Integrand size = 12, antiderivative size = 115 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=-\frac {4 a^3 (1-a x)}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {14 a^2 \sqrt {1-a^2 x^2}}{3 x}+\frac {11}{2} a^3 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \] Output:
-4*a^3*(-a*x+1)/(-a^2*x^2+1)^(1/2)-1/3*(-a^2*x^2+1)^(1/2)/x^3+3/2*a*(-a^2* x^2+1)^(1/2)/x^2-14/3*a^2*(-a^2*x^2+1)^(1/2)/x+11/2*a^3*arctanh((-a^2*x^2+ 1)^(1/2))
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.71 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=\frac {1}{6} \left (-\frac {\sqrt {1-a^2 x^2} \left (2-7 a x+19 a^2 x^2+52 a^3 x^3\right )}{x^3 (1+a x)}-33 a^3 \log (x)+33 a^3 \log \left (1+\sqrt {1-a^2 x^2}\right )\right ) \] Input:
Integrate[1/(E^(3*ArcTanh[a*x])*x^4),x]
Output:
(-((Sqrt[1 - a^2*x^2]*(2 - 7*a*x + 19*a^2*x^2 + 52*a^3*x^3))/(x^3*(1 + a*x ))) - 33*a^3*Log[x] + 33*a^3*Log[1 + Sqrt[1 - a^2*x^2]])/6
Time = 0.90 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6674, 2353, 2009}
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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 6674 |
\(\displaystyle \int \frac {(1-a x)^2}{x^4 (a x+1) \sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 2353 |
\(\displaystyle \int \left (\frac {4 a^2}{x^2 \sqrt {1-a^2 x^2}}+\frac {1}{x^4 \sqrt {1-a^2 x^2}}-\frac {3 a}{x^3 \sqrt {1-a^2 x^2}}+\frac {4 a^4}{(a x+1) \sqrt {1-a^2 x^2}}-\frac {4 a^3}{x \sqrt {1-a^2 x^2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {14 a^2 \sqrt {1-a^2 x^2}}{3 x}+\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {11}{2} a^3 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {4 a^3 \sqrt {1-a^2 x^2}}{a x+1}\) |
Input:
Int[1/(E^(3*ArcTanh[a*x])*x^4),x]
Output:
-1/3*Sqrt[1 - a^2*x^2]/x^3 + (3*a*Sqrt[1 - a^2*x^2])/(2*x^2) - (14*a^2*Sqr t[1 - a^2*x^2])/(3*x) - (4*a^3*Sqrt[1 - a^2*x^2])/(1 + a*x) + (11*a^3*ArcT anh[Sqrt[1 - a^2*x^2]])/2
Int[(Px_)*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2) ^(p_), x_Symbol] :> Int[ExpandIntegrand[Px*(e*x)^m*(c + d*x)^n*(a + b*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && PolyQ[Px, x] && (Integer Q[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0]))
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x )^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*x^2])), x] / ; FreeQ[{a, c, m}, x] && IntegerQ[(n - 1)/2]
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {28 a^{4} x^{4}-9 a^{3} x^{3}-26 a^{2} x^{2}+9 a x -2}{6 x^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {a^{3} \left (-11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {8 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a \left (x +\frac {1}{a}\right )}\right )}{2}\) | \(105\) |
default | \(-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{3 x^{3}}+\frac {16 a^{2} \left (-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{x}-4 a^{2} \left (\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}+\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 \sqrt {a^{2}}}\right )\right )}{3}+a \left (-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{3}}-2 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )\right )-3 a \left (-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{2 x^{2}}-\frac {3 a^{2} \left (\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3}+\sqrt {-a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}\right )-10 a^{3} \left (\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3}+\sqrt {-a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )+4 a^{2} \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )+10 a^{3} \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\) | \(674\) |
Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x,method=_RETURNVERBOSE)
Output:
1/6*(28*a^4*x^4-9*a^3*x^3-26*a^2*x^2+9*a*x-2)/x^3/(-a^2*x^2+1)^(1/2)-1/2*a ^3*(-11*arctanh(1/(-a^2*x^2+1)^(1/2))+8/a/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1 /a))^(1/2))
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=-\frac {24 \, a^{4} x^{4} + 24 \, a^{3} x^{3} + 33 \, {\left (a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + {\left (52 \, a^{3} x^{3} + 19 \, a^{2} x^{2} - 7 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a x^{4} + x^{3}\right )}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x, algorithm="fricas")
Output:
-1/6*(24*a^4*x^4 + 24*a^3*x^3 + 33*(a^4*x^4 + a^3*x^3)*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (52*a^3*x^3 + 19*a^2*x^2 - 7*a*x + 2)*sqrt(-a^2*x^2 + 1))/( a*x^4 + x^3)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{x^{4} \left (a x + 1\right )^{3}}\, dx \] Input:
integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/x**4,x)
Output:
Integral((-(a*x - 1)*(a*x + 1))**(3/2)/(x**4*(a*x + 1)**3), x)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x^{4}} \,d x } \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x, algorithm="maxima")
Output:
integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*x^4), x)
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (96) = 192\).
Time = 0.12 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.30 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=\frac {{\left (a^{4} - \frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{2}}{x} + \frac {48 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{x^{2}} + \frac {249 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{2} x^{3}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} + \frac {11 \, a^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} - \frac {\frac {57 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4}}{x} - \frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{2}}{x^{2}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{x^{3}}}{24 \, a^{2} {\left | a \right |}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x, algorithm="giac")
Output:
1/24*(a^4 - 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^2/x + 48*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/x^2 + 249*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^2*x^3))*a^ 6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/ (a^2*x) + 1)*abs(a)) + 11/2*a^4*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 1/24*(57*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4/ x - 9*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2/x^2 + (sqrt(-a^2*x^2 + 1)*abs( a) + a)^3/x^3)/(a^2*abs(a))
Time = 14.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=\frac {3\,a\,\sqrt {1-a^2\,x^2}}{2\,x^2}-\frac {\sqrt {1-a^2\,x^2}}{3\,x^3}-\frac {14\,a^2\,\sqrt {1-a^2\,x^2}}{3\,x}+\frac {4\,a^4\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {a^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,11{}\mathrm {i}}{2} \] Input:
int((1 - a^2*x^2)^(3/2)/(x^4*(a*x + 1)^3),x)
Output:
(3*a*(1 - a^2*x^2)^(1/2))/(2*x^2) - (1 - a^2*x^2)^(1/2)/(3*x^3) - (a^3*ata n((1 - a^2*x^2)^(1/2)*1i)*11i)/2 - (14*a^2*(1 - a^2*x^2)^(1/2))/(3*x) + (4 *a^4*(1 - a^2*x^2)^(1/2))/((x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)*(-a^2)^(1/2))
Time = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x^4} \, dx=\frac {a^{3} \left (-132 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}-132 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}+\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{7}-8 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{6}+48 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}+306 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}-48 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+8 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )}{24 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3} \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )+1\right )} \] Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x)
Output:
(a**3*( - 132*log(tan(asin(a*x)/2))*tan(asin(a*x)/2)**4 - 132*log(tan(asin (a*x)/2))*tan(asin(a*x)/2)**3 + tan(asin(a*x)/2)**7 - 8*tan(asin(a*x)/2)** 6 + 48*tan(asin(a*x)/2)**5 + 306*tan(asin(a*x)/2)**4 - 48*tan(asin(a*x)/2) **2 + 8*tan(asin(a*x)/2) - 1))/(24*tan(asin(a*x)/2)**3*(tan(asin(a*x)/2) + 1))