Integrand size = 23, antiderivative size = 101 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=-\frac {4 \sqrt {c-a c x}}{a^4}+\frac {14 (c-a c x)^{3/2}}{3 a^4 c}-\frac {18 (c-a c x)^{5/2}}{5 a^4 c^2}+\frac {10 (c-a c x)^{7/2}}{7 a^4 c^3}-\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4} \] Output:
-4*(-a*c*x+c)^(1/2)/a^4+14/3*(-a*c*x+c)^(3/2)/a^4/c-18/5*(-a*c*x+c)^(5/2)/ a^4/c^2+10/7*(-a*c*x+c)^(7/2)/a^4/c^3-2/9*(-a*c*x+c)^(9/2)/a^4/c^4
Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.48 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=-\frac {2 \sqrt {c-a c x} \left (272+136 a x+102 a^2 x^2+85 a^3 x^3+35 a^4 x^4\right )}{315 a^4} \] Input:
Integrate[E^(2*ArcTanh[a*x])*x^3*Sqrt[c - a*c*x],x]
Output:
(-2*Sqrt[c - a*c*x]*(272 + 136*a*x + 102*a^2*x^2 + 85*a^3*x^3 + 35*a^4*x^4 ))/(315*a^4)
Time = 0.37 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6680, 35, 86, 2009}
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.
\(\displaystyle \int x^3 e^{2 \text {arctanh}(a x)} \sqrt {c-a c x} \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {x^3 (a x+1) \sqrt {c-a c x}}{1-a x}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle c \int \frac {x^3 (a x+1)}{\sqrt {c-a c x}}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle c \int \left (\frac {(c-a c x)^{7/2}}{a^3 c^4}-\frac {5 (c-a c x)^{5/2}}{a^3 c^3}+\frac {9 (c-a c x)^{3/2}}{a^3 c^2}-\frac {7 \sqrt {c-a c x}}{a^3 c}+\frac {2}{a^3 \sqrt {c-a c x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c \left (-\frac {2 (c-a c x)^{9/2}}{9 a^4 c^5}+\frac {10 (c-a c x)^{7/2}}{7 a^4 c^4}-\frac {18 (c-a c x)^{5/2}}{5 a^4 c^3}+\frac {14 (c-a c x)^{3/2}}{3 a^4 c^2}-\frac {4 \sqrt {c-a c x}}{a^4 c}\right )\) |
Input:
Int[E^(2*ArcTanh[a*x])*x^3*Sqrt[c - a*c*x],x]
Output:
c*((-4*Sqrt[c - a*c*x])/(a^4*c) + (14*(c - a*c*x)^(3/2))/(3*a^4*c^2) - (18 *(c - a*c*x)^(5/2))/(5*a^4*c^3) + (10*(c - a*c*x)^(7/2))/(7*a^4*c^4) - (2* (c - a*c*x)^(9/2))/(9*a^4*c^5))
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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.45
method | result | size |
gosper | \(-\frac {2 \sqrt {-a c x +c}\, \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right )}{315 a^{4}}\) | \(45\) |
trager | \(-\frac {2 \sqrt {-a c x +c}\, \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right )}{315 a^{4}}\) | \(45\) |
pseudoelliptic | \(-\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right )}{315 a^{4}}\) | \(46\) |
risch | \(\frac {2 c \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right ) \left (a x -1\right )}{315 a^{4} \sqrt {-c \left (a x -1\right )}}\) | \(52\) |
orering | \(\frac {2 \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right ) \left (a x -1\right ) \left (a x +1\right ) \sqrt {-a c x +c}}{315 a^{4} \left (-a^{2} x^{2}+1\right )}\) | \(67\) |
derivativedivides | \(-\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {9}{2}}}{9}-\frac {5 c \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {9 c^{2} \left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {7 c^{3} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+2 \sqrt {-a c x +c}\, c^{4}\right )}{c^{4} a^{4}}\) | \(75\) |
default | \(-\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {9}{2}}}{9}-\frac {5 c \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {9 c^{2} \left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {7 c^{3} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+2 \sqrt {-a c x +c}\, c^{4}\right )}{c^{4} a^{4}}\) | \(75\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a*c*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/315*(-a*c*x+c)^(1/2)*(35*a^4*x^4+85*a^3*x^3+102*a^2*x^2+136*a*x+272)/a^ 4
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.44 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=-\frac {2 \, {\left (35 \, a^{4} x^{4} + 85 \, a^{3} x^{3} + 102 \, a^{2} x^{2} + 136 \, a x + 272\right )} \sqrt {-a c x + c}}{315 \, a^{4}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a*c*x+c)^(1/2),x, algorithm="fricas ")
Output:
-2/315*(35*a^4*x^4 + 85*a^3*x^3 + 102*a^2*x^2 + 136*a*x + 272)*sqrt(-a*c*x + c)/a^4
Time = 4.43 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.35 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=\begin {cases} \frac {2 \left (- 2 c^{4} \sqrt {- a c x + c} + \frac {7 c^{3} \left (- a c x + c\right )^{\frac {3}{2}}}{3} - \frac {9 c^{2} \left (- a c x + c\right )^{\frac {5}{2}}}{5} + \frac {5 c \left (- a c x + c\right )^{\frac {7}{2}}}{7} - \frac {\left (- a c x + c\right )^{\frac {9}{2}}}{9}\right )}{a^{4} c^{4}} & \text {for}\: a c \neq 0 \\\sqrt {c} \left (- \frac {x^{4}}{4} - \frac {2 x^{3}}{3 a} - \frac {x^{2}}{a^{2}} - \frac {2 x}{a^{3}} - \frac {2 \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {\log {\left (a x - 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{a^{3}}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*x**3*(-a*c*x+c)**(1/2),x)
Output:
Piecewise((2*(-2*c**4*sqrt(-a*c*x + c) + 7*c**3*(-a*c*x + c)**(3/2)/3 - 9* c**2*(-a*c*x + c)**(5/2)/5 + 5*c*(-a*c*x + c)**(7/2)/7 - (-a*c*x + c)**(9/ 2)/9)/(a**4*c**4), Ne(a*c, 0)), (sqrt(c)*(-x**4/4 - 2*x**3/(3*a) - x**2/a* *2 - 2*x/a**3 - 2*Piecewise((-x, Eq(a, 0)), (log(a*x - 1)/a, True))/a**3), True))
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.73 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=-\frac {2 \, {\left (35 \, {\left (-a c x + c\right )}^{\frac {9}{2}} - 225 \, {\left (-a c x + c\right )}^{\frac {7}{2}} c + 567 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c^{2} - 735 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{3} + 630 \, \sqrt {-a c x + c} c^{4}\right )}}{315 \, a^{4} c^{4}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a*c*x+c)^(1/2),x, algorithm="maxima ")
Output:
-2/315*(35*(-a*c*x + c)^(9/2) - 225*(-a*c*x + c)^(7/2)*c + 567*(-a*c*x + c )^(5/2)*c^2 - 735*(-a*c*x + c)^(3/2)*c^3 + 630*sqrt(-a*c*x + c)*c^4)/(a^4* c^4)
Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (83) = 166\).
Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.87 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=-\frac {2 \, {\left (\frac {9 \, {\left (5 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} + 21 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} c - 35 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-a c x + c} c^{3}\right )}}{a^{3} c^{3}} + \frac {35 \, {\left (a c x - c\right )}^{4} \sqrt {-a c x + c} + 180 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} c + 378 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} c^{2} - 420 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {-a c x + c} c^{4}}{a^{3} c^{4}}\right )}}{315 \, a} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a*c*x+c)^(1/2),x, algorithm="giac")
Output:
-2/315*(9*(5*(a*c*x - c)^3*sqrt(-a*c*x + c) + 21*(a*c*x - c)^2*sqrt(-a*c*x + c)*c - 35*(-a*c*x + c)^(3/2)*c^2 + 35*sqrt(-a*c*x + c)*c^3)/(a^3*c^3) + (35*(a*c*x - c)^4*sqrt(-a*c*x + c) + 180*(a*c*x - c)^3*sqrt(-a*c*x + c)*c + 378*(a*c*x - c)^2*sqrt(-a*c*x + c)*c^2 - 420*(-a*c*x + c)^(3/2)*c^3 + 3 15*sqrt(-a*c*x + c)*c^4)/(a^3*c^4))/a
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {14\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a^4\,c}-\frac {4\,\sqrt {c-a\,c\,x}}{a^4}-\frac {18\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a^4\,c^2}+\frac {10\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a^4\,c^3}-\frac {2\,{\left (c-a\,c\,x\right )}^{9/2}}{9\,a^4\,c^4} \] Input:
int(-(x^3*(c - a*c*x)^(1/2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)
Output:
(14*(c - a*c*x)^(3/2))/(3*a^4*c) - (4*(c - a*c*x)^(1/2))/a^4 - (18*(c - a* c*x)^(5/2))/(5*a^4*c^2) + (10*(c - a*c*x)^(7/2))/(7*a^4*c^3) - (2*(c - a*c *x)^(9/2))/(9*a^4*c^4)
Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.44 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c}\, \sqrt {-a x +1}\, \left (-35 a^{4} x^{4}-85 a^{3} x^{3}-102 a^{2} x^{2}-136 a x -272\right )}{315 a^{4}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^3*(-a*c*x+c)^(1/2),x)
Output:
(2*sqrt(c)*sqrt( - a*x + 1)*( - 35*a**4*x**4 - 85*a**3*x**3 - 102*a**2*x** 2 - 136*a*x - 272))/(315*a**4)