Integrand size = 24, antiderivative size = 74 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {26}{275} \sqrt {1-2 x}-\frac {3}{25} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{275 (3+5 x)}-\frac {26 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]
-3/25*(1-2*x)^(3/2)-1/275*(1-2*x)^(3/2)/(3+5*x)-26/1375*arctanh(1/11*55^(1 /2)*(1-2*x)^(1/2))*55^(1/2)+26/275*(1-2*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {\sqrt {1-2 x} \left (-2+15 x+30 x^2\right )}{25 (3+5 x)}-\frac {26 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]
(Sqrt[1 - 2*x]*(-2 + 15*x + 30*x^2))/(25*(3 + 5*x)) - (26*ArcTanh[Sqrt[5/1 1]*Sqrt[1 - 2*x]])/(25*Sqrt[55])
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {100, 90, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^2}{(5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {1}{275} \int \frac {\sqrt {1-2 x} (495 x+362)}{5 x+3}dx-\frac {(1-2 x)^{3/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{275} \left (65 \int \frac {\sqrt {1-2 x}}{5 x+3}dx-33 (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{3/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{275} \left (65 \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-33 (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{3/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{275} \left (65 \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-33 (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{3/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{275} \left (65 \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-33 (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{3/2}}{275 (5 x+3)}\) |
-1/275*(1 - 2*x)^(3/2)/(3 + 5*x) + (-33*(1 - 2*x)^(3/2) + 65*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/275
3.19.41.3.1 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 + 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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 = 46, normalized size of antiderivative = 0.62
method | result | size |
risch | \(-\frac {60 x^{3}-19 x +2}{25 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {26 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\) | \(46\) |
pseudoelliptic | \(\frac {-26 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+55 \sqrt {1-2 x}\, \left (30 x^{2}+15 x -2\right )}{4125+6875 x}\) | \(52\) |
derivativedivides | \(-\frac {3 \left (1-2 x \right )^{\frac {3}{2}}}{25}+\frac {12 \sqrt {1-2 x}}{125}+\frac {2 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}-\frac {26 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\) | \(54\) |
default | \(-\frac {3 \left (1-2 x \right )^{\frac {3}{2}}}{25}+\frac {12 \sqrt {1-2 x}}{125}+\frac {2 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}-\frac {26 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\) | \(54\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (30 x^{2}+15 x -2\right )}{75+125 x}-\frac {13 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1375}\) | \(72\) |
-1/25*(60*x^3-19*x+2)/(3+5*x)/(1-2*x)^(1/2)-26/1375*arctanh(1/11*55^(1/2)* (1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {13 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (30 \, x^{2} + 15 \, x - 2\right )} \sqrt {-2 \, x + 1}}{1375 \, {\left (5 \, x + 3\right )}} \]
1/1375*(13*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(30*x^2 + 15*x - 2)*sqrt(-2*x + 1))/(5*x + 3)
Time = 32.02 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.50 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx=- \frac {3 \left (1 - 2 x\right )^{\frac {3}{2}}}{25} + \frac {12 \sqrt {1 - 2 x}}{125} + \frac {64 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{6875} - \frac {44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} \]
-3*(1 - 2*x)**(3/2)/25 + 12*sqrt(1 - 2*x)/125 + 64*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/6875 - 44*Piecewise( (sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2 *x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqr t(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/125
Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {3}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {13}{1375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {12}{125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]
-3/25*(-2*x + 1)^(3/2) + 13/1375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1 ))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 12/125*sqrt(-2*x + 1) - 1/125*sqrt(-2* x + 1)/(5*x + 3)
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {3}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {13}{1375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {12}{125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \]
-3/25*(-2*x + 1)^(3/2) + 13/1375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqr t(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 12/125*sqrt(-2*x + 1) - 1/12 5*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {12\,\sqrt {1-2\,x}}{125}-\frac {2\,\sqrt {1-2\,x}}{625\,\left (2\,x+\frac {6}{5}\right )}-\frac {3\,{\left (1-2\,x\right )}^{3/2}}{25}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,26{}\mathrm {i}}{1375} \]