\(\int e^{\text {sech}^{-1}(a x^3)} x^m \, dx\) [60]

Optimal result

Integrand size = 12, antiderivative size = 85 \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=-\frac {x^{-2+m}}{a (2-m)}+\frac {\sqrt {-1+\frac {1}{a x^3}} x^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{6} (-1-m),\frac {5-m}{6},\frac {1}{a^2 x^6}\right )}{(1+m) \sqrt {1-\frac {1}{a x^3}}} \] Output:

-x^(-2+m)/a/(2-m)+(-1+1/a/x^3)^(1/2)*x^(1+m)*hypergeom([-1/2, -1/6-1/6*m], 
[5/6-1/6*m],1/a^2/x^6)/(1+m)/(1-1/a/x^3)^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(187\) vs. \(2(85)=170\).

Time = 0.68 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.20 \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\frac {2^{\frac {1+m}{3}} e^{\text {sech}^{-1}\left (a x^3\right )} \left (\frac {e^{\text {sech}^{-1}\left (a x^3\right )}}{1+e^{2 \text {sech}^{-1}\left (a x^3\right )}}\right )^{\frac {1+m}{3}} \left (1+e^{2 \text {sech}^{-1}\left (a x^3\right )}\right )^{\frac {1+m}{3}} x^{1+m} \left (a x^3\right )^{\frac {1}{3} (-1-m)} \left ((10+m) \operatorname {Hypergeometric2F1}\left (\frac {4+m}{6},\frac {4+m}{3},\frac {10+m}{6},-e^{2 \text {sech}^{-1}\left (a x^3\right )}\right )-e^{2 \text {sech}^{-1}\left (a x^3\right )} (4+m) \operatorname {Hypergeometric2F1}\left (\frac {4+m}{3},\frac {10+m}{6},\frac {16+m}{6},-e^{2 \text {sech}^{-1}\left (a x^3\right )}\right )\right )}{(4+m) (10+m)} \] Input:

Integrate[E^ArcSech[a*x^3]*x^m,x]
 

Output:

(2^((1 + m)/3)*E^ArcSech[a*x^3]*(E^ArcSech[a*x^3]/(1 + E^(2*ArcSech[a*x^3] 
)))^((1 + m)/3)*(1 + E^(2*ArcSech[a*x^3]))^((1 + m)/3)*x^(1 + m)*(a*x^3)^( 
(-1 - m)/3)*((10 + m)*Hypergeometric2F1[(4 + m)/6, (4 + m)/3, (10 + m)/6, 
-E^(2*ArcSech[a*x^3])] - E^(2*ArcSech[a*x^3])*(4 + m)*Hypergeometric2F1[(4 
 + m)/3, (10 + m)/6, (16 + m)/6, -E^(2*ArcSech[a*x^3])]))/((4 + m)*(10 + m 
))
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6889, 15, 791, 888}

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.

\(\displaystyle \int x^m e^{\text {sech}^{-1}\left (a x^3\right )} \, dx\)

\(\Big \downarrow \) 6889

\(\displaystyle \frac {3 \int x^{m-3}dx}{a (m+1)}+\frac {3 \sqrt {\frac {1}{a x^3+1}} \sqrt {a x^3+1} \int \frac {x^{m-3}}{\sqrt {1-a x^3} \sqrt {a x^3+1}}dx}{a (m+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^3\right )}}{m+1}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {3 \sqrt {\frac {1}{a x^3+1}} \sqrt {a x^3+1} \int \frac {x^{m-3}}{\sqrt {1-a x^3} \sqrt {a x^3+1}}dx}{a (m+1)}-\frac {3 x^{m-2}}{a (2-m) (m+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^3\right )}}{m+1}\)

\(\Big \downarrow \) 791

\(\displaystyle \frac {3 \sqrt {\frac {1}{a x^3+1}} \sqrt {a x^3+1} \int \frac {x^{m-3}}{\sqrt {1-a^2 x^6}}dx}{a (m+1)}-\frac {3 x^{m-2}}{a (2-m) (m+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^3\right )}}{m+1}\)

\(\Big \downarrow \) 888

\(\displaystyle -\frac {3 \sqrt {\frac {1}{a x^3+1}} \sqrt {a x^3+1} x^{m-2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m-2}{6},\frac {m+4}{6},a^2 x^6\right )}{a (2-m) (m+1)}-\frac {3 x^{m-2}}{a (2-m) (m+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^3\right )}}{m+1}\)

Input:

Int[E^ArcSech[a*x^3]*x^m,x]
 

Output:

(-3*x^(-2 + m))/(a*(2 - m)*(1 + m)) + (E^ArcSech[a*x^3]*x^(1 + m))/(1 + m) 
 - (3*x^(-2 + m)*Sqrt[(1 + a*x^3)^(-1)]*Sqrt[1 + a*x^3]*Hypergeometric2F1[ 
1/2, (-2 + m)/6, (4 + m)/6, a^2*x^6])/(a*(2 - m)*(1 + m))
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 791

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_) 
^(n_))^(p_), x_Symbol] :> Int[(c*x)^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; Free 
Q[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] 
 || (GtQ[a1, 0] && GtQ[a2, 0]))
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 6889

Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ 
ArcSech[a*x^p]/(m + 1)), x] + (Simp[p/(a*(m + 1))   Int[x^(m - p), x], x] + 
 Simp[p*(Sqrt[1 + a*x^p]/(a*(m + 1)))*Sqrt[1/(1 + a*x^p)]   Int[x^(m - p)/( 
Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, 
-1]
 

Maple [F]

\[\int \left (\frac {1}{a \,x^{3}}+\sqrt {-1+\frac {1}{a \,x^{3}}}\, \sqrt {\frac {1}{a \,x^{3}}+1}\right ) x^{m}d x\]

Input:

int((1/a/x^3+(-1+1/a/x^3)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x)
 

Output:

int((1/a/x^3+(-1+1/a/x^3)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x)
 

Fricas [F]

\[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\int { x^{m} {\left (\sqrt {\frac {1}{a x^{3}} + 1} \sqrt {\frac {1}{a x^{3}} - 1} + \frac {1}{a x^{3}}\right )} \,d x } \] Input:

integrate((1/a/x^3+(-1+1/a/x^3)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x, algorithm= 
"fricas")
 

Output:

integral((a*x^3*x^m*sqrt((a*x^3 + 1)/(a*x^3))*sqrt(-(a*x^3 - 1)/(a*x^3)) + 
 x^m)/(a*x^3), x)
 

Sympy [F]

\[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\frac {\int \frac {x^{m}}{x^{3}}\, dx + \int a x^{m} \sqrt {-1 + \frac {1}{a x^{3}}} \sqrt {1 + \frac {1}{a x^{3}}}\, dx}{a} \] Input:

integrate((1/a/x**3+(-1+1/a/x**3)**(1/2)*(1/a/x**3+1)**(1/2))*x**m,x)
 

Output:

(Integral(x**m/x**3, x) + Integral(a*x**m*sqrt(-1 + 1/(a*x**3))*sqrt(1 + 1 
/(a*x**3)), x))/a
 

Maxima [F(-2)]

Exception generated. \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\text {Exception raised: ValueError} \] Input:

integrate((1/a/x^3+(-1+1/a/x^3)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x, algorithm= 
"maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(m-3>0)', see `assume?` for more 
details)Is
 

Giac [F]

\[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\int { x^{m} {\left (\sqrt {\frac {1}{a x^{3}} + 1} \sqrt {\frac {1}{a x^{3}} - 1} + \frac {1}{a x^{3}}\right )} \,d x } \] Input:

integrate((1/a/x^3+(-1+1/a/x^3)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x, algorithm= 
"giac")
 

Output:

integrate(x^m*(sqrt(1/(a*x^3) + 1)*sqrt(1/(a*x^3) - 1) + 1/(a*x^3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\int x^m\,\left (\sqrt {\frac {1}{a\,x^3}-1}\,\sqrt {\frac {1}{a\,x^3}+1}+\frac {1}{a\,x^3}\right ) \,d x \] Input:

int(x^m*((1/(a*x^3) - 1)^(1/2)*(1/(a*x^3) + 1)^(1/2) + 1/(a*x^3)),x)
 

Output:

int(x^m*((1/(a*x^3) - 1)^(1/2)*(1/(a*x^3) + 1)^(1/2) + 1/(a*x^3)), x)
 

Reduce [F]

\[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\frac {x^{m} \sqrt {a \,x^{3}+1}\, \sqrt {-a \,x^{3}+1}\, m -2 x^{m} \sqrt {a \,x^{3}+1}\, \sqrt {-a \,x^{3}+1}+x^{m} m +x^{m}-3 \left (\int \frac {x^{m} \sqrt {a \,x^{3}+1}\, \sqrt {-a \,x^{3}+1}}{a^{2} m \,x^{9}+a^{2} x^{9}-m \,x^{3}-x^{3}}d x \right ) m^{2} x^{2}+3 \left (\int \frac {x^{m} \sqrt {a \,x^{3}+1}\, \sqrt {-a \,x^{3}+1}}{a^{2} m \,x^{9}+a^{2} x^{9}-m \,x^{3}-x^{3}}d x \right ) m \,x^{2}+6 \left (\int \frac {x^{m} \sqrt {a \,x^{3}+1}\, \sqrt {-a \,x^{3}+1}}{a^{2} m \,x^{9}+a^{2} x^{9}-m \,x^{3}-x^{3}}d x \right ) x^{2}}{a \,x^{2} \left (m^{2}-m -2\right )} \] Input:

int((1/a/x^3+(-1+1/a/x^3)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x)
 

Output:

(x**m*sqrt(a*x**3 + 1)*sqrt( - a*x**3 + 1)*m - 2*x**m*sqrt(a*x**3 + 1)*sqr 
t( - a*x**3 + 1) + x**m*m + x**m - 3*int((x**m*sqrt(a*x**3 + 1)*sqrt( - a* 
x**3 + 1))/(a**2*m*x**9 + a**2*x**9 - m*x**3 - x**3),x)*m**2*x**2 + 3*int( 
(x**m*sqrt(a*x**3 + 1)*sqrt( - a*x**3 + 1))/(a**2*m*x**9 + a**2*x**9 - m*x 
**3 - x**3),x)*m*x**2 + 6*int((x**m*sqrt(a*x**3 + 1)*sqrt( - a*x**3 + 1))/ 
(a**2*m*x**9 + a**2*x**9 - m*x**3 - x**3),x)*x**2)/(a*x**2*(m**2 - m - 2))