Integrand size = 18, antiderivative size = 115 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {64 a^2 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{105 (c-a c x)^{3/2}}+\frac {16 a^2 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{35 \sqrt {c-a c x}}+\frac {2}{7} a^2 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \sqrt {c-a c x} \]
64/105*a^2*c^4*(1-1/a^2/x^2)^(3/2)*x^3/(-a*c*x+c)^(3/2)+16/35*a^2*c^3*(1-1 /a^2/x^2)^(3/2)*x^3/(-a*c*x+c)^(1/2)+2/7*a^2*c^2*(1-1/a^2/x^2)^(3/2)*x^3*( -a*c*x+c)^(1/2)
Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.61 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 c^2 \sqrt {1+\frac {1}{a x}} \sqrt {c-a c x} \left (71+17 a x-39 a^2 x^2+15 a^3 x^3\right )}{105 a \sqrt {1-\frac {1}{a x}}} \]
(2*c^2*Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x]*(71 + 17*a*x - 39*a^2*x^2 + 15*a^ 3*x^3))/(105*a*Sqrt[1 - 1/(a*x)])
Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6727, 27, 100, 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.
\(\displaystyle \int (c-a c x)^{5/2} e^{\coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6727 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \int \frac {\left (a-\frac {1}{x}\right )^2 \sqrt {1+\frac {1}{a x}}}{a^2 \left (\frac {1}{x}\right )^{9/2}}d\frac {1}{x}}{\left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \int \frac {\left (a-\frac {1}{x}\right )^2 \sqrt {1+\frac {1}{a x}}}{\left (\frac {1}{x}\right )^{9/2}}d\frac {1}{x}}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \left (\frac {2}{7} \int -\frac {\left (18 a-\frac {7}{x}\right ) \sqrt {1+\frac {1}{a x}}}{2 \left (\frac {1}{x}\right )^{7/2}}d\frac {1}{x}-\frac {2 a^2 \left (\frac {1}{a x}+1\right )^{3/2}}{7 \left (\frac {1}{x}\right )^{7/2}}\right )}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \left (-\frac {1}{7} \int \frac {\left (18 a-\frac {7}{x}\right ) \sqrt {1+\frac {1}{a x}}}{\left (\frac {1}{x}\right )^{7/2}}d\frac {1}{x}-\frac {2 a^2 \left (\frac {1}{a x}+1\right )^{3/2}}{7 \left (\frac {1}{x}\right )^{7/2}}\right )}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2} \left (\frac {1}{7} \left (\frac {71}{5} \int \frac {\sqrt {1+\frac {1}{a x}}}{\left (\frac {1}{x}\right )^{5/2}}d\frac {1}{x}+\frac {36 a \left (\frac {1}{a x}+1\right )^{3/2}}{5 \left (\frac {1}{x}\right )^{5/2}}\right )-\frac {2 a^2 \left (\frac {1}{a x}+1\right )^{3/2}}{7 \left (\frac {1}{x}\right )^{7/2}}\right )}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{5/2} \left (\frac {1}{7} \left (\frac {36 a \left (\frac {1}{a x}+1\right )^{3/2}}{5 \left (\frac {1}{x}\right )^{5/2}}-\frac {142 \left (\frac {1}{a x}+1\right )^{3/2}}{15 \left (\frac {1}{x}\right )^{3/2}}\right )-\frac {2 a^2 \left (\frac {1}{a x}+1\right )^{3/2}}{7 \left (\frac {1}{x}\right )^{7/2}}\right ) (c-a c x)^{5/2}}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\) |
-(((((36*a*(1 + 1/(a*x))^(3/2))/(5*(x^(-1))^(5/2)) - (142*(1 + 1/(a*x))^(3 /2))/(15*(x^(-1))^(3/2)))/7 - (2*a^2*(1 + 1/(a*x))^(3/2))/(7*(x^(-1))^(7/2 )))*(x^(-1))^(5/2)*(c - a*c*x)^(5/2))/(a^2*(1 - 1/(a*x))^(5/2)))
3.3.28.3.1 Defintions of rubi rules used
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[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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 = 0.41 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.46
method | result | size |
default | \(\frac {2 \sqrt {-c \left (a x -1\right )}\, c^{2} \left (a x +1\right ) \left (15 a^{2} x^{2}-54 a x +71\right )}{105 \sqrt {\frac {a x -1}{a x +1}}\, a}\) | \(53\) |
gosper | \(\frac {2 \left (a x +1\right ) \left (15 a^{2} x^{2}-54 a x +71\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{105 a \left (a x -1\right )^{2} \sqrt {\frac {a x -1}{a x +1}}}\) | \(56\) |
risch | \(-\frac {2 c^{3} \left (a x -1\right ) \left (15 a^{3} x^{3}-39 a^{2} x^{2}+17 a x +71\right )}{105 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x -1\right )}\, a}\) | \(61\) |
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (15 \, a^{4} c^{2} x^{4} - 24 \, a^{3} c^{2} x^{3} - 22 \, a^{2} c^{2} x^{2} + 88 \, a c^{2} x + 71 \, c^{2}\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{105 \, {\left (a^{2} x - a\right )}} \]
2/105*(15*a^4*c^2*x^4 - 24*a^3*c^2*x^3 - 22*a^2*c^2*x^2 + 88*a*c^2*x + 71* c^2)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^2*x - a)
Timed out. \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.58 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (15 \, a^{3} \sqrt {-c} c^{2} x^{3} - 39 \, a^{2} \sqrt {-c} c^{2} x^{2} + 17 \, a \sqrt {-c} c^{2} x + 71 \, \sqrt {-c} c^{2}\right )} \sqrt {a x + 1}}{105 \, a} \]
2/105*(15*a^3*sqrt(-c)*c^2*x^3 - 39*a^2*sqrt(-c)*c^2*x^2 + 17*a*sqrt(-c)*c ^2*x + 71*sqrt(-c)*c^2)*sqrt(a*x + 1)/a
Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (64 \, \sqrt {2} \sqrt {-c} c - \frac {15 \, {\left (a c x + c\right )}^{3} \sqrt {-a c x - c} - 84 \, {\left (a c x + c\right )}^{2} \sqrt {-a c x - c} c - 140 \, {\left (-a c x - c\right )}^{\frac {3}{2}} c^{2}}{c^{2}}\right )} c^{2}}{105 \, a {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )} \]
2/105*(64*sqrt(2)*sqrt(-c)*c - (15*(a*c*x + c)^3*sqrt(-a*c*x - c) - 84*(a* c*x + c)^2*sqrt(-a*c*x - c)*c - 140*(-a*c*x - c)^(3/2)*c^2)/c^2)*c^2/(a*ab s(c)*sgn(a*x + 1))
Time = 4.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.52 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2\,c^2\,\sqrt {c-a\,c\,x}\,{\left (a\,x+1\right )}^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (15\,a^2\,x^2-54\,a\,x+71\right )}{105\,a\,\left (a\,x-1\right )} \]