Integrand size = 17, antiderivative size = 113 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=\frac {c^3 x \sqrt {1-a^2 x^2}}{4 a}-\frac {1}{2} a c^3 x^3 \sqrt {1-a^2 x^2}-\frac {7 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 a^2}-\frac {1}{5} c^3 x^2 \left (1-a^2 x^2\right )^{3/2}-\frac {c^3 \arcsin (a x)}{4 a^2} \] Output:
1/4*c^3*x*(-a^2*x^2+1)^(1/2)/a-1/2*a*c^3*x^3*(-a^2*x^2+1)^(1/2)-7/15*c^3*( -a^2*x^2+1)^(3/2)/a^2-1/5*c^3*x^2*(-a^2*x^2+1)^(3/2)-1/4*c^3*arcsin(a*x)/a ^2
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=\frac {c^3 \left (\sqrt {1-a^2 x^2} \left (-28+15 a x+16 a^2 x^2-30 a^3 x^3+12 a^4 x^4\right )+30 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{60 a^2} \] Input:
Integrate[E^ArcTanh[a*x]*x*(c - a*c*x)^3,x]
Output:
(c^3*(Sqrt[1 - a^2*x^2]*(-28 + 15*a*x + 16*a^2*x^2 - 30*a^3*x^3 + 12*a^4*x ^4) + 30*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(60*a^2)
Time = 0.52 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {6678, 27, 541, 25, 27, 533, 27, 455, 211, 223}
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.
\(\displaystyle \int x e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int c^2 x (1-a x)^2 \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^3 \int x (1-a x)^2 \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 541 |
\(\displaystyle c^3 \left (-\frac {\int -a^2 x (7-10 a x) \sqrt {1-a^2 x^2}dx}{5 a^2}-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle c^3 \left (\frac {\int a^2 x (7-10 a x) \sqrt {1-a^2 x^2}dx}{5 a^2}-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^3 \left (\frac {1}{5} \int x (7-10 a x) \sqrt {1-a^2 x^2}dx-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 533 |
\(\displaystyle c^3 \left (\frac {1}{5} \left (\frac {\int -2 a (5-14 a x) \sqrt {1-a^2 x^2}dx}{4 a^2}+\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{2 a}\right )-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^3 \left (\frac {1}{5} \left (\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{2 a}-\frac {\int (5-14 a x) \sqrt {1-a^2 x^2}dx}{2 a}\right )-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c^3 \left (\frac {1}{5} \left (\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{2 a}-\frac {5 \int \sqrt {1-a^2 x^2}dx+\frac {14 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{2 a}\right )-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle c^3 \left (\frac {1}{5} \left (\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{2 a}-\frac {5 \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {14 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{2 a}\right )-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c^3 \left (\frac {1}{5} \left (\frac {5 x \left (1-a^2 x^2\right )^{3/2}}{2 a}-\frac {5 \left (\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {14 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{2 a}\right )-\frac {1}{5} x^2 \left (1-a^2 x^2\right )^{3/2}\right )\) |
Input:
Int[E^ArcTanh[a*x]*x*(c - a*c*x)^3,x]
Output:
c^3*(-1/5*(x^2*(1 - a^2*x^2)^(3/2)) + ((5*x*(1 - a^2*x^2)^(3/2))/(2*a) - ( (14*(1 - a^2*x^2)^(3/2))/(3*a) + 5*((x*Sqrt[1 - a^2*x^2])/2 + ArcSin[a*x]/ (2*a)))/(2*a))/5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x ] + Simp[1/(b*(m + n + 2*p + 1)) Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) *x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {\left (12 a^{4} x^{4}-30 a^{3} x^{3}+16 a^{2} x^{2}+15 a x -28\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{60 a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{3}}{4 a \sqrt {a^{2}}}\) | \(94\) |
default | \(-c^{3} \left (a^{4} \left (-\frac {x^{4} \sqrt {-a^{2} x^{2}+1}}{5 a^{2}}+\frac {-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a^{2}}\right )+\frac {\sqrt {-a^{2} x^{2}+1}}{a^{2}}+2 a \left (-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\right )-2 a^{3} \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )\right )\) | \(223\) |
meijerg | \(\frac {c^{3} \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}+8 a^{2} x^{2}+16\right ) \sqrt {-a^{2} x^{2}+1}}{15}\right )}{2 a^{2} \sqrt {\pi }}+\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}\right )}{a \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}} \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{a \sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {c^{3} \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {\pi }}\) | \(227\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x*(-a*c*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/60*(12*a^4*x^4-30*a^3*x^3+16*a^2*x^2+15*a*x-28)*(a^2*x^2-1)/a^2/(-a^2*x ^2+1)^(1/2)*c^3-1/4/a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2)) *c^3
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=\frac {30 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (12 \, a^{4} c^{3} x^{4} - 30 \, a^{3} c^{3} x^{3} + 16 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x - 28 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{60 \, a^{2}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x*(-a*c*x+c)^3,x, algorithm="fricas")
Output:
1/60*(30*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (12*a^4*c^3*x^4 - 30 *a^3*c^3*x^3 + 16*a^2*c^3*x^2 + 15*a*c^3*x - 28*c^3)*sqrt(-a^2*x^2 + 1))/a ^2
Time = 0.93 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.37 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (\frac {a^{2} c^{3} x^{4}}{5} - \frac {a c^{3} x^{3}}{2} + \frac {4 c^{3} x^{2}}{15} + \frac {c^{3} x}{4 a} - \frac {7 c^{3}}{15 a^{2}}\right ) - \frac {c^{3} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{4 a \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\- \frac {a^{4} c^{3} x^{6}}{6} + \frac {2 a^{3} c^{3} x^{5}}{5} - \frac {2 a c^{3} x^{3}}{3} + \frac {c^{3} x^{2}}{2} & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x*(-a*c*x+c)**3,x)
Output:
Piecewise((sqrt(-a**2*x**2 + 1)*(a**2*c**3*x**4/5 - a*c**3*x**3/2 + 4*c**3 *x**2/15 + c**3*x/(4*a) - 7*c**3/(15*a**2)) - c**3*log(-2*a**2*x + 2*sqrt( -a**2)*sqrt(-a**2*x**2 + 1))/(4*a*sqrt(-a**2)), Ne(a**2, 0)), (-a**4*c**3* x**6/6 + 2*a**3*c**3*x**5/5 - 2*a*c**3*x**3/3 + c**3*x**2/2, True))
Time = 0.13 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=\frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3} x^{4} - \frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} a c^{3} x^{3} + \frac {4}{15} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x^{2} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{3} x}{4 \, a} - \frac {c^{3} \arcsin \left (a x\right )}{4 \, a^{2}} - \frac {7 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{15 \, a^{2}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x*(-a*c*x+c)^3,x, algorithm="maxima")
Output:
1/5*sqrt(-a^2*x^2 + 1)*a^2*c^3*x^4 - 1/2*sqrt(-a^2*x^2 + 1)*a*c^3*x^3 + 4/ 15*sqrt(-a^2*x^2 + 1)*c^3*x^2 + 1/4*sqrt(-a^2*x^2 + 1)*c^3*x/a - 1/4*c^3*a rcsin(a*x)/a^2 - 7/15*sqrt(-a^2*x^2 + 1)*c^3/a^2
Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=-\frac {c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{4 \, a {\left | a \right |}} + \frac {1}{60} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (\frac {15 \, c^{3}}{a} + 2 \, {\left (8 \, c^{3} + 3 \, {\left (2 \, a^{2} c^{3} x - 5 \, a c^{3}\right )} x\right )} x\right )} x - \frac {28 \, c^{3}}{a^{2}}\right )} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x*(-a*c*x+c)^3,x, algorithm="giac")
Output:
-1/4*c^3*arcsin(a*x)*sgn(a)/(a*abs(a)) + 1/60*sqrt(-a^2*x^2 + 1)*((15*c^3/ a + 2*(8*c^3 + 3*(2*a^2*c^3*x - 5*a*c^3)*x)*x)*x - 28*c^3/a^2)
Time = 23.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=\frac {c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}{4\,a^3}-\frac {\frac {2\,c^3\,{\left (1-a^2\,x^2\right )}^{3/2}}{3}-\frac {c^3\,{\left (1-a^2\,x^2\right )}^{5/2}}{5}}{a^2}-\frac {\frac {c^3\,x\,\sqrt {1-a^2\,x^2}}{4}-\frac {c^3\,x\,{\left (1-a^2\,x^2\right )}^{3/2}}{2}}{a} \] Input:
int((x*(c - a*c*x)^3*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
(c^3*asinh(x*(-a^2)^(1/2))*(-a^2)^(1/2))/(4*a^3) - ((2*c^3*(1 - a^2*x^2)^( 3/2))/3 - (c^3*(1 - a^2*x^2)^(5/2))/5)/a^2 - ((c^3*x*(1 - a^2*x^2)^(1/2))/ 4 - (c^3*x*(1 - a^2*x^2)^(3/2))/2)/a
Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int e^{\text {arctanh}(a x)} x (c-a c x)^3 \, dx=\frac {c^{3} \left (-15 \mathit {asin} \left (a x \right )+12 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}-30 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+16 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+15 \sqrt {-a^{2} x^{2}+1}\, a x -28 \sqrt {-a^{2} x^{2}+1}+28\right )}{60 a^{2}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x*(-a*c*x+c)^3,x)
Output:
(c**3*( - 15*asin(a*x) + 12*sqrt( - a**2*x**2 + 1)*a**4*x**4 - 30*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 16*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 15*sqrt( - a**2*x**2 + 1)*a*x - 28*sqrt( - a**2*x**2 + 1) + 28))/(60*a**2)