Integrand size = 24, antiderivative size = 61 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx=-\frac {9}{25} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{275 (3+5 x)}-\frac {134 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]
-134/15125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-9/25*(1-2*x)^(1/2 )-1/275*(1-2*x)^(1/2)/(3+5*x)
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x} (298+495 x)}{275 (3+5 x)}-\frac {134 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]
-1/275*(Sqrt[1 - 2*x]*(298 + 495*x))/(3 + 5*x) - (134*ArcTanh[Sqrt[5/11]*S qrt[1 - 2*x]])/(275*Sqrt[55])
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {100, 90, 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.
\(\displaystyle \int \frac {(3 x+2)^2}{\sqrt {1-2 x} (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {1}{275} \int \frac {495 x+364}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{275} \left (67 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-99 \sqrt {1-2 x}\right )-\frac {\sqrt {1-2 x}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{275} \left (-67 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-99 \sqrt {1-2 x}\right )-\frac {\sqrt {1-2 x}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{275} \left (-\frac {134 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}}-99 \sqrt {1-2 x}\right )-\frac {\sqrt {1-2 x}}{275 (5 x+3)}\) |
-1/275*Sqrt[1 - 2*x]/(3 + 5*x) + (-99*Sqrt[1 - 2*x] - (134*ArcTanh[Sqrt[5/ 11]*Sqrt[1 - 2*x]])/Sqrt[55])/275
3.21.51.3.1 Defintions of rubi rules used
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 = 1.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(-\frac {9 \sqrt {1-2 x}}{25}+\frac {2 \sqrt {1-2 x}}{1375 \left (-\frac {6}{5}-2 x \right )}-\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(45\) |
default | \(-\frac {9 \sqrt {1-2 x}}{25}+\frac {2 \sqrt {1-2 x}}{1375 \left (-\frac {6}{5}-2 x \right )}-\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(45\) |
risch | \(\frac {990 x^{2}+101 x -298}{275 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(46\) |
pseudoelliptic | \(\frac {-134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}-55 \sqrt {1-2 x}\, \left (298+495 x \right )}{45375+75625 x}\) | \(47\) |
trager | \(-\frac {\left (298+495 x \right ) \sqrt {1-2 x}}{275 \left (3+5 x \right )}-\frac {67 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{15125}\) | \(67\) |
-9/25*(1-2*x)^(1/2)+2/1375*(1-2*x)^(1/2)/(-6/5-2*x)-134/15125*arctanh(1/11 *55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx=\frac {67 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (495 \, x + 298\right )} \sqrt {-2 \, x + 1}}{15125 \, {\left (5 \, x + 3\right )}} \]
1/15125*(67*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5* x + 3)) - 55*(495*x + 298)*sqrt(-2*x + 1))/(5*x + 3)
Time = 31.39 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.84 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx=- \frac {9 \sqrt {1 - 2 x}}{25} + \frac {6 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{1375} - \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} \]
-9*sqrt(1 - 2*x)/25 + 6*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sq rt(1 - 2*x) + sqrt(55)/5))/1375 - 4*Piecewise((sqrt(55)*(-log(sqrt(55)*sqr t(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(5 5)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (s qrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/25
Time = 0.32 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx=\frac {67}{15125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {9}{25} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{275 \, {\left (5 \, x + 3\right )}} \]
67/15125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) - 9/25*sqrt(-2*x + 1) - 1/275*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx=\frac {67}{15125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {9}{25} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{275 \, {\left (5 \, x + 3\right )}} \]
67/15125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 9/25*sqrt(-2*x + 1) - 1/275*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.44 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx=-\frac {134\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {2\,\sqrt {1-2\,x}}{1375\,\left (2\,x+\frac {6}{5}\right )}-\frac {9\,\sqrt {1-2\,x}}{25} \]