Integrand size = 18, antiderivative size = 61 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {1}{2} c^2 x \sqrt {1-a^2 x^2}-\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {c^2 \arcsin (a x)}{2 a} \] Output:
1/2*c^2*x*(-a^2*x^2+1)^(1/2)-1/3*c^2*(-a^2*x^2+1)^(3/2)/a+1/2*c^2*arcsin(a *x)/a
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {c^2 \left (\sqrt {1-a^2 x^2} \left (-2+3 a x+2 a^2 x^2\right )-6 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{6 a} \] Input:
Integrate[E^(3*ArcTanh[a*x])*(c - a*c*x)^2,x]
Output:
(c^2*(Sqrt[1 - a^2*x^2]*(-2 + 3*a*x + 2*a^2*x^2) - 6*ArcSin[Sqrt[1 - a*x]/ Sqrt[2]]))/(6*a)
Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6677, 27, 466, 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.
\(\displaystyle \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{c (1-a x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^2 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{1-a x}dx\) |
\(\Big \downarrow \) 466 |
\(\displaystyle c^2 \left (\int \sqrt {1-a^2 x^2}dx-\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle c^2 \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c^2 \left (-\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )\) |
Input:
Int[E^(3*ArcTanh[a*x])*(c - a*c*x)^2,x]
Output:
c^2*((x*Sqrt[1 - a^2*x^2])/2 - (1 - a^2*x^2)^(3/2)/(3*a) + ArcSin[a*x]/(2* a))
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_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 2*(n + 2*p + 1))) Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 ] || EqQ[n + p + 1, 0]) && NeQ[n + 2*p + 1, 0] && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23
method | result | size |
risch | \(-\frac {\left (2 a^{2} x^{2}+3 a x -2\right ) \left (a^{2} x^{2}-1\right ) c^{2}}{6 a \sqrt {-a^{2} x^{2}+1}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{2}}{2 \sqrt {a^{2}}}\) | \(75\) |
default | \(c^{2} \left (\frac {x}{\sqrt {-a^{2} x^{2}+1}}+\frac {1}{a \sqrt {-a^{2} x^{2}+1}}+a^{4} \left (-\frac {x^{3}}{2 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {\frac {3 x}{2 a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )+a^{5} \left (-\frac {x^{4}}{3 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {-\frac {4 x^{2}}{3 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {8}{3 a^{4} \sqrt {-a^{2} x^{2}+1}}}{a^{2}}\right )-2 a^{2} \left (\frac {x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}\right )-2 a^{3} \left (-\frac {x^{2}}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {-a^{2} x^{2}+1}}\right )\right )\) | \(278\) |
meijerg | \(\frac {2 c^{2} \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {c^{2} \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}+\frac {c^{2} x}{\sqrt {-a^{2} x^{2}+1}}-\frac {c^{2} \left (\frac {8 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-2 a^{4} x^{4}-8 a^{2} x^{2}+16\right )}{6 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}+\frac {c^{2} \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (-5 a^{2} x^{2}+15\right )}{10 a^{4} \sqrt {-a^{2} x^{2}+1}}-\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{2 a^{5}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {2 c^{2} \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{4 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}\) | \(281\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-1/6*(2*a^2*x^2+3*a*x-2)*(a^2*x^2-1)/a/(-a^2*x^2+1)^(1/2)*c^2+1/2/(a^2)^(1 /2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c^2
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=-\frac {6 \, c^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (2 \, a^{2} c^{2} x^{2} + 3 \, a c^{2} x - 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, a} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^2,x, algorithm="fricas")
Output:
-1/6*(6*c^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (2*a^2*c^2*x^2 + 3*a* c^2*x - 2*c^2)*sqrt(-a^2*x^2 + 1))/a
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (48) = 96\).
Time = 5.02 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.95 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=\begin {cases} \frac {c^{2} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{2 \sqrt {- a^{2}}} + \sqrt {- a^{2} x^{2} + 1} \left (\frac {a c^{2} x^{2}}{3} + \frac {c^{2} x}{2} - \frac {c^{2}}{3 a}\right ) & \text {for}\: a^{2} \neq 0 \\- \frac {a^{3} c^{2} x^{4}}{4} - \frac {a^{2} c^{2} x^{3}}{3} + \frac {a c^{2} x^{2}}{2} + c^{2} x & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(-a*c*x+c)**2,x)
Output:
Piecewise((c**2*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(2*sqr t(-a**2)) + sqrt(-a**2*x**2 + 1)*(a*c**2*x**2/3 + c**2*x/2 - c**2/(3*a)), Ne(a**2, 0)), (-a**3*c**2*x**4/4 - a**2*c**2*x**3/3 + a*c**2*x**2/2 + c**2 *x, True))
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (51) = 102\).
Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.93 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=-\frac {a^{3} c^{2} x^{4}}{3 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {a^{2} c^{2} x^{3}}{2 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {2 \, a c^{2} x^{2}}{3 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {c^{2} x}{2 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {c^{2} \arcsin \left (a x\right )}{2 \, a} - \frac {c^{2}}{3 \, \sqrt {-a^{2} x^{2} + 1} a} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^2,x, algorithm="maxima")
Output:
-1/3*a^3*c^2*x^4/sqrt(-a^2*x^2 + 1) - 1/2*a^2*c^2*x^3/sqrt(-a^2*x^2 + 1) + 2/3*a*c^2*x^2/sqrt(-a^2*x^2 + 1) + 1/2*c^2*x/sqrt(-a^2*x^2 + 1) + 1/2*c^2 *arcsin(a*x)/a - 1/3*c^2/(sqrt(-a^2*x^2 + 1)*a)
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, {\left | a \right |}} + \frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a c^{2} x + 3 \, c^{2}\right )} x - \frac {2 \, c^{2}}{a}\right )} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^2,x, algorithm="giac")
Output:
1/2*c^2*arcsin(a*x)*sgn(a)/abs(a) + 1/6*sqrt(-a^2*x^2 + 1)*((2*a*c^2*x + 3 *c^2)*x - 2*c^2/a)
Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.34 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {c^2\,x\,\sqrt {1-a^2\,x^2}}{2}+\frac {c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}}-\frac {c^2\,\sqrt {1-a^2\,x^2}}{3\,a}+\frac {a\,c^2\,x^2\,\sqrt {1-a^2\,x^2}}{3} \] Input:
int(((c - a*c*x)^2*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
Output:
(c^2*x*(1 - a^2*x^2)^(1/2))/2 + (c^2*asinh(x*(-a^2)^(1/2)))/(2*(-a^2)^(1/2 )) - (c^2*(1 - a^2*x^2)^(1/2))/(3*a) + (a*c^2*x^2*(1 - a^2*x^2)^(1/2))/3
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {c^{2} \left (3 \mathit {asin} \left (a x \right )+2 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+3 \sqrt {-a^{2} x^{2}+1}\, a x -2 \sqrt {-a^{2} x^{2}+1}+2\right )}{6 a} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^2,x)
Output:
(c**2*(3*asin(a*x) + 2*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 3*sqrt( - a**2*x **2 + 1)*a*x - 2*sqrt( - a**2*x**2 + 1) + 2))/(6*a)