Integrand size = 16, antiderivative size = 106 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b}+\frac {5}{8} x^{3/2} \sqrt {2-b x}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{3/2}} \]
5/12*x^(3/2)*(-b*x+2)^(3/2)+1/4*x^(3/2)*(-b*x+2)^(5/2)+5/4*arcsin(1/2*b^(1 /2)*x^(1/2)*2^(1/2))/b^(3/2)+5/8*x^(3/2)*(-b*x+2)^(1/2)-5/8*x^(1/2)*(-b*x+ 2)^(1/2)/b
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\frac {\sqrt {x} \sqrt {2-b x} \left (-15+59 b x-34 b^2 x^2+6 b^3 x^3\right )}{24 b}-\frac {5 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{2 b^{3/2}} \]
(Sqrt[x]*Sqrt[2 - b*x]*(-15 + 59*b*x - 34*b^2*x^2 + 6*b^3*x^3))/(24*b) - ( 5*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])/(2*b^(3/2))
Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {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.
\(\displaystyle \int \sqrt {x} (2-b x)^{5/2} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{4} \int \sqrt {x} (2-b x)^{3/2}dx+\frac {1}{4} x^{3/2} (2-b x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{4} \left (\int \sqrt {x} \sqrt {2-b x}dx+\frac {1}{3} x^{3/2} (2-b x)^{3/2}\right )+\frac {1}{4} x^{3/2} (2-b x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \int \frac {\sqrt {x}}{\sqrt {2-b x}}dx+\frac {1}{3} x^{3/2} (2-b x)^{3/2}+\frac {1}{2} x^{3/2} \sqrt {2-b x}\right )+\frac {1}{4} x^{3/2} (2-b x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {2-b x}}dx}{b}-\frac {\sqrt {x} \sqrt {2-b x}}{b}\right )+\frac {1}{3} x^{3/2} (2-b x)^{3/2}+\frac {1}{2} x^{3/2} \sqrt {2-b x}\right )+\frac {1}{4} x^{3/2} (2-b x)^{5/2}\) |
\(\Big \downarrow \) 63 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {2 \int \frac {1}{\sqrt {2-b x}}d\sqrt {x}}{b}-\frac {\sqrt {x} \sqrt {2-b x}}{b}\right )+\frac {1}{3} x^{3/2} (2-b x)^{3/2}+\frac {1}{2} x^{3/2} \sqrt {2-b x}\right )+\frac {1}{4} x^{3/2} (2-b x)^{5/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}-\frac {\sqrt {x} \sqrt {2-b x}}{b}\right )+\frac {1}{3} x^{3/2} (2-b x)^{3/2}+\frac {1}{2} x^{3/2} \sqrt {2-b x}\right )+\frac {1}{4} x^{3/2} (2-b x)^{5/2}\) |
(x^(3/2)*(2 - b*x)^(5/2))/4 + (5*((x^(3/2)*Sqrt[2 - b*x])/2 + (x^(3/2)*(2 - b*x)^(3/2))/3 + (-((Sqrt[x]*Sqrt[2 - b*x])/b) + (2*ArcSin[(Sqrt[b]*Sqrt[ x])/Sqrt[2]])/b^(3/2))/2))/4
3.6.65.3.1 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 = 89, normalized size of antiderivative = 0.84
method | result | size |
meijerg | \(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {3}{2}} \left (-6 b^{3} x^{3}+34 b^{2} x^{2}-59 b x +15\right ) \sqrt {-\frac {b x}{2}+1}}{24 b}-\frac {5 \sqrt {\pi }\, \left (-b \right )^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {3}{2}}}}{\sqrt {-b}\, \sqrt {\pi }\, b}\) | \(89\) |
default | \(\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {5}{2}}}{4}+\frac {5 x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {3}{2}}}{12}+\frac {5 x^{\frac {3}{2}} \sqrt {-b x +2}}{8}-\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{8 b}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\) | \(107\) |
risch | \(-\frac {\left (6 b^{3} x^{3}-34 b^{2} x^{2}+59 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{24 b \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\) | \(115\) |
30/(-b)^(1/2)/Pi^(1/2)/b*(1/720*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(3/2)*(-6*b^ 3*x^3+34*b^2*x^2-59*b*x+15)/b*(-1/2*b*x+1)^(1/2)-1/24*Pi^(1/2)*(-b)^(3/2)/ b^(3/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))
Time = 0.24 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.33 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\left [\frac {{\left (6 \, b^{4} x^{3} - 34 \, b^{3} x^{2} + 59 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{24 \, b^{2}}, \frac {{\left (6 \, b^{4} x^{3} - 34 \, b^{3} x^{2} + 59 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{24 \, b^{2}}\right ] \]
[1/24*((6*b^4*x^3 - 34*b^3*x^2 + 59*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^2, 1/24*(( 6*b^4*x^3 - 34*b^3*x^2 + 59*b^2*x - 15*b)*sqrt(-b*x + 2)*sqrt(x) - 30*sqrt (b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^2]
Result contains complex when optimal does not.
Time = 9.92 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.39 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\begin {cases} \frac {i b^{3} x^{\frac {9}{2}}}{4 \sqrt {b x - 2}} - \frac {23 i b^{2} x^{\frac {7}{2}}}{12 \sqrt {b x - 2}} + \frac {127 i b x^{\frac {5}{2}}}{24 \sqrt {b x - 2}} - \frac {133 i x^{\frac {3}{2}}}{24 \sqrt {b x - 2}} + \frac {5 i \sqrt {x}}{4 b \sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {3}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b^{3} x^{\frac {9}{2}}}{4 \sqrt {- b x + 2}} + \frac {23 b^{2} x^{\frac {7}{2}}}{12 \sqrt {- b x + 2}} - \frac {127 b x^{\frac {5}{2}}}{24 \sqrt {- b x + 2}} + \frac {133 x^{\frac {3}{2}}}{24 \sqrt {- b x + 2}} - \frac {5 \sqrt {x}}{4 b \sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((I*b**3*x**(9/2)/(4*sqrt(b*x - 2)) - 23*I*b**2*x**(7/2)/(12*sqrt (b*x - 2)) + 127*I*b*x**(5/2)/(24*sqrt(b*x - 2)) - 133*I*x**(3/2)/(24*sqrt (b*x - 2)) + 5*I*sqrt(x)/(4*b*sqrt(b*x - 2)) - 5*I*acosh(sqrt(2)*sqrt(b)*s qrt(x)/2)/(4*b**(3/2)), Abs(b*x) > 2), (-b**3*x**(9/2)/(4*sqrt(-b*x + 2)) + 23*b**2*x**(7/2)/(12*sqrt(-b*x + 2)) - 127*b*x**(5/2)/(24*sqrt(-b*x + 2) ) + 133*x**(3/2)/(24*sqrt(-b*x + 2)) - 5*sqrt(x)/(4*b*sqrt(-b*x + 2)) + 5* asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(3/2)), True))
Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.37 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\frac {\frac {15 \, \sqrt {-b x + 2} b^{3}}{\sqrt {x}} + \frac {55 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} + \frac {73 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}}{12 \, {\left (b^{5} - \frac {4 \, {\left (b x - 2\right )} b^{4}}{x} + \frac {6 \, {\left (b x - 2\right )}^{2} b^{3}}{x^{2}} - \frac {4 \, {\left (b x - 2\right )}^{3} b^{2}}{x^{3}} + \frac {{\left (b x - 2\right )}^{4} b}{x^{4}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{4 \, b^{\frac {3}{2}}} \]
1/12*(15*sqrt(-b*x + 2)*b^3/sqrt(x) + 55*(-b*x + 2)^(3/2)*b^2/x^(3/2) + 73 *(-b*x + 2)^(5/2)*b/x^(5/2) - 15*(-b*x + 2)^(7/2)/x^(7/2))/(b^5 - 4*(b*x - 2)*b^4/x + 6*(b*x - 2)^2*b^3/x^2 - 4*(b*x - 2)^3*b^2/x^3 + (b*x - 2)^4*b/ x^4) - 5/4*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (73) = 146\).
Time = 23.34 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.35 \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\frac {{\left ({\left ({\left (b x - 2\right )} {\left (2 \, {\left (b x - 2\right )} {\left (\frac {3 \, {\left (b x - 2\right )}}{b^{3}} + \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} + \frac {279}{b^{3}}\right )} \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b^{2}}\right )} b {\left | b \right |} - 24 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left ({\left (b x - 2\right )} {\left (\frac {2 \, {\left (b x - 2\right )}}{b^{2}} + \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} - \frac {30 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b}\right )} {\left | b \right |} + \frac {144 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} {\left (b x + 3\right )} \sqrt {-b x + 2} - \frac {6 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} {\left | b \right |}}{b^{2}} + \frac {192 \, {\left (\frac {2 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}} - \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2}\right )} {\left | b \right |}}{b^{2}}}{24 \, b} \]
1/24*((((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))*b*abs(b) - 24*(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) + 144*(sqrt((b*x - 2)*b + 2*b)*(b*x + 3)*sqrt(-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^2 + 192*(2*b*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/sqrt(-b) - sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2))*abs(b)/b ^2)/b
Timed out. \[ \int \sqrt {x} (2-b x)^{5/2} \, dx=\int \sqrt {x}\,{\left (2-b\,x\right )}^{5/2} \,d x \]