Integrand size = 17, antiderivative size = 76 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx=-\frac {22}{25} \sqrt {1-2 x}-\frac {2}{15} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{5 (3+5 x)}+\frac {22}{25} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-2/15*(1-2*x)^(3/2)-1/5*(1-2*x)^(5/2)/(3+5*x)+22/125*arctanh(1/11*55^(1/2) *(1-2*x)^(1/2))*55^(1/2)-22/25*(1-2*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx=\frac {1}{375} \left (\frac {5 \sqrt {1-2 x} \left (-243-260 x+40 x^2\right )}{3+5 x}+66 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
((5*Sqrt[1 - 2*x]*(-243 - 260*x + 40*x^2))/(3 + 5*x) + 66*Sqrt[55]*ArcTanh [Sqrt[5/11]*Sqrt[1 - 2*x]])/375
Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {51, 60, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2}}{(5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\int \frac {(1-2 x)^{3/2}}{5 x+3}dx-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\) |
(-2*(1 - 2*x)^(3/2))/15 - (1 - 2*x)^(5/2)/(5*(3 + 5*x)) - (11*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/5
3.20.84.3.1 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 = 0.67
method | result | size |
risch | \(-\frac {80 x^{3}-560 x^{2}-226 x +243}{75 \left (3+5 x \right ) \sqrt {1-2 x}}+\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(51\) |
pseudoelliptic | \(\frac {66 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+5 \sqrt {1-2 x}\, \left (40 x^{2}-260 x -243\right )}{1125+1875 x}\) | \(52\) |
derivativedivides | \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{75}-\frac {88 \sqrt {1-2 x}}{125}+\frac {242 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}+\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(54\) |
default | \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{75}-\frac {88 \sqrt {1-2 x}}{125}+\frac {242 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}+\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(54\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (40 x^{2}-260 x -243\right )}{225+375 x}-\frac {11 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{125}\) | \(72\) |
-1/75*(80*x^3-560*x^2-226*x+243)/(3+5*x)/(1-2*x)^(1/2)+22/125*arctanh(1/11 *55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx=\frac {33 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 5 \, {\left (40 \, x^{2} - 260 \, x - 243\right )} \sqrt {-2 \, x + 1}}{375 \, {\left (5 \, x + 3\right )}} \]
1/375*(33*sqrt(11)*sqrt(5)*(5*x + 3)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 5*(40*x^2 - 260*x - 243)*sqrt(-2*x + 1))/(5*x + 3 )
Result contains complex when optimal does not.
Time = 1.94 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.58 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx=\begin {cases} \frac {8 \sqrt {5} i \left (x + \frac {3}{5}\right ) \sqrt {10 x - 5}}{375} - \frac {308 \sqrt {5} i \sqrt {10 x - 5}}{1875} - \frac {22 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} - \frac {121 \sqrt {5} i \sqrt {10 x - 5}}{3125 \left (x + \frac {3}{5}\right )} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {11}{10} \\\frac {8 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )}{375} - \frac {308 \sqrt {5} \sqrt {5 - 10 x}}{1875} - \frac {121 \sqrt {5} \sqrt {5 - 10 x}}{3125 \left (x + \frac {3}{5}\right )} - \frac {11 \sqrt {55} \log {\left (x + \frac {3}{5} \right )}}{125} + \frac {22 \sqrt {55} \log {\left (\sqrt {\frac {5}{11} - \frac {10 x}{11}} + 1 \right )}}{125} & \text {otherwise} \end {cases} \]
Piecewise((8*sqrt(5)*I*(x + 3/5)*sqrt(10*x - 5)/375 - 308*sqrt(5)*I*sqrt(1 0*x - 5)/1875 - 22*sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/125 - 121 *sqrt(5)*I*sqrt(10*x - 5)/(3125*(x + 3/5)), Abs(x + 3/5) > 11/10), (8*sqrt (5)*sqrt(5 - 10*x)*(x + 3/5)/375 - 308*sqrt(5)*sqrt(5 - 10*x)/1875 - 121*s qrt(5)*sqrt(5 - 10*x)/(3125*(x + 3/5)) - 11*sqrt(55)*log(x + 3/5)/125 + 22 *sqrt(55)*log(sqrt(5/11 - 10*x/11) + 1)/125, True))
Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx=-\frac {4}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {11}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {88}{125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]
-4/75*(-2*x + 1)^(3/2) - 11/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1) )/(sqrt(55) + 5*sqrt(-2*x + 1))) - 88/125*sqrt(-2*x + 1) - 121/125*sqrt(-2 *x + 1)/(5*x + 3)
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx=-\frac {4}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {11}{125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {88}{125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]
-4/75*(-2*x + 1)^(3/2) - 11/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt (-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 88/125*sqrt(-2*x + 1) - 121/1 25*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.44 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx=-\frac {242\,\sqrt {1-2\,x}}{625\,\left (2\,x+\frac {6}{5}\right )}-\frac {88\,\sqrt {1-2\,x}}{125}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{75}-\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,22{}\mathrm {i}}{125} \]