Integrand size = 20, antiderivative size = 115 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^5} \, dx=-\frac {\sqrt {1-x} \sqrt {1+x}}{4 x^4}-\frac {2 \sqrt {1-x} \sqrt {1+x}}{3 x^3}-\frac {7 \sqrt {1-x} \sqrt {1+x}}{8 x^2}-\frac {4 \sqrt {1-x} \sqrt {1+x}}{3 x}-\frac {7}{8} \text {arctanh}\left (\sqrt {1-x} \sqrt {1+x}\right ) \]
-7/8*arctanh((1-x)^(1/2)*(1+x)^(1/2))-1/4*(1-x)^(1/2)*(1+x)^(1/2)/x^4-2/3* (1-x)^(1/2)*(1+x)^(1/2)/x^3-7/8*(1-x)^(1/2)*(1+x)^(1/2)/x^2-4/3*(1-x)^(1/2 )*(1+x)^(1/2)/x
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.57 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^5} \, dx=-\frac {\sqrt {1-x} \left (6+22 x+37 x^2+53 x^3+32 x^4\right )}{24 x^4 \sqrt {1+x}}-\frac {7}{4} \text {arctanh}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \]
-1/24*(Sqrt[1 - x]*(6 + 22*x + 37*x^2 + 53*x^3 + 32*x^4))/(x^4*Sqrt[1 + x] ) - (7*ArcTanh[Sqrt[1 - x]/Sqrt[1 + x]])/4
Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {109, 25, 168, 25, 168, 25, 168, 27, 103, 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.
\(\displaystyle \int \frac {(x+1)^{3/2}}{\sqrt {1-x} x^5} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle -\frac {1}{4} \int -\frac {7 x+8}{\sqrt {1-x} x^4 \sqrt {x+1}}dx-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \int \frac {7 x+8}{\sqrt {1-x} x^4 \sqrt {x+1}}dx-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{3} \int -\frac {16 x+21}{\sqrt {1-x} x^3 \sqrt {x+1}}dx-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {16 x+21}{\sqrt {1-x} x^3 \sqrt {x+1}}dx-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (-\frac {1}{2} \int -\frac {21 x+32}{\sqrt {1-x} x^2 \sqrt {x+1}}dx-\frac {21 \sqrt {1-x} \sqrt {x+1}}{2 x^2}\right )-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {21 x+32}{\sqrt {1-x} x^2 \sqrt {x+1}}dx-\frac {21 \sqrt {1-x} \sqrt {x+1}}{2 x^2}\right )-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-\int -\frac {21}{\sqrt {1-x} x \sqrt {x+1}}dx-\frac {32 \sqrt {1-x} \sqrt {x+1}}{x}\right )-\frac {21 \sqrt {1-x} \sqrt {x+1}}{2 x^2}\right )-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (21 \int \frac {1}{\sqrt {1-x} x \sqrt {x+1}}dx-\frac {32 \sqrt {1-x} \sqrt {x+1}}{x}\right )-\frac {21 \sqrt {1-x} \sqrt {x+1}}{2 x^2}\right )-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 103 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-21 \int \frac {1}{1-(1-x) (x+1)}d\left (\sqrt {1-x} \sqrt {x+1}\right )-\frac {32 \sqrt {1-x} \sqrt {x+1}}{x}\right )-\frac {21 \sqrt {1-x} \sqrt {x+1}}{2 x^2}\right )-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-21 \text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )-\frac {32 \sqrt {1-x} \sqrt {x+1}}{x}\right )-\frac {21 \sqrt {1-x} \sqrt {x+1}}{2 x^2}\right )-\frac {8 \sqrt {1-x} \sqrt {x+1}}{3 x^3}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{4 x^4}\) |
-1/4*(Sqrt[1 - x]*Sqrt[1 + x])/x^4 + ((-8*Sqrt[1 - x]*Sqrt[1 + x])/(3*x^3) + ((-21*Sqrt[1 - x]*Sqrt[1 + x])/(2*x^2) + ((-32*Sqrt[1 - x]*Sqrt[1 + x]) /x - 21*ArcTanh[Sqrt[1 - x]*Sqrt[1 + x]])/2)/3)/4
3.8.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
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])
Time = 0.57 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\left (-1+x \right ) \sqrt {1+x}\, \left (32 x^{3}+21 x^{2}+16 x +6\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{24 x^{4} \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}-\frac {7 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{8 \sqrt {1-x}\, \sqrt {1+x}}\) | \(93\) |
default | \(-\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (21 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{4}+32 x^{3} \sqrt {-x^{2}+1}+21 x^{2} \sqrt {-x^{2}+1}+16 x \sqrt {-x^{2}+1}+6 \sqrt {-x^{2}+1}\right )}{24 x^{4} \sqrt {-x^{2}+1}}\) | \(94\) |
1/24*(-1+x)*(1+x)^(1/2)*(32*x^3+21*x^2+16*x+6)/x^4/(-(-1+x)*(1+x))^(1/2)*( (1+x)*(1-x))^(1/2)/(1-x)^(1/2)-7/8*arctanh(1/(-x^2+1)^(1/2))*((1+x)*(1-x)) ^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)
Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.52 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^5} \, dx=\frac {21 \, x^{4} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - {\left (32 \, x^{3} + 21 \, x^{2} + 16 \, x + 6\right )} \sqrt {x + 1} \sqrt {-x + 1}}{24 \, x^{4}} \]
1/24*(21*x^4*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - (32*x^3 + 21*x^2 + 16 *x + 6)*sqrt(x + 1)*sqrt(-x + 1))/x^4
\[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^5} \, dx=\int \frac {\left (x + 1\right )^{\frac {3}{2}}}{x^{5} \sqrt {1 - x}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.71 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^5} \, dx=-\frac {4 \, \sqrt {-x^{2} + 1}}{3 \, x} - \frac {7 \, \sqrt {-x^{2} + 1}}{8 \, x^{2}} - \frac {2 \, \sqrt {-x^{2} + 1}}{3 \, x^{3}} - \frac {\sqrt {-x^{2} + 1}}{4 \, x^{4}} - \frac {7}{8} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]
-4/3*sqrt(-x^2 + 1)/x - 7/8*sqrt(-x^2 + 1)/x^2 - 2/3*sqrt(-x^2 + 1)/x^3 - 1/4*sqrt(-x^2 + 1)/x^4 - 7/8*log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (85) = 170\).
Time = 0.35 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.83 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^5} \, dx=-\frac {21 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{7} - 308 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{5} + 1328 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{3} - \frac {4800 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}{\sqrt {x + 1}} + \frac {4800 \, \sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}}{6 \, {\left ({\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{2} - 4\right )}^{4}} - \frac {7}{8} \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} + 2 \right |}\right ) + \frac {7}{8} \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \right |}\right ) \]
-1/6*(21*((sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sq rt(-x + 1)))^7 - 308*((sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/( sqrt(2) - sqrt(-x + 1)))^5 + 1328*((sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))^3 - 4800*(sqrt(2) - sqrt(-x + 1))/sq rt(x + 1) + 4800*sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))/(((sqrt(2) - sqrt(- x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))^2 - 4)^4 - 7/8 *log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - sq rt(-x + 1)) + 2)) + 7/8*log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sq rt(x + 1)/(sqrt(2) - sqrt(-x + 1)) - 2))
Timed out. \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^5} \, dx=\int \frac {{\left (x+1\right )}^{3/2}}{x^5\,\sqrt {1-x}} \,d x \]