Integrand size = 21, antiderivative size = 97 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {4 \sqrt {c-a c x}}{a^2}+\frac {2 (c-a c x)^{3/2}}{3 a^2 c}+\frac {2 (c-a c x)^{5/2}}{5 a^2 c^2}-\frac {4 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a^2} \] Output:
4*(-a*c*x+c)^(1/2)/a^2+2/3*(-a*c*x+c)^(3/2)/a^2/c+2/5*(-a*c*x+c)^(5/2)/a^2 /c^2-4*2^(1/2)*c^(1/2)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))/a^2
Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c-a c x} \left (38-11 a x+3 a^2 x^2\right )-60 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{15 a^2} \] Input:
Integrate[(x*Sqrt[c - a*c*x])/E^(2*ArcCoth[a*x]),x]
Output:
(2*Sqrt[c - a*c*x]*(38 - 11*a*x + 3*a^2*x^2) - 60*Sqrt[2]*Sqrt[c]*ArcTanh[ Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/(15*a^2)
Time = 0.67 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6717, 6680, 35, 90, 60, 60, 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.
\(\displaystyle \int x \sqrt {c-a c x} e^{-2 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int e^{-2 \text {arctanh}(a x)} x \sqrt {c-a c x}dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle -\int \frac {x (1-a x) \sqrt {c-a c x}}{a x+1}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle -\frac {\int \frac {x (c-a c x)^{3/2}}{a x+1}dx}{c}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle -\frac {-\frac {\int \frac {(c-a c x)^{3/2}}{a x+1}dx}{a}-\frac {2 (c-a c x)^{5/2}}{5 a^2 c}}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {-\frac {2 c \int \frac {\sqrt {c-a c x}}{a x+1}dx+\frac {2 (c-a c x)^{3/2}}{3 a}}{a}-\frac {2 (c-a c x)^{5/2}}{5 a^2 c}}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {-\frac {2 c \left (2 c \int \frac {1}{(a x+1) \sqrt {c-a c x}}dx+\frac {2 \sqrt {c-a c x}}{a}\right )+\frac {2 (c-a c x)^{3/2}}{3 a}}{a}-\frac {2 (c-a c x)^{5/2}}{5 a^2 c}}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {-\frac {2 c \left (\frac {2 \sqrt {c-a c x}}{a}-\frac {4 \int \frac {1}{2-\frac {c-a c x}{c}}d\sqrt {c-a c x}}{a}\right )+\frac {2 (c-a c x)^{3/2}}{3 a}}{a}-\frac {2 (c-a c x)^{5/2}}{5 a^2 c}}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {-\frac {2 (c-a c x)^{5/2}}{5 a^2 c}-\frac {2 c \left (\frac {2 \sqrt {c-a c x}}{a}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}\right )+\frac {2 (c-a c x)^{3/2}}{3 a}}{a}}{c}\) |
Input:
Int[(x*Sqrt[c - a*c*x])/E^(2*ArcCoth[a*x]),x]
Output:
-(((-2*(c - a*c*x)^(5/2))/(5*a^2*c) - ((2*(c - a*c*x)^(3/2))/(3*a) + 2*c*( (2*Sqrt[c - a*c*x])/a - (2*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2 ]*Sqrt[c])])/a))/a)/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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(\frac {-60 \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+\left (6 a^{2} x^{2}-22 a x +76\right ) \sqrt {-c \left (a x -1\right )}}{15 a^{2}}\) | \(59\) |
risch | \(-\frac {2 \left (3 a^{2} x^{2}-11 a x +38\right ) \left (a x -1\right ) c}{15 a^{2} \sqrt {-c \left (a x -1\right )}}-\frac {4 \sqrt {2}\, \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a^{2}}\) | \(66\) |
derivativedivides | \(\frac {\frac {2 \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-a c x +c \right )^{\frac {3}{2}} c}{3}+4 \sqrt {-a c x +c}\, c^{2}-4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a^{2} c^{2}}\) | \(73\) |
default | \(\frac {\frac {2 \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-a c x +c \right )^{\frac {3}{2}} c}{3}+4 \sqrt {-a c x +c}\, c^{2}-4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a^{2} c^{2}}\) | \(73\) |
Input:
int(x*(-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1),x,method=_RETURNVERBOSE)
Output:
1/15*(-60*c^(1/2)*2^(1/2)*arctanh(1/2*(-c*(a*x-1))^(1/2)*2^(1/2)/c^(1/2))+ (6*a^2*x^2-22*a*x+76)*(-c*(a*x-1))^(1/2))/a^2
Time = 0.10 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.47 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\left [\frac {2 \, {\left (15 \, \sqrt {2} \sqrt {c} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + {\left (3 \, a^{2} x^{2} - 11 \, a x + 38\right )} \sqrt {-a c x + c}\right )}}{15 \, a^{2}}, -\frac {2 \, {\left (30 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) - {\left (3 \, a^{2} x^{2} - 11 \, a x + 38\right )} \sqrt {-a c x + c}\right )}}{15 \, a^{2}}\right ] \] Input:
integrate(x*(-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1),x, algorithm="fricas")
Output:
[2/15*(15*sqrt(2)*sqrt(c)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (3*a^2*x^2 - 11*a*x + 38)*sqrt(-a*c*x + c))/a^2, -2/15 *(30*sqrt(2)*sqrt(-c)*arctan(sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c) ) - (3*a^2*x^2 - 11*a*x + 38)*sqrt(-a*c*x + c))/a^2]
Time = 3.00 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\begin {cases} \frac {2 \cdot \left (\frac {2 \sqrt {2} c^{3} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} + 2 c^{2} \sqrt {- a c x + c} + \frac {c \left (- a c x + c\right )^{\frac {3}{2}}}{3} + \frac {\left (- a c x + c\right )^{\frac {5}{2}}}{5}\right )}{a^{2} c^{2}} & \text {for}\: a c \neq 0 \\\sqrt {c} \left (\frac {x^{2}}{2} - \frac {2 x}{a} + \frac {2 \left (\begin {cases} x & \text {for}\: a = 0 \\\frac {\log {\left (a x + 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{a}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x*(-a*c*x+c)**(1/2)*(a*x-1)/(a*x+1),x)
Output:
Piecewise((2*(2*sqrt(2)*c**3*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/s qrt(-c) + 2*c**2*sqrt(-a*c*x + c) + c*(-a*c*x + c)**(3/2)/3 + (-a*c*x + c) **(5/2)/5)/(a**2*c**2), Ne(a*c, 0)), (sqrt(c)*(x**2/2 - 2*x/a + 2*Piecewis e((x, Eq(a, 0)), (log(a*x + 1)/a, True))/a), True))
Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 \, {\left (15 \, \sqrt {2} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 3 \, {\left (-a c x + c\right )}^{\frac {5}{2}} + 5 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 30 \, \sqrt {-a c x + c} c^{2}\right )}}{15 \, a^{2} c^{2}} \] Input:
integrate(x*(-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1),x, algorithm="maxima")
Output:
2/15*(15*sqrt(2)*c^(5/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2 )*sqrt(c) + sqrt(-a*c*x + c))) + 3*(-a*c*x + c)^(5/2) + 5*(-a*c*x + c)^(3/ 2)*c + 30*sqrt(-a*c*x + c)*c^2)/(a^2*c^2)
Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {4 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {2 \, {\left (3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{8} c^{8} + 5 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{8} c^{9} + 30 \, \sqrt {-a c x + c} a^{8} c^{10}\right )}}{15 \, a^{10} c^{10}} \] Input:
integrate(x*(-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1),x, algorithm="giac")
Output:
4*sqrt(2)*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a^2*sqrt(-c)) + 2/15*(3*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^8*c^8 + 5*(-a*c*x + c)^(3/2)*a^8 *c^9 + 30*sqrt(-a*c*x + c)*a^8*c^10)/(a^10*c^10)
Time = 13.66 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {4\,\sqrt {c-a\,c\,x}}{a^2}+\frac {2\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a^2\,c}+\frac {2\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a^2\,c^2}+\frac {\sqrt {2}\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,4{}\mathrm {i}}{a^2} \] Input:
int((x*(c - a*c*x)^(1/2)*(a*x - 1))/(a*x + 1),x)
Output:
(4*(c - a*c*x)^(1/2))/a^2 + (2*(c - a*c*x)^(3/2))/(3*a^2*c) + (2*(c - a*c* x)^(5/2))/(5*a^2*c^2) + (2^(1/2)*c^(1/2)*atan((2^(1/2)*(c - a*c*x)^(1/2)*1 i)/(2*c^(1/2)))*4i)/a^2
Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int e^{-2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c}\, \left (3 \sqrt {-a x +1}\, a^{2} x^{2}-11 \sqrt {-a x +1}\, a x +38 \sqrt {-a x +1}+15 \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )-15 \sqrt {2}\, \mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )\right )}{15 a^{2}} \] Input:
int(x*(-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1),x)
Output:
(2*sqrt(c)*(3*sqrt( - a*x + 1)*a**2*x**2 - 11*sqrt( - a*x + 1)*a*x + 38*sq rt( - a*x + 1) + 15*sqrt(2)*log(sqrt( - a*x + 1) - sqrt(2)) - 15*sqrt(2)*l og(sqrt( - a*x + 1) + sqrt(2))))/(15*a**2)