Integrand size = 16, antiderivative size = 67 \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\frac {2 x^{3/2}}{3 b (2-b x)^{3/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {2-b x}}+\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \]
2/3*x^(3/2)/b/(-b*x+2)^(3/2)+2*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2) -2*x^(1/2)/b^2/(-b*x+2)^(1/2)
Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\frac {4 \sqrt {x} (-3+2 b x)}{3 b^2 (2-b x)^{3/2}}-\frac {4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{5/2}} \]
(4*Sqrt[x]*(-3 + 2*b*x))/(3*b^2*(2 - b*x)^(3/2)) - (4*ArcTan[(Sqrt[b]*Sqrt [x])/(Sqrt[2] - Sqrt[2 - b*x])])/b^(5/2)
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {57, 57, 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.
\(\displaystyle \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {2 x^{3/2}}{3 b (2-b x)^{3/2}}-\frac {\int \frac {\sqrt {x}}{(2-b x)^{3/2}}dx}{b}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {2 x^{3/2}}{3 b (2-b x)^{3/2}}-\frac {\frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {\int \frac {1}{\sqrt {x} \sqrt {2-b x}}dx}{b}}{b}\) |
\(\Big \downarrow \) 63 |
\(\displaystyle \frac {2 x^{3/2}}{3 b (2-b x)^{3/2}}-\frac {\frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \int \frac {1}{\sqrt {2-b x}}d\sqrt {x}}{b}}{b}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {2 x^{3/2}}{3 b (2-b x)^{3/2}}-\frac {\frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}}{b}\) |
(2*x^(3/2))/(3*b*(2 - b*x)^(3/2)) - ((2*Sqrt[x])/(b*Sqrt[2 - b*x]) - (2*Ar cSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2))/b
3.7.42.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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 = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09
method | result | size |
meijerg | \(-\frac {4 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {5}{2}} \left (-10 b x +15\right )}{20 b^{2} \left (-\frac {b x}{2}+1\right )^{\frac {3}{2}}}+\frac {3 \sqrt {\pi }\, \left (-b \right )^{\frac {5}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{2 b^{\frac {5}{2}}}\right )}{3 \left (-b \right )^{\frac {3}{2}} \sqrt {\pi }\, b}\) | \(73\) |
-4/3/(-b)^(3/2)/Pi^(1/2)/b*(-1/20*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(5/2)*(-10 *b*x+15)/b^2/(-1/2*b*x+1)^(3/2)+3/2*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 = 173, normalized size of antiderivative = 2.58 \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\left [-\frac {3 \, {\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) - 4 \, {\left (2 \, b^{2} x - 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, {\left (b^{5} x^{2} - 4 \, b^{4} x + 4 \, b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) - 2 \, {\left (2 \, b^{2} x - 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x}\right )}}{3 \, {\left (b^{5} x^{2} - 4 \, b^{4} x + 4 \, b^{3}\right )}}\right ] \]
[-1/3*(3*(b^2*x^2 - 4*b*x + 4)*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b) *sqrt(x) + 1) - 4*(2*b^2*x - 3*b)*sqrt(-b*x + 2)*sqrt(x))/(b^5*x^2 - 4*b^4 *x + 4*b^3), -2/3*(3*(b^2*x^2 - 4*b*x + 4)*sqrt(b)*arctan(sqrt(-b*x + 2)/( sqrt(b)*sqrt(x))) - 2*(2*b^2*x - 3*b)*sqrt(-b*x + 2)*sqrt(x))/(b^5*x^2 - 4 *b^4*x + 4*b^3)]
Result contains complex when optimal does not.
Time = 2.95 (sec) , antiderivative size = 648, normalized size of antiderivative = 9.67 \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\begin {cases} \frac {8 i b^{\frac {11}{2}} x^{8}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} - \frac {12 i b^{\frac {9}{2}} x^{7}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} - \frac {6 i b^{5} x^{\frac {15}{2}} \sqrt {b x - 2} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} + \frac {3 \pi b^{5} x^{\frac {15}{2}} \sqrt {b x - 2}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} + \frac {12 i b^{4} x^{\frac {13}{2}} \sqrt {b x - 2} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} - \frac {6 \pi b^{4} x^{\frac {13}{2}} \sqrt {b x - 2}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {8 b^{\frac {11}{2}} x^{8}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} + \frac {12 b^{\frac {9}{2}} x^{7}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} + \frac {6 b^{5} x^{\frac {15}{2}} \sqrt {- b x + 2} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} - \frac {12 b^{4} x^{\frac {13}{2}} \sqrt {- b x + 2} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} & \text {otherwise} \end {cases} \]
Piecewise((8*I*b**(11/2)*x**8/(3*b**(15/2)*x**(15/2)*sqrt(b*x - 2) - 6*b** (13/2)*x**(13/2)*sqrt(b*x - 2)) - 12*I*b**(9/2)*x**7/(3*b**(15/2)*x**(15/2 )*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqrt(b*x - 2)) - 6*I*b**5*x**(15/2 )*sqrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)*sq rt(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqrt(b*x - 2)) + 3*pi*b**5*x**(15/2)*s qrt(b*x - 2)/(3*b**(15/2)*x**(15/2)*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2)* sqrt(b*x - 2)) + 12*I*b**4*x**(13/2)*sqrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*s qrt(x)/2)/(3*b**(15/2)*x**(15/2)*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqr t(b*x - 2)) - 6*pi*b**4*x**(13/2)*sqrt(b*x - 2)/(3*b**(15/2)*x**(15/2)*sqr t(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqrt(b*x - 2)), Abs(b*x) > 2), (-8*b**( 11/2)*x**8/(3*b**(15/2)*x**(15/2)*sqrt(-b*x + 2) - 6*b**(13/2)*x**(13/2)*s qrt(-b*x + 2)) + 12*b**(9/2)*x**7/(3*b**(15/2)*x**(15/2)*sqrt(-b*x + 2) - 6*b**(13/2)*x**(13/2)*sqrt(-b*x + 2)) + 6*b**5*x**(15/2)*sqrt(-b*x + 2)*as in(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)*sqrt(-b*x + 2) - 6*b* *(13/2)*x**(13/2)*sqrt(-b*x + 2)) - 12*b**4*x**(13/2)*sqrt(-b*x + 2)*asin( sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)*sqrt(-b*x + 2) - 6*b**(1 3/2)*x**(13/2)*sqrt(-b*x + 2)), True))
Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.75 \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\frac {2 \, {\left (b + \frac {3 \, {\left (b x - 2\right )}}{x}\right )} x^{\frac {3}{2}}}{3 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {5}{2}}} \]
2/3*(b + 3*(b*x - 2)/x)*x^(3/2)/((-b*x + 2)^(3/2)*b^2) - 2*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. 178 vs. \(2 (50) = 100\).
Time = 1.53 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.66 \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\frac {{\left (\frac {3 \, \log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b}} + \frac {16 \, {\left (3 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{4} \sqrt {-b} - 6 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} \sqrt {-b} b + 8 \, \sqrt {-b} b^{2}\right )}}{{\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )}^{3}}\right )} {\left | b \right |}}{3 \, b^{3}} \]
1/3*(3*log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2)/sqrt(-b) + 16*(3*(sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^4*sqrt(-b) - 6*(sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2*sqrt(-b)*b + 8*sqr t(-b)*b^2)/((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)^3 )*abs(b)/b^3
Timed out. \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\int \frac {x^{3/2}}{{\left (2-b\,x\right )}^{5/2}} \,d x \]
Time = 0.00 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.51 \[ \int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx=\frac {-2 \sqrt {b}\, \sqrt {-b x +2}\, \mathrm {log}\left (\frac {\sqrt {-b x +2}+\sqrt {x}\, \sqrt {b}\, i}{\sqrt {2}}\right ) b i x +4 \sqrt {b}\, \sqrt {-b x +2}\, \mathrm {log}\left (\frac {\sqrt {-b x +2}+\sqrt {x}\, \sqrt {b}\, i}{\sqrt {2}}\right ) i -\frac {8 \sqrt {x}\, b^{2} x}{3}+4 \sqrt {x}\, b}{\sqrt {-b x +2}\, b^{3} \left (b x -2\right )} \]