Integrand size = 20, antiderivative size = 57 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {1}{a c^2 \sqrt {c-a c x}}-\frac {\text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}} \] 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 = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.63 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1-a x)\right )}{a c^2 \sqrt {c-a c x}} \] Input:
Integrate[1/(E^(2*ArcTanh[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 = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.
\(\displaystyle \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 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*ArcTanh[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])]/(Sqr t[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])
Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {\frac {1}{c \sqrt {-a c x +c}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 c^{\frac {3}{2}}}}{a c}\) | \(50\) |
default | \(-\frac {2 \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{2 c \sqrt {-a c x +c}}\right )}{c a}\) | \(50\) |
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(1/(a*x+1)^2*(-a^2*x^2+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 = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {c} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) - 4 \, \sqrt {-a c x + c}}{4 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}, -\frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + 2 \, \sqrt {-a c x + c}}{2 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}\right ] \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+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*(s qrt(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 = 5.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\begin {cases} - \frac {2 \left (- \frac {1}{2 c \sqrt {- a c x + c}} - \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{4 c \sqrt {- c}}\right )}{a c} & \text {for}\: a c \neq 0 \\\frac {- x + 2 \left (\begin {cases} x & \text {for}\: a = 0 \\\frac {\log {\left (a x + 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(a*x+1)**2*(-a**2*x**2+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)), ((-x + 2*Piecewis e((x, Eq(a, 0)), (log(a*x + 1)/a, True)))/c**(5/2), True))
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\frac {\sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right )}{c^{\frac {3}{2}}} + \frac {4}{\sqrt {-a c x + c} c}}{4 \, a c} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+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 = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{2 \, a \sqrt {-c} c^{2}} + \frac {1}{\sqrt {-a c x + c} a c^{2}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+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 = 46, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {1}{a\,c^2\,\sqrt {c-a\,c\,x}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )}{2\,a\,c^{5/2}} \] Input:
int(-(a^2*x^2 - 1)/((c - a*c*x)^(5/2)*(a*x + 1)^2),x)
Output:
1/(a*c^2*(c - a*c*x)^(1/2)) - (2^(1/2)*atanh((2^(1/2)*(c - a*c*x)^(1/2))/( 2*c^(1/2))))/(2*a*c^(5/2))
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (\sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )-\sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )+4\right )}{4 \sqrt {-a x +1}\, a \,c^{3}} \] Input:
int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(5/2),x)
Output:
(sqrt(c)*(sqrt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) - sqrt(2)) - sqrt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) + sqrt(2)) + 4))/(4*sqrt( - a*x + 1)*a*c**3)