Integrand size = 25, antiderivative size = 66 \[ \int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\frac {x^2 (1-a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {(4-3 a x) \sqrt {1-a^2 x^2}}{2 a^4}+\frac {3 \arcsin (a x)}{2 a^4} \]
3/2*arcsin(a*x)/a^4+x^2*(-a*x+1)/a^2/(-a^2*x^2+1)^(1/2)+1/2*(-3*a*x+4)*(-a ^2*x^2+1)^(1/2)/a^4
Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.05 \[ \int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (4+a x-a^2 x^2\right )}{2 a^4 (1+a x)}+\frac {3 \arctan \left (\frac {a x}{-1+\sqrt {1-a^2 x^2}}\right )}{a^4} \]
(Sqrt[1 - a^2*x^2]*(4 + a*x - a^2*x^2))/(2*a^4*(1 + a*x)) + (3*ArcTan[(a*x )/(-1 + Sqrt[1 - a^2*x^2])])/a^4
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {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.
| \(\displaystyle \int \frac {x^3}{(a x+1) \sqrt {1-a^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 563 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2}}{a^4 (a x+1)}-\frac {\int -\frac {a^2 x^2-a x+1}{\sqrt {1-a^2 x^2}}dx}{a^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a^2 x^2-a x+1}{\sqrt {1-a^2 x^2}}dx}{a^3}+\frac {\sqrt {1-a^2 x^2}}{a^4 (a x+1)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {-\frac {\int -\frac {a^2 (3-2 a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^3}+\frac {\sqrt {1-a^2 x^2}}{a^4 (a x+1)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (3-2 a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^3}+\frac {\sqrt {1-a^2 x^2}}{a^4 (a x+1)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {3-2 a x}{\sqrt {1-a^2 x^2}}dx-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^3}+\frac {\sqrt {1-a^2 x^2}}{a^4 (a x+1)}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {2 \sqrt {1-a^2 x^2}}{a}\right )-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^3}+\frac {\sqrt {1-a^2 x^2}}{a^4 (a x+1)}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2}}{a^4 (a x+1)}+\frac {\frac {1}{2} \left (\frac {2 \sqrt {1-a^2 x^2}}{a}+\frac {3 \arcsin (a x)}{a}\right )-\frac {1}{2} x \sqrt {1-a^2 x^2}}{a^3}\) |
Sqrt[1 - a^2*x^2]/(a^4*(1 + a*x)) + (-1/2*(x*Sqrt[1 - a^2*x^2]) + ((2*Sqrt [1 - a^2*x^2])/a + (3*ArcSin[a*x])/a)/2)/a^3
3.2.50.3.1 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]
Time = 0.37 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.47
| method | result | size |
| risch | \(\frac {\left (a x -2\right ) \left (a^{2} x^{2}-1\right )}{2 a^{4} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{3} \sqrt {a^{2}}}+\frac {\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2}+2 \left (x +\frac {1}{a}\right ) a}}{a^{5} \left (x +\frac {1}{a}\right )}\) | \(97\) |
| default | \(\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {a^{2}}}+\frac {-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}}{a}+\frac {\sqrt {-a^{2} x^{2}+1}}{a^{4}}+\frac {\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2}+2 \left (x +\frac {1}{a}\right ) a}}{a^{5} \left (x +\frac {1}{a}\right )}\) | \(134\) |
1/2*(a*x-2)*(a^2*x^2-1)/a^4/(-a^2*x^2+1)^(1/2)+3/2/a^3/(a^2)^(1/2)*arctan( (a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+1/a^5/(x+1/a)*(-(x+1/a)^2*a^2+2*(x+1/a)* a)^(1/2)
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.14 \[ \int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\frac {4 \, a x - 6 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (a^{2} x^{2} - a x - 4\right )} \sqrt {-a^{2} x^{2} + 1} + 4}{2 \, {\left (a^{5} x + a^{4}\right )}} \]
1/2*(4*a*x - 6*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (a^2*x^2 - a*x - 4)*sqrt(-a^2*x^2 + 1) + 4)/(a^5*x + a^4)
\[ \int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\int \frac {x^{3}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03 \[ \int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\frac {\sqrt {-a^{2} x^{2} + 1}}{a^{5} x + a^{4}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{3}} + \frac {3 \, \arcsin \left (a x\right )}{2 \, a^{4}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4}} \]
sqrt(-a^2*x^2 + 1)/(a^5*x + a^4) - 1/2*sqrt(-a^2*x^2 + 1)*x/a^3 + 3/2*arcs in(a*x)/a^4 + sqrt(-a^2*x^2 + 1)/a^4
Exception generated. \[ \int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\text {Exception raised: TypeError} \]
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 = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.76 \[ \int \frac {x^3}{(1+a x) \sqrt {1-a^2 x^2}} \, dx=\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^3\,\sqrt {-a^2}}-\frac {\left (\frac {1}{a^2\,\sqrt {-a^2}}+\frac {x\,\sqrt {-a^2}}{2\,a^3}\right )\,\sqrt {1-a^2\,x^2}}{\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a^3\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]