3+144*a^3*x^3-9*ln(a*x+1)*x^2*a^2+441*a^2*ln(a*x- 1)*x^2+42*a^2*x^2+9*ln(a*x+1)*x*a-441*a*ln(a*x-1)*x-270*a*x-3*ln(a*x+1)+14 7*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 = 138, normalized size of antiderivative = 0.52 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {{\left (48 \, a^{4} x^{4} - 144 \, a^{3} x^{3} - 42 \, a^{2} x^{2} + 270 \, a x - 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x + 1\right ) + 147 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 140\right )} \sqrt {a^{2} c}}{48 \, {\left (a^{5} c^{3} x^{3} - 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x - a^{2} c^{3}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="fric as")
Output:
1/48*(48*a^4*x^4 - 144*a^3*x^3 - 42*a^2*x^2 + 270*a*x - 3*(a^3*x^3 - 3*a^2 *x^2 + 3*a*x - 1)*log(a*x + 1) + 147*(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(1/((a*x-1)/(a*x+1))**(3/2)/(c-c/a**2/x**2)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="maxi ma")
Output:
integrate(1/((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2)), 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(1/((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. \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {1}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \] Input:
int(1/((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2)),x)
Output:
int(1/((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2)), x)
Time = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.55 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (147 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-441 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+441 \,\mathrm {log}\left (a x -1\right ) a x -147 \,\mathrm {log}\left (a x -1\right )-3 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+9 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-9 \,\mathrm {log}\left (a x +1\right ) a x +3 \,\mathrm {log}\left (a x +1\right )+48 a^{4} x^{4}-158 a^{3} x^{3}+228 a x -126\right )}{48 a \,c^{3} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x)
Output:
(sqrt(c)*(147*log(a*x - 1)*a**3*x**3 - 441*log(a*x - 1)*a**2*x**2 + 441*lo g(a*x - 1)*a*x - 147*log(a*x - 1) - 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) + 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 ))