Integrand size = 24, antiderivative size = 94 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {357 \sqrt {1-2 x}}{1375}+\frac {119 (1-2 x)^{3/2}}{3025}-\frac {(1-2 x)^{5/2}}{550 (3+5 x)^2}-\frac {131 (1-2 x)^{5/2}}{6050 (3+5 x)}-\frac {357 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{125 \sqrt {55}} \]
119/3025*(1-2*x)^(3/2)-1/550*(1-2*x)^(5/2)/(3+5*x)^2-131/6050*(1-2*x)^(5/2 )/(3+5*x)-357/6875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+357/1375* (1-2*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {\sqrt {1-2 x} \left (656+2105 x+1320 x^2-600 x^3\right )}{250 (3+5 x)^2}-\frac {357 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{125 \sqrt {55}} \]
(Sqrt[1 - 2*x]*(656 + 2105*x + 1320*x^2 - 600*x^3))/(250*(3 + 5*x)^2) - (3 57*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(125*Sqrt[55])
Time = 0.20 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {100, 27, 87, 60, 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.
| \(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)^2}{(5 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {1}{550} \int \frac {5 (1-2 x)^{3/2} (198 x+145)}{(5 x+3)^2}dx-\frac {(1-2 x)^{5/2}}{550 (5 x+3)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{110} \int \frac {(1-2 x)^{3/2} (198 x+145)}{(5 x+3)^2}dx-\frac {(1-2 x)^{5/2}}{550 (5 x+3)^2}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {1}{110} \left (\frac {357}{11} \int \frac {(1-2 x)^{3/2}}{5 x+3}dx-\frac {131 (1-2 x)^{5/2}}{55 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{550 (5 x+3)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{110} \left (\frac {357}{11} \left (\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {131 (1-2 x)^{5/2}}{55 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{550 (5 x+3)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{110} \left (\frac {357}{11} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {131 (1-2 x)^{5/2}}{55 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{550 (5 x+3)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{110} \left (\frac {357}{11} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {131 (1-2 x)^{5/2}}{55 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{550 (5 x+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{110} \left (\frac {357}{11} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {131 (1-2 x)^{5/2}}{55 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{550 (5 x+3)^2}\) |
-1/550*(1 - 2*x)^(5/2)/(3 + 5*x)^2 + ((-131*(1 - 2*x)^(5/2))/(55*(3 + 5*x) ) + (357*((2*(1 - 2*x)^(3/2))/15 + (11*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5 ]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/5))/11)/110
3.20.21.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 1.00 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.60
| method | result | size |
| risch | \(\frac {1200 x^{4}-3240 x^{3}-2890 x^{2}+793 x +656}{250 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {357 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{6875}\) | \(56\) |
| pseudoelliptic | \(\frac {-714 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}-55 \sqrt {1-2 x}\, \left (600 x^{3}-1320 x^{2}-2105 x -656\right )}{13750 \left (3+5 x \right )^{2}}\) | \(60\) |
| derivativedivides | \(\frac {6 \left (1-2 x \right )^{\frac {3}{2}}}{125}+\frac {174 \sqrt {1-2 x}}{625}+\frac {\frac {127 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {1419 \sqrt {1-2 x}}{625}}{\left (-6-10 x \right )^{2}}-\frac {357 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{6875}\) | \(66\) |
| default | \(\frac {6 \left (1-2 x \right )^{\frac {3}{2}}}{125}+\frac {174 \sqrt {1-2 x}}{625}+\frac {\frac {127 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {1419 \sqrt {1-2 x}}{625}}{\left (-6-10 x \right )^{2}}-\frac {357 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{6875}\) | \(66\) |
| trager | \(-\frac {\left (600 x^{3}-1320 x^{2}-2105 x -656\right ) \sqrt {1-2 x}}{250 \left (3+5 x \right )^{2}}-\frac {357 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{13750}\) | \(77\) |
1/250*(1200*x^4-3240*x^3-2890*x^2+793*x+656)/(3+5*x)^2/(1-2*x)^(1/2)-357/6 875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {357 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (600 \, x^{3} - 1320 \, x^{2} - 2105 \, x - 656\right )} \sqrt {-2 \, x + 1}}{13750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/13750*(357*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(600*x^3 - 1320*x^2 - 2105*x - 656)*sqrt(-2*x + 1) )/(25*x^2 + 30*x + 9)
Time = 106.96 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.77 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {6 \left (1 - 2 x\right )^{\frac {3}{2}}}{125} + \frac {174 \sqrt {1 - 2 x}}{625} + \frac {829 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{34375} - \frac {2728 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} + \frac {968 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \]
6*(1 - 2*x)**(3/2)/125 + 174*sqrt(1 - 2*x)/625 + 829*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/34375 - 2728*Piece wise((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 )*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2 *x) < sqrt(55)/5)))/625 + 968*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(5 5)*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) ))/625
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {6}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {357}{13750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {174}{625} \, \sqrt {-2 \, x + 1} + \frac {635 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1419 \, \sqrt {-2 \, x + 1}}{625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
6/125*(-2*x + 1)^(3/2) + 357/13750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 174/625*sqrt(-2*x + 1) + 1/625*(635* (-2*x + 1)^(3/2) - 1419*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {6}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {357}{13750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {174}{625} \, \sqrt {-2 \, x + 1} + \frac {635 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1419 \, \sqrt {-2 \, x + 1}}{2500 \, {\left (5 \, x + 3\right )}^{2}} \]
6/125*(-2*x + 1)^(3/2) + 357/13750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*s qrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 174/625*sqrt(-2*x + 1) + 1 /2500*(635*(-2*x + 1)^(3/2) - 1419*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 1.39 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {174\,\sqrt {1-2\,x}}{625}+\frac {6\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {\frac {1419\,\sqrt {1-2\,x}}{15625}-\frac {127\,{\left (1-2\,x\right )}^{3/2}}{3125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,357{}\mathrm {i}}{6875} \]