Integrand size = 20, antiderivative size = 41 \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=-\frac {3}{8} x \sqrt {1-x^2}-\frac {1}{4} x^3 \sqrt {1-x^2}+\frac {3 \arcsin (x)}{8} \] Output:
-3/8*x*(-x^2+1)^(1/2)-1/4*x^3*(-x^2+1)^(1/2)+3/8*arcsin(x)
Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=-\frac {1}{8} x \sqrt {1-x^2} \left (3+2 x^2\right )+\frac {3}{4} \arctan \left (\frac {x}{-1+\sqrt {1-x^2}}\right ) \] Input:
Integrate[x^4/(Sqrt[1 - x]*Sqrt[1 + x]),x]
Output:
-1/8*(x*Sqrt[1 - x^2]*(3 + 2*x^2)) + (3*ArcTan[x/(-1 + Sqrt[1 - x^2])])/4
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.37, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {111, 27, 101, 25, 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.
\(\displaystyle \int \frac {x^4}{\sqrt {1-x} \sqrt {x+1}} \, dx\) |
\(\Big \downarrow \) 111 |
\(\displaystyle -\frac {1}{4} \int -\frac {3 x^2}{\sqrt {1-x} \sqrt {x+1}}dx-\frac {1}{4} \sqrt {1-x} \sqrt {x+1} x^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{4} \int \frac {x^2}{\sqrt {1-x} \sqrt {x+1}}dx-\frac {1}{4} \sqrt {1-x} x^3 \sqrt {x+1}\) |
\(\Big \downarrow \) 101 |
\(\displaystyle \frac {3}{4} \left (-\frac {1}{2} \int -\frac {1}{\sqrt {1-x} \sqrt {x+1}}dx-\frac {1}{2} \sqrt {1-x} \sqrt {x+1} x\right )-\frac {1}{4} \sqrt {1-x} x^3 \sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-x} \sqrt {x+1}}dx-\frac {1}{2} \sqrt {1-x} x \sqrt {x+1}\right )-\frac {1}{4} \sqrt {1-x} x^3 \sqrt {x+1}\) |
\(\Big \downarrow \) 39 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}}dx-\frac {1}{2} \sqrt {1-x} x \sqrt {x+1}\right )-\frac {1}{4} \sqrt {1-x} x^3 \sqrt {x+1}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {3}{4} \left (\frac {\arcsin (x)}{2}-\frac {1}{2} \sqrt {1-x} x \sqrt {x+1}\right )-\frac {1}{4} \sqrt {1-x} x^3 \sqrt {x+1}\) |
Input:
Int[x^4/(Sqrt[1 - x]*Sqrt[1 + x]),x]
Output:
-1/4*(Sqrt[1 - x]*x^3*Sqrt[1 + x]) + (3*(-1/2*(Sqrt[1 - x]*x*Sqrt[1 + x]) + ArcSin[x]/2))/4
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_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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 = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.34
method | result | size |
default | \(\frac {\sqrt {1-x}\, \sqrt {1+x}\, \left (-2 x^{3} \sqrt {-x^{2}+1}-3 x \sqrt {-x^{2}+1}+3 \arcsin \left (x \right )\right )}{8 \sqrt {-x^{2}+1}}\) | \(55\) |
risch | \(\frac {x \left (2 x^{2}+3\right ) \sqrt {1+x}\, \left (-1+x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{8 \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}}+\frac {3 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{8 \sqrt {1-x}\, \sqrt {1+x}}\) | \(75\) |
Input:
int(x^4/(1-x)^(1/2)/(1+x)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*(1-x)^(1/2)*(1+x)^(1/2)*(-2*x^3*(-x^2+1)^(1/2)-3*x*(-x^2+1)^(1/2)+3*ar csin(x))/(-x^2+1)^(1/2)
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=-\frac {1}{8} \, {\left (2 \, x^{3} + 3 \, x\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {3}{4} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \] Input:
integrate(x^4/(1-x)^(1/2)/(1+x)^(1/2),x, algorithm="fricas")
Output:
-1/8*(2*x^3 + 3*x)*sqrt(x + 1)*sqrt(-x + 1) - 3/4*arctan((sqrt(x + 1)*sqrt (-x + 1) - 1)/x)
Timed out. \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=\text {Timed out} \] Input:
integrate(x**4/(1-x)**(1/2)/(1+x)**(1/2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=-\frac {1}{4} \, \sqrt {-x^{2} + 1} x^{3} - \frac {3}{8} \, \sqrt {-x^{2} + 1} x + \frac {3}{8} \, \arcsin \left (x\right ) \] Input:
integrate(x^4/(1-x)^(1/2)/(1+x)^(1/2),x, algorithm="maxima")
Output:
-1/4*sqrt(-x^2 + 1)*x^3 - 3/8*sqrt(-x^2 + 1)*x + 3/8*arcsin(x)
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.07 \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=-\frac {1}{8} \, {\left ({\left (2 \, {\left (x + 1\right )} {\left (x - 2\right )} + 9\right )} {\left (x + 1\right )} - 5\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {3}{4} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \] Input:
integrate(x^4/(1-x)^(1/2)/(1+x)^(1/2),x, algorithm="giac")
Output:
-1/8*((2*(x + 1)*(x - 2) + 9)*(x + 1) - 5)*sqrt(x + 1)*sqrt(-x + 1) + 3/4* arcsin(1/2*sqrt(2)*sqrt(x + 1))
Time = 3.80 (sec) , antiderivative size = 380, normalized size of antiderivative = 9.27 \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=\frac {\frac {3\,\left (\sqrt {1-x}-1\right )}{2\,\left (\sqrt {x+1}-1\right )}+\frac {23\,{\left (\sqrt {1-x}-1\right )}^3}{2\,{\left (\sqrt {x+1}-1\right )}^3}-\frac {333\,{\left (\sqrt {1-x}-1\right )}^5}{2\,{\left (\sqrt {x+1}-1\right )}^5}+\frac {671\,{\left (\sqrt {1-x}-1\right )}^7}{2\,{\left (\sqrt {x+1}-1\right )}^7}-\frac {671\,{\left (\sqrt {1-x}-1\right )}^9}{2\,{\left (\sqrt {x+1}-1\right )}^9}+\frac {333\,{\left (\sqrt {1-x}-1\right )}^{11}}{2\,{\left (\sqrt {x+1}-1\right )}^{11}}-\frac {23\,{\left (\sqrt {1-x}-1\right )}^{13}}{2\,{\left (\sqrt {x+1}-1\right )}^{13}}-\frac {3\,{\left (\sqrt {1-x}-1\right )}^{15}}{2\,{\left (\sqrt {x+1}-1\right )}^{15}}}{\frac {8\,{\left (\sqrt {1-x}-1\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}+\frac {28\,{\left (\sqrt {1-x}-1\right )}^4}{{\left (\sqrt {x+1}-1\right )}^4}+\frac {56\,{\left (\sqrt {1-x}-1\right )}^6}{{\left (\sqrt {x+1}-1\right )}^6}+\frac {70\,{\left (\sqrt {1-x}-1\right )}^8}{{\left (\sqrt {x+1}-1\right )}^8}+\frac {56\,{\left (\sqrt {1-x}-1\right )}^{10}}{{\left (\sqrt {x+1}-1\right )}^{10}}+\frac {28\,{\left (\sqrt {1-x}-1\right )}^{12}}{{\left (\sqrt {x+1}-1\right )}^{12}}+\frac {8\,{\left (\sqrt {1-x}-1\right )}^{14}}{{\left (\sqrt {x+1}-1\right )}^{14}}+\frac {{\left (\sqrt {1-x}-1\right )}^{16}}{{\left (\sqrt {x+1}-1\right )}^{16}}+1}-\frac {3\,\mathrm {atan}\left (\frac {\sqrt {1-x}-1}{\sqrt {x+1}-1}\right )}{2} \] Input:
int(x^4/((1 - x)^(1/2)*(x + 1)^(1/2)),x)
Output:
((3*((1 - x)^(1/2) - 1))/(2*((x + 1)^(1/2) - 1)) + (23*((1 - x)^(1/2) - 1) ^3)/(2*((x + 1)^(1/2) - 1)^3) - (333*((1 - x)^(1/2) - 1)^5)/(2*((x + 1)^(1 /2) - 1)^5) + (671*((1 - x)^(1/2) - 1)^7)/(2*((x + 1)^(1/2) - 1)^7) - (671 *((1 - x)^(1/2) - 1)^9)/(2*((x + 1)^(1/2) - 1)^9) + (333*((1 - x)^(1/2) - 1)^11)/(2*((x + 1)^(1/2) - 1)^11) - (23*((1 - x)^(1/2) - 1)^13)/(2*((x + 1 )^(1/2) - 1)^13) - (3*((1 - x)^(1/2) - 1)^15)/(2*((x + 1)^(1/2) - 1)^15))/ ((8*((1 - x)^(1/2) - 1)^2)/((x + 1)^(1/2) - 1)^2 + (28*((1 - x)^(1/2) - 1) ^4)/((x + 1)^(1/2) - 1)^4 + (56*((1 - x)^(1/2) - 1)^6)/((x + 1)^(1/2) - 1) ^6 + (70*((1 - x)^(1/2) - 1)^8)/((x + 1)^(1/2) - 1)^8 + (56*((1 - x)^(1/2) - 1)^10)/((x + 1)^(1/2) - 1)^10 + (28*((1 - x)^(1/2) - 1)^12)/((x + 1)^(1 /2) - 1)^12 + (8*((1 - x)^(1/2) - 1)^14)/((x + 1)^(1/2) - 1)^14 + ((1 - x) ^(1/2) - 1)^16/((x + 1)^(1/2) - 1)^16 + 1) - (3*atan(((1 - x)^(1/2) - 1)/( (x + 1)^(1/2) - 1)))/2
Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.05 \[ \int \frac {x^4}{\sqrt {1-x} \sqrt {1+x}} \, dx=-\frac {3 \mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{4}-\frac {\sqrt {x +1}\, \sqrt {1-x}\, x^{3}}{4}-\frac {3 \sqrt {x +1}\, \sqrt {1-x}\, x}{8} \] Input:
int(x^4/(1-x)^(1/2)/(1+x)^(1/2),x)
Output:
( - 6*asin(sqrt( - x + 1)/sqrt(2)) - 2*sqrt(x + 1)*sqrt( - x + 1)*x**3 - 3 *sqrt(x + 1)*sqrt( - x + 1)*x)/8