Integrand size = 23, antiderivative size = 136 \[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {152 c \sqrt {1-a^2 x^2}}{105 a^3 \sqrt {c-a c x}}+\frac {38 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{105 a^3}-\frac {4 (c-a c x)^{3/2} \sqrt {1-a^2 x^2}}{35 a^3 c}+\frac {2 (c-a c x)^{5/2} \sqrt {1-a^2 x^2}}{7 a^3 c^2} \] Output:
152/105*c*(-a^2*x^2+1)^(1/2)/a^3/(-a*c*x+c)^(1/2)+38/105*(-a*c*x+c)^(1/2)* (-a^2*x^2+1)^(1/2)/a^3-4/35*(-a*c*x+c)^(3/2)*(-a^2*x^2+1)^(1/2)/a^3/c+2/7* (-a*c*x+c)^(5/2)*(-a^2*x^2+1)^(1/2)/a^3/c^2
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.40 \[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=-\frac {2 c \sqrt {1-a^2 x^2} \left (-104+52 a x-39 a^2 x^2+15 a^3 x^3\right )}{105 a^3 \sqrt {c-a c x}} \] Input:
Integrate[(x^2*Sqrt[c - a*c*x])/E^ArcTanh[a*x],x]
Output:
(-2*c*Sqrt[1 - a^2*x^2]*(-104 + 52*a*x - 39*a^2*x^2 + 15*a^3*x^3))/(105*a^ 3*Sqrt[c - a*c*x])
Time = 0.43 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6678, 574, 581, 27, 672, 458}
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.
\(\displaystyle \int x^2 e^{-\text {arctanh}(a x)} \sqrt {c-a c x} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle \frac {\int \frac {x^2 (c-a c x)^{3/2}}{\sqrt {1-a^2 x^2}}dx}{c}\) |
\(\Big \downarrow \) 574 |
\(\displaystyle \frac {\frac {13}{7} c \int \frac {x^2 \sqrt {c-a c x}}{\sqrt {1-a^2 x^2}}dx-\frac {2 c^2 x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}}{c}\) |
\(\Big \downarrow \) 581 |
\(\displaystyle \frac {\frac {13}{7} c \left (\frac {2 \int \frac {c^2 (2 a x+3) \sqrt {c-a c x}}{2 \sqrt {1-a^2 x^2}}dx}{5 a^2 c^2}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a^3 c}\right )-\frac {2 c^2 x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {13}{7} c \left (\frac {\int \frac {(2 a x+3) \sqrt {c-a c x}}{\sqrt {1-a^2 x^2}}dx}{5 a^2}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a^3 c}\right )-\frac {2 c^2 x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}}{c}\) |
\(\Big \downarrow \) 672 |
\(\displaystyle \frac {\frac {13}{7} c \left (\frac {\frac {7}{3} \int \frac {\sqrt {c-a c x}}{\sqrt {1-a^2 x^2}}dx-\frac {4 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{3 a}}{5 a^2}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a^3 c}\right )-\frac {2 c^2 x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}}{c}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {\frac {13}{7} c \left (\frac {\frac {14 c \sqrt {1-a^2 x^2}}{3 a \sqrt {c-a c x}}-\frac {4 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{3 a}}{5 a^2}+\frac {2 \sqrt {1-a^2 x^2} (c-a c x)^{3/2}}{5 a^3 c}\right )-\frac {2 c^2 x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}}{c}\) |
Input:
Int[(x^2*Sqrt[c - a*c*x])/E^ArcTanh[a*x],x]
Output:
((-2*c^2*x^3*Sqrt[1 - a^2*x^2])/(7*Sqrt[c - a*c*x]) + (13*c*((2*(c - a*c*x )^(3/2)*Sqrt[1 - a^2*x^2])/(5*a^3*c) + ((14*c*Sqrt[1 - a^2*x^2])/(3*a*Sqrt [c - a*c*x]) - (4*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(3*a))/(5*a^2)))/7)/c
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
Int[((e_.)*(x_))^(n_)*((c_) + (d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d^2*(e*x)^(n + 1)*(c + d*x)^(m - 2)*((a + b*x^2)^(p + 1)/ (b*e*(n + p + 2))), x] + Simp[c*((2*n + p + 3)/(n + p + 2)) Int[(e*x)^n*( c + d*x)^(m - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x ] && EqQ[b*c^2 + a*d^2, 0] && EqQ[m + p - 1, 0] && !LtQ[n, -1] && IntegerQ [2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b*x^ 2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m + c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p )*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & & IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] )
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)) Int[(d + e*x )^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^ 2 + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
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 = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.41
method | result | size |
gosper | \(\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-a c x +c}\, \left (15 a^{3} x^{3}-39 a^{2} x^{2}+52 a x -104\right )}{105 \left (a x -1\right ) a^{3}}\) | \(56\) |
orering | \(\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-a c x +c}\, \left (15 a^{3} x^{3}-39 a^{2} x^{2}+52 a x -104\right )}{105 \left (a x -1\right ) a^{3}}\) | \(56\) |
default | \(\frac {2 \sqrt {-c \left (a x -1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (15 a^{3} x^{3}-39 a^{2} x^{2}+52 a x -104\right )}{105 \left (a x -1\right ) a^{3}}\) | \(57\) |
risch | \(\frac {2 \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c \left (15 a^{3} x^{3}-39 a^{2} x^{2}+52 a x -104\right ) \left (a x +1\right )}{105 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, a^{3} \sqrt {c \left (a x +1\right )}}\) | \(92\) |
Input:
int(x^2*(-a*c*x+c)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOS E)
Output:
2/105*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)*(15*a^3*x^3-39*a^2*x^2+52*a*x-10 4)/(a*x-1)/a^3
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.43 \[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2 \, {\left (15 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 52 \, a x - 104\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{105 \, {\left (a^{4} x - a^{3}\right )}} \] Input:
integrate(x^2*(-a*c*x+c)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fr icas")
Output:
2/105*(15*a^3*x^3 - 39*a^2*x^2 + 52*a*x - 104)*sqrt(-a^2*x^2 + 1)*sqrt(-a* c*x + c)/(a^4*x - a^3)
\[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\int \frac {x^{2} \sqrt {- c \left (a x - 1\right )} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, dx \] Input:
integrate(x**2*(-a*c*x+c)**(1/2)/(a*x+1)*(-a**2*x**2+1)**(1/2),x)
Output:
Integral(x**2*sqrt(-c*(a*x - 1))*sqrt(-(a*x - 1)*(a*x + 1))/(a*x + 1), x)
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.46 \[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=-\frac {2 \, {\left (15 \, a^{3} \sqrt {c} x^{3} - 39 \, a^{2} \sqrt {c} x^{2} + 52 \, a \sqrt {c} x - 104 \, \sqrt {c}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{105 \, {\left (a^{4} x - a^{3}\right )}} \] Input:
integrate(x^2*(-a*c*x+c)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="ma xima")
Output:
-2/105*(15*a^3*sqrt(c)*x^3 - 39*a^2*sqrt(c)*x^2 + 52*a*sqrt(c)*x - 104*sqr t(c))*sqrt(a*x + 1)*(a*x - 1)/(a^4*x - a^3)
Exception generated. \[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^2*(-a*c*x+c)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="gi ac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 14.37 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54 \[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}\,\left (15\,a^2\,x^2-24\,a\,x+28\right )}{105\,a^3}-\frac {152\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{105\,a^3\,\left (a\,x-1\right )} \] Input:
int((x^2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2))/(a*x + 1),x)
Output:
(2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2)*(15*a^2*x^2 - 24*a*x + 28))/(105* a^3) - (152*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2))/(105*a^3*(a*x - 1))
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.26 \[ \int e^{-\text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c}\, \sqrt {a x +1}\, \left (-15 a^{3} x^{3}+39 a^{2} x^{2}-52 a x +104\right )}{105 a^{3}} \] Input:
int(x^2*(-a*c*x+c)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x)
Output:
(2*sqrt(c)*sqrt(a*x + 1)*( - 15*a**3*x**3 + 39*a**2*x**2 - 52*a*x + 104))/ (105*a**3)