Integrand size = 18, antiderivative size = 66 \[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=\frac {4 c (c-a c x)^{-1+p}}{a (1-p)}+\frac {4 (c-a c x)^p}{a p}-\frac {(c-a c x)^{1+p}}{a c (1+p)} \] Output:
4*c*(-a*c*x+c)^(-1+p)/a/(1-p)+4*(-a*c*x+c)^p/a/p-(-a*c*x+c)^(p+1)/a/c/(p+1 )
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=\frac {(c-a c x)^p \left (\frac {4+3 p}{p (1+p)}+\frac {a x}{1+p}+\frac {4}{(-1+p) (-1+a x)}\right )}{a} \] Input:
Integrate[E^(4*ArcTanh[a*x])*(c - a*c*x)^p,x]
Output:
((c - a*c*x)^p*((4 + 3*p)/(p*(1 + p)) + (a*x)/(1 + p) + 4/((-1 + p)*(-1 + a*x))))/a
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6680, 35, 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.
\(\displaystyle \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {(a x+1)^2 (c-a c x)^p}{(1-a x)^2}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle c^2 \int (a x+1)^2 (c-a c x)^{p-2}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle c^2 \int \left (4 (c-a c x)^{p-2}-\frac {4 (c-a c x)^{p-1}}{c}+\frac {(c-a c x)^p}{c^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c^2 \left (-\frac {(c-a c x)^{p+1}}{a c^3 (p+1)}+\frac {4 (c-a c x)^p}{a c^2 p}+\frac {4 (c-a c x)^{p-1}}{a c (1-p)}\right )\) |
Input:
Int[E^(4*ArcTanh[a*x])*(c - a*c*x)^p,x]
Output:
c^2*((4*(c - a*c*x)^(-1 + p))/(a*c*(1 - p)) + (4*(c - a*c*x)^p)/(a*c^2*p) - (c - a*c*x)^(1 + p)/(a*c^3*(1 + p)))
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
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_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Int[u*(c + d*x)^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c , d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.12
method | result | size |
gosper | \(\frac {\left (a^{2} p^{2} x^{2}-a^{2} x^{2} p +2 a \,p^{2} x +2 a p x -4 a x +p^{2}+3 p +4\right ) \left (-a c x +c \right )^{p}}{\left (a x -1\right ) a p \left (p^{2}-1\right )}\) | \(74\) |
risch | \(\frac {\left (a^{2} p^{2} x^{2}-a^{2} x^{2} p +2 a \,p^{2} x +2 a p x -4 a x +p^{2}+3 p +4\right ) \left (-a c x +c \right )^{p}}{a p \left (p +1\right ) \left (p -1\right ) \left (a x -1\right )}\) | \(77\) |
orering | \(\frac {\left (a^{2} p^{2} x^{2}-a^{2} x^{2} p +2 a \,p^{2} x +2 a p x -4 a x +p^{2}+3 p +4\right ) \left (a x +1\right )^{2} \left (a x -1\right ) \left (-a c x +c \right )^{p}}{a p \left (p^{2}-1\right ) \left (-a^{2} x^{2}+1\right )^{2}}\) | \(91\) |
norman | \(\frac {\frac {a^{2} x^{3} {\mathrm e}^{p \ln \left (-a c x +c \right )}}{p +1}+\frac {\left (3 p +5\right ) x \,{\mathrm e}^{p \ln \left (-a c x +c \right )}}{p^{2}-1}+\frac {\left (p^{2}+3 p +4\right ) {\mathrm e}^{p \ln \left (-a c x +c \right )}}{a p \left (p^{2}-1\right )}+\frac {\left (3 p +4\right ) a \,x^{2} {\mathrm e}^{p \ln \left (-a c x +c \right )}}{p \left (p +1\right )}}{a^{2} x^{2}-1}\) | \(124\) |
parallelrisch | \(\frac {x^{2} \left (-a c x +c \right )^{p} a^{2} p^{2}-x^{2} \left (-a c x +c \right )^{p} a^{2} p +2 x \left (-a c x +c \right )^{p} a \,p^{2}+2 x \left (-a c x +c \right )^{p} a p -4 a \left (-a c x +c \right )^{p} x +\left (-a c x +c \right )^{p} p^{2}+3 \left (-a c x +c \right )^{p} p +4 \left (-a c x +c \right )^{p}}{\left (a x -1\right ) a p \left (p^{2}-1\right )}\) | \(139\) |
Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c)^p,x,method=_RETURNVERBOSE)
Output:
(a^2*p^2*x^2-a^2*p*x^2+2*a*p^2*x+2*a*p*x-4*a*x+p^2+3*p+4)*(-a*c*x+c)^p/(a* x-1)/a/p/(p^2-1)
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23 \[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=-\frac {{\left ({\left (a^{2} p^{2} - a^{2} p\right )} x^{2} + p^{2} + 2 \, {\left (a p^{2} + a p - 2 \, a\right )} x + 3 \, p + 4\right )} {\left (-a c x + c\right )}^{p}}{a p^{3} - a p - {\left (a^{2} p^{3} - a^{2} p\right )} x} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c)^p,x, algorithm="fricas")
Output:
-((a^2*p^2 - a^2*p)*x^2 + p^2 + 2*(a*p^2 + a*p - 2*a)*x + 3*p + 4)*(-a*c*x + c)^p/(a*p^3 - a*p - (a^2*p^3 - a^2*p)*x)
Leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (48) = 96\).
Time = 1.19 (sec) , antiderivative size = 541, normalized size of antiderivative = 8.20 \[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=\begin {cases} c^{p} x & \text {for}\: a = 0 \\- \frac {a^{2} x^{2} \log {\left (x - \frac {1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {2 a x \log {\left (x - \frac {1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {4 a x}{a^{3} c x^{2} - 2 a^{2} c x + a c} - \frac {\log {\left (x - \frac {1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} - \frac {2}{a^{3} c x^{2} - 2 a^{2} c x + a c} & \text {for}\: p = -1 \\\frac {a^{2} x^{2}}{a^{2} x - a} + \frac {4 a x \log {\left (x - \frac {1}{a} \right )}}{a^{2} x - a} - \frac {a x}{a^{2} x - a} - \frac {4 \log {\left (x - \frac {1}{a} \right )}}{a^{2} x - a} - \frac {4}{a^{2} x - a} & \text {for}\: p = 0 \\- \frac {a c x^{2}}{2} - 3 c x - \frac {4 c \log {\left (x - \frac {1}{a} \right )}}{a} & \text {for}\: p = 1 \\\frac {a^{2} p^{2} x^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} - \frac {a^{2} p x^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {2 a p^{2} x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {2 a p x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} - \frac {4 a x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {p^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {3 p \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {4 \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)**4/(-a**2*x**2+1)**2*(-a*c*x+c)**p,x)
Output:
Piecewise((c**p*x, Eq(a, 0)), (-a**2*x**2*log(x - 1/a)/(a**3*c*x**2 - 2*a* *2*c*x + a*c) + 2*a*x*log(x - 1/a)/(a**3*c*x**2 - 2*a**2*c*x + a*c) + 4*a* x/(a**3*c*x**2 - 2*a**2*c*x + a*c) - log(x - 1/a)/(a**3*c*x**2 - 2*a**2*c* x + a*c) - 2/(a**3*c*x**2 - 2*a**2*c*x + a*c), Eq(p, -1)), (a**2*x**2/(a** 2*x - a) + 4*a*x*log(x - 1/a)/(a**2*x - a) - a*x/(a**2*x - a) - 4*log(x - 1/a)/(a**2*x - a) - 4/(a**2*x - a), Eq(p, 0)), (-a*c*x**2/2 - 3*c*x - 4*c* log(x - 1/a)/a, Eq(p, 1)), (a**2*p**2*x**2*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) - a**2*p*x**2*(-a*c*x + c)**p/(a**2*p**3*x - a**2 *p*x - a*p**3 + a*p) + 2*a*p**2*x*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) + 2*a*p*x*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) - 4*a*x*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) + p**2*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) + 3*p*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) + 4*(-a*c*x + c)**p/(a**2 *p**3*x - a**2*p*x - a*p**3 + a*p), True))
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.17 \[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=\frac {{\left ({\left (p^{2} - p\right )} a^{2} c^{p} x^{2} + 2 \, {\left (p^{2} + p - 2\right )} a c^{p} x + {\left (p^{2} + 3 \, p + 4\right )} c^{p}\right )} {\left (-a x + 1\right )}^{p}}{{\left (p^{3} - p\right )} a^{2} x - {\left (p^{3} - p\right )} a} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c)^p,x, algorithm="maxima")
Output:
((p^2 - p)*a^2*c^p*x^2 + 2*(p^2 + p - 2)*a*c^p*x + (p^2 + 3*p + 4)*c^p)*(- a*x + 1)^p/((p^3 - p)*a^2*x - (p^3 - p)*a)
\[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=\int { \frac {{\left (a x + 1\right )}^{4} {\left (-a c x + c\right )}^{p}}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c)^p,x, algorithm="giac")
Output:
integrate((a*x + 1)^4*(-a*c*x + c)^p/(a^2*x^2 - 1)^2, x)
Time = 14.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86 \[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=\frac {4\,{\left (c-a\,c\,x\right )}^p}{a\,\left (a\,x-1\right )\,\left (p-1\right )}+\frac {{\left (c-a\,c\,x\right )}^p\,\left (3\,p+a\,p\,x+4\right )}{a\,p\,\left (p+1\right )} \] Input:
int(((c - a*c*x)^p*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)
Output:
(4*(c - a*c*x)^p)/(a*(a*x - 1)*(p - 1)) + ((c - a*c*x)^p*(3*p + a*p*x + 4) )/(a*p*(p + 1))
Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.18 \[ \int e^{4 \text {arctanh}(a x)} (c-a c x)^p \, dx=\frac {\left (-a c x +c \right )^{p} \left (a^{2} p^{2} x^{2}-a^{2} p \,x^{2}+2 a \,p^{2} x +2 a p x -4 a x +p^{2}+3 p +4\right )}{a p \left (a \,p^{2} x -a x -p^{2}+1\right )} \] Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c)^p,x)
Output:
(( - a*c*x + c)**p*(a**2*p**2*x**2 - a**2*p*x**2 + 2*a*p**2*x + 2*a*p*x - 4*a*x + p**2 + 3*p + 4))/(a*p*(a*p**2*x - a*x - p**2 + 1))