Integrand size = 14, antiderivative size = 136 \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\frac {25 a^2 \sqrt [4]{1-a x}}{2 \sqrt [4]{1+a x}}+\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{1+a x}}-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}+\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac {25}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right ) \] Output:
25/2*a^2*(-a*x+1)^(1/4)/(a*x+1)^(1/4)+5/4*a*(-a*x+1)^(5/4)/x/(a*x+1)^(1/4) -1/2*(-a*x+1)^(9/4)/x^2/(a*x+1)^(1/4)+25/4*a^2*arctan((a*x+1)^(1/4)/(-a*x+ 1)^(1/4))-25/4*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 = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.51 \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\frac {\sqrt [4]{1-a x} \left (-2+9 a x+43 a^2 x^2-50 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1-a x}{1+a x}\right )\right )}{4 x^2 \sqrt [4]{1+a x}} \] Input:
Integrate[1/(E^((5*ArcTanh[a*x])/2)*x^3),x]
Output:
((1 - a*x)^(1/4)*(-2 + 9*a*x + 43*a^2*x^2 - 50*a^2*x^2*Hypergeometric2F1[1 /4, 1, 5/4, (1 - a*x)/(1 + a*x)]))/(4*x^2*(1 + a*x)^(1/4))
Time = 0.44 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6676, 107, 105, 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.
\(\displaystyle \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 6676 |
\(\displaystyle \int \frac {(1-a x)^{5/4}}{x^3 (a x+1)^{5/4}}dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle -\frac {5}{4} a \int \frac {(1-a x)^{5/4}}{x^2 (a x+1)^{5/4}}dx-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle -\frac {5}{4} a \left (-\frac {5}{2} a \int \frac {\sqrt [4]{1-a x}}{x (a x+1)^{5/4}}dx-\frac {(1-a x)^{5/4}}{x \sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle -\frac {5}{4} a \left (-\frac {5}{2} a \left (\int \frac {1}{x (1-a x)^{3/4} \sqrt [4]{a x+1}}dx+\frac {4 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{5/4}}{x \sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {5}{4} a \left (-\frac {5}{2} a \left (4 \int -\frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}+\frac {4 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{5/4}}{x \sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {5}{4} a \left (-\frac {5}{2} a \left (\frac {4 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}-4 \int \frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {(1-a x)^{5/4}}{x \sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {5}{4} a \left (-\frac {5}{2} a \left (4 \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {a x+1}}{\sqrt {1-a x}}+1}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )+\frac {4 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{5/4}}{x \sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {5}{4} a \left (-\frac {5}{2} a \left (4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )+\frac {4 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{5/4}}{x \sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {5}{4} a \left (-\frac {5}{2} a \left (4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )\right )+\frac {4 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{5/4}}{x \sqrt [4]{a x+1}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}\) |
Input:
Int[1/(E^((5*ArcTanh[a*x])/2)*x^3),x]
Output:
-1/2*(1 - a*x)^(9/4)/(x^2*(1 + a*x)^(1/4)) - (5*a*(-((1 - a*x)^(5/4)/(x*(1 + a*x)^(1/4))) - (5*a*((4*(1 - a*x)^(1/4))/(1 + a*x)^(1/4) + 4*(ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2 - ArcTanh[(1 + a*x)^(1/4)/(1 - a*x)^(1/4) ]/2)))/2))/4
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]
\[\int \frac {1}{{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x^{3}}d x\]
Input:
int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)
Output:
int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)
Time = 0.10 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\frac {2 \, {\left (43 \, a^{2} x^{2} + 9 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 50 \, {\left (a^{3} x^{3} + a^{2} x^{2}\right )} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 25 \, {\left (a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 25 \, {\left (a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right )}{8 \, {\left (a x^{3} + x^{2}\right )}} \] Input:
integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x, algorithm="fricas")
Output:
1/8*(2*(43*a^2*x^2 + 9*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-sqrt(-a^2*x^2 + 1 )/(a*x - 1)) + 50*(a^3*x^3 + a^2*x^2)*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) - 25*(a^3*x^3 + a^2*x^2)*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) + 25*(a^3*x^3 + a^2*x^2)*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1)) /(a*x^3 + x^2)
\[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int \frac {1}{x^{3} \left (\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/((a*x+1)/(-a**2*x**2+1)**(1/2))**(5/2)/x**3,x)
Output:
Integral(1/(x**3*((a*x + 1)/sqrt(-a**2*x**2 + 1))**(5/2)), x)
\[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int { \frac {1}{x^{3} \left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x, algorithm="maxima")
Output:
integrate(1/(x^3*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)
\[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int { \frac {1}{x^{3} \left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x, algorithm="giac")
Output:
integrate(1/(x^3*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)
Timed out. \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int \frac {1}{x^3\,{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2}} \,d x \] Input:
int(1/(x^3*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2)),x)
Output:
int(1/(x^3*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2)), x)
\[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int \frac {1}{{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x^{3}}d x \] Input:
int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)
Output:
int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)