Integrand size = 20, antiderivative size = 110 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {3}{4} \sqrt {1-x} \sqrt {1+x}-\frac {1}{4} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3 \arcsin (x)}{4} \]
3/4*arcsin(x)-1/4*(1-x)^(1/2)*(1+x)^(3/2)-1/10*(1-x)^(1/2)*(1+x)^(5/2)-1/5 *x^2*(1+x)^(5/2)*(1-x)^(1/2)-1/10*(1+x)^(7/2)*(1-x)^(1/2)-3/4*(1-x)^(1/2)* (1+x)^(1/2)
Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {\sqrt {1-x} \left (24+39 x+27 x^2+22 x^3+14 x^4+4 x^5\right )}{20 \sqrt {1+x}}-\frac {3}{2} \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \]
-1/20*(Sqrt[1 - x]*(24 + 39*x + 27*x^2 + 22*x^3 + 14*x^4 + 4*x^5))/Sqrt[1 + x] - (3*ArcTan[Sqrt[1 - x]/Sqrt[1 + x]])/2
Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {111, 27, 90, 60, 60, 50, 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.
\(\displaystyle \int \frac {x^3 (x+1)^{3/2}}{\sqrt {1-x}} \, dx\) |
\(\Big \downarrow \) 111 |
\(\displaystyle -\frac {1}{5} \int -\frac {2 x (x+1)^{5/2}}{\sqrt {1-x}}dx-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \int \frac {x (x+1)^{5/2}}{\sqrt {1-x}}dx-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {2}{5} \left (\frac {3}{4} \int \frac {(x+1)^{5/2}}{\sqrt {1-x}}dx-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\right )-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2}{5} \left (\frac {3}{4} \left (\frac {5}{3} \int \frac {(x+1)^{3/2}}{\sqrt {1-x}}dx-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\right )-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2}{5} \left (\frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int \frac {\sqrt {x+1}}{\sqrt {1-x}}dx-\frac {1}{2} \sqrt {1-x} (x+1)^{3/2}\right )-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\right )-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}\) |
\(\Big \downarrow \) 50 |
\(\displaystyle \frac {2}{5} \left (\frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {1-x^2}}dx-\sqrt {1-x^2}\right )-\frac {1}{2} \sqrt {1-x} (x+1)^{3/2}\right )-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\right )-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {2}{5} \left (\frac {3}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\arcsin (x)-\sqrt {1-x^2}\right )-\frac {1}{2} \sqrt {1-x} (x+1)^{3/2}\right )-\frac {1}{3} \sqrt {1-x} (x+1)^{5/2}\right )-\frac {1}{4} \sqrt {1-x} (x+1)^{7/2}\right )-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}\) |
-1/5*(Sqrt[1 - x]*x^2*(1 + x)^(5/2)) + (2*(-1/4*(Sqrt[1 - x]*(1 + x)^(7/2) ) + (3*(-1/3*(Sqrt[1 - x]*(1 + x)^(5/2)) + (5*(-1/2*(Sqrt[1 - x]*(1 + x)^( 3/2)) + (3*(-Sqrt[1 - x^2] + ArcSin[x]))/2))/3))/4))/5
3.8.24.3.1 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[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a *c + b*d*x^2)^m/(2*d*m), x] + Simp[a Int[(a*c + b*d*x^2)^n, x], x] /; Fre eQ[{a, b, c, d, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 1] && GtQ[m, 0 ] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
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]
Time = 1.65 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {\left (4 x^{4}+10 x^{3}+12 x^{2}+15 x +24\right ) \left (-1+x \right ) \sqrt {1+x}\, \sqrt {\left (1+x \right ) \left (1-x \right )}}{20 \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}+\frac {3 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{4 \sqrt {1-x}\, \sqrt {1+x}}\) | \(87\) |
default | \(\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (-4 x^{4} \sqrt {-x^{2}+1}-10 x^{3} \sqrt {-x^{2}+1}-12 x^{2} \sqrt {-x^{2}+1}-15 x \sqrt {-x^{2}+1}+15 \arcsin \left (x \right )-24 \sqrt {-x^{2}+1}\right )}{20 \sqrt {-x^{2}+1}}\) | \(94\) |
1/20*(4*x^4+10*x^3+12*x^2+15*x+24)*(-1+x)*(1+x)^(1/2)/(-(-1+x)*(1+x))^(1/2 )*((1+x)*(1-x))^(1/2)/(1-x)^(1/2)+3/4*arcsin(x)*((1+x)*(1-x))^(1/2)/(1-x)^ (1/2)/(1+x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.52 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{20} \, {\left (4 \, x^{4} + 10 \, x^{3} + 12 \, x^{2} + 15 \, x + 24\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {3}{2} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]
-1/20*(4*x^4 + 10*x^3 + 12*x^2 + 15*x + 24)*sqrt(x + 1)*sqrt(-x + 1) - 3/2 *arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)
\[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\int \frac {x^{3} \left (x + 1\right )^{\frac {3}{2}}}{\sqrt {1 - x}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{5} \, \sqrt {-x^{2} + 1} x^{4} - \frac {1}{2} \, \sqrt {-x^{2} + 1} x^{3} - \frac {3}{5} \, \sqrt {-x^{2} + 1} x^{2} - \frac {3}{4} \, \sqrt {-x^{2} + 1} x - \frac {6}{5} \, \sqrt {-x^{2} + 1} + \frac {3}{4} \, \arcsin \left (x\right ) \]
-1/5*sqrt(-x^2 + 1)*x^4 - 1/2*sqrt(-x^2 + 1)*x^3 - 3/5*sqrt(-x^2 + 1)*x^2 - 3/4*sqrt(-x^2 + 1)*x - 6/5*sqrt(-x^2 + 1) + 3/4*arcsin(x)
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.47 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, {\left ({\left (2 \, x - 1\right )} {\left (x + 1\right )} + 3\right )} {\left (x + 1\right )} + 5\right )} {\left (x + 1\right )} + 15\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {3}{2} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]
-1/20*((2*((2*x - 1)*(x + 1) + 3)*(x + 1) + 5)*(x + 1) + 15)*sqrt(x + 1)*s qrt(-x + 1) + 3/2*arcsin(1/2*sqrt(2)*sqrt(x + 1))
Timed out. \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\int \frac {x^3\,{\left (x+1\right )}^{3/2}}{\sqrt {1-x}} \,d x \]