Integrand size = 20, antiderivative size = 78 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {13 x^2 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 x^3 \sqrt {1+x}}{3 (1-x)^{3/2}}-\frac {1}{6} \sqrt {1-x} \sqrt {1+x} (52+33 x)+\frac {11 \arcsin (x)}{2} \]
11/2*arcsin(x)+2/3*x^3*(1+x)^(1/2)/(1-x)^(3/2)-13/3*x^2*(1+x)^(1/2)/(1-x)^ (1/2)-1/6*(52+33*x)*(1-x)^(1/2)*(1+x)^(1/2)
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {\sqrt {1+x} \left (52-71 x+12 x^2+3 x^3\right )}{6 (1-x)^{3/2}}+11 \arctan \left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \]
-1/6*(Sqrt[1 + x]*(52 - 71*x + 12*x^2 + 3*x^3))/(1 - x)^(3/2) + 11*ArcTan[ Sqrt[1 + x]/Sqrt[1 - x]]
Time = 0.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {108, 27, 167, 25, 164, 39, 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.
\(\displaystyle \int \frac {x^3 \sqrt {x+1}}{(1-x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 108 |
\(\displaystyle \frac {2 x^3 \sqrt {x+1}}{3 (1-x)^{3/2}}-\frac {2}{3} \int \frac {x^2 (7 x+6)}{2 (1-x)^{3/2} \sqrt {x+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 x^3 \sqrt {x+1}}{3 (1-x)^{3/2}}-\frac {1}{3} \int \frac {x^2 (7 x+6)}{(1-x)^{3/2} \sqrt {x+1}}dx\) |
\(\Big \downarrow \) 167 |
\(\displaystyle \frac {1}{3} \left (-\int -\frac {x (33 x+26)}{\sqrt {1-x} \sqrt {x+1}}dx-\frac {13 \sqrt {x+1} x^2}{\sqrt {1-x}}\right )+\frac {2 \sqrt {x+1} x^3}{3 (1-x)^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (\int \frac {x (33 x+26)}{\sqrt {1-x} \sqrt {x+1}}dx-\frac {13 x^2 \sqrt {x+1}}{\sqrt {1-x}}\right )+\frac {2 \sqrt {x+1} x^3}{3 (1-x)^{3/2}}\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {1}{3} \left (\frac {33}{2} \int \frac {1}{\sqrt {1-x} \sqrt {x+1}}dx-\frac {13 \sqrt {x+1} x^2}{\sqrt {1-x}}-\frac {1}{2} \sqrt {1-x} \sqrt {x+1} (33 x+52)\right )+\frac {2 \sqrt {x+1} x^3}{3 (1-x)^{3/2}}\) |
\(\Big \downarrow \) 39 |
\(\displaystyle \frac {1}{3} \left (\frac {33}{2} \int \frac {1}{\sqrt {1-x^2}}dx-\frac {13 \sqrt {x+1} x^2}{\sqrt {1-x}}-\frac {1}{2} \sqrt {1-x} \sqrt {x+1} (33 x+52)\right )+\frac {2 \sqrt {x+1} x^3}{3 (1-x)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{3} \left (\frac {33 \arcsin (x)}{2}-\frac {13 \sqrt {x+1} x^2}{\sqrt {1-x}}-\frac {1}{2} \sqrt {1-x} \sqrt {x+1} (33 x+52)\right )+\frac {2 \sqrt {x+1} x^3}{3 (1-x)^{3/2}}\) |
(2*x^3*Sqrt[1 + x])/(3*(1 - x)^(3/2)) + ((-13*x^2*Sqrt[1 + x])/Sqrt[1 - x] - (Sqrt[1 - x]*Sqrt[1 + x]*(52 + 33*x))/2 + (33*ArcSin[x])/2)/3
3.9.28.3.1 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[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[( a*c + b*d*x^2)^m, x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 0] && ( IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
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 = 0.60 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {\left (3 x^{4}+15 x^{3}-59 x^{2}-19 x +52\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{6 \left (-1+x \right ) \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}+\frac {11 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {1-x}\, \sqrt {1+x}}\) | \(89\) |
default | \(\frac {\left (-3 x^{3} \sqrt {-x^{2}+1}+33 \arcsin \left (x \right ) x^{2}-12 x^{2} \sqrt {-x^{2}+1}-66 \arcsin \left (x \right ) x +71 x \sqrt {-x^{2}+1}+33 \arcsin \left (x \right )-52 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{6 \left (-1+x \right )^{2} \sqrt {-x^{2}+1}}\) | \(97\) |
1/6*(3*x^4+15*x^3-59*x^2-19*x+52)/(-1+x)/(-(-1+x)*(1+x))^(1/2)*((1+x)*(1-x ))^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)+11/2*arcsin(x)*((1+x)*(1-x))^(1/2)/(1-x)^ (1/2)/(1+x)^(1/2)
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {52 \, x^{2} + {\left (3 \, x^{3} + 12 \, x^{2} - 71 \, x + 52\right )} \sqrt {x + 1} \sqrt {-x + 1} + 66 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 104 \, x + 52}{6 \, {\left (x^{2} - 2 \, x + 1\right )}} \]
-1/6*(52*x^2 + (3*x^3 + 12*x^2 - 71*x + 52)*sqrt(x + 1)*sqrt(-x + 1) + 66* (x^2 - 2*x + 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - 104*x + 52)/(x^ 2 - 2*x + 1)
\[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\int \frac {x^{3} \sqrt {x + 1}}{\left (1 - x\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.63 \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {{\left ({\left (3 \, {\left (x + 2\right )} {\left (x + 1\right )} - 86\right )} {\left (x + 1\right )} + 132\right )} \sqrt {x + 1} \sqrt {-x + 1}}{6 \, {\left (x - 1\right )}^{2}} + 11 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]
-1/6*((3*(x + 2)*(x + 1) - 86)*(x + 1) + 132)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 + 11*arcsin(1/2*sqrt(2)*sqrt(x + 1))
Timed out. \[ \int \frac {x^3 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\int \frac {x^3\,\sqrt {x+1}}{{\left (1-x\right )}^{5/2}} \,d x \]