Integrand size = 18, antiderivative size = 86 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {11 c^2 \sqrt {1-a^2 x^2}}{3 a}-\frac {3}{2} c^2 x \sqrt {1-a^2 x^2}+\frac {1}{3} a c^2 x^2 \sqrt {1-a^2 x^2}+\frac {5 c^2 \arcsin (a x)}{2 a} \] Output:
11/3*c^2*(-a^2*x^2+1)^(1/2)/a-3/2*c^2*x*(-a^2*x^2+1)^(1/2)+1/3*a*c^2*x^2*( -a^2*x^2+1)^(1/2)+5/2*c^2*arcsin(a*x)/a
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {c^2 \left (\frac {\sqrt {1+a x} \left (22-31 a x+11 a^2 x^2-2 a^3 x^3\right )}{\sqrt {1-a x}}-30 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{6 a} \] Input:
Integrate[(c - a*c*x)^2/E^ArcTanh[a*x],x]
Output:
(c^2*((Sqrt[1 + a*x]*(22 - 31*a*x + 11*a^2*x^2 - 2*a^3*x^3))/Sqrt[1 - a*x] - 30*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(6*a)
Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6677, 27, 469, 469, 455, 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.
\(\displaystyle \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle \frac {\int \frac {c^3 (1-a x)^3}{\sqrt {1-a^2 x^2}}dx}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^2 \int \frac {(1-a x)^3}{\sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^2 \left (\frac {5}{3} \int \frac {(1-a x)^2}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^2 \left (\frac {5}{3} \left (\frac {3}{2} \int \frac {1-a x}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2} (1-a x)}{2 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c^2 \left (\frac {5}{3} \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)}{2 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c^2 \left (\frac {5}{3} \left (\frac {3}{2} \left (\frac {\sqrt {1-a^2 x^2}}{a}+\frac {\arcsin (a x)}{a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)}{2 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )\) |
Input:
Int[(c - a*c*x)^2/E^ArcTanh[a*x],x]
Output:
c^2*(((1 - a*x)^2*Sqrt[1 - a^2*x^2])/(3*a) + (5*(((1 - a*x)*Sqrt[1 - a^2*x ^2])/(2*a) + (3*(Sqrt[1 - a^2*x^2]/a + ArcSin[a*x]/a))/2))/3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* ((n + p)/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 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 = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87
method | result | size |
risch | \(-\frac {\left (2 a^{2} x^{2}-9 a x +22\right ) \left (a^{2} x^{2}-1\right ) c^{2}}{6 a \sqrt {-a^{2} x^{2}+1}}+\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{2}}{2 \sqrt {a^{2}}}\) | \(75\) |
default | \(c^{2} \left (-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a}-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{2}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}+\frac {4 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {4 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{a}\right )\) | \(132\) |
Input:
int((-a*c*x+c)^2/(a*x+1)*(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/6*(2*a^2*x^2-9*a*x+22)*(a^2*x^2-1)/a/(-a^2*x^2+1)^(1/2)*c^2+5/2/(a^2)^( 1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c^2
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=-\frac {30 \, c^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (2 \, a^{2} c^{2} x^{2} - 9 \, a c^{2} x + 22 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, a} \] Input:
integrate((-a*c*x+c)^2/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")
Output:
-1/6*(30*c^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (2*a^2*c^2*x^2 - 9*a *c^2*x + 22*c^2)*sqrt(-a^2*x^2 + 1))/a
\[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=c^{2} \left (\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a x + 1}\, dx + \int \left (- \frac {2 a x \sqrt {- a^{2} x^{2} + 1}}{a x + 1}\right )\, dx + \int \frac {a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a x + 1}\, dx\right ) \] Input:
integrate((-a*c*x+c)**2/(a*x+1)*(-a**2*x**2+1)**(1/2),x)
Output:
c**2*(Integral(sqrt(-a**2*x**2 + 1)/(a*x + 1), x) + Integral(-2*a*x*sqrt(- a**2*x**2 + 1)/(a*x + 1), x) + Integral(a**2*x**2*sqrt(-a**2*x**2 + 1)/(a* x + 1), x))
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=-\frac {3}{2} \, \sqrt {-a^{2} x^{2} + 1} c^{2} x - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c^{2}}{3 \, a} + \frac {5 \, c^{2} \arcsin \left (a x\right )}{2 \, a} + \frac {4 \, \sqrt {-a^{2} x^{2} + 1} c^{2}}{a} \] Input:
integrate((-a*c*x+c)^2/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")
Output:
-3/2*sqrt(-a^2*x^2 + 1)*c^2*x - 1/3*(-a^2*x^2 + 1)^(3/2)*c^2/a + 5/2*c^2*a rcsin(a*x)/a + 4*sqrt(-a^2*x^2 + 1)*c^2/a
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.63 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {5 \, c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, {\left | a \right |}} + \frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a c^{2} x - 9 \, c^{2}\right )} x + \frac {22 \, c^{2}}{a}\right )} \] Input:
integrate((-a*c*x+c)^2/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")
Output:
5/2*c^2*arcsin(a*x)*sgn(a)/abs(a) + 1/6*sqrt(-a^2*x^2 + 1)*((2*a*c^2*x - 9 *c^2)*x + 22*c^2/a)
Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {5\,c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}}-\frac {3\,c^2\,x\,\sqrt {1-a^2\,x^2}}{2}+\frac {11\,c^2\,\sqrt {1-a^2\,x^2}}{3\,a}+\frac {a\,c^2\,x^2\,\sqrt {1-a^2\,x^2}}{3} \] Input:
int(((1 - a^2*x^2)^(1/2)*(c - a*c*x)^2)/(a*x + 1),x)
Output:
(5*c^2*asinh(x*(-a^2)^(1/2)))/(2*(-a^2)^(1/2)) - (3*c^2*x*(1 - a^2*x^2)^(1 /2))/2 + (11*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 = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^2 \, dx=\frac {c^{2} \left (15 \mathit {asin} \left (a x \right )+2 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-9 \sqrt {-a^{2} x^{2}+1}\, a x +22 \sqrt {-a^{2} x^{2}+1}-22\right )}{6 a} \] Input:
int((-a*c*x+c)^2/(a*x+1)*(-a^2*x^2+1)^(1/2),x)
Output:
(c**2*(15*asin(a*x) + 2*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 9*sqrt( - a**2* x**2 + 1)*a*x + 22*sqrt( - a**2*x**2 + 1) - 22))/(6*a)