Integrand size = 24, antiderivative size = 81 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {409 \sqrt {1-2 x}}{3025}-\frac {(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac {133 (1-2 x)^{3/2}}{6050 (3+5 x)}-\frac {409 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]
-1/550*(1-2*x)^(3/2)/(3+5*x)^2-133/6050*(1-2*x)^(3/2)/(3+5*x)-409/15125*ar ctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+409/3025*(1-2*x)^(1/2)
Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {\sqrt {1-2 x} \left (632+2245 x+1980 x^2\right )}{550 (3+5 x)^2}-\frac {409 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]
(Sqrt[1 - 2*x]*(632 + 2245*x + 1980*x^2))/(550*(3 + 5*x)^2) - (409*ArcTanh [Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])
Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {100, 87, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^2}{(5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {1}{550} \int \frac {\sqrt {1-2 x} (990 x+727)}{(5 x+3)^2}dx-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{550} \left (\frac {2045}{11} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {133 (1-2 x)^{3/2}}{11 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}\) |
-1/550*(1 - 2*x)^(3/2)/(3 + 5*x)^2 + ((-133*(1 - 2*x)^(3/2))/(11*(3 + 5*x) ) + (2045*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/11)/550
3.19.50.3.1 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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.99 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.63
method | result | size |
risch | \(-\frac {3960 x^{3}+2510 x^{2}-981 x -632}{550 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {409 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(51\) |
pseudoelliptic | \(\frac {-818 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+55 \sqrt {1-2 x}\, \left (1980 x^{2}+2245 x +632\right )}{30250 \left (3+5 x \right )^{2}}\) | \(55\) |
derivativedivides | \(\frac {18 \sqrt {1-2 x}}{125}+\frac {\frac {131 \left (1-2 x \right )^{\frac {3}{2}}}{275}-\frac {133 \sqrt {1-2 x}}{125}}{\left (-6-10 x \right )^{2}}-\frac {409 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(57\) |
default | \(\frac {18 \sqrt {1-2 x}}{125}+\frac {\frac {131 \left (1-2 x \right )^{\frac {3}{2}}}{275}-\frac {133 \sqrt {1-2 x}}{125}}{\left (-6-10 x \right )^{2}}-\frac {409 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(57\) |
trager | \(\frac {\left (1980 x^{2}+2245 x +632\right ) \sqrt {1-2 x}}{550 \left (3+5 x \right )^{2}}-\frac {409 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{30250}\) | \(72\) |
-1/550*(3960*x^3+2510*x^2-981*x-632)/(3+5*x)^2/(1-2*x)^(1/2)-409/15125*arc tanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {409 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (1980 \, x^{2} + 2245 \, x + 632\right )} \sqrt {-2 \, x + 1}}{30250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/30250*(409*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(1980*x^2 + 2245*x + 632)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Time = 106.33 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.22 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {18 \sqrt {1 - 2 x}}{125} + \frac {87 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{6875} - \frac {256 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} + \frac {88 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} \]
18*sqrt(1 - 2*x)/125 + 87*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log( sqrt(1 - 2*x) + sqrt(55)/5))/6875 - 256*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*(sq rt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605 , (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/125 + 88* Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt( 55)*sqrt(1 - 2*x)/11 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/ (16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/125
Time = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {409}{30250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {18}{125} \, \sqrt {-2 \, x + 1} + \frac {655 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1463 \, \sqrt {-2 \, x + 1}}{1375 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
409/30250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(- 2*x + 1))) + 18/125*sqrt(-2*x + 1) + 1/1375*(655*(-2*x + 1)^(3/2) - 1463*s qrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {409}{30250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {18}{125} \, \sqrt {-2 \, x + 1} + \frac {655 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1463 \, \sqrt {-2 \, x + 1}}{5500 \, {\left (5 \, x + 3\right )}^{2}} \]
409/30250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/125*sqrt(-2*x + 1) + 1/5500*(655*(-2*x + 1)^(3/2 ) - 1463*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {18\,\sqrt {1-2\,x}}{125}-\frac {409\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {\frac {133\,\sqrt {1-2\,x}}{3125}-\frac {131\,{\left (1-2\,x\right )}^{3/2}}{6875}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]