Integrand size = 10, antiderivative size = 98 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx=\sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x-\frac {3 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}+\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \]
(1-1/a/x)^(1/4)*(1+1/a/x)^(3/4)*x-3*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4) )/a+3*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx=\frac {8 e^{\frac {3}{2} \coth ^{-1}(a x)} \left (1+\left (-1+e^{2 \coth ^{-1}(a x)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},2,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )\right )}{a \left (-1+e^{2 \coth ^{-1}(a x)}\right )} \]
(8*E^((3*ArcCoth[a*x])/2)*(1 + (-1 + E^(2*ArcCoth[a*x]))*Hypergeometric2F1 [3/4, 2, 7/4, E^(2*ArcCoth[a*x])]))/(a*(-1 + E^(2*ArcCoth[a*x])))
Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6720, 105, 104, 25, 827, 216, 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.
\(\displaystyle \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6720 |
\(\displaystyle -\int \frac {\left (1+\frac {1}{a x}\right )^{3/4} x^2}{\left (1-\frac {1}{a x}\right )^{3/4}}d\frac {1}{x}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {3 \int \frac {x}{\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}}d\frac {1}{x}}{2 a}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {6 \int -\frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {6 \int \frac {1}{\left (1-\frac {1}{x^4}\right ) x^2}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}}{a}+x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {6 \left (\frac {1}{2} \int \frac {1}{1+\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {6 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {1}{x^2}}d\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {6 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )\right )}{a}\) |
(1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4)*x - (6*(ArcTan[(1 + 1/(a*x))^(1/4) /(1 - 1/(a*x))^(1/4)]/2 - ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)] /2))/a
3.1.72.3.1 Defintions of rubi rules used
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_)), x_Symbol] :> -Subst[Int[(1 + x/a)^(n/2)/( x^2*(1 - x/a)^(n/2)), x], x, 1/x] /; FreeQ[{a, n}, x] && !IntegerQ[n]
\[\int \frac {1}{\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{4}}}d x\]
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx=\frac {2 \, {\left (a x + 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 6 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + 3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - 3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{2 \, a} \]
1/2*(2*(a*x + 1)*((a*x - 1)/(a*x + 1))^(1/4) + 6*arctan(((a*x - 1)/(a*x + 1))^(1/4)) + 3*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) - 3*log(((a*x - 1)/(a* x + 1))^(1/4) - 1))/a
\[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx=\int \frac {1}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx=-\frac {1}{2} \, a {\left (\frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} - \frac {6 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} - \frac {3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} + \frac {3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{2}}\right )} \]
-1/2*a*(4*((a*x - 1)/(a*x + 1))^(1/4)/((a*x - 1)*a^2/(a*x + 1) - a^2) - 6* arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^2 - 3*log(((a*x - 1)/(a*x + 1))^(1/4 ) + 1)/a^2 + 3*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^2)
Time = 0.33 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.11 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx=\frac {1}{2} \, a {\left (\frac {6 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} + \frac {3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} - \frac {3 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{2}} - \frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{2} {\left (\frac {a x - 1}{a x + 1} - 1\right )}}\right )} \]
1/2*a*(6*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^2 + 3*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^2 - 3*log(abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^2 - 4* ((a*x - 1)/(a*x + 1))^(1/4)/(a^2*((a*x - 1)/(a*x + 1) - 1)))
Time = 4.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int e^{\frac {3}{2} \coth ^{-1}(a x)} \, dx=\frac {2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}+\frac {3\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a}+\frac {3\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a} \]