Integrand size = 20, antiderivative size = 104 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\frac {1}{5 a c^2 (c-a c x)^{5/2}}+\frac {1}{6 a c^3 (c-a c x)^{3/2}}+\frac {1}{4 a c^4 \sqrt {c-a c x}}-\frac {\text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a c^{9/2}} \] Output:
1/5/a/c^2/(-a*c*x+c)^(5/2)+1/6/a/c^3/(-a*c*x+c)^(3/2)+1/4/a/c^4/(-a*c*x+c) ^(1/2)-1/8*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a/c^(9/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.38 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {1}{2} (1-a x)\right )}{5 a c^2 (c-a c x)^{5/2}} \] Input:
Integrate[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^(9/2)),x]
Output:
Hypergeometric2F1[-5/2, 1, -3/2, (1 - a*x)/2]/(5*a*c^2*(c - a*c*x)^(5/2))
Time = 0.47 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6680, 35, 61, 61, 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.
\(\displaystyle \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {1-a x}{(a x+1) (c-a c x)^{9/2}}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle \frac {\int \frac {1}{(a x+1) (c-a c x)^{7/2}}dx}{c}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\int \frac {1}{(a x+1) (c-a c x)^{5/2}}dx}{2 c}+\frac {1}{5 a c (c-a c x)^{5/2}}}{c}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{(a x+1) (c-a c x)^{3/2}}dx}{2 c}+\frac {1}{3 a c (c-a c x)^{3/2}}}{2 c}+\frac {1}{5 a c (c-a c x)^{5/2}}}{c}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {1}{(a x+1) \sqrt {c-a c x}}dx}{2 c}+\frac {1}{a c \sqrt {c-a c x}}}{2 c}+\frac {1}{3 a c (c-a c x)^{3/2}}}{2 c}+\frac {1}{5 a c (c-a c x)^{5/2}}}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {\frac {\frac {1}{a c \sqrt {c-a c x}}-\frac {\int \frac {1}{2-\frac {c-a c x}{c}}d\sqrt {c-a c x}}{a c^2}}{2 c}+\frac {1}{3 a c (c-a c x)^{3/2}}}{2 c}+\frac {1}{5 a c (c-a c x)^{5/2}}}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {\frac {1}{a c \sqrt {c-a c x}}-\frac {\text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{3/2}}}{2 c}+\frac {1}{3 a c (c-a c x)^{3/2}}}{2 c}+\frac {1}{5 a c (c-a c x)^{5/2}}}{c}\) |
Input:
Int[1/(E^(2*ArcTanh[a*x])*(c - a*c*x)^(9/2)),x]
Output:
(1/(5*a*c*(c - a*c*x)^(5/2)) + (1/(3*a*c*(c - a*c*x)^(3/2)) + (1/(a*c*Sqrt [c - a*c*x]) - ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])]/(Sqrt[2]*a*c^(3/ 2)))/(2*c))/(2*c))/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 = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {7}{2}}}+\frac {1}{4 c^{3} \sqrt {-a c x +c}}+\frac {1}{6 c^{2} \left (-a c x +c \right )^{\frac {3}{2}}}+\frac {1}{5 c \left (-a c x +c \right )^{\frac {5}{2}}}}{a c}\) | \(78\) |
default | \(-\frac {2 \left (-\frac {1}{8 c^{3} \sqrt {-a c x +c}}-\frac {1}{12 c^{2} \left (-a c x +c \right )^{\frac {3}{2}}}-\frac {1}{10 c \left (-a c x +c \right )^{\frac {5}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 c^{\frac {7}{2}}}\right )}{c a}\) | \(78\) |
pseudoelliptic | \(-\frac {\sqrt {2}\, \sqrt {-c \left (a x -1\right )}\, \left (a x -1\right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-2 \sqrt {c}\, \left (a^{2} x^{2}-\frac {8}{3} a x +\frac {37}{15}\right )}{8 \sqrt {-c \left (a x -1\right )}\, c^{\frac {9}{2}} \left (a x -1\right )^{2} a}\) | \(85\) |
Input:
int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(9/2),x,method=_RETURNVERBOSE)
Output:
2/c/a*(1/8/c^3/(-a*c*x+c)^(1/2)+1/12/c^2/(-a*c*x+c)^(3/2)+1/10/c/(-a*c*x+c )^(5/2)-1/16/c^(7/2)*2^(1/2)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2)) )
Time = 0.08 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.48 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\left [\frac {15 \, \sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {c} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) - 4 \, {\left (15 \, a^{2} x^{2} - 40 \, a x + 37\right )} \sqrt {-a c x + c}}{240 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}}, -\frac {15 \, \sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + 2 \, {\left (15 \, a^{2} x^{2} - 40 \, a x + 37\right )} \sqrt {-a c x + c}}{120 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}}\right ] \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(9/2),x, algorithm="fricas")
Output:
[1/240*(15*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(c)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) - 4*(15*a^2*x^2 - 40* a*x + 37)*sqrt(-a*c*x + c))/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a *c^5), -1/120*(15*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(-c)*arcta n(sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + 2*(15*a^2*x^2 - 40*a*x + 37)*sqrt(-a*c*x + c))/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a*c^5 )]
Time = 7.10 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\begin {cases} - \frac {2 \left (- \frac {1}{10 c \left (- a c x + c\right )^{\frac {5}{2}}} - \frac {1}{12 c^{2} \left (- a c x + c\right )^{\frac {3}{2}}} - \frac {1}{8 c^{3} \sqrt {- a c x + c}} - \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{16 c^{3} \sqrt {- c}}\right )}{a c} & \text {for}\: a c \neq 0 \\\frac {- x + 2 \left (\begin {cases} x & \text {for}\: a = 0 \\\frac {\log {\left (a x + 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{c^{\frac {9}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(a*x+1)**2*(-a**2*x**2+1)/(-a*c*x+c)**(9/2),x)
Output:
Piecewise((-2*(-1/(10*c*(-a*c*x + c)**(5/2)) - 1/(12*c**2*(-a*c*x + c)**(3 /2)) - 1/(8*c**3*sqrt(-a*c*x + c)) - sqrt(2)*atan(sqrt(2)*sqrt(-a*c*x + c) /(2*sqrt(-c)))/(16*c**3*sqrt(-c)))/(a*c), Ne(a*c, 0)), ((-x + 2*Piecewise( (x, Eq(a, 0)), (log(a*x + 1)/a, True)))/c**(9/2), True))
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\frac {\frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right )}{c^{\frac {7}{2}}} + \frac {4 \, {\left (15 \, {\left (a c x - c\right )}^{2} - 10 \, {\left (a c x - c\right )} c + 12 \, c^{2}\right )}}{{\left (-a c x + c\right )}^{\frac {5}{2}} c^{3}}}{240 \, a c} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(9/2),x, algorithm="maxima")
Output:
1/240*(15*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt( c) + sqrt(-a*c*x + c)))/c^(7/2) + 4*(15*(a*c*x - c)^2 - 10*(a*c*x - c)*c + 12*c^2)/((-a*c*x + c)^(5/2)*c^3))/(a*c)
Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{8 \, a \sqrt {-c} c^{4}} + \frac {15 \, {\left (a c x - c\right )}^{2} - 10 \, {\left (a c x - c\right )} c + 12 \, c^{2}}{60 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a c^{4}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(9/2),x, algorithm="giac")
Output:
1/8*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)*c^4) + 1/60*(15*(a*c*x - c)^2 - 10*(a*c*x - c)*c + 12*c^2)/((a*c*x - c)^2*sqrt (-a*c*x + c)*a*c^4)
Time = 22.72 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\frac {\frac {c-a\,c\,x}{6\,c^2}+\frac {1}{5\,c}+\frac {{\left (c-a\,c\,x\right )}^2}{4\,c^3}}{a\,c\,{\left (c-a\,c\,x\right )}^{5/2}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right )}{8\,a\,c^{9/2}} \] Input:
int(-(a^2*x^2 - 1)/((c - a*c*x)^(9/2)*(a*x + 1)^2),x)
Output:
((c - a*c*x)/(6*c^2) + 1/(5*c) + (c - a*c*x)^2/(4*c^3))/(a*c*(c - a*c*x)^( 5/2)) - (2^(1/2)*atanh((2^(1/2)*(c - a*c*x)^(1/2))/(2*c^(1/2))))/(8*a*c^(9 /2))
Time = 0.16 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.94 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{(c-a c x)^{9/2}} \, dx=\frac {\sqrt {c}\, \left (15 \sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}-30 \sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +15 \sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )-15 \sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}+30 \sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x -15 \sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )+60 a^{2} x^{2}-160 a x +148\right )}{240 \sqrt {-a x +1}\, a \,c^{5} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a*c*x+c)^(9/2),x)
Output:
(sqrt(c)*(15*sqrt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) - sqrt(2))*a**2 *x**2 - 30*sqrt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) - sqrt(2))*a*x + 15*sqrt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) - sqrt(2)) - 15*sqrt( - a *x + 1)*sqrt(2)*log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 + 30*sqrt( - a*x + 1)*sqrt(2)*log(sqrt( - a*x + 1) + sqrt(2))*a*x - 15*sqrt( - a*x + 1)*sq rt(2)*log(sqrt( - a*x + 1) + sqrt(2)) + 60*a**2*x**2 - 160*a*x + 148))/(24 0*sqrt( - a*x + 1)*a*c**5*(a**2*x**2 - 2*a*x + 1))