Integrand size = 15, antiderivative size = 49 \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\frac {5}{8} x \sqrt {-1+x^2}+\frac {1}{12} (8+3 x) \left (-1+x^2\right )^{3/2}-\frac {5}{8} \text {arctanh}\left (\frac {x}{\sqrt {-1+x^2}}\right ) \] Output:
5/8*x*(x^2-1)^(1/2)+1/12*(8+3*x)*(x^2-1)^(3/2)-5/8*arctanh(x/(x^2-1)^(1/2) )
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\frac {1}{24} \sqrt {-1+x^2} \left (-16+9 x+16 x^2+6 x^3\right )-\frac {5}{4} \text {arctanh}\left (\frac {\sqrt {-1+x^2}}{-1+x}\right ) \] Input:
Integrate[(1 + x)^2*Sqrt[-1 + x^2],x]
Output:
(Sqrt[-1 + x^2]*(-16 + 9*x + 16*x^2 + 6*x^3))/24 - (5*ArcTanh[Sqrt[-1 + x^ 2]/(-1 + x)])/4
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {469, 455, 211, 224, 219}
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.
\(\displaystyle \int (x+1)^2 \sqrt {x^2-1} \, dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {5}{4} \int (x+1) \sqrt {x^2-1}dx+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {5}{4} \left (\int \sqrt {x^2-1}dx+\frac {1}{3} \left (x^2-1\right )^{3/2}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {5}{4} \left (-\frac {1}{2} \int \frac {1}{\sqrt {x^2-1}}dx+\frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {5}{4} \left (-\frac {1}{2} \int \frac {1}{1-\frac {x^2}{x^2-1}}d\frac {x}{\sqrt {x^2-1}}+\frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{4} \left (-\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt {x^2-1}}\right )+\frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\) |
Input:
Int[(1 + x)^2*Sqrt[-1 + x^2],x]
Output:
((1 + x)*(-1 + x^2)^(3/2))/4 + (5*((x*Sqrt[-1 + x^2])/2 + (-1 + x^2)^(3/2) /3 - ArcTanh[x/Sqrt[-1 + x^2]]/2))/4
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* ((n + p)/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* p + 1, 0] && IntegerQ[2*p]
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76
method | result | size |
trager | \(\left (\frac {1}{4} x^{3}+\frac {2}{3} x^{2}+\frac {3}{8} x -\frac {2}{3}\right ) \sqrt {x^{2}-1}-\frac {5 \ln \left (x +\sqrt {x^{2}-1}\right )}{8}\) | \(37\) |
risch | \(\frac {\left (6 x^{3}+16 x^{2}+9 x -16\right ) \sqrt {x^{2}-1}}{24}-\frac {5 \ln \left (x +\sqrt {x^{2}-1}\right )}{8}\) | \(38\) |
default | \(\frac {5 x \sqrt {x^{2}-1}}{8}-\frac {5 \ln \left (x +\sqrt {x^{2}-1}\right )}{8}+\frac {x \left (x^{2}-1\right )^{\frac {3}{2}}}{4}+\frac {2 \left (x^{2}-1\right )^{\frac {3}{2}}}{3}\) | \(43\) |
meijerg | \(\frac {i \sqrt {\operatorname {signum}\left (x^{2}-1\right )}\, \left (-2 i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-2 i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{4 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}}-\frac {i \sqrt {\operatorname {signum}\left (x^{2}-1\right )}\, \left (-\frac {i \sqrt {\pi }\, x \left (-6 x^{2}+3\right ) \sqrt {-x^{2}+1}}{6}+\frac {i \sqrt {\pi }\, \arcsin \left (x \right )}{2}\right )}{4 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}}+\frac {\sqrt {\operatorname {signum}\left (x^{2}-1\right )}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {2 \sqrt {\pi }\, \left (-2 x^{2}+2\right ) \sqrt {-x^{2}+1}}{3}\right )}{2 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}}\) | \(157\) |
Input:
int((x+1)^2*(x^2-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
(1/4*x^3+2/3*x^2+3/8*x-2/3)*(x^2-1)^(1/2)-5/8*ln(x+(x^2-1)^(1/2))
Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\frac {1}{24} \, {\left (6 \, x^{3} + 16 \, x^{2} + 9 \, x - 16\right )} \sqrt {x^{2} - 1} + \frac {5}{8} \, \log \left (-x + \sqrt {x^{2} - 1}\right ) \] Input:
integrate((1+x)^2*(x^2-1)^(1/2),x, algorithm="fricas")
Output:
1/24*(6*x^3 + 16*x^2 + 9*x - 16)*sqrt(x^2 - 1) + 5/8*log(-x + sqrt(x^2 - 1 ))
Time = 0.36 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\sqrt {x^{2} - 1} \left (\frac {x^{3}}{4} + \frac {2 x^{2}}{3} + \frac {3 x}{8} - \frac {2}{3}\right ) - \frac {5 \log {\left (2 x + 2 \sqrt {x^{2} - 1} \right )}}{8} \] Input:
integrate((1+x)**2*(x**2-1)**(1/2),x)
Output:
sqrt(x**2 - 1)*(x**3/4 + 2*x**2/3 + 3*x/8 - 2/3) - 5*log(2*x + 2*sqrt(x**2 - 1))/8
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\frac {1}{4} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} x + \frac {2}{3} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} + \frac {5}{8} \, \sqrt {x^{2} - 1} x - \frac {5}{8} \, \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \] Input:
integrate((1+x)^2*(x^2-1)^(1/2),x, algorithm="maxima")
Output:
1/4*(x^2 - 1)^(3/2)*x + 2/3*(x^2 - 1)^(3/2) + 5/8*sqrt(x^2 - 1)*x - 5/8*lo g(2*x + 2*sqrt(x^2 - 1))
Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\frac {1}{24} \, {\left ({\left (2 \, {\left (3 \, x + 8\right )} x + 9\right )} x - 16\right )} \sqrt {x^{2} - 1} + \frac {5}{8} \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right ) \] Input:
integrate((1+x)^2*(x^2-1)^(1/2),x, algorithm="giac")
Output:
1/24*((2*(3*x + 8)*x + 9)*x - 16)*sqrt(x^2 - 1) + 5/8*log(abs(-x + sqrt(x^ 2 - 1)))
Timed out. \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\int \sqrt {x^2-1}\,{\left (x+1\right )}^2 \,d x \] Input:
int((x^2 - 1)^(1/2)*(x + 1)^2,x)
Output:
int((x^2 - 1)^(1/2)*(x + 1)^2, x)
Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int (1+x)^2 \sqrt {-1+x^2} \, dx=\frac {\sqrt {x^{2}-1}\, x^{3}}{4}+\frac {2 \sqrt {x^{2}-1}\, x^{2}}{3}+\frac {3 \sqrt {x^{2}-1}\, x}{8}-\frac {2 \sqrt {x^{2}-1}}{3}-\frac {5 \,\mathrm {log}\left (\sqrt {x^{2}-1}+x \right )}{8} \] Input:
int((1+x)^2*(x^2-1)^(1/2),x)
Output:
(6*sqrt(x**2 - 1)*x**3 + 16*sqrt(x**2 - 1)*x**2 + 9*sqrt(x**2 - 1)*x - 16* sqrt(x**2 - 1) - 15*log(sqrt(x**2 - 1) + x))/24