Integrand size = 20, antiderivative size = 82 \[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {2^{1+\frac {n}{2}} (1-a x)^{\frac {1}{2} (-2-n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3-n),-\frac {n}{2},\frac {1}{2} (-1-n),\frac {1}{2} (1-a x)\right )}{a c^2 (3+n) \sqrt {c-a c x}} \] Output:
2^(1+1/2*n)*(-a*x+1)^(-1-1/2*n)*hypergeom([-1/2*n, -3/2-1/2*n],[-1/2-1/2*n ],-1/2*a*x+1/2)/a/c^2/(3+n)/(-a*c*x+c)^(1/2)
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {2^{1+\frac {n}{2}} (1-a x)^{-1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2}-\frac {n}{2},-\frac {n}{2},-\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {a x}{2}\right )}{a c^2 (3+n) \sqrt {c-a c x}} \] Input:
Integrate[E^(n*ArcTanh[a*x])/(c - a*c*x)^(5/2),x]
Output:
(2^(1 + n/2)*(1 - a*x)^(-1 - n/2)*Hypergeometric2F1[-3/2 - n/2, -1/2*n, -1 /2 - n/2, 1/2 - (a*x)/2])/(a*c^2*(3 + n)*Sqrt[c - a*c*x])
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6680, 37, 79}
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.
\(\displaystyle \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {(1-a x)^{-n/2} (a x+1)^{n/2}}{(c-a c x)^{5/2}}dx\) |
\(\Big \downarrow \) 37 |
\(\displaystyle \frac {(1-a x)^{5/2} \int (1-a x)^{\frac {1}{2} (-n-5)} (a x+1)^{n/2}dx}{(c-a c x)^{5/2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {2^{\frac {n}{2}+1} (1-a x)^{\frac {1}{2} (-n-3)+\frac {5}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-n-3),-\frac {n}{2},\frac {1}{2} (-n-1),\frac {1}{2} (1-a x)\right )}{a (n+3) (c-a c x)^{5/2}}\) |
Input:
Int[E^(n*ArcTanh[a*x])/(c - a*c*x)^(5/2),x]
Output:
(2^(1 + n/2)*(1 - a*x)^(5/2 + (-3 - n)/2)*Hypergeometric2F1[(-3 - n)/2, -1 /2*n, (-1 - n)/2, (1 - a*x)/2])/(a*(3 + n)*(c - a*c*x)^(5/2))
Int[(u_.)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> S imp[(a + b*x)^m/(c + d*x)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] && !SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 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])
\[\int \frac {{\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{\left (-a c x +c \right )^{\frac {5}{2}}}d x\]
Input:
int(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x)
Output:
int(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x)
\[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a c x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x, algorithm="fricas")
Output:
integral(-sqrt(-a*c*x + c)*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a^3*c^3*x^3 - 3 *a^2*c^3*x^2 + 3*a*c^3*x - c^3), x)
\[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(exp(n*atanh(a*x))/(-a*c*x+c)**(5/2),x)
Output:
Integral(exp(n*atanh(a*x))/(-c*(a*x - 1))**(5/2), x)
\[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a c x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x, algorithm="maxima")
Output:
integrate((-(a*x + 1)/(a*x - 1))^(1/2*n)/(-a*c*x + c)^(5/2), x)
\[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a c x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(exp(n*arctanh(a*x))/(-a*c*x+c)^(5/2),x, algorithm="giac")
Output:
integrate((-(a*x + 1)/(a*x - 1))^(1/2*n)/(-a*c*x + c)^(5/2), x)
Timed out. \[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{{\left (c-a\,c\,x\right )}^{5/2}} \,d x \] Input:
int(exp(n*atanh(a*x))/(c - a*c*x)^(5/2),x)
Output:
int(exp(n*atanh(a*x))/(c - a*c*x)^(5/2), x)
\[ \int \frac {e^{n \text {arctanh}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\int \frac {e^{\mathit {atanh} \left (a x \right ) n}}{\sqrt {-a x +1}\, a^{2} x^{2}-2 \sqrt {-a x +1}\, a x +\sqrt {-a x +1}}d x}{\sqrt {c}\, c^{2}} \] Input:
int(exp(n*atanh(a*x))/(-a*c*x+c)^(5/2),x)
Output:
int(e**(atanh(a*x)*n)/(sqrt( - a*x + 1)*a**2*x**2 - 2*sqrt( - a*x + 1)*a*x + sqrt( - a*x + 1)),x)/(sqrt(c)*c**2)