Integrand size = 27, antiderivative size = 105 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2}}{a^2 c \sqrt {1-a x} \sqrt [4]{c-a^2 c x^2}}+\frac {\sqrt [4]{1-a^2 x^2} \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^2 c \sqrt [4]{c-a^2 c x^2}} \]
1/2*(-a^2*x^2+1)^(1/4)*arctanh(1/2*(-a*x+1)^(1/2)*2^(1/2))/a^2/c/(-a^2*c*x ^2+c)^(1/4)*2^(1/2)+(-a^2*x^2+1)^(1/4)/a^2/c/(-a^2*c*x^2+c)^(1/4)/(-a*x+1) ^(1/2)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2} \left (\frac {1}{a^2 \sqrt {1-a x}}+\frac {\text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^2}\right )}{c \sqrt [4]{c-a^2 c x^2}} \]
((1 - a^2*x^2)^(1/4)*(1/(a^2*Sqrt[1 - a*x]) + ArcTanh[Sqrt[1 - a*x]/Sqrt[2 ]]/(Sqrt[2]*a^2)))/(c*(c - a^2*c*x^2)^(1/4))
Time = 0.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6703, 6700, 87, 73, 219}
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.
\(\displaystyle \int \frac {x e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (1-a^2 x^2\right )^{5/4}}dx}{c \sqrt [4]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \int \frac {x}{(1-a x)^{3/2} (a x+1)}dx}{c \sqrt [4]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \left (\frac {1}{a^2 \sqrt {1-a x}}-\frac {\int \frac {1}{\sqrt {1-a x} (a x+1)}dx}{2 a}\right )}{c \sqrt [4]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \left (\frac {\int \frac {1}{a x+1}d\sqrt {1-a x}}{a^2}+\frac {1}{a^2 \sqrt {1-a x}}\right )}{c \sqrt [4]{c-a^2 c x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \left (\frac {\text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^2}+\frac {1}{a^2 \sqrt {1-a x}}\right )}{c \sqrt [4]{c-a^2 c x^2}}\) |
((1 - a^2*x^2)^(1/4)*(1/(a^2*Sqrt[1 - a*x]) + ArcTanh[Sqrt[1 - a*x]/Sqrt[2 ]]/(Sqrt[2]*a^2)))/(c*(c - a^2*c*x^2)^(1/4))
3.13.99.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
\[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, x}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{4}}}d x\]
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\text {Timed out} \]
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\int { \frac {x \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\int { \frac {x \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\int \frac {x\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{5/4}} \,d x \]