Integrand size = 27, antiderivative size = 133 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\frac {4 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{3 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac {8 \sqrt [8]{2} (1-a x)^{5/8} \sqrt [8]{1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {7}{8},\frac {13}{8},\frac {1}{2} (1-a x)\right )}{15 a^2 c \sqrt [8]{c-a^2 c x^2}} \]
4/3*(a*x+1)^(1/8)*(-a^2*x^2+1)^(1/8)/a^2/c/(-a*x+1)^(3/8)/(-a^2*c*x^2+c)^( 1/8)+8/15*2^(1/8)*(-a*x+1)^(5/8)*(-a^2*x^2+1)^(1/8)*hypergeom([5/8, 7/8],[ 13/8],-1/2*a*x+1/2)/a^2/c/(-a^2*c*x^2+c)^(1/8)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=-\frac {4 \sqrt [8]{1-a^2 x^2} \left (-5 \sqrt [8]{1+a x}+2 \sqrt [8]{2} (-1+a x) \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {7}{8},\frac {13}{8},\frac {1}{2} (1-a x)\right )\right )}{15 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]
(-4*(1 - a^2*x^2)^(1/8)*(-5*(1 + a*x)^(1/8) + 2*2^(1/8)*(-1 + a*x)*Hyperge ometric2F1[5/8, 7/8, 13/8, (1 - a*x)/2]))/(15*a^2*c*(1 - a*x)^(3/8)*(c - a ^2*c*x^2)^(1/8))
Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.77, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6703, 6700, 87, 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.
\(\displaystyle \int \frac {x e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (1-a^2 x^2\right )^{9/8}}dx}{c \sqrt [8]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \int \frac {x}{(1-a x)^{11/8} (a x+1)^{7/8}}dx}{c \sqrt [8]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \left (\frac {4 \sqrt [8]{a x+1}}{3 a^2 (1-a x)^{3/8}}-\frac {2 \int \frac {1}{(1-a x)^{3/8} (a x+1)^{7/8}}dx}{3 a}\right )}{c \sqrt [8]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \left (\frac {8 \sqrt [8]{2} (1-a x)^{5/8} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {7}{8},\frac {13}{8},\frac {1}{2} (1-a x)\right )}{15 a^2}+\frac {4 \sqrt [8]{a x+1}}{3 a^2 (1-a x)^{3/8}}\right )}{c \sqrt [8]{c-a^2 c x^2}}\) |
((1 - a^2*x^2)^(1/8)*((4*(1 + a*x)^(1/8))/(3*a^2*(1 - a*x)^(3/8)) + (8*2^( 1/8)*(1 - a*x)^(5/8)*Hypergeometric2F1[5/8, 7/8, 13/8, (1 - a*x)/2])/(15*a ^2)))/(c*(c - a^2*c*x^2)^(1/8))
3.14.5.3.1 Defintions of rubi rules used
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[((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[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar t[p]) Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && !I ntegerQ[n/2]
\[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, x}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {9}{8}}}d x\]
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\text {Timed out} \]
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int { \frac {x \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{8}}} \,d x } \]
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int { \frac {x \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{8}}} \,d x } \]
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int \frac {x\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{9/8}} \,d x \]