Integrand size = 18, antiderivative size = 106 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=\frac {64 c^4 \left (1-a^2 x^2\right )^{3/2}}{105 a (c-a c x)^{3/2}}+\frac {16 c^3 \left (1-a^2 x^2\right )^{3/2}}{35 a \sqrt {c-a c x}}+\frac {2 c^2 \sqrt {c-a c x} \left (1-a^2 x^2\right )^{3/2}}{7 a} \] Output:
64/105*c^4*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(3/2)+16/35*c^3*(-a^2*x^2+1)^(3 /2)/a/(-a*c*x+c)^(1/2)+2/7*c^2*(-a*c*x+c)^(1/2)*(-a^2*x^2+1)^(3/2)/a
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.51 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 c^2 (1+a x)^{3/2} \sqrt {c-a c x} \left (71-54 a x+15 a^2 x^2\right )}{105 a \sqrt {1-a x}} \] Input:
Integrate[E^ArcTanh[a*x]*(c - a*c*x)^(5/2),x]
Output:
(2*c^2*(1 + a*x)^(3/2)*Sqrt[c - a*c*x]*(71 - 54*a*x + 15*a^2*x^2))/(105*a* Sqrt[1 - a*x])
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6677, 459, 459, 458}
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.
\(\displaystyle \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle c \int (c-a c x)^{3/2} \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 459 |
\(\displaystyle c \left (\frac {8}{7} c \int \sqrt {c-a c x} \sqrt {1-a^2 x^2}dx+\frac {2 c \left (1-a^2 x^2\right )^{3/2} \sqrt {c-a c x}}{7 a}\right )\) |
\(\Big \downarrow \) 459 |
\(\displaystyle c \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}dx+\frac {2 c \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt {c-a c x}}\right )+\frac {2 c \left (1-a^2 x^2\right )^{3/2} \sqrt {c-a c x}}{7 a}\right )\) |
\(\Big \downarrow \) 458 |
\(\displaystyle c \left (\frac {8}{7} c \left (\frac {8 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 a (c-a c x)^{3/2}}+\frac {2 c \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt {c-a c x}}\right )+\frac {2 c \left (1-a^2 x^2\right )^{3/2} \sqrt {c-a c x}}{7 a}\right )\) |
Input:
Int[E^ArcTanh[a*x]*(c - a*c*x)^(5/2),x]
Output:
c*((2*c*Sqrt[c - a*c*x]*(1 - a^2*x^2)^(3/2))/(7*a) + (8*c*((8*c^2*(1 - a^2 *x^2)^(3/2))/(15*a*(c - a*c*x)^(3/2)) + (2*c*(1 - a^2*x^2)^(3/2))/(5*a*Sqr t[c - a*c*x])))/7)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* (Simplify[n + p]/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[Simplif y[n + p], 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(\frac {2 \left (a x +1\right )^{2} \left (15 a^{2} x^{2}-54 a x +71\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{105 a \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}\) | \(55\) |
orering | \(\frac {2 \left (a x +1\right )^{2} \left (15 a^{2} x^{2}-54 a x +71\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{105 a \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}\) | \(55\) |
default | \(-\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, c^{2} \left (a x +1\right ) \left (15 a^{2} x^{2}-54 a x +71\right )}{105 \left (a x -1\right ) a}\) | \(57\) |
risch | \(-\frac {2 \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c^{3} \left (15 a^{3} x^{3}-39 a^{2} x^{2}+17 a x +71\right ) \left (a x +1\right )}{105 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, a \sqrt {c \left (a x +1\right )}}\) | \(94\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/105*(a*x+1)^2*(15*a^2*x^2-54*a*x+71)*(-a*c*x+c)^(5/2)/a/(a*x-1)^2/(-a^2* x^2+1)^(1/2)
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.65 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2 \, {\left (15 \, a^{3} c^{2} x^{3} - 39 \, a^{2} c^{2} x^{2} + 17 \, a c^{2} x + 71 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{105 \, {\left (a^{2} x - a\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(5/2),x, algorithm="fricas ")
Output:
-2/105*(15*a^3*c^2*x^3 - 39*a^2*c^2*x^2 + 17*a*c^2*x + 71*c^2)*sqrt(-a^2*x ^2 + 1)*sqrt(-a*c*x + c)/(a^2*x - a)
\[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=\int \frac {\left (- c \left (a x - 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)**(5/2),x)
Output:
Integral((-c*(a*x - 1))**(5/2)*(a*x + 1)/sqrt(-(a*x - 1)*(a*x + 1)), x)
Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (3 \, a^{4} c^{\frac {5}{2}} x^{4} - 9 \, a^{3} c^{\frac {5}{2}} x^{3} + 11 \, a^{2} c^{\frac {5}{2}} x^{2} - 23 \, a c^{\frac {5}{2}} x - 46 \, c^{\frac {5}{2}}\right )}}{21 \, \sqrt {a x + 1} a} + \frac {2 \, {\left (3 \, a^{3} c^{\frac {5}{2}} x^{3} - 11 \, a^{2} c^{\frac {5}{2}} x^{2} + 29 \, a c^{\frac {5}{2}} x + 43 \, c^{\frac {5}{2}}\right )}}{15 \, \sqrt {a x + 1} a} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(5/2),x, algorithm="maxima ")
Output:
2/21*(3*a^4*c^(5/2)*x^4 - 9*a^3*c^(5/2)*x^3 + 11*a^2*c^(5/2)*x^2 - 23*a*c^ (5/2)*x - 46*c^(5/2))/(sqrt(a*x + 1)*a) + 2/15*(3*a^3*c^(5/2)*x^3 - 11*a^2 *c^(5/2)*x^2 + 29*a*c^(5/2)*x + 43*c^(5/2))/(sqrt(a*x + 1)*a)
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2 \, {\left (64 \, \sqrt {2} c^{\frac {3}{2}} - \frac {15 \, {\left (a c x + c\right )}^{\frac {7}{2}} - 84 \, {\left (a c x + c\right )}^{\frac {5}{2}} c + 140 \, {\left (a c x + c\right )}^{\frac {3}{2}} c^{2}}{c^{2}}\right )} c^{2}}{105 \, a {\left | c \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(5/2),x, algorithm="giac")
Output:
-2/105*(64*sqrt(2)*c^(3/2) - (15*(a*c*x + c)^(7/2) - 84*(a*c*x + c)^(5/2)* c + 140*(a*c*x + c)^(3/2)*c^2)/c^2)*c^2/(a*abs(c))
Time = 14.62 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=\frac {\sqrt {c-a\,c\,x}\,\left (\frac {176\,c^2\,x}{105}+\frac {142\,c^2}{105\,a}-\frac {44\,a\,c^2\,x^2}{105}-\frac {16\,a^2\,c^2\,x^3}{35}+\frac {2\,a^3\,c^2\,x^4}{7}\right )}{\sqrt {1-a^2\,x^2}} \] Input:
int(((c - a*c*x)^(5/2)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
((c - a*c*x)^(1/2)*((176*c^2*x)/105 + (142*c^2)/(105*a) - (44*a*c^2*x^2)/1 05 - (16*a^2*c^2*x^3)/35 + (2*a^3*c^2*x^4)/7))/(1 - a^2*x^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.36 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \sqrt {c}\, \sqrt {a x +1}\, c^{2} \left (15 a^{3} x^{3}-39 a^{2} x^{2}+17 a x +71\right )}{105 a} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(5/2),x)
Output:
(2*sqrt(c)*sqrt(a*x + 1)*c**2*(15*a**3*x**3 - 39*a**2*x**2 + 17*a*x + 71)) /(105*a)