Integrand size = 15, antiderivative size = 67 \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\frac {7}{8} x \sqrt {-1+x^2}+\frac {1}{5} (1+x)^2 \left (-1+x^2\right )^{3/2}+\frac {7}{60} (8+3 x) \left (-1+x^2\right )^{3/2}-\frac {7}{8} \text {arctanh}\left (\frac {x}{\sqrt {-1+x^2}}\right ) \] Output:
7/8*x*(x^2-1)^(1/2)+1/5*(1+x)^2*(x^2-1)^(3/2)+7/60*(8+3*x)*(x^2-1)^(3/2)-7 /8*arctanh(x/(x^2-1)^(1/2))
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\frac {1}{120} \sqrt {-1+x^2} \left (-136+15 x+112 x^2+90 x^3+24 x^4\right )-\frac {7}{4} \text {arctanh}\left (\frac {\sqrt {-1+x^2}}{-1+x}\right ) \] Input:
Integrate[(1 + x)^3*Sqrt[-1 + x^2],x]
Output:
(Sqrt[-1 + x^2]*(-136 + 15*x + 112*x^2 + 90*x^3 + 24*x^4))/120 - (7*ArcTan h[Sqrt[-1 + x^2]/(-1 + x)])/4
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {469, 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.
\(\displaystyle \int (x+1)^3 \sqrt {x^2-1} \, dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {7}{5} \int (x+1)^2 \sqrt {x^2-1}dx+\frac {1}{5} \left (x^2-1\right )^{3/2} (x+1)^2\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \int (x+1) \sqrt {x^2-1}dx+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\right )+\frac {1}{5} \left (x^2-1\right )^{3/2} (x+1)^2\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \left (\int \sqrt {x^2-1}dx+\frac {1}{3} \left (x^2-1\right )^{3/2}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\right )+\frac {1}{5} \left (x^2-1\right )^{3/2} (x+1)^2\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \left (-\frac {1}{2} \int \frac {1}{\sqrt {x^2-1}}dx+\frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\right )+\frac {1}{5} \left (x^2-1\right )^{3/2} (x+1)^2\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \left (-\frac {1}{2} \int \frac {1}{1-\frac {x^2}{x^2-1}}d\frac {x}{\sqrt {x^2-1}}+\frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\right )+\frac {1}{5} \left (x^2-1\right )^{3/2} (x+1)^2\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \left (-\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt {x^2-1}}\right )+\frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}\right )+\frac {1}{4} (x+1) \left (x^2-1\right )^{3/2}\right )+\frac {1}{5} \left (x^2-1\right )^{3/2} (x+1)^2\) |
Input:
Int[(1 + x)^3*Sqrt[-1 + x^2],x]
Output:
((1 + x)^2*(-1 + x^2)^(3/2))/5 + (7*(((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) )/5
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 = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.63
method | result | size |
trager | \(\left (\frac {1}{5} x^{4}+\frac {3}{4} x^{3}+\frac {14}{15} x^{2}+\frac {1}{8} x -\frac {17}{15}\right ) \sqrt {x^{2}-1}-\frac {7 \ln \left (x +\sqrt {x^{2}-1}\right )}{8}\) | \(42\) |
risch | \(\frac {\left (24 x^{4}+90 x^{3}+112 x^{2}+15 x -136\right ) \sqrt {x^{2}-1}}{120}-\frac {7 \ln \left (x +\sqrt {x^{2}-1}\right )}{8}\) | \(43\) |
default | \(\frac {7 x \sqrt {x^{2}-1}}{8}-\frac {7 \ln \left (x +\sqrt {x^{2}-1}\right )}{8}+\frac {x^{2} \left (x^{2}-1\right )^{\frac {3}{2}}}{5}+\frac {17 \left (x^{2}-1\right )^{\frac {3}{2}}}{15}+\frac {3 x \left (x^{2}-1\right )^{\frac {3}{2}}}{4}\) | \(55\) |
meijerg | \(\frac {i \sqrt {\operatorname {signum}\left (x^{2}-1\right )}\, \left (-2 i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-2 i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{4 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}}-\frac {\sqrt {\operatorname {signum}\left (x^{2}-1\right )}\, \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (-x^{2}+1\right )^{\frac {3}{2}} \left (3 x^{2}+2\right )}{15}\right )}{4 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}}-\frac {3 i \sqrt {\operatorname {signum}\left (x^{2}-1\right )}\, \left (-\frac {i \sqrt {\pi }\, x \left (-6 x^{2}+3\right ) \sqrt {-x^{2}+1}}{6}+\frac {i \sqrt {\pi }\, \arcsin \left (x \right )}{2}\right )}{4 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}}+\frac {3 \sqrt {\operatorname {signum}\left (x^{2}-1\right )}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {2 \sqrt {\pi }\, \left (-2 x^{2}+2\right ) \sqrt {-x^{2}+1}}{3}\right )}{4 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{2}-1\right )}}\) | \(207\) |
Input:
int((x+1)^3*(x^2-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
(1/5*x^4+3/4*x^3+14/15*x^2+1/8*x-17/15)*(x^2-1)^(1/2)-7/8*ln(x+(x^2-1)^(1/ 2))
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66 \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\frac {1}{120} \, {\left (24 \, x^{4} + 90 \, x^{3} + 112 \, x^{2} + 15 \, x - 136\right )} \sqrt {x^{2} - 1} + \frac {7}{8} \, \log \left (-x + \sqrt {x^{2} - 1}\right ) \] Input:
integrate((1+x)^3*(x^2-1)^(1/2),x, algorithm="fricas")
Output:
1/120*(24*x^4 + 90*x^3 + 112*x^2 + 15*x - 136)*sqrt(x^2 - 1) + 7/8*log(-x + sqrt(x^2 - 1))
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.22 \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\frac {x^{4} \sqrt {x^{2} - 1}}{5} + \frac {3 x^{3} \sqrt {x^{2} - 1}}{4} + \frac {14 x^{2} \sqrt {x^{2} - 1}}{15} + \frac {x \sqrt {x^{2} - 1}}{8} - \frac {17 \sqrt {x^{2} - 1}}{15} - \frac {7 \log {\left (x + \sqrt {x^{2} - 1} \right )}}{8} \] Input:
integrate((1+x)**3*(x**2-1)**(1/2),x)
Output:
x**4*sqrt(x**2 - 1)/5 + 3*x**3*sqrt(x**2 - 1)/4 + 14*x**2*sqrt(x**2 - 1)/1 5 + x*sqrt(x**2 - 1)/8 - 17*sqrt(x**2 - 1)/15 - 7*log(x + sqrt(x**2 - 1))/ 8
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.87 \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\frac {1}{5} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} x^{2} + \frac {3}{4} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} x + \frac {17}{15} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} + \frac {7}{8} \, \sqrt {x^{2} - 1} x - \frac {7}{8} \, \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \] Input:
integrate((1+x)^3*(x^2-1)^(1/2),x, algorithm="maxima")
Output:
1/5*(x^2 - 1)^(3/2)*x^2 + 3/4*(x^2 - 1)^(3/2)*x + 17/15*(x^2 - 1)^(3/2) + 7/8*sqrt(x^2 - 1)*x - 7/8*log(2*x + 2*sqrt(x^2 - 1))
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66 \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\frac {1}{120} \, {\left ({\left (2 \, {\left (3 \, {\left (4 \, x + 15\right )} x + 56\right )} x + 15\right )} x - 136\right )} \sqrt {x^{2} - 1} + \frac {7}{8} \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right ) \] Input:
integrate((1+x)^3*(x^2-1)^(1/2),x, algorithm="giac")
Output:
1/120*((2*(3*(4*x + 15)*x + 56)*x + 15)*x - 136)*sqrt(x^2 - 1) + 7/8*log(a bs(-x + sqrt(x^2 - 1)))
Timed out. \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\int \sqrt {x^2-1}\,{\left (x+1\right )}^3 \,d x \] Input:
int((x^2 - 1)^(1/2)*(x + 1)^3,x)
Output:
int((x^2 - 1)^(1/2)*(x + 1)^3, x)
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93 \[ \int (1+x)^3 \sqrt {-1+x^2} \, dx=\frac {\sqrt {x^{2}-1}\, x^{4}}{5}+\frac {3 \sqrt {x^{2}-1}\, x^{3}}{4}+\frac {14 \sqrt {x^{2}-1}\, x^{2}}{15}+\frac {\sqrt {x^{2}-1}\, x}{8}-\frac {17 \sqrt {x^{2}-1}}{15}-\frac {7 \,\mathrm {log}\left (\sqrt {x^{2}-1}+x \right )}{8} \] Input:
int((1+x)^3*(x^2-1)^(1/2),x)
Output:
(24*sqrt(x**2 - 1)*x**4 + 90*sqrt(x**2 - 1)*x**3 + 112*sqrt(x**2 - 1)*x**2 + 15*sqrt(x**2 - 1)*x - 136*sqrt(x**2 - 1) - 105*log(sqrt(x**2 - 1) + x)) /120