Integrand size = 25, antiderivative size = 174 \[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {c x^{1+m} \sqrt {c-a^2 c x^2}}{(1+m) \sqrt {1-a^2 x^2}}+\frac {a c x^{2+m} \sqrt {c-a^2 c x^2}}{(2+m) \sqrt {1-a^2 x^2}}-\frac {a^2 c x^{3+m} \sqrt {c-a^2 c x^2}}{(3+m) \sqrt {1-a^2 x^2}}-\frac {a^3 c x^{4+m} \sqrt {c-a^2 c x^2}}{(4+m) \sqrt {1-a^2 x^2}} \] Output:
c*x^(1+m)*(-a^2*c*x^2+c)^(1/2)/(1+m)/(-a^2*x^2+1)^(1/2)+a*c*x^(2+m)*(-a^2* c*x^2+c)^(1/2)/(2+m)/(-a^2*x^2+1)^(1/2)-a^2*c*x^(3+m)*(-a^2*c*x^2+c)^(1/2) /(3+m)/(-a^2*x^2+1)^(1/2)-a^3*c*x^(4+m)*(-a^2*c*x^2+c)^(1/2)/(4+m)/(-a^2*x ^2+1)^(1/2)
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.48 \[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {c x^{1+m} \sqrt {c-a^2 c x^2} \left (-(1+a x)^3+(5+2 m) \left (\frac {1}{1+m}+\frac {2 a x}{2+m}+\frac {a^2 x^2}{3+m}\right )\right )}{(4+m) \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]*x^m*(c - a^2*c*x^2)^(3/2),x]
Output:
(c*x^(1 + m)*Sqrt[c - a^2*c*x^2]*(-(1 + a*x)^3 + (5 + 2*m)*((1 + m)^(-1) + (2*a*x)/(2 + m) + (a^2*x^2)/(3 + m))))/((4 + m)*Sqrt[1 - a^2*x^2])
Time = 0.79 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 84, 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.
\(\displaystyle \int x^m e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {c \sqrt {c-a^2 c x^2} \int e^{\text {arctanh}(a x)} x^m \left (1-a^2 x^2\right )^{3/2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {c \sqrt {c-a^2 c x^2} \int x^m (1-a x) (a x+1)^2dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {c \sqrt {c-a^2 c x^2} \int \left (x^m+a x^{m+1}-a^2 x^{m+2}-a^3 x^{m+3}\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \sqrt {c-a^2 c x^2} \left (-\frac {a^3 x^{m+4}}{m+4}-\frac {a^2 x^{m+3}}{m+3}+\frac {a x^{m+2}}{m+2}+\frac {x^{m+1}}{m+1}\right )}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[E^ArcTanh[a*x]*x^m*(c - a^2*c*x^2)^(3/2),x]
Output:
(c*Sqrt[c - a^2*c*x^2]*(x^(1 + m)/(1 + m) + (a*x^(2 + m))/(2 + m) - (a^2*x ^(3 + m))/(3 + m) - (a^3*x^(4 + m))/(4 + m)))/Sqrt[1 - a^2*x^2]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 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])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar t[p]) Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && !I ntegerQ[n/2]
Time = 0.12 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.03
method | result | size |
orering | \(\frac {x \left (a^{3} m^{3} x^{3}+6 a^{3} m^{2} x^{3}+11 a^{3} m \,x^{3}+a^{2} x^{2} m^{3}+6 a^{3} x^{3}+7 a^{2} m^{2} x^{2}+14 a^{2} m \,x^{2}-a \,m^{3} x +8 a^{2} x^{2}-8 a x \,m^{2}-19 a m x -m^{3}-12 a x -9 m^{2}-26 m -24\right ) x^{m} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right ) \left (a x +1\right ) \left (a x -1\right ) \sqrt {-a^{2} x^{2}+1}}\) | \(179\) |
gosper | \(\frac {x^{1+m} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}} \left (a^{3} m^{3} x^{3}+6 a^{3} m^{2} x^{3}+11 a^{3} m \,x^{3}+a^{2} x^{2} m^{3}+6 a^{3} x^{3}+7 a^{2} m^{2} x^{2}+14 a^{2} m \,x^{2}-a \,m^{3} x +8 a^{2} x^{2}-8 a x \,m^{2}-19 a m x -m^{3}-12 a x -9 m^{2}-26 m -24\right )}{\left (1+m \right ) \left (2+m \right ) \left (3+m \right ) \left (4+m \right ) \left (a x -1\right ) \left (a x +1\right ) \sqrt {-a^{2} x^{2}+1}}\) | \(180\) |
risch | \(\frac {\sqrt {-\frac {c \left (-a^{2} x^{2}+1\right )}{a^{2} x^{2}-1}}\, \left (a^{2} x^{2}-1\right ) c^{\frac {3}{2}} \left (a^{3} m^{3} x^{3}+6 a^{3} m^{2} x^{3}+11 a^{3} m \,x^{3}+a^{2} x^{2} m^{3}+6 a^{3} x^{3}+7 a^{2} m^{2} x^{2}+14 a^{2} m \,x^{2}-a \,m^{3} x +8 a^{2} x^{2}-8 a x \,m^{2}-19 a m x -m^{3}-12 a x -9 m^{2}-26 m -24\right ) x \,x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(204\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m*(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVE RBOSE)
Output:
x*(a^3*m^3*x^3+6*a^3*m^2*x^3+11*a^3*m*x^3+a^2*m^3*x^2+6*a^3*x^3+7*a^2*m^2* x^2+14*a^2*m*x^2-a*m^3*x+8*a^2*x^2-8*a*m^2*x-19*a*m*x-m^3-12*a*x-9*m^2-26* m-24)*x^m*(-a^2*c*x^2+c)^(3/2)/(4+m)/(3+m)/(2+m)/(1+m)/(a*x+1)/(a*x-1)/(-a ^2*x^2+1)^(1/2)
Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.21 \[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {\sqrt {-a^{2} c x^{2} + c} {\left ({\left (a^{3} c m^{3} + 6 \, a^{3} c m^{2} + 11 \, a^{3} c m + 6 \, a^{3} c\right )} x^{4} + {\left (a^{2} c m^{3} + 7 \, a^{2} c m^{2} + 14 \, a^{2} c m + 8 \, a^{2} c\right )} x^{3} - {\left (a c m^{3} + 8 \, a c m^{2} + 19 \, a c m + 12 \, a c\right )} x^{2} - {\left (c m^{3} + 9 \, c m^{2} + 26 \, c m + 24 \, c\right )} x\right )} \sqrt {-a^{2} x^{2} + 1} x^{m}}{m^{4} + 10 \, m^{3} - {\left (a^{2} m^{4} + 10 \, a^{2} m^{3} + 35 \, a^{2} m^{2} + 50 \, a^{2} m + 24 \, a^{2}\right )} x^{2} + 35 \, m^{2} + 50 \, m + 24} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m*(-a^2*c*x^2+c)^(3/2),x, algorithm ="fricas")
Output:
-sqrt(-a^2*c*x^2 + c)*((a^3*c*m^3 + 6*a^3*c*m^2 + 11*a^3*c*m + 6*a^3*c)*x^ 4 + (a^2*c*m^3 + 7*a^2*c*m^2 + 14*a^2*c*m + 8*a^2*c)*x^3 - (a*c*m^3 + 8*a* c*m^2 + 19*a*c*m + 12*a*c)*x^2 - (c*m^3 + 9*c*m^2 + 26*c*m + 24*c)*x)*sqrt (-a^2*x^2 + 1)*x^m/(m^4 + 10*m^3 - (a^2*m^4 + 10*a^2*m^3 + 35*a^2*m^2 + 50 *a^2*m + 24*a^2)*x^2 + 35*m^2 + 50*m + 24)
\[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=\int \frac {x^{m} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**m*(-a**2*c*x**2+c)**(3/2),x)
Output:
Integral(x**m*(-c*(a*x - 1)*(a*x + 1))**(3/2)*(a*x + 1)/sqrt(-(a*x - 1)*(a *x + 1)), x)
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.46 \[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {{\left (a^{2} c^{\frac {3}{2}} {\left (m + 2\right )} x^{4} - c^{\frac {3}{2}} {\left (m + 4\right )} x^{2}\right )} a x^{m}}{m^{2} + 6 \, m + 8} - \frac {{\left (a^{2} c^{\frac {3}{2}} {\left (m + 1\right )} x^{3} - c^{\frac {3}{2}} {\left (m + 3\right )} x\right )} x^{m}}{m^{2} + 4 \, m + 3} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m*(-a^2*c*x^2+c)^(3/2),x, algorithm ="maxima")
Output:
-(a^2*c^(3/2)*(m + 2)*x^4 - c^(3/2)*(m + 4)*x^2)*a*x^m/(m^2 + 6*m + 8) - ( a^2*c^(3/2)*(m + 1)*x^3 - c^(3/2)*(m + 3)*x)*x^m/(m^2 + 4*m + 3)
\[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a x + 1\right )} x^{m}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m*(-a^2*c*x^2+c)^(3/2),x, algorithm ="giac")
Output:
integrate((-a^2*c*x^2 + c)^(3/2)*(a*x + 1)*x^m/sqrt(-a^2*x^2 + 1), x)
Time = 23.94 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.31 \[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {x^m\,\left (\frac {c\,x\,\sqrt {c-a^2\,c\,x^2}\,\left (m^3+9\,m^2+26\,m+24\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {a\,c\,x^2\,\sqrt {c-a^2\,c\,x^2}\,\left (m^3+8\,m^2+19\,m+12\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}-\frac {a^3\,c\,x^4\,\sqrt {c-a^2\,c\,x^2}\,\left (m^3+6\,m^2+11\,m+6\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}-\frac {a^2\,c\,x^3\,\sqrt {c-a^2\,c\,x^2}\,\left (m^3+7\,m^2+14\,m+8\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}\right )}{\sqrt {1-a^2\,x^2}} \] Input:
int((x^m*(c - a^2*c*x^2)^(3/2)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
(x^m*((c*x*(c - a^2*c*x^2)^(1/2)*(26*m + 9*m^2 + m^3 + 24))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (a*c*x^2*(c - a^2*c*x^2)^(1/2)*(19*m + 8*m^2 + m^3 + 12))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) - (a^3*c*x^4*(c - a^2*c*x^2)^( 1/2)*(11*m + 6*m^2 + m^3 + 6))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) - (a^2* c*x^3*(c - a^2*c*x^2)^(1/2)*(14*m + 7*m^2 + m^3 + 8))/(50*m + 35*m^2 + 10* m^3 + m^4 + 24)))/(1 - a^2*x^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.81 \[ \int e^{\text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {x^{m} \sqrt {c}\, c x \left (-a^{3} m^{3} x^{3}-6 a^{3} m^{2} x^{3}-11 a^{3} m \,x^{3}-a^{2} m^{3} x^{2}-6 a^{3} x^{3}-7 a^{2} m^{2} x^{2}-14 a^{2} m \,x^{2}+a \,m^{3} x -8 a^{2} x^{2}+8 a \,m^{2} x +19 a m x +m^{3}+12 a x +9 m^{2}+26 m +24\right )}{m^{4}+10 m^{3}+35 m^{2}+50 m +24} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m*(-a^2*c*x^2+c)^(3/2),x)
Output:
(x**m*sqrt(c)*c*x*( - a**3*m**3*x**3 - 6*a**3*m**2*x**3 - 11*a**3*m*x**3 - 6*a**3*x**3 - a**2*m**3*x**2 - 7*a**2*m**2*x**2 - 14*a**2*m*x**2 - 8*a**2 *x**2 + a*m**3*x + 8*a*m**2*x + 19*a*m*x + 12*a*x + m**3 + 9*m**2 + 26*m + 24))/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24)