Integrand size = 19, antiderivative size = 114 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{c-a c x} \, dx=\frac {(1+a x)^2}{a^4 c \sqrt {1-a^2 x^2}}+\frac {11 \sqrt {1-a^2 x^2}}{3 a^4 c}+\frac {x \sqrt {1-a^2 x^2}}{a^3 c}+\frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2 c}-\frac {3 \arcsin (a x)}{a^4 c} \]
-3*arcsin(a*x)/a^4/c+(a*x+1)^2/a^4/c/(-a^2*x^2+1)^(1/2)+11/3*(-a^2*x^2+1)^ (1/2)/a^4/c+x*(-a^2*x^2+1)^(1/2)/a^3/c+1/3*x^2*(-a^2*x^2+1)^(1/2)/a^2/c
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.63 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{c-a c x} \, dx=\frac {-\frac {\sqrt {1+a x} \left (-14+5 a x+2 a^2 x^2+a^3 x^3\right )}{\sqrt {1-a x}}+18 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{3 a^4 c} \]
(-((Sqrt[1 + a*x]*(-14 + 5*a*x + 2*a^2*x^2 + a^3*x^3))/Sqrt[1 - a*x]) + 18 *ArcSin[Sqrt[1 - a*x]/Sqrt[2]])/(3*a^4*c)
Time = 0.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {6678, 27, 563, 2346, 27, 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.
\(\displaystyle \int \frac {x^3 e^{\text {arctanh}(a x)}}{c-a c x} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {x^3 \sqrt {1-a^2 x^2}}{c^2 (1-a x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x^3 \sqrt {1-a^2 x^2}}{(1-a x)^2}dx}{c}\) |
\(\Big \downarrow \) 563 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {\int \frac {a^3 x^3+2 a^2 x^2+2 a x+2}{\sqrt {1-a^2 x^2}}dx}{a^3}}{c}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {-\frac {\int -\frac {2 \left (3 x^2 a^4+4 x a^3+3 a^2\right )}{\sqrt {1-a^2 x^2}}dx}{3 a^2}-\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^3}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {\frac {2 \int \frac {3 x^2 a^4+4 x a^3+3 a^2}{\sqrt {1-a^2 x^2}}dx}{3 a^2}-\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^3}}{c}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {\frac {2 \left (-\frac {\int -\frac {a^4 (8 a x+9)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {3}{2} a^2 x \sqrt {1-a^2 x^2}\right )}{3 a^2}-\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^3}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {\frac {2 \left (\frac {\int \frac {a^4 (8 a x+9)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {3}{2} a^2 x \sqrt {1-a^2 x^2}\right )}{3 a^2}-\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^3}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {\frac {2 \left (\frac {1}{2} a^2 \int \frac {8 a x+9}{\sqrt {1-a^2 x^2}}dx-\frac {3}{2} a^2 x \sqrt {1-a^2 x^2}\right )}{3 a^2}-\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^3}}{c}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {\frac {2 \left (\frac {1}{2} a^2 \left (9 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {8 \sqrt {1-a^2 x^2}}{a}\right )-\frac {3}{2} a^2 x \sqrt {1-a^2 x^2}\right )}{3 a^2}-\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^3}}{c}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^4 (1-a x)}-\frac {\frac {2 \left (\frac {1}{2} a^2 \left (\frac {9 \arcsin (a x)}{a}-\frac {8 \sqrt {1-a^2 x^2}}{a}\right )-\frac {3}{2} a^2 x \sqrt {1-a^2 x^2}\right )}{3 a^2}-\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^3}}{c}\) |
((2*Sqrt[1 - a^2*x^2])/(a^4*(1 - a*x)) - (-1/3*(a*x^2*Sqrt[1 - a^2*x^2]) + (2*((-3*a^2*x*Sqrt[1 - a^2*x^2])/2 + (a^2*((-8*Sqrt[1 - a^2*x^2])/a + (9* ArcSin[a*x])/a))/2))/(3*a^2))/a^3)/c
3.4.28.3.1 Defintions of rubi rules used
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 = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.06
method | result | size |
risch | \(-\frac {\left (a^{2} x^{2}+3 a x +8\right ) \left (a^{2} x^{2}-1\right )}{3 a^{4} \sqrt {-a^{2} x^{2}+1}\, c}-\frac {\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {a^{2}}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{a^{5} \left (x -\frac {1}{a}\right )}}{c}\) | \(121\) |
default | \(-\frac {-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}+\frac {2 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {a^{2}}}+\frac {-\frac {x \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}}{a}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{a^{5} \left (x -\frac {1}{a}\right )}}{c}\) | \(169\) |
-1/3*(a^2*x^2+3*a*x+8)*(a^2*x^2-1)/a^4/(-a^2*x^2+1)^(1/2)/c-(3/a^3/(a^2)^( 1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2/a^5/(x-1/a)*(-(x-1/a)^2*a^ 2-2*(x-1/a)*a)^(1/2))/c
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{c-a c x} \, dx=\frac {14 \, a x + 18 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (a^{3} x^{3} + 2 \, a^{2} x^{2} + 5 \, a x - 14\right )} \sqrt {-a^{2} x^{2} + 1} - 14}{3 \, {\left (a^{5} c x - a^{4} c\right )}} \]
1/3*(14*a*x + 18*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a^3*x ^3 + 2*a^2*x^2 + 5*a*x - 14)*sqrt(-a^2*x^2 + 1) - 14)/(a^5*c*x - a^4*c)
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{c-a c x} \, dx=- \frac {\int \frac {x^{3}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{4}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \]
-(Integral(x**3/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + In tegral(a*x**4/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{c-a c x} \, dx=-\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{5} c x - a^{4} c} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{3 \, a^{2} c} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{3} c} - \frac {3 \, \arcsin \left (a x\right )}{a^{4} c} + \frac {8 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, a^{4} c} \]
-2*sqrt(-a^2*x^2 + 1)/(a^5*c*x - a^4*c) + 1/3*sqrt(-a^2*x^2 + 1)*x^2/(a^2* c) + sqrt(-a^2*x^2 + 1)*x/(a^3*c) - 3*arcsin(a*x)/(a^4*c) + 8/3*sqrt(-a^2* x^2 + 1)/(a^4*c)
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{c-a c x} \, dx=\frac {1}{3} \, \sqrt {-a^{2} x^{2} + 1} {\left (x {\left (\frac {x}{a^{2} c} + \frac {3}{a^{3} c}\right )} + \frac {8}{a^{4} c}\right )} - \frac {3 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{a^{3} c {\left | a \right |}} + \frac {4}{a^{3} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \]
1/3*sqrt(-a^2*x^2 + 1)*(x*(x/(a^2*c) + 3/(a^3*c)) + 8/(a^4*c)) - 3*arcsin( a*x)*sgn(a)/(a^3*c*abs(a)) + 4/(a^3*c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^ 2*x) - 1)*abs(a))
Time = 3.71 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{c-a c x} \, dx=\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2}{3\,c\,{\left (-a^2\right )}^{3/2}}-\frac {2}{a^2\,c\,\sqrt {-a^2}}+\frac {a^2\,x^2}{3\,c\,{\left (-a^2\right )}^{3/2}}+\frac {x\,\sqrt {-a^2}}{a^3\,c}\right )}{\sqrt {-a^2}}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^3\,c\,\sqrt {-a^2}}+\frac {2\,\sqrt {1-a^2\,x^2}}{a^3\,c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]