Integrand size = 16, antiderivative size = 128 \[ \int x^{3/2} (2-b x)^{5/2} \, dx=-\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2-b x}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {3 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}} \]
1/4*x^(5/2)*(-b*x+2)^(3/2)+1/5*x^(5/2)*(-b*x+2)^(5/2)+3/4*arcsin(1/2*b^(1/ 2)*x^(1/2)*2^(1/2))/b^(5/2)-1/8*x^(3/2)*(-b*x+2)^(1/2)/b+1/4*x^(5/2)*(-b*x +2)^(1/2)-3/8*x^(1/2)*(-b*x+2)^(1/2)/b^2
Time = 0.38 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.73 \[ \int x^{3/2} (2-b x)^{5/2} \, dx=\frac {\sqrt {x} \sqrt {2-b x} \left (-15-5 b x+62 b^2 x^2-42 b^3 x^3+8 b^4 x^4\right )}{40 b^2}-\frac {3 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{2 b^{5/2}} \]
(Sqrt[x]*Sqrt[2 - b*x]*(-15 - 5*b*x + 62*b^2*x^2 - 42*b^3*x^3 + 8*b^4*x^4) )/(40*b^2) - (3*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])/(2*b^ (5/2))
Time = 0.19 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {60, 60, 60, 60, 60, 63, 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.
| \(\displaystyle \int x^{3/2} (2-b x)^{5/2} \, dx\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \int x^{3/2} (2-b x)^{3/2}dx+\frac {1}{5} x^{5/2} (2-b x)^{5/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{4} \int x^{3/2} \sqrt {2-b x}dx+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{4} \left (\frac {1}{3} \int \frac {x^{3/2}}{\sqrt {2-b x}}dx+\frac {1}{3} x^{5/2} \sqrt {2-b x}\right )+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{4} \left (\frac {1}{3} \left (\frac {3 \int \frac {\sqrt {x}}{\sqrt {2-b x}}dx}{2 b}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}\right )+\frac {1}{3} x^{5/2} \sqrt {2-b x}\right )+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{4} \left (\frac {1}{3} \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {2-b x}}dx}{b}-\frac {\sqrt {x} \sqrt {2-b x}}{b}\right )}{2 b}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}\right )+\frac {1}{3} x^{5/2} \sqrt {2-b x}\right )+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}\) |
| \(\Big \downarrow \) 63 |
| \(\displaystyle \frac {3}{4} \left (\frac {1}{3} \left (\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {2-b x}}d\sqrt {x}}{b}-\frac {\sqrt {x} \sqrt {2-b x}}{b}\right )}{2 b}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}\right )+\frac {1}{3} x^{5/2} \sqrt {2-b x}\right )+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {3}{4} \left (\frac {1}{3} \left (\frac {3 \left (\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}-\frac {\sqrt {x} \sqrt {2-b x}}{b}\right )}{2 b}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}\right )+\frac {1}{3} x^{5/2} \sqrt {2-b x}\right )+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}\) |
(x^(5/2)*(2 - b*x)^(3/2))/4 + (x^(5/2)*(2 - b*x)^(5/2))/5 + (3*((x^(5/2)*S qrt[2 - b*x])/3 + (-1/2*(x^(3/2)*Sqrt[2 - b*x])/b + (3*(-((Sqrt[x]*Sqrt[2 - b*x])/b) + (2*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2)))/(2*b))/3))/4
3.6.64.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b S ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x ] && GtQ[c, 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 = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.76
| method | result | size |
| meijerg | \(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {5}{2}} \left (-8 b^{4} x^{4}+42 b^{3} x^{3}-62 b^{2} x^{2}+5 b x +15\right ) \sqrt {-\frac {b x}{2}+1}}{40 b^{2}}-\frac {3 \sqrt {\pi }\, \left (-b \right )^{\frac {5}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {5}{2}}}}{\left (-b \right )^{\frac {3}{2}} \sqrt {\pi }\, b}\) | \(97\) |
| risch | \(-\frac {\left (8 b^{4} x^{4}-42 b^{3} x^{3}+62 b^{2} x^{2}-5 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{40 b^{2} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {3 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {-b x +2}}\) | \(123\) |
| default | \(-\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {7}{2}}}{5 b}+\frac {-\frac {3 \sqrt {x}\, \left (-b x +2\right )^{\frac {7}{2}}}{20 b}+\frac {3 \left (\frac {\left (-b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{2}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\right )}{20 b}}{b}\) | \(135\) |
60/(-b)^(3/2)/Pi^(1/2)/b*(1/2400*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(5/2)*(-8*b ^4*x^4+42*b^3*x^3-62*b^2*x^2+5*b*x+15)/b^2*(-1/2*b*x+1)^(1/2)-1/80*Pi^(1/2 )*(-b)^(5/2)/b^(5/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))
Time = 0.24 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.23 \[ \int x^{3/2} (2-b x)^{5/2} \, dx=\left [\frac {{\left (8 \, b^{5} x^{4} - 42 \, b^{4} x^{3} + 62 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{40 \, b^{3}}, \frac {{\left (8 \, b^{5} x^{4} - 42 \, b^{4} x^{3} + 62 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{40 \, b^{3}}\right ] \]
[1/40*((8*b^5*x^4 - 42*b^4*x^3 + 62*b^3*x^2 - 5*b^2*x - 15*b)*sqrt(-b*x + 2)*sqrt(x) - 15*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1))/ b^3, 1/40*((8*b^5*x^4 - 42*b^4*x^3 + 62*b^3*x^2 - 5*b^2*x - 15*b)*sqrt(-b* x + 2)*sqrt(x) - 30*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^3]
Result contains complex when optimal does not.
Time = 41.45 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.28 \[ \int x^{3/2} (2-b x)^{5/2} \, dx=\begin {cases} \frac {i b^{3} x^{\frac {11}{2}}}{5 \sqrt {b x - 2}} - \frac {29 i b^{2} x^{\frac {9}{2}}}{20 \sqrt {b x - 2}} + \frac {73 i b x^{\frac {7}{2}}}{20 \sqrt {b x - 2}} - \frac {129 i x^{\frac {5}{2}}}{40 \sqrt {b x - 2}} - \frac {i x^{\frac {3}{2}}}{8 b \sqrt {b x - 2}} + \frac {3 i \sqrt {x}}{4 b^{2} \sqrt {b x - 2}} - \frac {3 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {5}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b^{3} x^{\frac {11}{2}}}{5 \sqrt {- b x + 2}} + \frac {29 b^{2} x^{\frac {9}{2}}}{20 \sqrt {- b x + 2}} - \frac {73 b x^{\frac {7}{2}}}{20 \sqrt {- b x + 2}} + \frac {129 x^{\frac {5}{2}}}{40 \sqrt {- b x + 2}} + \frac {x^{\frac {3}{2}}}{8 b \sqrt {- b x + 2}} - \frac {3 \sqrt {x}}{4 b^{2} \sqrt {- b x + 2}} + \frac {3 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((I*b**3*x**(11/2)/(5*sqrt(b*x - 2)) - 29*I*b**2*x**(9/2)/(20*sqr t(b*x - 2)) + 73*I*b*x**(7/2)/(20*sqrt(b*x - 2)) - 129*I*x**(5/2)/(40*sqrt (b*x - 2)) - I*x**(3/2)/(8*b*sqrt(b*x - 2)) + 3*I*sqrt(x)/(4*b**2*sqrt(b*x - 2)) - 3*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(5/2)), Abs(b*x) > 2), (-b**3*x**(11/2)/(5*sqrt(-b*x + 2)) + 29*b**2*x**(9/2)/(20*sqrt(-b*x + 2) ) - 73*b*x**(7/2)/(20*sqrt(-b*x + 2)) + 129*x**(5/2)/(40*sqrt(-b*x + 2)) + x**(3/2)/(8*b*sqrt(-b*x + 2)) - 3*sqrt(x)/(4*b**2*sqrt(-b*x + 2)) + 3*asi n(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(5/2)), True))
Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (89) = 178\).
Time = 0.32 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.40 \[ \int x^{3/2} (2-b x)^{5/2} \, dx=\frac {\frac {15 \, \sqrt {-b x + 2} b^{4}}{\sqrt {x}} + \frac {70 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{3}}{x^{\frac {3}{2}}} + \frac {128 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b^{2}}{x^{\frac {5}{2}}} - \frac {70 \, {\left (-b x + 2\right )}^{\frac {7}{2}} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {9}{2}}}{x^{\frac {9}{2}}}}{20 \, {\left (b^{7} - \frac {5 \, {\left (b x - 2\right )} b^{6}}{x} + \frac {10 \, {\left (b x - 2\right )}^{2} b^{5}}{x^{2}} - \frac {10 \, {\left (b x - 2\right )}^{3} b^{4}}{x^{3}} + \frac {5 \, {\left (b x - 2\right )}^{4} b^{3}}{x^{4}} - \frac {{\left (b x - 2\right )}^{5} b^{2}}{x^{5}}\right )}} - \frac {3 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{4 \, b^{\frac {5}{2}}} \]
1/20*(15*sqrt(-b*x + 2)*b^4/sqrt(x) + 70*(-b*x + 2)^(3/2)*b^3/x^(3/2) + 12 8*(-b*x + 2)^(5/2)*b^2/x^(5/2) - 70*(-b*x + 2)^(7/2)*b/x^(7/2) - 15*(-b*x + 2)^(9/2)/x^(9/2))/(b^7 - 5*(b*x - 2)*b^6/x + 10*(b*x - 2)^2*b^5/x^2 - 10 *(b*x - 2)^3*b^4/x^3 + 5*(b*x - 2)^4*b^3/x^4 - (b*x - 2)^5*b^2/x^5) - 3/4* arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (89) = 178\).
Time = 23.39 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.20 \[ \int x^{3/2} (2-b x)^{5/2} \, dx=\frac {{\left ({\left ({\left (2 \, {\left (b x - 2\right )} {\left ({\left (b x - 2\right )} {\left (\frac {4 \, {\left (b x - 2\right )}}{b^{4}} + \frac {41}{b^{4}}\right )} + \frac {171}{b^{4}}\right )} + \frac {745}{b^{4}}\right )} {\left (b x - 2\right )} + \frac {965}{b^{4}}\right )} \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} - \frac {630 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b^{3}}\right )} b {\left | b \right |} - 10 \, {\left ({\left ({\left (b x - 2\right )} {\left (2 \, {\left (b x - 2\right )} {\left (\frac {3 \, {\left (b x - 2\right )}}{b^{3}} + \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} + \frac {279}{b^{3}}\right )} \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b^{2}}\right )} {\left | b \right |} + \frac {80 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left ({\left (b x - 2\right )} {\left (\frac {2 \, {\left (b x - 2\right )}}{b^{2}} + \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} - \frac {30 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b}\right )} {\left | b \right |}}{b} - \frac {160 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} {\left (b x + 3\right )} \sqrt {-b x + 2} - \frac {6 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} {\left | b \right |}}{b^{3}}}{40 \, b} \]
1/40*((((2*(b*x - 2)*((b*x - 2)*(4*(b*x - 2)/b^4 + 41/b^4) + 171/b^4) + 74 5/b^4)*(b*x - 2) + 965/b^4)*sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2) - 630*l og(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/(sqrt(-b)*b^3) )*b*abs(b) - 10*(((b*x - 2)*(2*(b*x - 2)*(3*(b*x - 2)/b^3 + 25/b^3) + 163/ b^3) + 279/b^3)*sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2) - 210*log(abs(-sqrt (-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/(sqrt(-b)*b^2))*abs(b) + 8 0*(sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)*(2*(b*x - 2)/b^2 + 13 /b^2) + 33/b^2) - 30*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/(sqrt(-b)*b))*abs(b)/b - 160*(sqrt((b*x - 2)*b + 2*b)*(b*x + 3)*sq rt(-b*x + 2) - 6*b*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2 *b)))/sqrt(-b))*abs(b)/b^3)/b
Timed out. \[ \int x^{3/2} (2-b x)^{5/2} \, dx=\int x^{3/2}\,{\left (2-b\,x\right )}^{5/2} \,d x \]