Integrand size = 17, antiderivative size = 76 \[ \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {10}{363 (1-2 x)^{3/2}}+\frac {50}{1331 \sqrt {1-2 x}}-\frac {1}{11 (1-2 x)^{3/2} (3+5 x)}-\frac {50 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331} \]
10/363/(1-2*x)^(3/2)-1/11/(1-2*x)^(3/2)/(3+5*x)-50/14641*arctanh(1/11*55^( 1/2)*(1-2*x)^(1/2))*55^(1/2)+50/1331/(1-2*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {2 \left (-\frac {11 \left (-417-400 x+1500 x^2\right )}{2 (1-2 x)^{3/2} (3+5 x)}-75 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )}{43923} \]
(2*((-11*(-417 - 400*x + 1500*x^2))/(2*(1 - 2*x)^(3/2)*(3 + 5*x)) - 75*Sqr t[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]))/43923
Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {52, 61, 61, 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.
\(\displaystyle \int \frac {1}{(1-2 x)^{5/2} (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5}{11} \int \frac {1}{(1-2 x)^{5/2} (5 x+3)}dx-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5}{11} \left (\frac {5}{11} \int \frac {1}{(1-2 x)^{3/2} (5 x+3)}dx+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5}{11} \left (\frac {5}{11} \left (\frac {5}{11} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{11 \sqrt {1-2 x}}\right )+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5}{11} \left (\frac {5}{11} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {5}{11} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{11} \left (\frac {5}{11} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {2}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\) |
-1/11*1/((1 - 2*x)^(3/2)*(3 + 5*x)) + (5*(2/(33*(1 - 2*x)^(3/2)) + (5*(2/( 11*Sqrt[1 - 2*x]) - (2*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/11))/ 11))/11
3.22.86.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 = 1.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {1500 x^{2}-400 x -417}{3993 \left (-1+2 x \right ) \sqrt {1-2 x}\, \left (3+5 x \right )}-\frac {50 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}\) | \(53\) |
derivativedivides | \(\frac {10 \sqrt {1-2 x}}{1331 \left (-\frac {6}{5}-2 x \right )}-\frac {50 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {4}{363 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {40}{1331 \sqrt {1-2 x}}\) | \(54\) |
default | \(\frac {10 \sqrt {1-2 x}}{1331 \left (-\frac {6}{5}-2 x \right )}-\frac {50 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {4}{363 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {40}{1331 \sqrt {1-2 x}}\) | \(54\) |
pseudoelliptic | \(\frac {150 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (10 x^{2}+x -3\right ) \sqrt {55}-16500 x^{2}+4400 x +4587}{\left (1-2 x \right )^{\frac {3}{2}} \left (131769+219615 x \right )}\) | \(60\) |
trager | \(-\frac {\left (1500 x^{2}-400 x -417\right ) \sqrt {1-2 x}}{3993 \left (-1+2 x \right )^{2} \left (3+5 x \right )}+\frac {25 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{14641}\) | \(79\) |
1/3993*(1500*x^2-400*x-417)/(-1+2*x)/(1-2*x)^(1/2)/(3+5*x)-50/14641*arctan h(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {75 \, \sqrt {11} \sqrt {5} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 11 \, {\left (1500 \, x^{2} - 400 \, x - 417\right )} \sqrt {-2 \, x + 1}}{43923 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]
1/43923*(75*sqrt(11)*sqrt(5)*(20*x^3 - 8*x^2 - 7*x + 3)*log((sqrt(11)*sqrt (5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 11*(1500*x^2 - 400*x - 417)*sqr t(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)
Result contains complex when optimal does not.
Time = 2.56 (sec) , antiderivative size = 2285, normalized size of antiderivative = 30.07 \[ \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\text {Too large to display} \]
Piecewise((15000*sqrt(5)*I*(x + 3/5)**3*asin(sqrt(110)/(10*sqrt(x + 3/5))) /(399300*sqrt(11)*(x + 3/5)**3 - 878460*sqrt(11)*(x + 3/5)**2 + 483153*sqr t(11)*(x + 3/5)) - 7500*sqrt(5)*(x + 3/5)**3*log(110)/(399300*sqrt(11)*(x + 3/5)**3 - 878460*sqrt(11)*(x + 3/5)**2 + 483153*sqrt(11)*(x + 3/5)) - 75 00*sqrt(5)*(x + 3/5)**3*log(11)/(399300*sqrt(11)*(x + 3/5)**3 - 878460*sqr t(11)*(x + 3/5)**2 + 483153*sqrt(11)*(x + 3/5)) - 15000*sqrt(5)*(x + 3/5)* *3*log(2)/(399300*sqrt(11)*(x + 3/5)**3 - 878460*sqrt(11)*(x + 3/5)**2 + 4 83153*sqrt(11)*(x + 3/5)) + 7500*sqrt(5)*(x + 3/5)**3*log(10)/(399300*sqrt (11)*(x + 3/5)**3 - 878460*sqrt(11)*(x + 3/5)**2 + 483153*sqrt(11)*(x + 3/ 5)) + 15000*sqrt(5)*(x + 3/5)**3*log(22)/(399300*sqrt(11)*(x + 3/5)**3 - 8 78460*sqrt(11)*(x + 3/5)**2 + 483153*sqrt(11)*(x + 3/5)) - 1500*sqrt(55)*I *(x + 3/5)**2*sqrt(10*x - 5)/(399300*sqrt(11)*(x + 3/5)**3 - 878460*sqrt(1 1)*(x + 3/5)**2 + 483153*sqrt(11)*(x + 3/5)) - 33000*sqrt(5)*I*(x + 3/5)** 2*asin(sqrt(110)/(10*sqrt(x + 3/5)))/(399300*sqrt(11)*(x + 3/5)**3 - 87846 0*sqrt(11)*(x + 3/5)**2 + 483153*sqrt(11)*(x + 3/5)) - 33000*sqrt(5)*(x + 3/5)**2*log(22)/(399300*sqrt(11)*(x + 3/5)**3 - 878460*sqrt(11)*(x + 3/5)* *2 + 483153*sqrt(11)*(x + 3/5)) - 16500*sqrt(5)*(x + 3/5)**2*log(10)/(3993 00*sqrt(11)*(x + 3/5)**3 - 878460*sqrt(11)*(x + 3/5)**2 + 483153*sqrt(11)* (x + 3/5)) + 33000*sqrt(5)*(x + 3/5)**2*log(2)/(399300*sqrt(11)*(x + 3/5)* *3 - 878460*sqrt(11)*(x + 3/5)**2 + 483153*sqrt(11)*(x + 3/5)) + 16500*...
Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {25}{14641} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2 \, {\left (375 \, {\left (2 \, x - 1\right )}^{2} + 1100 \, x - 792\right )}}{3993 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]
25/14641*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) + 2/3993*(375*(2*x - 1)^2 + 1100*x - 792)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {25}{14641} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {4 \, {\left (60 \, x - 41\right )}}{3993 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {25 \, \sqrt {-2 \, x + 1}}{1331 \, {\left (5 \, x + 3\right )}} \]
25/14641*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 4/3993*(60*x - 41)/((2*x - 1)*sqrt(-2*x + 1)) - 25/1 331*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=-\frac {\frac {40\,x}{363}+\frac {50\,{\left (2\,x-1\right )}^2}{1331}-\frac {48}{605}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {50\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{14641} \]