Integrand size = 23, antiderivative size = 42 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=\frac {c x^{1+m}}{1+m}+\frac {2 a c x^{2+m}}{2+m}+\frac {a^2 c x^{3+m}}{3+m} \] Output:
c*x^(1+m)/(1+m)+2*a*c*x^(2+m)/(2+m)+a^2*c*x^(3+m)/(3+m)
Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=c x^{1+m} \left (\frac {1}{1+m}+\frac {2 a x}{2+m}+\frac {a^2 x^2}{3+m}\right ) \] Input:
Integrate[E^(2*ArcTanh[a*x])*x^m*(c - a^2*c*x^2),x]
Output:
c*x^(1 + m)*((1 + m)^(-1) + (2*a*x)/(2 + m) + (a^2*x^2)/(3 + m))
Time = 0.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6700, 53, 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.
\(\displaystyle \int x^m e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right ) \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle c \int x^m (a x+1)^2dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle c \int \left (x^m+2 a x^{m+1}+a^2 x^{m+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c \left (\frac {a^2 x^{m+3}}{m+3}+\frac {2 a x^{m+2}}{m+2}+\frac {x^{m+1}}{m+1}\right )\) |
Input:
Int[E^(2*ArcTanh[a*x])*x^m*(c - a^2*c*x^2),x]
Output:
c*(x^(1 + m)/(1 + m) + (2*a*x^(2 + m))/(2 + m) + (a^2*x^(3 + m))/(3 + m))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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])
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19
method | result | size |
norman | \(\frac {c x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}+\frac {a^{2} c \,x^{3} {\mathrm e}^{m \ln \left (x \right )}}{3+m}+\frac {2 a c \,x^{2} {\mathrm e}^{m \ln \left (x \right )}}{2+m}\) | \(50\) |
risch | \(\frac {c \,x^{m} \left (a^{2} m^{2} x^{2}+3 a^{2} m \,x^{2}+2 a^{2} x^{2}+2 a x \,m^{2}+8 a m x +6 a x +m^{2}+5 m +6\right ) x}{\left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(73\) |
gosper | \(\frac {c \,x^{1+m} \left (a^{2} m^{2} x^{2}+3 a^{2} m \,x^{2}+2 a^{2} x^{2}+2 a x \,m^{2}+8 a m x +6 a x +m^{2}+5 m +6\right )}{\left (1+m \right ) \left (2+m \right ) \left (3+m \right )}\) | \(74\) |
orering | \(\frac {\left (a^{2} m^{2} x^{2}+3 a^{2} m \,x^{2}+2 a^{2} x^{2}+2 a x \,m^{2}+8 a m x +6 a x +m^{2}+5 m +6\right ) x \,x^{m} \left (-a^{2} c \,x^{2}+c \right )}{\left (3+m \right ) \left (2+m \right ) \left (1+m \right ) \left (-a^{2} x^{2}+1\right )}\) | \(95\) |
parallelrisch | \(\frac {x^{3} x^{m} a^{2} c \,m^{2}+3 x^{3} x^{m} a^{2} c m +2 x^{3} x^{m} a^{2} c +2 x^{2} x^{m} a c \,m^{2}+8 x^{2} x^{m} a c m +6 x^{2} x^{m} a c +x \,x^{m} c \,m^{2}+5 x \,x^{m} c m +6 c \,x^{m} x}{\left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(115\) |
meijerg | \(-\frac {\left (-a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} c \left (-\frac {2 x^{1+m} \left (-a^{2}\right )^{\frac {5}{2}+\frac {m}{2}} \left (a^{2} m \,x^{2}+a^{2} x^{2}+m +3\right )}{a^{4} \left (3+m \right ) \left (1+m \right )}+\frac {x^{1+m} \left (-a^{2}\right )^{\frac {5}{2}+\frac {m}{2}} \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}+\frac {1}{2}\right )}{a^{4}}\right )}{2}-\frac {\left (-a^{2}\right )^{-\frac {m}{2}} c \left (-\frac {2 x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \left (a^{2} m \,x^{2}+m +2\right )}{\left (2+m \right ) m}+x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}\right )\right )}{a}-\frac {\left (-a^{2}\right )^{-\frac {m}{2}} c \left (-\frac {2 x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \left (-2-m \right )}{\left (2+m \right ) m}-x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}\right )\right )}{a}+\frac {c \,x^{1+m} \left (\frac {m}{2}+\frac {1}{2}\right ) \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}+\frac {1}{2}\right )}{1+m}\) | \(274\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c),x,method=_RETURNVERBOSE)
Output:
c/(1+m)*x*exp(m*ln(x))+a^2*c/(3+m)*x^3*exp(m*ln(x))+2*a*c/(2+m)*x^2*exp(m* ln(x))
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.95 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=\frac {{\left ({\left (a^{2} c m^{2} + 3 \, a^{2} c m + 2 \, a^{2} c\right )} x^{3} + 2 \, {\left (a c m^{2} + 4 \, a c m + 3 \, a c\right )} x^{2} + {\left (c m^{2} + 5 \, c m + 6 \, c\right )} x\right )} x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c),x, algorithm="fricas")
Output:
((a^2*c*m^2 + 3*a^2*c*m + 2*a^2*c)*x^3 + 2*(a*c*m^2 + 4*a*c*m + 3*a*c)*x^2 + (c*m^2 + 5*c*m + 6*c)*x)*x^m/(m^3 + 6*m^2 + 11*m + 6)
Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (36) = 72\).
Time = 0.39 (sec) , antiderivative size = 299, normalized size of antiderivative = 7.12 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=\begin {cases} a^{2} c \log {\left (x \right )} - \frac {2 a c}{x} - \frac {c}{2 x^{2}} & \text {for}\: m = -3 \\a^{2} c x + 2 a c \log {\left (x \right )} - \frac {c}{x} & \text {for}\: m = -2 \\\frac {a^{2} c x^{2}}{2} + 2 a c x + c \log {\left (x \right )} & \text {for}\: m = -1 \\\frac {a^{2} c m^{2} x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {3 a^{2} c m x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {2 a^{2} c x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {2 a c m^{2} x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {8 a c m x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {6 a c x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {c m^{2} x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {5 c m x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {6 c x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*x**m*(-a**2*c*x**2+c),x)
Output:
Piecewise((a**2*c*log(x) - 2*a*c/x - c/(2*x**2), Eq(m, -3)), (a**2*c*x + 2 *a*c*log(x) - c/x, Eq(m, -2)), (a**2*c*x**2/2 + 2*a*c*x + c*log(x), Eq(m, -1)), (a**2*c*m**2*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 3*a**2*c*m*x**3* x**m/(m**3 + 6*m**2 + 11*m + 6) + 2*a**2*c*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 2*a*c*m**2*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + 8*a*c*m*x**2*x** m/(m**3 + 6*m**2 + 11*m + 6) + 6*a*c*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + c*m**2*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + 5*c*m*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + 6*c*x*x**m/(m**3 + 6*m**2 + 11*m + 6), True))
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.48 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=\frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{2} c x^{3} + 2 \, {\left (m^{2} + 4 \, m + 3\right )} a c x^{2} + {\left (m^{2} + 5 \, m + 6\right )} c x\right )} x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c),x, algorithm="maxima")
Output:
((m^2 + 3*m + 2)*a^2*c*x^3 + 2*(m^2 + 4*m + 3)*a*c*x^2 + (m^2 + 5*m + 6)*c *x)*x^m/(m^3 + 6*m^2 + 11*m + 6)
\[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=\int { \frac {{\left (a^{2} c x^{2} - c\right )} {\left (a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} - 1} \,d x } \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c),x, algorithm="giac")
Output:
integrate((a^2*c*x^2 - c)*(a*x + 1)^2*x^m/(a^2*x^2 - 1), x)
Time = 23.54 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.19 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=x^m\,\left (\frac {c\,x\,\left (m^2+5\,m+6\right )}{m^3+6\,m^2+11\,m+6}+\frac {2\,a\,c\,x^2\,\left (m^2+4\,m+3\right )}{m^3+6\,m^2+11\,m+6}+\frac {a^2\,c\,x^3\,\left (m^2+3\,m+2\right )}{m^3+6\,m^2+11\,m+6}\right ) \] Input:
int(-(x^m*(c - a^2*c*x^2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)
Output:
x^m*((c*x*(5*m + m^2 + 6))/(11*m + 6*m^2 + m^3 + 6) + (2*a*c*x^2*(4*m + m^ 2 + 3))/(11*m + 6*m^2 + m^3 + 6) + (a^2*c*x^3*(3*m + m^2 + 2))/(11*m + 6*m ^2 + m^3 + 6))
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.71 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right ) \, dx=\frac {x^{m} c x \left (a^{2} m^{2} x^{2}+3 a^{2} m \,x^{2}+2 a^{2} x^{2}+2 a \,m^{2} x +8 a m x +6 a x +m^{2}+5 m +6\right )}{m^{3}+6 m^{2}+11 m +6} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c),x)
Output:
(x**m*c*x*(a**2*m**2*x**2 + 3*a**2*m*x**2 + 2*a**2*x**2 + 2*a*m**2*x + 8*a *m*x + 6*a*x + m**2 + 5*m + 6))/(m**3 + 6*m**2 + 11*m + 6)