Integrand size = 20, antiderivative size = 89 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {18 \left (1+\frac {1}{a x}\right )^{5/2} (c-a c x)^{5/2}}{35 a \left (1-\frac {1}{a x}\right )^{5/2}}+\frac {2 \left (1+\frac {1}{a x}\right )^{5/2} x (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}} \] Output:
-18/35*(1+1/a/x)^(5/2)*(-a*c*x+c)^(5/2)/a/(1-1/a/x)^(5/2)+2/7*(1+1/a/x)^(5 /2)*x*(-a*c*x+c)^(5/2)/(1-1/a/x)^(5/2)
Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \sqrt {1+\frac {1}{a x}} (-9+5 a x) \sqrt {c-a c x} (c+a c x)^2}{35 a \sqrt {1-\frac {1}{a x}}} \] Input:
Integrate[E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]
Output:
(2*Sqrt[1 + 1/(a*x)]*(-9 + 5*a*x)*Sqrt[c - a*c*x]*(c + a*c*x)^2)/(35*a*Sqr t[1 - 1/(a*x)])
Time = 0.41 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6727, 27, 87, 48}
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.
\(\displaystyle \int (c-a c x)^{5/2} e^{3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6727 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \int \frac {\left (a-\frac {1}{x}\right ) \left (1+\frac {1}{a x}\right )^{3/2}}{a \left (\frac {1}{x}\right )^{9/2}}d\frac {1}{x}}{\left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \int \frac {\left (a-\frac {1}{x}\right ) \left (1+\frac {1}{a x}\right )^{3/2}}{\left (\frac {1}{x}\right )^{9/2}}d\frac {1}{x}}{a \left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \left (-\frac {9}{7} \int \frac {\left (1+\frac {1}{a x}\right )^{3/2}}{\left (\frac {1}{x}\right )^{7/2}}d\frac {1}{x}-\frac {2 a \left (\frac {1}{a x}+1\right )^{5/2}}{7 \left (\frac {1}{x}\right )^{7/2}}\right )}{a \left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} \left (\frac {18 \left (\frac {1}{a x}+1\right )^{5/2}}{35 \left (\frac {1}{x}\right )^{5/2}}-\frac {2 a \left (\frac {1}{a x}+1\right )^{5/2}}{7 \left (\frac {1}{x}\right )^{7/2}}\right ) (c-a c x)^{5/2}}{a \left (1-\frac {1}{a x}\right )^{5/2}}\) |
Input:
Int[E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]
Output:
-((((-2*a*(1 + 1/(a*x))^(5/2))/(7*(x^(-1))^(7/2)) + (18*(1 + 1/(a*x))^(5/2 ))/(35*(x^(-1))^(5/2)))*(x^(-1))^(5/2)*(c - a*c*x)^(5/2))/(a*(1 - 1/(a*x)) ^(5/2)))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Si mp[(-(1/x)^p)*((c + d*x)^p/(1 + c/(d*x))^p) Subst[Int[((1 + c*(x/d))^p*(( 1 + x/a)^(n/2)/x^(p + 2)))/(1 - x/a)^(n/2), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && !IntegerQ[p]
Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(\frac {2 \left (a x +1\right ) \left (5 a x -9\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{35 a \left (a x -1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(48\) |
orering | \(\frac {2 \left (a x +1\right ) \left (5 a x -9\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{35 a \left (a x -1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(48\) |
default | \(\frac {2 \left (a x -1\right ) \left (a x +1\right ) \sqrt {-c \left (a x -1\right )}\, c^{2} \left (5 a x -9\right )}{35 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\) | \(50\) |
risch | \(-\frac {2 c^{3} \left (a x -1\right ) \left (5 a^{3} x^{3}+a^{2} x^{2}-13 a x -9\right )}{35 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x -1\right )}\, a}\) | \(60\) |
Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/35*(a*x+1)*(5*a*x-9)*(-a*c*x+c)^(5/2)/a/(a*x-1)/((a*x-1)/(a*x+1))^(3/2)
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (5 \, a^{4} c^{2} x^{4} + 6 \, a^{3} c^{2} x^{3} - 12 \, a^{2} c^{2} x^{2} - 22 \, a c^{2} x - 9 \, c^{2}\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{2} x - a\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="fricas" )
Output:
2/35*(5*a^4*c^2*x^4 + 6*a^3*c^2*x^3 - 12*a^2*c^2*x^2 - 22*a*c^2*x - 9*c^2) *sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^2*x - a)
Timed out. \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(3/2)*(-a*c*x+c)**(5/2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (5 \, a^{3} \sqrt {-c} c^{2} x^{3} - 9 \, a^{2} \sqrt {-c} c^{2} x^{2} - 5 \, a \sqrt {-c} c^{2} x + 9 \, \sqrt {-c} c^{2}\right )} {\left (a x + 1\right )}^{\frac {3}{2}}}{35 \, {\left (a x - 1\right )} a} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="maxima" )
Output:
2/35*(5*a^3*sqrt(-c)*c^2*x^3 - 9*a^2*sqrt(-c)*c^2*x^2 - 5*a*sqrt(-c)*c^2*x + 9*sqrt(-c)*c^2)*(a*x + 1)^(3/2)/((a*x - 1)*a)
Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2 \, {\left (16 \, \sqrt {2} \sqrt {-c} c + \frac {5 \, {\left (a c x + c\right )}^{3} \sqrt {-a c x - c} - 14 \, {\left (a c x + c\right )}^{2} \sqrt {-a c x - c} c}{c^{2}}\right )} c^{2}}{35 \, a {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="giac")
Output:
-2/35*(16*sqrt(2)*sqrt(-c)*c + (5*(a*c*x + c)^3*sqrt(-a*c*x - c) - 14*(a*c *x + c)^2*sqrt(-a*c*x - c)*c)/c^2)*c^2/(a*abs(c)*sgn(a*x + 1))
Time = 14.01 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2\,c^2\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (-5\,a^3\,x^3-11\,a^2\,x^2+a\,x+23\right )}{35\,a}-\frac {64\,c^2\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{35\,a\,\left (a\,x-1\right )} \] Input:
int((c - a*c*x)^(5/2)/((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
- (2*c^2*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(a*x - 11*a^2*x^2 - 5*a^3*x^3 + 23))/(35*a) - (64*c^2*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1)) ^(1/2))/(35*a*(a*x - 1))
Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.44 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \sqrt {c}\, \sqrt {a x +1}\, c^{2} i \left (-5 a^{3} x^{3}-a^{2} x^{2}+13 a x +9\right )}{35 a} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(5/2),x)
Output:
(2*sqrt(c)*sqrt(a*x + 1)*c**2*i*( - 5*a**3*x**3 - a**2*x**2 + 13*a*x + 9)) /(35*a)