Integrand size = 14, antiderivative size = 110 \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=-\frac {3 a \sqrt [4]{1-a x} (1+a x)^{3/4}}{4 x}-\frac {\sqrt [4]{1-a x} (1+a x)^{7/4}}{2 x^2}+\frac {9}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac {9}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right ) \]
-3/4*a*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/x-1/2*(-a*x+1)^(1/4)*(a*x+1)^(7/4)/x^2 +9/4*a^2*arctan((a*x+1)^(1/4)/(-a*x+1)^(1/4))-9/4*a^2*arctanh((a*x+1)^(1/4 )/(-a*x+1)^(1/4))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64 \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=-\frac {\sqrt [4]{1-a x} \left (2+7 a x+5 a^2 x^2+18 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1-a x}{1+a x}\right )\right )}{4 x^2 \sqrt [4]{1+a x}} \]
-1/4*((1 - a*x)^(1/4)*(2 + 7*a*x + 5*a^2*x^2 + 18*a^2*x^2*Hypergeometric2F 1[1/4, 1, 5/4, (1 - a*x)/(1 + a*x)]))/(x^2*(1 + a*x)^(1/4))
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6676, 107, 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.
\(\displaystyle \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 6676 |
\(\displaystyle \int \frac {(a x+1)^{3/4}}{x^3 (1-a x)^{3/4}}dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle \frac {3}{4} a \int \frac {(a x+1)^{3/4}}{x^2 (1-a x)^{3/4}}dx-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {3}{4} a \left (\frac {3}{2} a \int \frac {1}{x (1-a x)^{3/4} \sqrt [4]{a x+1}}dx-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {3}{4} a \left (6 a \int -\frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {3}{4} a \left (-6 a \int \frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {3}{4} a \left (6 a \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {a x+1}}{\sqrt {1-a x}}+1}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {3}{4} a \left (6 a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3}{4} a \left (6 a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\) |
-1/2*((1 - a*x)^(1/4)*(1 + a*x)^(7/4))/x^2 + (3*a*(-(((1 - a*x)^(1/4)*(1 + a*x)^(3/4))/x) + 6*a*(ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2 - ArcTanh [(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2)))/4
3.1.77.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[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^(ArcTanh[(a_.)*(x_)]*(n_))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x) ^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c, m, n}, x] && !Int egerQ[(n - 1)/2]
\[\int \frac {{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}^{\frac {3}{2}}}{x^{3}}d x\]
Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=\frac {18 \, a^{2} x^{2} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 9 \, a^{2} x^{2} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 9 \, a^{2} x^{2} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} {\left (5 \, a x + 2\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{8 \, x^{2}} \]
1/8*(18*a^2*x^2*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) - 9*a^2*x^2*lo g(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) + 9*a^2*x^2*log(sqrt(-sqrt(-a^2 *x^2 + 1)/(a*x - 1)) - 1) - 2*sqrt(-a^2*x^2 + 1)*(5*a*x + 2)*sqrt(-sqrt(-a ^2*x^2 + 1)/(a*x - 1)))/x^2
\[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int \frac {\left (\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x^{3}}\, dx \]
\[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int { \frac {\left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x^{3}} \,d x } \]
\[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int { \frac {\left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int \frac {{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{3/2}}{x^3} \,d x \]