Integrand size = 26, antiderivative size = 74 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\frac {4 \sqrt [8]{2} \sqrt [8]{1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},\frac {7}{8},\frac {5}{8},\frac {1}{2} (1-a x)\right )}{3 a c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \] Output:
4/3*2^(1/8)*(-a^2*x^2+1)^(1/8)*hypergeom([-3/8, 7/8],[5/8],-1/2*a*x+1/2)/a /c/(-a*x+1)^(3/8)/(-a^2*c*x^2+c)^(1/8)
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\frac {4 \sqrt [8]{2-2 a^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},\frac {7}{8},\frac {5}{8},\frac {1}{2} (1-a x)\right )}{3 a c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \] Input:
Integrate[E^(ArcTanh[a*x]/2)/(c - a^2*c*x^2)^(9/8),x]
Output:
(4*(2 - 2*a^2*x^2)^(1/8)*Hypergeometric2F1[-3/8, 7/8, 5/8, (1 - a*x)/2])/( 3*a*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8))
Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6693, 6690, 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.
\(\displaystyle \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^{9/8}}dx}{c \sqrt [8]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \int \frac {1}{(1-a x)^{11/8} (a x+1)^{7/8}}dx}{c \sqrt [8]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {4 \sqrt [8]{2} \sqrt [8]{1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},\frac {7}{8},\frac {5}{8},\frac {1}{2} (1-a x)\right )}{3 a c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}\) |
Input:
Int[E^(ArcTanh[a*x]/2)/(c - a^2*c*x^2)^(9/8),x]
Output:
(4*2^(1/8)*(1 - a^2*x^2)^(1/8)*Hypergeometric2F1[-3/8, 7/8, 5/8, (1 - a*x) /2])/(3*a*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8))
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_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p Int[(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a , c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPart[p]) Int [(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0])
\[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {9}{8}}}d x\]
Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*c*x^2+c)^(9/8),x)
Output:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*c*x^2+c)^(9/8),x)
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\text {Timed out} \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*c*x^2+c)^(9/8),x, algor ithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\text {Timed out} \] Input:
integrate(((a*x+1)/(-a**2*x**2+1)**(1/2))**(1/2)/(-a**2*c*x**2+c)**(9/8),x )
Output:
Timed out
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int { \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{8}}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*c*x^2+c)^(9/8),x, algor ithm="maxima")
Output:
integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(9/8), x)
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int { \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{8}}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*c*x^2+c)^(9/8),x, algor ithm="giac")
Output:
integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(9/8), x)
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int \frac {\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{9/8}} \,d x \] Input:
int(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2)/(c - a^2*c*x^2)^(9/8),x)
Output:
int(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2)/(c - a^2*c*x^2)^(9/8), x)
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\frac {\int \frac {\sqrt {a x +1}\, \left (-a^{2} x^{2}+1\right )^{\frac {5}{8}}}{a^{4} x^{4}-2 a^{2} x^{2}+1}d x}{c^{\frac {5}{8}} \sqrt {c}} \] Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*c*x^2+c)^(9/8),x)
Output:
(c**(3/8)*int((sqrt(a*x + 1)*( - a**2*x**2 + 1)**(5/8))/(a**4*x**4 - 2*a** 2*x**2 + 1),x))/(sqrt(c)*c)