Integrand size = 20, antiderivative size = 90 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{2 a c (c-a c x)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{2 \sqrt {2} a c^{5/2}} \] Output:
1/2*(-a^2*x^2+1)^(1/2)/a/c/(-a*c*x+c)^(3/2)+1/4*arctanh(1/2*c^(1/2)*(-a^2* x^2+1)^(1/2)*2^(1/2)/(-a*c*x+c)^(1/2))*2^(1/2)/a/c^(5/2)
Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {-2 \sqrt {1+a x}+\sqrt {2} (-1+a x) \text {arctanh}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )}{4 a c^2 \sqrt {1-a x} \sqrt {c-a c x}} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - a*c*x)^(5/2)),x]
Output:
-1/4*(-2*Sqrt[1 + a*x] + Sqrt[2]*(-1 + a*x)*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]] )/(a*c^2*Sqrt[1 - a*x]*Sqrt[c - a*c*x])
Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6677, 470, 471, 221}
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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle \frac {\int \frac {1}{(c-a c x)^{3/2} \sqrt {1-a^2 x^2}}dx}{c}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}}dx}{4 c}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}}{c}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle \frac {\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}-\frac {1}{2} a \int \frac {1}{\frac {a^2 c^2 \left (1-a^2 x^2\right )}{c-a c x}-2 a^2 c}d\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{2 \sqrt {2} a c^{3/2}}+\frac {\sqrt {1-a^2 x^2}}{2 a (c-a c x)^{3/2}}}{c}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - a*c*x)^(5/2)),x]
Output:
(Sqrt[1 - a^2*x^2]/(2*a*(c - a*c*x)^(3/2)) + ArcTanh[(Sqrt[c]*Sqrt[1 - a^2 *x^2])/(Sqrt[2]*Sqrt[c - a*c*x])]/(2*Sqrt[2]*a*c^(3/2)))/c
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 2*p + 2)/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a c x -\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -2 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{4 c^{\frac {7}{2}} \left (a x -1\right )^{2} \sqrt {c \left (a x +1\right )}\, a}\) | \(111\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(7/2)*(2^(1/2)*arctanh(1/2*(c *(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*a*c*x-2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/ 2)*2^(1/2)/c^(1/2))*c-2*(c*(a*x+1))^(1/2)*c^(1/2))/(a*x-1)^2/(c*(a*x+1))^( 1/2)/a
Time = 0.09 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.83 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{8 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}}, \frac {\sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{4 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}}\right ] \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(5/2),x, algorithm="fric as")
Output:
[1/8*(sqrt(2)*(a^2*x^2 - 2*a*x + 1)*sqrt(c)*log(-(a^2*c*x^2 + 2*a*c*x - 2* sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a^2*x^2 - 2*a* x + 1)) + 4*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^3*c^3*x^2 - 2*a^2*c^3* x + a*c^3), 1/4*(sqrt(2)*(a^2*x^2 - 2*a*x + 1)*sqrt(-c)*arctan(1/2*sqrt(2) *sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + 2*sqrt(-a^2*x ^2 + 1)*sqrt(-a*c*x + c))/(a^3*c^3*x^2 - 2*a^2*c^3*x + a*c^3)]
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**(5/2),x)
Output:
Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(a*x - 1))**(5/2)*(a*x + 1)), x)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (-a c x + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(5/2),x, algorithm="maxi ma")
Output:
integrate(sqrt(-a^2*x^2 + 1)/((-a*c*x + c)^(5/2)*(a*x + 1)), x)
Exception generated. \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(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. \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-a\,c\,x\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \] Input:
int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^(5/2)*(a*x + 1)),x)
Output:
int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^(5/2)*(a*x + 1)), x)
Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-2 \sqrt {a x +1}-\sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x +\sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )\right )}{4 a \,c^{3} \left (a x -1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(5/2),x)
Output:
(sqrt(c)*( - 2*sqrt(a*x + 1) - sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt( 2))/2))*a*x + sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2))))/(4*a*c* *3*(a*x - 1))