Integrand size = 24, antiderivative size = 70 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{2+m} \] Output:
x^(1+m)*hypergeom([2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)+a*x^(2+m)*hype rgeom([2, 1+1/2*m],[2+1/2*m],a^2*x^2)/(2+m)
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx=x^{1+m} \left (\frac {a x \operatorname {Hypergeometric2F1}\left (2,1+\frac {m}{2},2+\frac {m}{2},a^2 x^2\right )}{2+m}+\frac {\operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}\right ) \] Input:
Integrate[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(3/2),x]
Output:
x^(1 + m)*((a*x*Hypergeometric2F1[2, 1 + m/2, 2 + m/2, a^2*x^2])/(2 + m) + Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, a^2*x^2]/(1 + m))
Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6700, 92, 82, 278}
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.
\(\displaystyle \int \frac {x^m e^{\text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \int \frac {x^m}{(1-a x)^2 (a x+1)}dx\) |
\(\Big \downarrow \) 92 |
\(\displaystyle a \int \frac {x^{m+1}}{(1-a x)^2 (a x+1)^2}dx+\int \frac {x^m}{(1-a x)^2 (a x+1)^2}dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle a \int \frac {x^{m+1}}{\left (1-a^2 x^2\right )^2}dx+\int \frac {x^m}{\left (1-a^2 x^2\right )^2}dx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (2,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\) |
Input:
Int[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(3/2),x]
Output:
(x^(1 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(2 + m)
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> Simp[a Int[(a + b*x)^n*(c + d*x)^n*(f*x)^p, x], x] + Simp[b/f In t[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] && !RationalQ[p] && !IGtQ[m, 0] && NeQ[m + n + p + 2, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 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])
Result contains higher order function than in optimal. Order 9 vs. order 5.
Time = 0.22 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.53
method | result | size |
meijerg | \(-\frac {\left (-a^{2}\right )^{-\frac {m}{2}} \left (\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \left (-2-m \right )}{\left (2+m \right ) \left (-a^{2} x^{2}+1\right )}+\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} m \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}\right )}{2}\right )}{2 a}+\frac {\left (-a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (-\frac {2 x^{1+m} \left (-a^{2}\right )^{\frac {m}{2}+\frac {1}{2}} \left (-1-m \right )}{\left (1+m \right ) \left (-2 a^{2} x^{2}+2\right )}+\frac {2 x^{1+m} \left (-a^{2}\right )^{\frac {m}{2}+\frac {1}{2}} \left (-\frac {m^{2}}{4}+\frac {1}{4}\right ) \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}+\frac {1}{2}\right )}{1+m}\right )}{2}\) | \(177\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^2*x^m,x,method=_RETURNVERBOSE)
Output:
-1/2/a*(-a^2)^(-1/2*m)*(1/(2+m)*x^m*(-a^2)^(1/2*m)*(-2-m)/(-a^2*x^2+1)+1/2 *x^m*(-a^2)^(1/2*m)*m*LerchPhi(a^2*x^2,1,1/2*m))+1/2*(-a^2)^(-1/2-1/2*m)*( -2/(1+m)*x^(1+m)*(-a^2)^(1/2*m+1/2)*(-1-m)/(-2*a^2*x^2+2)+2/(1+m)*x^(1+m)* (-a^2)^(1/2*m+1/2)*(-1/4*m^2+1/4)*LerchPhi(a^2*x^2,1,1/2*m+1/2))
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^2*x^m,x, algorithm="fricas")
Output:
integral(x^m/(a^3*x^3 - a^2*x^2 - a*x + 1), x)
Result contains complex when optimal does not.
Time = 3.32 (sec) , antiderivative size = 673, normalized size of antiderivative = 9.61 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**2*x**m,x)
Output:
-a**2*m**2*x**2*x**(m + 1)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a**2*x**2*x**(m + 1)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2) *gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a* (-a**2*m**2*x**2*x**(m + 2)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*a** 2*m*x**2*x**(m + 2)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamm a(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) + m**2*x**(m + 2)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a** 2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) + 2*m*x**(m + 2)*lerchphi(a**2*x **2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*m*x**(m + 2)*gamma(m/2 + 1)/(8*a**2*x**2*gamma (m/2 + 2) - 8*gamma(m/2 + 2)) - 4*x**(m + 2)*gamma(m/2 + 1)/(8*a**2*x**2*g amma(m/2 + 2) - 8*gamma(m/2 + 2))) + m**2*x**(m + 1)*lerchphi(a**2*x**2*ex p_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3 /2) - 8*gamma(m/2 + 3/2)) - 2*m*x**(m + 1)*gamma(m/2 + 1/2)/(8*a**2*x**2*g amma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - x**(m + 1)*lerchphi(a**2*x**2*exp_ polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2 ) - 8*gamma(m/2 + 3/2)) - 2*x**(m + 1)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma (m/2 + 3/2) - 8*gamma(m/2 + 3/2))
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^2*x^m,x, algorithm="maxima")
Output:
integrate((a*x + 1)*x^m/(a^2*x^2 - 1)^2, x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^2*x^m,x, algorithm="giac")
Output:
integrate((a*x + 1)*x^m/(a^2*x^2 - 1)^2, x)
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int \frac {x^m\,\left (a\,x+1\right )}{{\left (a^2\,x^2-1\right )}^2} \,d x \] Input:
int((x^m*(a*x + 1))/(a^2*x^2 - 1)^2,x)
Output:
int((x^m*(a*x + 1))/(a^2*x^2 - 1)^2, x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{m}}{a^{3} x^{3}-a^{2} x^{2}-a x +1}d x \] Input:
int((a*x+1)/(-a^2*x^2+1)^2*x^m,x)
Output:
int(x**m/(a**3*x**3 - a**2*x**2 - a*x + 1),x)