Integrand size = 20, antiderivative size = 37 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {1-x} \sqrt {1+x}+2 \arcsin (x)-\text {arctanh}\left (\sqrt {1-x^2}\right ) \] Output:
-(1-x)^(1/2)*(1+x)^(1/2)+2*arcsin(x)-arctanh((-x^2+1)^(1/2))
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.46 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {1-x^2}-4 \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right )-2 \text {arctanh}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \] Input:
Integrate[(1 + x)^(3/2)/(Sqrt[1 - x]*x),x]
Output:
-Sqrt[1 - x^2] - 4*ArcTan[Sqrt[1 - x]/Sqrt[1 + x]] - 2*ArcTanh[Sqrt[1 - x] /Sqrt[1 + x]]
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {113, 25, 175, 39, 103, 219, 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.
\(\displaystyle \int \frac {(x+1)^{3/2}}{\sqrt {1-x} x} \, dx\) |
\(\Big \downarrow \) 113 |
\(\displaystyle -\int -\frac {2 x+1}{\sqrt {1-x} x \sqrt {x+1}}dx-\sqrt {1-x} \sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {2 x+1}{\sqrt {1-x} x \sqrt {x+1}}dx-\sqrt {1-x} \sqrt {x+1}\) |
\(\Big \downarrow \) 175 |
\(\displaystyle 2 \int \frac {1}{\sqrt {1-x} \sqrt {x+1}}dx+\int \frac {1}{\sqrt {1-x} x \sqrt {x+1}}dx-\sqrt {1-x} \sqrt {x+1}\) |
\(\Big \downarrow \) 39 |
\(\displaystyle 2 \int \frac {1}{\sqrt {1-x^2}}dx+\int \frac {1}{\sqrt {1-x} x \sqrt {x+1}}dx-\sqrt {1-x} \sqrt {x+1}\) |
\(\Big \downarrow \) 103 |
\(\displaystyle 2 \int \frac {1}{\sqrt {1-x^2}}dx-\int \frac {1}{1-(1-x) (x+1)}d\left (\sqrt {1-x} \sqrt {x+1}\right )-\sqrt {1-x} \sqrt {x+1}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \int \frac {1}{\sqrt {1-x^2}}dx-\text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )-\sqrt {1-x} \sqrt {x+1}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle 2 \arcsin (x)-\text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )-\sqrt {1-x} \sqrt {x+1}\) |
Input:
Int[(1 + x)^(3/2)/(Sqrt[1 - x]*x),x]
Output:
-(Sqrt[1 - x]*Sqrt[1 + x]) + 2*ArcSin[x] - ArcTanh[Sqrt[1 - x]*Sqrt[1 + 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[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ ))), x_] :> Simp[b*f Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq rt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d *e - f*(b*c + a*d), 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] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
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] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.38
method | result | size |
default | \(\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (-\sqrt {-x^{2}+1}+2 \arcsin \left (x \right )-\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )\right )}{\sqrt {-x^{2}+1}}\) | \(51\) |
Input:
int((1+x)^(3/2)/(1-x)^(1/2)/x,x,method=_RETURNVERBOSE)
Output:
(1+x)^(1/2)*(1-x)^(1/2)/(-x^2+1)^(1/2)*(-(-x^2+1)^(1/2)+2*arcsin(x)-arctan h(1/(-x^2+1)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.54 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {x + 1} \sqrt {-x + 1} - 4 \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \] Input:
integrate((1+x)^(3/2)/(1-x)^(1/2)/x,x, algorithm="fricas")
Output:
-sqrt(x + 1)*sqrt(-x + 1) - 4*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + l og((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)
\[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=\int \frac {\left (x + 1\right )^{\frac {3}{2}}}{x \sqrt {1 - x}}\, dx \] Input:
integrate((1+x)**(3/2)/(1-x)**(1/2)/x,x)
Output:
Integral((x + 1)**(3/2)/(x*sqrt(1 - x)), x)
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {-x^{2} + 1} + 2 \, \arcsin \left (x\right ) - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \] Input:
integrate((1+x)^(3/2)/(1-x)^(1/2)/x,x, algorithm="maxima")
Output:
-sqrt(-x^2 + 1) + 2*arcsin(x) - log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (31) = 62\).
Time = 0.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 4.27 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=2 \, \pi - \sqrt {x + 1} \sqrt {-x + 1} + 4 \, \arctan \left (\frac {\sqrt {x + 1} {\left (\frac {{\left (\sqrt {2} - \sqrt {-x + 1}\right )}^{2}}{x + 1} - 1\right )}}{2 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}\right ) - \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} + 2 \right |}\right ) + \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \right |}\right ) \] Input:
integrate((1+x)^(3/2)/(1-x)^(1/2)/x,x, algorithm="giac")
Output:
2*pi - sqrt(x + 1)*sqrt(-x + 1) + 4*arctan(1/2*sqrt(x + 1)*((sqrt(2) - sqr t(-x + 1))^2/(x + 1) - 1)/(sqrt(2) - sqrt(-x + 1))) - log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)) + 2)) + l og(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - sqrt (-x + 1)) - 2))
Timed out. \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=\int \frac {{\left (x+1\right )}^{3/2}}{x\,\sqrt {1-x}} \,d x \] Input:
int((x + 1)^(3/2)/(x*(1 - x)^(1/2)),x)
Output:
int((x + 1)^(3/2)/(x*(1 - x)^(1/2)), x)
Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.11 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-4 \mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )-\sqrt {x +1}\, \sqrt {1-x}-\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )-1\right )+\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )+1\right )-\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )-1\right )+\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )+1\right ) \] Input:
int((1+x)^(3/2)/(1-x)^(1/2)/x,x)
Output:
- 4*asin(sqrt( - x + 1)/sqrt(2)) - sqrt(x + 1)*sqrt( - x + 1) - log( - sq rt(2) + tan(asin(sqrt( - x + 1)/sqrt(2))/2) - 1) + log( - sqrt(2) + tan(as in(sqrt( - x + 1)/sqrt(2))/2) + 1) - log(sqrt(2) + tan(asin(sqrt( - x + 1) /sqrt(2))/2) - 1) + log(sqrt(2) + tan(asin(sqrt( - x + 1)/sqrt(2))/2) + 1)