Integrand size = 19, antiderivative size = 92 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=\frac {2 \sqrt {1-a^2 x^2}}{a^3 c}+\frac {x \sqrt {1-a^2 x^2}}{2 a^2 c}+\frac {2 \sqrt {1-a^2 x^2}}{a^3 c (1-a x)}-\frac {5 \arcsin (a x)}{2 a^3 c} \] Output:
2*(-a^2*x^2+1)^(1/2)/a^3/c+1/2*x*(-a^2*x^2+1)^(1/2)/a^2/c+2*(-a^2*x^2+1)^( 1/2)/a^3/c/(-a*x+1)-5/2*arcsin(a*x)/a^3/c
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=\frac {-\frac {\sqrt {1+a x} \left (-8+3 a x+a^2 x^2\right )}{\sqrt {1-a x}}+10 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{2 a^3 c} \] Input:
Integrate[(E^ArcTanh[a*x]*x^2)/(c - a*c*x),x]
Output:
(-((Sqrt[1 + a*x]*(-8 + 3*a*x + a^2*x^2))/Sqrt[1 - a*x]) + 10*ArcSin[Sqrt[ 1 - a*x]/Sqrt[2]])/(2*a^3*c)
Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {6678, 27, 563, 25, 2346, 25, 27, 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.
\(\displaystyle \int \frac {x^2 e^{\text {arctanh}(a x)}}{c-a c x} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {x^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x^2 \sqrt {1-a^2 x^2}}{(1-a x)^2}dx}{c}\) |
\(\Big \downarrow \) 563 |
\(\displaystyle \frac {\frac {\int -\frac {a^2 x^2+2 a x+2}{\sqrt {1-a^2 x^2}}dx}{a^2}+\frac {2 \sqrt {1-a^2 x^2}}{a^3 (1-a x)}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^3 (1-a x)}-\frac {\int \frac {a^2 x^2+2 a x+2}{\sqrt {1-a^2 x^2}}dx}{a^2}}{c}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^3 (1-a x)}-\frac {-\frac {\int -\frac {a^2 (4 a x+5)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^2}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^3 (1-a x)}-\frac {\frac {\int \frac {a^2 (4 a x+5)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^2}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^3 (1-a x)}-\frac {\frac {1}{2} \int \frac {4 a x+5}{\sqrt {1-a^2 x^2}}dx-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^2}}{c}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^3 (1-a x)}-\frac {\frac {1}{2} \left (5 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {4 \sqrt {1-a^2 x^2}}{a}\right )-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^2}}{c}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^3 (1-a x)}-\frac {\frac {1}{2} \left (\frac {5 \arcsin (a x)}{a}-\frac {4 \sqrt {1-a^2 x^2}}{a}\right )-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^2}}{c}\) |
Input:
Int[(E^ArcTanh[a*x]*x^2)/(c - a*c*x),x]
Output:
((2*Sqrt[1 - a^2*x^2])/(a^3*(1 - a*x)) - (-1/2*(x*Sqrt[1 - a^2*x^2]) + ((- 4*Sqrt[1 - a^2*x^2])/a + (5*ArcSin[a*x])/a)/2)/a^2)/c
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[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)* b^(n + 2)*(c + d*x))), x] - Simp[d^(2*n - m + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(-n - 1)*(-c)^(m - n - 1) - d^m*x^m*(-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2 , 0] && IGtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
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 = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{a^{3}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{4} \left (x -\frac {1}{a}\right )}}{c}\) | \(113\) |
risch | \(-\frac {\left (a x +4\right ) \left (a^{2} x^{2}-1\right )}{2 a^{3} \sqrt {-a^{2} x^{2}+1}\, c}-\frac {\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{4} \left (x -\frac {1}{a}\right )}}{c}\) | \(113\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a*c*x+c),x,method=_RETURNVERBOSE)
Output:
-1/c*(-1/2*x*(-a^2*x^2+1)^(1/2)/a^2+5/2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2) *x/(-a^2*x^2+1)^(1/2))-2*(-a^2*x^2+1)^(1/2)/a^3+2/a^4/(x-1/a)*(-(x-1/a)^2* a^2-2*a*(x-1/a))^(1/2))
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=\frac {8 \, a x + 10 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (a^{2} x^{2} + 3 \, a x - 8\right )} \sqrt {-a^{2} x^{2} + 1} - 8}{2 \, {\left (a^{4} c x - a^{3} c\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a*c*x+c),x, algorithm="fricas")
Output:
1/2*(8*a*x + 10*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a^2*x^ 2 + 3*a*x - 8)*sqrt(-a^2*x^2 + 1) - 8)/(a^4*c*x - a^3*c)
\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=- \frac {\int \frac {x^{2}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{3}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**2/(-a*c*x+c),x)
Output:
-(Integral(x**2/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + In tegral(a*x**3/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=-\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4} c x - a^{3} c} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{2} c} - \frac {5 \, \arcsin \left (a x\right )}{2 \, a^{3} c} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{3} c} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a*c*x+c),x, algorithm="maxima")
Output:
-2*sqrt(-a^2*x^2 + 1)/(a^4*c*x - a^3*c) + 1/2*sqrt(-a^2*x^2 + 1)*x/(a^2*c) - 5/2*arcsin(a*x)/(a^3*c) + 2*sqrt(-a^2*x^2 + 1)/(a^3*c)
Exception generated. \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a*c*x+c),x, algorithm="giac")
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 = 0.03 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2\,\sqrt {-a^2}}{a^3\,c}+\frac {x\,\sqrt {-a^2}}{2\,a^2\,c}\right )}{\sqrt {-a^2}}-\frac {5\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^2\,c\,\sqrt {-a^2}}+\frac {2\,\sqrt {1-a^2\,x^2}}{a^2\,c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \] Input:
int((x^2*(a*x + 1))/((1 - a^2*x^2)^(1/2)*(c - a*c*x)),x)
Output:
((1 - a^2*x^2)^(1/2)*((2*(-a^2)^(1/2))/(a^3*c) + (x*(-a^2)^(1/2))/(2*a^2*c )))/(-a^2)^(1/2) - (5*asinh(x*(-a^2)^(1/2)))/(2*a^2*c*(-a^2)^(1/2)) + (2*( 1 - a^2*x^2)^(1/2))/(a^2*c*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)*(-a^2)^(1/2))
Time = 0.15 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{c-a c x} \, dx=\frac {-5 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-5 \mathit {asin} \left (a x \right ) a x +5 \mathit {asin} \left (a x \right )+\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+3 \sqrt {-a^{2} x^{2}+1}\, a x -10 \sqrt {-a^{2} x^{2}+1}-a^{3} x^{3}-4 a^{2} x^{2}+3 a x +10}{2 a^{3} c \left (\sqrt {-a^{2} x^{2}+1}+a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a*c*x+c),x)
Output:
( - 5*sqrt( - a**2*x**2 + 1)*asin(a*x) - 5*asin(a*x)*a*x + 5*asin(a*x) + s qrt( - a**2*x**2 + 1)*a**2*x**2 + 3*sqrt( - a**2*x**2 + 1)*a*x - 10*sqrt( - a**2*x**2 + 1) - a**3*x**3 - 4*a**2*x**2 + 3*a*x + 10)/(2*a**3*c*(sqrt( - a**2*x**2 + 1) + a*x - 1))