Integrand size = 17, antiderivative size = 63 \[ \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx=-\frac {6}{25} \sqrt {1-2 x}-\frac {(1-2 x)^{3/2}}{5 (3+5 x)}+\frac {6}{25} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-1/5*(1-2*x)^(3/2)/(3+5*x)+6/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^( 1/2)-6/25*(1-2*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx=\frac {1}{125} \left (-\frac {5 \sqrt {1-2 x} (23+20 x)}{3+5 x}+6 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
((-5*Sqrt[1 - 2*x]*(23 + 20*x))/(3 + 5*x) + 6*Sqrt[55]*ArcTanh[Sqrt[5/11]* Sqrt[1 - 2*x]])/125
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {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.
\(\displaystyle \int \frac {(1-2 x)^{3/2}}{(5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {3}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {3}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {3}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {3}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\) |
-1/5*(1 - 2*x)^(3/2)/(3 + 5*x) - (3*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*A rcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/5
3.20.13.3.1 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 = 0.98 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(-\frac {4 \sqrt {1-2 x}}{25}+\frac {22 \sqrt {1-2 x}}{125 \left (-\frac {6}{5}-2 x \right )}+\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(45\) |
default | \(-\frac {4 \sqrt {1-2 x}}{25}+\frac {22 \sqrt {1-2 x}}{125 \left (-\frac {6}{5}-2 x \right )}+\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(45\) |
risch | \(\frac {40 x^{2}+26 x -23}{25 \left (3+5 x \right ) \sqrt {1-2 x}}+\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(46\) |
pseudoelliptic | \(\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}-5 \sqrt {1-2 x}\, \left (23+20 x \right )}{375+625 x}\) | \(47\) |
trager | \(-\frac {\left (23+20 x \right ) \sqrt {1-2 x}}{25 \left (3+5 x \right )}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{125}\) | \(67\) |
-4/25*(1-2*x)^(1/2)+22/125*(1-2*x)^(1/2)/(-6/5-2*x)+6/125*arctanh(1/11*55^ (1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.05 \[ \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx=\frac {3 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) - 5 \, {\left (20 \, x + 23\right )} \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]
1/125*(3*sqrt(11)*sqrt(5)*(5*x + 3)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) - 5*(20*x + 23)*sqrt(-2*x + 1))/(5*x + 3)
Result contains complex when optimal does not.
Time = 1.30 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.78 \[ \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx=\begin {cases} \frac {6 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} + \frac {4 \sqrt {2} \sqrt {x + \frac {3}{5}}}{25 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}}} - \frac {11 \sqrt {2}}{125 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {121 \sqrt {2}}{1250 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} & \text {for}\: \frac {1}{\left |{x + \frac {3}{5}}\right |} > \frac {10}{11} \\- \frac {6 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} - \frac {4 \sqrt {2} i \sqrt {x + \frac {3}{5}}}{25 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}} + \frac {11 \sqrt {2} i}{125 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {121 \sqrt {2} i}{1250 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((6*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/125 + 4*sqrt(2)* sqrt(x + 3/5)/(25*sqrt(-1 + 11/(10*(x + 3/5)))) - 11*sqrt(2)/(125*sqrt(-1 + 11/(10*(x + 3/5)))*sqrt(x + 3/5)) - 121*sqrt(2)/(1250*sqrt(-1 + 11/(10*( x + 3/5)))*(x + 3/5)**(3/2)), 1/Abs(x + 3/5) > 10/11), (-6*sqrt(55)*I*asin (sqrt(110)/(10*sqrt(x + 3/5)))/125 - 4*sqrt(2)*I*sqrt(x + 3/5)/(25*sqrt(1 - 11/(10*(x + 3/5)))) + 11*sqrt(2)*I/(125*sqrt(1 - 11/(10*(x + 3/5)))*sqrt (x + 3/5)) + 121*sqrt(2)*I/(1250*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(3 /2)), True))
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx=-\frac {3}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {4}{25} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{25 \, {\left (5 \, x + 3\right )}} \]
-3/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 4/25*sqrt(-2*x + 1) - 11/25*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03 \[ \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx=-\frac {3}{125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {4}{25} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{25 \, {\left (5 \, x + 3\right )}} \]
-3/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5 *sqrt(-2*x + 1))) - 4/25*sqrt(-2*x + 1) - 11/25*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx=\frac {6\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}-\frac {22\,\sqrt {1-2\,x}}{125\,\left (2\,x+\frac {6}{5}\right )}-\frac {4\,\sqrt {1-2\,x}}{25} \]