Integrand size = 20, antiderivative size = 57 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {1}{a c^2 \sqrt {c-a c x}}+\frac {\text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}} \] Output:
-1/a/c^2/(-a*c*x+c)^(1/2)+1/2*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2) )*2^(1/2)/a/c^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.65 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1-a x)\right )}{a c^2 \sqrt {c-a c x}} \] Input:
Integrate[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^(5/2)),x]
Output:
-(Hypergeometric2F1[-1/2, 1, 1/2, (1 - a*x)/2]/(a*c^2*Sqrt[c - a*c*x]))
Time = 0.55 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6717, 6680, 35, 61, 73, 219}
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.
\(\displaystyle \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}}dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle -\int \frac {1-a x}{(a x+1) (c-a c x)^{5/2}}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle -\frac {\int \frac {1}{(a x+1) (c-a c x)^{3/2}}dx}{c}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {\frac {\int \frac {1}{(a x+1) \sqrt {c-a c x}}dx}{2 c}+\frac {1}{a c \sqrt {c-a c x}}}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\frac {1}{a c \sqrt {c-a c x}}-\frac {\int \frac {1}{2-\frac {c-a c x}{c}}d\sqrt {c-a c x}}{a c^2}}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {1}{a c \sqrt {c-a c x}}-\frac {\text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{3/2}}}{c}\) |
Input:
Int[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^(5/2)),x]
Output:
-((1/(a*c*Sqrt[c - a*c*x]) - ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]/(S qrt[2]*a*c^(3/2)))/c)
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Int[u*(c + d*x)^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c , d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && !(IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(-\frac {2 \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}+\frac {1}{2 c \sqrt {-a c x +c}}\right )}{c a}\) | \(50\) |
default | \(\frac {-\frac {1}{c \sqrt {-a c x +c}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 c^{\frac {3}{2}}}}{a c}\) | \(50\) |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {-c \left (a x -1\right )}-2 \sqrt {c}}{2 c^{\frac {5}{2}} \sqrt {-c \left (a x -1\right )}\, a}\) | \(58\) |
Input:
int((a*x-1)/(a*x+1)/(-a*c*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/c/a*(-1/4/c^(3/2)*2^(1/2)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2)) +1/2/c/(-a*c*x+c)^(1/2))
Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {c} \log \left (\frac {a c x - 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + 4 \, \sqrt {-a c x + c}}{4 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}, \frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + 2 \, \sqrt {-a c x + c}}{2 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}\right ] \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(5/2),x, algorithm="fricas")
Output:
[1/4*(sqrt(2)*(a*x - 1)*sqrt(c)*log((a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*sq rt(c) - 3*c)/(a*x + 1)) + 4*sqrt(-a*c*x + c))/(a^2*c^3*x - a*c^3), 1/2*(sq rt(2)*(a*x - 1)*sqrt(-c)*arctan(sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + 2*sqrt(-a*c*x + c))/(a^2*c^3*x - a*c^3)]
Time = 2.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\begin {cases} - \frac {2 \cdot \left (\frac {1}{2 c \sqrt {- a c x + c}} + \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{4 c \sqrt {- c}}\right )}{a c} & \text {for}\: a c \neq 0 \\\frac {\begin {cases} - x & \text {for}\: a = 0 \\\frac {a x - 2 \log {\left (a x + 1 \right )} + 1}{a} & \text {otherwise} \end {cases}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)**(5/2),x)
Output:
Piecewise((-2*(1/(2*c*sqrt(-a*c*x + c)) + sqrt(2)*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/(4*c*sqrt(-c)))/(a*c), Ne(a*c, 0)), (Piecewise((-x, Eq (a, 0)), ((a*x - 2*log(a*x + 1) + 1)/a, True))/c**(5/2), True))
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {\frac {\sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right )}{c^{\frac {3}{2}}} + \frac {4}{\sqrt {-a c x + c} c}}{4 \, a c} \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(5/2),x, algorithm="maxima")
Output:
-1/4*(sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c)))/c^(3/2) + 4/(sqrt(-a*c*x + c)*c))/(a*c)
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{2 \, a \sqrt {-c} c^{2}} - \frac {1}{\sqrt {-a c x + c} a c^{2}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^(5/2),x, algorithm="giac")
Output:
-1/2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)*c^2 ) - 1/(sqrt(-a*c*x + c)*a*c^2)
Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )}{2\,a\,c^{5/2}}-\frac {1}{a\,c^2\,\sqrt {c-a\,c\,x}} \] Input:
int((a*x - 1)/((c - a*c*x)^(5/2)*(a*x + 1)),x)
Output:
(2^(1/2)*atanh((2^(1/2)*(c - a*c*x)^(1/2))/(2*c^(1/2))))/(2*a*c^(5/2)) - 1 /(a*c^2*(c - a*c*x)^(1/2))
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-\sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )+\sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-4\right )}{4 \sqrt {-a x +1}\, a \,c^{3}} \] Input:
int((a*x-1)/(a*x+1)/(-a*c*x+c)^(5/2),x)
Output:
(sqrt(c)*( - sqrt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) - sqrt(2)) + sq rt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) + sqrt(2)) - 4))/(4*sqrt( - a* x + 1)*a*c**3)