Integrand size = 17, antiderivative size = 69 \[ \int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx=-\frac {3}{8} \sqrt {-1+x} \sqrt {1+x}+\frac {1}{24} (7-2 x) (-1+x)^{3/2} \sqrt {1+x}+\frac {1}{4} (-1+x)^{3/2} x^2 \sqrt {1+x}+\frac {3 \text {arccosh}(x)}{8} \]
3/8*arccosh(x)+1/24*(7-2*x)*(-1+x)^(3/2)*(1+x)^(1/2)+1/4*(-1+x)^(3/2)*x^2* (1+x)^(1/2)-3/8*(-1+x)^(1/2)*(1+x)^(1/2)
Time = 0.00 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx=\frac {\sqrt {\frac {-1+x}{1+x}} \left (\sqrt {-1+x} \left (-16-7 x+x^2-2 x^3+6 x^4\right )-18 \sqrt {1+x} \log \left (\sqrt {-1+x}-\sqrt {1+x}\right )\right )}{24 \sqrt {-1+x}} \]
(Sqrt[(-1 + x)/(1 + x)]*(Sqrt[-1 + x]*(-16 - 7*x + x^2 - 2*x^3 + 6*x^4) - 18*Sqrt[1 + x]*Log[Sqrt[-1 + x] - Sqrt[1 + x]]))/(24*Sqrt[-1 + x])
Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2050, 111, 164, 60, 43}
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.
\(\displaystyle \int x^3 \sqrt {\frac {x-1}{x+1}} \, dx\) |
\(\Big \downarrow \) 2050 |
\(\displaystyle \int \frac {\sqrt {x-1} x^3}{\sqrt {x+1}}dx\) |
\(\Big \downarrow \) 111 |
\(\displaystyle \frac {1}{4} \int \frac {(2-x) \sqrt {x-1} x}{\sqrt {x+1}}dx+\frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{6} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{2} \int \frac {\sqrt {x-1}}{\sqrt {x+1}}dx\right )+\frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{6} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{2} \left (\sqrt {x-1} \sqrt {x+1}-\int \frac {1}{\sqrt {x-1} \sqrt {x+1}}dx\right )\right )+\frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2\) |
\(\Big \downarrow \) 43 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{6} (7-2 x) (x-1)^{3/2} \sqrt {x+1}-\frac {3}{2} \left (\sqrt {x-1} \sqrt {x+1}-\text {arccosh}(x)\right )\right )+\frac {1}{4} (x-1)^{3/2} \sqrt {x+1} x^2\) |
((-1 + x)^(3/2)*x^2*Sqrt[1 + x])/4 + (((7 - 2*x)*(-1 + x)^(3/2)*Sqrt[1 + x ])/6 - (3*(Sqrt[-1 + x]*Sqrt[1 + x] - ArcCosh[x]))/2)/4
3.8.36.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ ArcCosh[b*(x/a)]/(b*Sqrt[d/b]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a *d, 0] && GtQ[a, 0] && GtQ[d/b, 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_))^(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[((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[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p _), x_Symbol] :> Int[u*((a*e + b*e*x^n)^p/(c + d*x^n)^p), x] /; FreeQ[{a, b , c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - a*(d/b), 0]
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01
method | result | size |
risch | \(\frac {\left (6 x^{3}-8 x^{2}+9 x -16\right ) \left (x +1\right ) \sqrt {\frac {x -1}{x +1}}}{24}+\frac {3 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\frac {x -1}{x +1}}\, \sqrt {\left (x -1\right ) \left (x +1\right )}}{8 \left (x -1\right )}\) | \(70\) |
trager | \(\frac {\left (x +1\right ) \left (6 x^{3}-8 x^{2}+9 x -16\right ) \sqrt {-\frac {1-x}{x +1}}}{24}-\frac {3 \ln \left (-\sqrt {-\frac {1-x}{x +1}}\, x -\sqrt {-\frac {1-x}{x +1}}+x \right )}{8}\) | \(74\) |
default | \(\frac {\sqrt {\frac {x -1}{x +1}}\, \left (x +1\right ) \left (6 x \left (x^{2}-1\right )^{\frac {3}{2}}-8 \left (\left (x -1\right ) \left (x +1\right )\right )^{\frac {3}{2}}+15 x \sqrt {x^{2}-1}-24 \sqrt {x^{2}-1}+9 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{24 \sqrt {\left (x -1\right ) \left (x +1\right )}}\) | \(79\) |
1/24*(6*x^3-8*x^2+9*x-16)*(x+1)*((x-1)/(x+1))^(1/2)+3/8*ln(x+(x^2-1)^(1/2) )*((x-1)/(x+1))^(1/2)*((x-1)*(x+1))^(1/2)/(x-1)
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx=\frac {1}{24} \, {\left (6 \, x^{4} - 2 \, x^{3} + x^{2} - 7 \, x - 16\right )} \sqrt {\frac {x - 1}{x + 1}} + \frac {3}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {3}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]
1/24*(6*x^4 - 2*x^3 + x^2 - 7*x - 16)*sqrt((x - 1)/(x + 1)) + 3/8*log(sqrt ((x - 1)/(x + 1)) + 1) - 3/8*log(sqrt((x - 1)/(x + 1)) - 1)
\[ \int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx=\int x^{3} \sqrt {\frac {x - 1}{x + 1}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (49) = 98\).
Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.00 \[ \int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx=-\frac {39 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {7}{2}} - 31 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} + 49 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} - 9 \, \sqrt {\frac {x - 1}{x + 1}}}{12 \, {\left (\frac {4 \, {\left (x - 1\right )}}{x + 1} - \frac {6 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {4 \, {\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac {{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} + \frac {3}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {3}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]
-1/12*(39*((x - 1)/(x + 1))^(7/2) - 31*((x - 1)/(x + 1))^(5/2) + 49*((x - 1)/(x + 1))^(3/2) - 9*sqrt((x - 1)/(x + 1)))/(4*(x - 1)/(x + 1) - 6*(x - 1 )^2/(x + 1)^2 + 4*(x - 1)^3/(x + 1)^3 - (x - 1)^4/(x + 1)^4 - 1) + 3/8*log (sqrt((x - 1)/(x + 1)) + 1) - 3/8*log(sqrt((x - 1)/(x + 1)) - 1)
Time = 0.34 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90 \[ \int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx=-\frac {3}{8} \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (x + 1\right ) + \frac {1}{24} \, {\left ({\left (2 \, {\left (3 \, x \mathrm {sgn}\left (x + 1\right ) - 4 \, \mathrm {sgn}\left (x + 1\right )\right )} x + 9 \, \mathrm {sgn}\left (x + 1\right )\right )} x - 16 \, \mathrm {sgn}\left (x + 1\right )\right )} \sqrt {x^{2} - 1} \]
-3/8*log(abs(-x + sqrt(x^2 - 1)))*sgn(x + 1) + 1/24*((2*(3*x*sgn(x + 1) - 4*sgn(x + 1))*x + 9*sgn(x + 1))*x - 16*sgn(x + 1))*sqrt(x^2 - 1)
Time = 0.05 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.72 \[ \int x^3 \sqrt {\frac {-1+x}{1+x}} \, dx=\frac {3\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )}{4}-\frac {\frac {3\,\sqrt {\frac {x-1}{x+1}}}{4}-\frac {49\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{12}+\frac {31\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{12}-\frac {13\,{\left (\frac {x-1}{x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {4\,\left (x-1\right )}{x+1}-\frac {4\,{\left (x-1\right )}^3}{{\left (x+1\right )}^3}+\frac {{\left (x-1\right )}^4}{{\left (x+1\right )}^4}+1} \]