Integrand size = 24, antiderivative size = 100 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {\sqrt {1-2 x} (2+3 x)^3}{10 (3+5 x)^2}-\frac {49 \sqrt {1-2 x} (2+3 x)^2}{275 (3+5 x)}+\frac {21 \sqrt {1-2 x} (44+75 x)}{2750}-\frac {1267 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1375 \sqrt {55}} \]
-1267/75625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/10*(2+3*x)^3*( 1-2*x)^(1/2)/(3+5*x)^2-49/275*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)+21/2750*(44+ 75*x)*(1-2*x)^(1/2)
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {\frac {55 \sqrt {1-2 x} \left (236+4555 x+12870 x^2+9900 x^3\right )}{(3+5 x)^2}-2534 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{151250} \]
((55*Sqrt[1 - 2*x]*(236 + 4555*x + 12870*x^2 + 9900*x^3))/(3 + 5*x)^2 - 25 34*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/151250
Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {108, 27, 166, 164, 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.
| \(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^3}{(5 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{10} \int \frac {7 (1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)^2}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7}{10} \int \frac {(1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)^2}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {7}{10} \left (\frac {1}{55} \int \frac {(46-225 x) (3 x+2)}{\sqrt {1-2 x} (5 x+3)}dx-\frac {14 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {7}{10} \left (\frac {1}{55} \left (\frac {181}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {3}{5} \sqrt {1-2 x} (75 x+44)\right )-\frac {14 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {7}{10} \left (\frac {1}{55} \left (\frac {3}{5} \sqrt {1-2 x} (75 x+44)-\frac {181}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {14 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {7}{10} \left (\frac {1}{55} \left (\frac {3}{5} \sqrt {1-2 x} (75 x+44)-\frac {362 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}\right )-\frac {14 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}\) |
-1/10*(Sqrt[1 - 2*x]*(2 + 3*x)^3)/(3 + 5*x)^2 + (7*((-14*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(55*(3 + 5*x)) + ((3*Sqrt[1 - 2*x]*(44 + 75*x))/5 - (362*ArcTanh [Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55]))/55))/10
3.19.49.3.1 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] :> 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
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 = 56, normalized size of antiderivative = 0.56
| method | result | size |
| risch | \(-\frac {19800 x^{4}+15840 x^{3}-3760 x^{2}-4083 x -236}{2750 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {1267 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{75625}\) | \(56\) |
| pseudoelliptic | \(\frac {-2534 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+55 \sqrt {1-2 x}\, \left (9900 x^{3}+12870 x^{2}+4555 x +236\right )}{151250 \left (3+5 x \right )^{2}}\) | \(60\) |
| derivativedivides | \(-\frac {9 \left (1-2 x \right )^{\frac {3}{2}}}{125}+\frac {54 \sqrt {1-2 x}}{625}+\frac {\frac {197 \left (1-2 x \right )^{\frac {3}{2}}}{1375}-\frac {199 \sqrt {1-2 x}}{625}}{\left (-6-10 x \right )^{2}}-\frac {1267 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{75625}\) | \(66\) |
| default | \(-\frac {9 \left (1-2 x \right )^{\frac {3}{2}}}{125}+\frac {54 \sqrt {1-2 x}}{625}+\frac {\frac {197 \left (1-2 x \right )^{\frac {3}{2}}}{1375}-\frac {199 \sqrt {1-2 x}}{625}}{\left (-6-10 x \right )^{2}}-\frac {1267 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{75625}\) | \(66\) |
| trager | \(\frac {\left (9900 x^{3}+12870 x^{2}+4555 x +236\right ) \sqrt {1-2 x}}{2750 \left (3+5 x \right )^{2}}+\frac {1267 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{151250}\) | \(77\) |
-1/2750*(19800*x^4+15840*x^3-3760*x^2-4083*x-236)/(3+5*x)^2/(1-2*x)^(1/2)- 1267/75625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {1267 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (9900 \, x^{3} + 12870 \, x^{2} + 4555 \, x + 236\right )} \sqrt {-2 \, x + 1}}{151250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/151250*(1267*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(9900*x^3 + 12870*x^2 + 4555*x + 236)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Time = 136.44 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.54 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx=- \frac {9 \left (1 - 2 x\right )^{\frac {3}{2}}}{125} + \frac {54 \sqrt {1 - 2 x}}{625} + \frac {279 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{34375} - \frac {388 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} + \frac {88 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \]
-9*(1 - 2*x)**(3/2)/125 + 54*sqrt(1 - 2*x)/625 + 279*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/34375 - 388*Piecew ise((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 + 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/(1 6*(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 = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {9}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1267}{151250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {54}{625} \, \sqrt {-2 \, x + 1} + \frac {985 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2189 \, \sqrt {-2 \, x + 1}}{6875 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
-9/125*(-2*x + 1)^(3/2) + 1267/151250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2* x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 54/625*sqrt(-2*x + 1) + 1/6875*(9 85*(-2*x + 1)^(3/2) - 2189*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {9}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1267}{151250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {54}{625} \, \sqrt {-2 \, x + 1} + \frac {985 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2189 \, \sqrt {-2 \, x + 1}}{27500 \, {\left (5 \, x + 3\right )}^{2}} \]
-9/125*(-2*x + 1)^(3/2) + 1267/151250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 1 0*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 54/625*sqrt(-2*x + 1) + 1/27500*(985*(-2*x + 1)^(3/2) - 2189*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {54\,\sqrt {1-2\,x}}{625}-\frac {9\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {\frac {199\,\sqrt {1-2\,x}}{15625}-\frac {197\,{\left (1-2\,x\right )}^{3/2}}{34375}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,1267{}\mathrm {i}}{75625} \]