Integrand size = 17, antiderivative size = 83 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx=\frac {3}{25} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{10 (3+5 x)^2}+\frac {(1-2 x)^{3/2}}{10 (3+5 x)}-\frac {3}{25} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-1/10*(1-2*x)^(5/2)/(3+5*x)^2+1/10*(1-2*x)^(3/2)/(3+5*x)-3/125*arctanh(1/1 1*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+3/25*(1-2*x)^(1/2)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx=\frac {1}{250} \left (\frac {5 \sqrt {1-2 x} \left (64+195 x+80 x^2\right )}{(3+5 x)^2}-6 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
((5*Sqrt[1 - 2*x]*(64 + 195*x + 80*x^2))/(3 + 5*x)^2 - 6*Sqrt[55]*ArcTanh[ Sqrt[5/11]*Sqrt[1 - 2*x]])/250
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {51, 51, 60, 73, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2}}{(5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {1}{2} \int \frac {(1-2 x)^{3/2}}{(5 x+3)^2}dx-\frac {(1-2 x)^{5/2}}{10 (5 x+3)^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{10 (5 x+3)^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{10 (5 x+3)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{10 (5 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{10 (5 x+3)^2}\) |
-1/10*(1 - 2*x)^(5/2)/(3 + 5*x)^2 + ((1 - 2*x)^(3/2)/(5*(3 + 5*x)) + (3*(( 2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/ 5)/2
3.20.95.3.1 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, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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])
Time = 1.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61
method | result | size |
risch | \(-\frac {160 x^{3}+310 x^{2}-67 x -64}{50 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(51\) |
pseudoelliptic | \(\frac {-6 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+5 \sqrt {1-2 x}\, \left (80 x^{2}+195 x +64\right )}{250 \left (3+5 x \right )^{2}}\) | \(55\) |
derivativedivides | \(\frac {8 \sqrt {1-2 x}}{125}+\frac {-\frac {99 \left (1-2 x \right )^{\frac {3}{2}}}{25}+\frac {847 \sqrt {1-2 x}}{125}}{\left (-6-10 x \right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(57\) |
default | \(\frac {8 \sqrt {1-2 x}}{125}+\frac {-\frac {99 \left (1-2 x \right )^{\frac {3}{2}}}{25}+\frac {847 \sqrt {1-2 x}}{125}}{\left (-6-10 x \right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(57\) |
trager | \(\frac {\left (80 x^{2}+195 x +64\right ) \sqrt {1-2 x}}{50 \left (3+5 x \right )^{2}}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{250}\) | \(72\) |
-1/50*(160*x^3+310*x^2-67*x-64)/(3+5*x)^2/(1-2*x)^(1/2)-3/125*arctanh(1/11 *55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx=\frac {3 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (80 \, x^{2} + 195 \, x + 64\right )} \sqrt {-2 \, x + 1}}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/250*(3*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log((sqrt(11)*sqrt(5)*sqrt(- 2*x + 1) + 5*x - 8)/(5*x + 3)) + 5*(80*x^2 + 195*x + 64)*sqrt(-2*x + 1))/( 25*x^2 + 30*x + 9)
Result contains complex when optimal does not.
Time = 2.26 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.59 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx=\begin {cases} - \frac {3 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} - \frac {8 \sqrt {2} \sqrt {x + \frac {3}{5}}}{125 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}}} - \frac {11 \sqrt {2}}{1250 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {1331 \sqrt {2}}{12500 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} - \frac {1331 \sqrt {2}}{62500 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {for}\: \frac {1}{\left |{x + \frac {3}{5}}\right |} > \frac {10}{11} \\\frac {3 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} + \frac {8 \sqrt {2} i \sqrt {x + \frac {3}{5}}}{125 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}} + \frac {11 \sqrt {2} i}{1250 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {1331 \sqrt {2} i}{12500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} + \frac {1331 \sqrt {2} i}{62500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-3*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/125 - 8*sqrt(2) *sqrt(x + 3/5)/(125*sqrt(-1 + 11/(10*(x + 3/5)))) - 11*sqrt(2)/(1250*sqrt( -1 + 11/(10*(x + 3/5)))*sqrt(x + 3/5)) + 1331*sqrt(2)/(12500*sqrt(-1 + 11/ (10*(x + 3/5)))*(x + 3/5)**(3/2)) - 1331*sqrt(2)/(62500*sqrt(-1 + 11/(10*( x + 3/5)))*(x + 3/5)**(5/2)), 1/Abs(x + 3/5) > 10/11), (3*sqrt(55)*I*asin( sqrt(110)/(10*sqrt(x + 3/5)))/125 + 8*sqrt(2)*I*sqrt(x + 3/5)/(125*sqrt(1 - 11/(10*(x + 3/5)))) + 11*sqrt(2)*I/(1250*sqrt(1 - 11/(10*(x + 3/5)))*sqr t(x + 3/5)) - 1331*sqrt(2)*I/(12500*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)* *(3/2)) + 1331*sqrt(2)*I/(62500*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(5/ 2)), True))
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx=\frac {3}{250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {8}{125} \, \sqrt {-2 \, x + 1} - \frac {11 \, {\left (45 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{125 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
3/250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 8/125*sqrt(-2*x + 1) - 11/125*(45*(-2*x + 1)^(3/2) - 77*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx=\frac {3}{250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {8}{125} \, \sqrt {-2 \, x + 1} - \frac {11 \, {\left (45 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{500 \, {\left (5 \, x + 3\right )}^{2}} \]
3/250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5* sqrt(-2*x + 1))) + 8/125*sqrt(-2*x + 1) - 11/500*(45*(-2*x + 1)^(3/2) - 77 *sqrt(-2*x + 1))/(5*x + 3)^2
Time = 1.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx=\frac {8\,\sqrt {1-2\,x}}{125}-\frac {3\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}+\frac {\frac {847\,\sqrt {1-2\,x}}{3125}-\frac {99\,{\left (1-2\,x\right )}^{3/2}}{625}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]