Integrand size = 17, antiderivative size = 62 \[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=-\frac {4 c (c+a c x)^{-1+p}}{a (1-p)}-\frac {4 (c+a c x)^p}{a p}+\frac {(c+a c x)^{1+p}}{a c (1+p)} \] 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 = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=\frac {(c+a c x)^p \left (-\frac {4+3 p}{p (1+p)}+\frac {a x}{1+p}+\frac {4}{(-1+p) (1+a x)}\right )}{a} \] Input:
Integrate[(c + a*c*x)^p/E^(4*ArcTanh[a*x]),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 = 0.44 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, 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.
\(\displaystyle \int e^{-4 \text {arctanh}(a x)} (a c x+c)^p \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {(1-a x)^2 (a c x+c)^p}{(a x+1)^2}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle c^2 \int (1-a x)^2 (a x c+c)^{p-2}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle c^2 \int \left (4 (a x c+c)^{p-2}-\frac {4 (a x c+c)^{p-1}}{c}+\frac {(a x c+c)^p}{c^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c^2 \left (\frac {(a c x+c)^{p+1}}{a c^3 (p+1)}-\frac {4 (a c x+c)^p}{a c^2 p}-\frac {4 (a c x+c)^{p-1}}{a c (1-p)}\right )\) |
Input:
Int[(c + a*c*x)^p/E^(4*ArcTanh[a*x]),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 = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18
method | result | size |
gosper | \(\frac {\left (a c x +c \right )^{p} \left (a^{2} p^{2} x^{2}-a^{2} x^{2} p -2 a \,p^{2} x -2 a p x +4 a x +p^{2}+3 p +4\right )}{\left (p^{2}-1\right ) a p \left (a x +1\right )}\) | \(73\) |
risch | \(\frac {\left (a^{2} p^{2} x^{2}-a^{2} x^{2} p -2 a \,p^{2} x -2 a p x +4 a x +p^{2}+3 p +4\right ) \left (a c x +c \right )^{p}}{a p \left (p +1\right ) \left (p -1\right ) \left (a x +1\right )}\) | \(76\) |
orering | \(\frac {\left (a^{2} p^{2} x^{2}-a^{2} x^{2} p -2 a \,p^{2} x -2 a p x +4 a x +p^{2}+3 p +4\right ) \left (a c x +c \right )^{p} \left (-a^{2} x^{2}+1\right )^{2}}{a p \left (p^{2}-1\right ) \left (a x +1\right )^{3} \left (a x -1\right )^{2}}\) | \(92\) |
parallelrisch | \(\frac {x^{2} \left (a c x +c \right )^{p} a^{2} p^{2}-x^{2} \left (a c x +c \right )^{p} a^{2} p -2 x \left (a c x +c \right )^{p} a \,p^{2}-2 x \left (a c x +c \right )^{p} a p +4 \left (a c x +c \right )^{p} x a +\left (a c x +c \right )^{p} p^{2}+3 \left (a c x +c \right )^{p} p +4 \left (a c x +c \right )^{p}}{\left (a x +1\right ) a p \left (p^{2}-1\right )}\) | \(131\) |
norman | \(\frac {\frac {a^{3} x^{4} {\mathrm e}^{p \ln \left (a c x +c \right )}}{p +1}+\frac {\left (p^{2}+3 p +4\right ) {\mathrm e}^{p \ln \left (a c x +c \right )}}{a p \left (p^{2}-1\right )}-\frac {4 a^{2} x^{3} {\mathrm e}^{p \ln \left (a c x +c \right )}}{p \left (p +1\right )}+\frac {4 \left (3+p \right ) x \,{\mathrm e}^{p \ln \left (a c x +c \right )}}{p \left (p^{2}-1\right )}-\frac {2 \left (p^{2}+p -6\right ) a \,x^{2} {\mathrm e}^{p \ln \left (a c x +c \right )}}{p \left (p^{2}-1\right )}}{\left (a x +1\right )^{3}}\) | \(148\) |
Input:
int((a*c*x+c)^p/(a*x+1)^4*(-a^2*x^2+1)^2,x,method=_RETURNVERBOSE)
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)/(p^2 -1)/a/p/(a*x+1)
Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.26 \[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=\frac {{\left ({\left (a^{2} p^{2} - a^{2} p\right )} x^{2} + p^{2} - 2 \, {\left (a p^{2} + a p - 2 \, a\right )} x + 3 \, p + 4\right )} {\left (a c x + c\right )}^{p}}{a p^{3} - a p + {\left (a^{2} p^{3} - a^{2} p\right )} x} \] Input:
integrate((a*c*x+c)^p/(a*x+1)^4*(-a^2*x^2+1)^2,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 = 1.04 (sec) , antiderivative size = 541, normalized size of antiderivative = 8.73 \[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=\begin {cases} c^{p} x & \text {for}\: a = 0 \\\frac {a^{2} x^{2} \log {\left (x + \frac {1}{a} \right )}}{a^{3} c x^{2} + 2 a^{2} c x + a c} + \frac {2 a x \log {\left (x + \frac {1}{a} \right )}}{a^{3} c x^{2} + 2 a^{2} c x + a c} + \frac {4 a x}{a^{3} c x^{2} + 2 a^{2} c x + a c} + \frac {\log {\left (x + \frac {1}{a} \right )}}{a^{3} c x^{2} + 2 a^{2} c x + a c} + \frac {2}{a^{3} c x^{2} + 2 a^{2} c x + a c} & \text {for}\: p = -1 \\\frac {a^{2} x^{2}}{a^{2} x + a} - \frac {4 a x \log {\left (x + \frac {1}{a} \right )}}{a^{2} x + a} - \frac {2 a x}{a^{2} x + a} - \frac {4 \log {\left (x + \frac {1}{a} \right )}}{a^{2} x + a} - \frac {7}{a^{2} x + a} & \text {for}\: p = 0 \\\frac {a c x^{2}}{2} - 3 c x + \frac {4 c \log {\left (x + \frac {1}{a} \right )}}{a} & \text {for}\: p = 1 \\\frac {a^{2} p^{2} x^{2} \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} - \frac {a^{2} p x^{2} \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} - \frac {2 a p^{2} x \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} - \frac {2 a p x \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} + \frac {4 a x \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} + \frac {p^{2} \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} + \frac {3 p \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} + \frac {4 \left (a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x + a p^{3} - a p} & \text {otherwise} \end {cases} \] Input:
integrate((a*c*x+c)**p/(a*x+1)**4*(-a**2*x**2+1)**2,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) - 2*a*x/(a**2*x + a) - 4*log(x + 1/a)/(a**2*x + a) - 7/(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))
Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (60) = 120\).
Time = 0.06 (sec) , antiderivative size = 389, normalized size of antiderivative = 6.27 \[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=\frac {{\left ({\left (p^{4} - 6 \, p^{3} + 11 \, p^{2} - 6 \, p\right )} a^{4} c^{p} x^{4} - 4 \, {\left (p^{3} - 6 \, p^{2} + 11 \, p - 6\right )} a^{3} c^{p} x^{3} + 12 \, {\left (p^{2} - 5 \, p + 6\right )} a^{2} c^{p} x^{2} - 24 \, a c^{p} {\left (p - 3\right )} x + 24 \, c^{p}\right )} {\left (a x + 1\right )}^{p} a^{4}}{{\left (p^{5} - 5 \, p^{4} + 5 \, p^{3} + 5 \, p^{2} - 6 \, p\right )} a^{8} x^{3} + 3 \, {\left (p^{5} - 5 \, p^{4} + 5 \, p^{3} + 5 \, p^{2} - 6 \, p\right )} a^{7} x^{2} + 3 \, {\left (p^{5} - 5 \, p^{4} + 5 \, p^{3} + 5 \, p^{2} - 6 \, p\right )} a^{6} x + {\left (p^{5} - 5 \, p^{4} + 5 \, p^{3} + 5 \, p^{2} - 6 \, p\right )} a^{5}} - \frac {2 \, {\left ({\left (p^{2} - 5 \, p + 6\right )} a^{2} c^{p} x^{2} - 2 \, a c^{p} {\left (p - 3\right )} x + 2 \, c^{p}\right )} {\left (a x + 1\right )}^{p} a^{2}}{{\left (p^{3} - 6 \, p^{2} + 11 \, p - 6\right )} a^{6} x^{3} + 3 \, {\left (p^{3} - 6 \, p^{2} + 11 \, p - 6\right )} a^{5} x^{2} + 3 \, {\left (p^{3} - 6 \, p^{2} + 11 \, p - 6\right )} a^{4} x + {\left (p^{3} - 6 \, p^{2} + 11 \, p - 6\right )} a^{3}} + \frac {{\left (a x + 1\right )}^{p} c^{p}}{a^{4} {\left (p - 3\right )} x^{3} + 3 \, a^{3} {\left (p - 3\right )} x^{2} + 3 \, a^{2} {\left (p - 3\right )} x + a {\left (p - 3\right )}} \] Input:
integrate((a*c*x+c)^p/(a*x+1)^4*(-a^2*x^2+1)^2,x, algorithm="maxima")
Output:
((p^4 - 6*p^3 + 11*p^2 - 6*p)*a^4*c^p*x^4 - 4*(p^3 - 6*p^2 + 11*p - 6)*a^3 *c^p*x^3 + 12*(p^2 - 5*p + 6)*a^2*c^p*x^2 - 24*a*c^p*(p - 3)*x + 24*c^p)*( a*x + 1)^p*a^4/((p^5 - 5*p^4 + 5*p^3 + 5*p^2 - 6*p)*a^8*x^3 + 3*(p^5 - 5*p ^4 + 5*p^3 + 5*p^2 - 6*p)*a^7*x^2 + 3*(p^5 - 5*p^4 + 5*p^3 + 5*p^2 - 6*p)* a^6*x + (p^5 - 5*p^4 + 5*p^3 + 5*p^2 - 6*p)*a^5) - 2*((p^2 - 5*p + 6)*a^2* c^p*x^2 - 2*a*c^p*(p - 3)*x + 2*c^p)*(a*x + 1)^p*a^2/((p^3 - 6*p^2 + 11*p - 6)*a^6*x^3 + 3*(p^3 - 6*p^2 + 11*p - 6)*a^5*x^2 + 3*(p^3 - 6*p^2 + 11*p - 6)*a^4*x + (p^3 - 6*p^2 + 11*p - 6)*a^3) + (a*x + 1)^p*c^p/(a^4*(p - 3)* x^3 + 3*a^3*(p - 3)*x^2 + 3*a^2*(p - 3)*x + a*(p - 3))
\[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=\int { \frac {{\left (a^{2} x^{2} - 1\right )}^{2} {\left (a c x + c\right )}^{p}}{{\left (a x + 1\right )}^{4}} \,d x } \] Input:
integrate((a*c*x+c)^p/(a*x+1)^4*(-a^2*x^2+1)^2,x, algorithm="giac")
Output:
integrate((a^2*x^2 - 1)^2*(a*c*x + c)^p/(a*x + 1)^4, x)
Time = 22.99 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=\frac {4\,{\left (c+a\,c\,x\right )}^p}{a\,\left (a\,x+1\right )\,\left (p-1\right )}-\frac {{\left (c+a\,c\,x\right )}^p\,\left (3\,p-a\,p\,x+4\right )}{a\,p\,\left (p+1\right )} \] Input:
int(((a^2*x^2 - 1)^2*(c + a*c*x)^p)/(a*x + 1)^4,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 = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.21 \[ \int e^{-4 \text {arctanh}(a x)} (c+a c x)^p \, dx=\frac {\left (a c x +c \right )^{p} \left (a^{2} p^{2} x^{2}-a^{2} p \,x^{2}-2 a \,p^{2} x -2 a p x +4 a x +p^{2}+3 p +4\right )}{a p \left (a \,p^{2} x -a x +p^{2}-1\right )} \] Input:
int((a*c*x+c)^p/(a*x+1)^4*(-a^2*x^2+1)^2,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))