Integrand size = 26, antiderivative size = 139 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\frac {2203}{320} \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}+\frac {27}{16} \sqrt {1-2 x} (2+3 x)^3 \sqrt {3+5 x}+\frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}}+\frac {\sqrt {1-2 x} \sqrt {3+5 x} (11129753+4618500 x)}{51200}-\frac {92108287 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{51200 \sqrt {10}} \]
-92108287/512000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+(2+3*x)^4*(3 +5*x)^(1/2)/(1-2*x)^(1/2)+2203/320*(2+3*x)^2*(1-2*x)^(1/2)*(3+5*x)^(1/2)+2 7/16*(2+3*x)^3*(1-2*x)^(1/2)*(3+5*x)^(1/2)+1/51200*(11129753+4618500*x)*(1 -2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.56 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (-14050073+9587886 x+5020200 x^2+2283840 x^3+518400 x^4\right )+92108287 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{512000 \sqrt {1-2 x}} \]
(-10*Sqrt[3 + 5*x]*(-14050073 + 9587886*x + 5020200*x^2 + 2283840*x^3 + 51 8400*x^4) + 92108287*Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] )/(512000*Sqrt[1 - 2*x])
Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {108, 27, 170, 27, 170, 27, 164, 64, 223}
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.
| \(\displaystyle \int \frac {(3 x+2)^4 \sqrt {5 x+3}}{(1-2 x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {(3 x+2)^4 \sqrt {5 x+3}}{\sqrt {1-2 x}}-\int \frac {(3 x+2)^3 (135 x+82)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(3 x+2)^4 \sqrt {5 x+3}}{\sqrt {1-2 x}}-\frac {1}{2} \int \frac {(3 x+2)^3 (135 x+82)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{40} \int -\frac {5 (3 x+2)^2 (6609 x+4028)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {27}{8} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^3\right )+\frac {\sqrt {5 x+3} (3 x+2)^4}{\sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {27}{8} \sqrt {1-2 x} (3 x+2)^3 \sqrt {5 x+3}-\frac {1}{16} \int \frac {(3 x+2)^2 (6609 x+4028)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {\sqrt {5 x+3} (3 x+2)^4}{\sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{16} \left (\frac {1}{30} \int -\frac {3 (3 x+2) (384875 x+236022)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {2203}{10} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2\right )+\frac {27}{8} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^3\right )+\frac {\sqrt {5 x+3} (3 x+2)^4}{\sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{16} \left (\frac {2203}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}-\frac {1}{20} \int \frac {(3 x+2) (384875 x+236022)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {27}{8} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^3\right )+\frac {\sqrt {5 x+3} (3 x+2)^4}{\sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{16} \left (\frac {1}{20} \left (\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (4618500 x+11129753)-\frac {92108287}{160} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {2203}{10} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2\right )+\frac {27}{8} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^3\right )+\frac {\sqrt {5 x+3} (3 x+2)^4}{\sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{16} \left (\frac {1}{20} \left (\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (4618500 x+11129753)-\frac {92108287}{400} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {2203}{10} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2\right )+\frac {27}{8} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^3\right )+\frac {\sqrt {5 x+3} (3 x+2)^4}{\sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{16} \left (\frac {1}{20} \left (\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (4618500 x+11129753)-\frac {92108287 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{80 \sqrt {10}}\right )+\frac {2203}{10} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2\right )+\frac {27}{8} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^3\right )+\frac {\sqrt {5 x+3} (3 x+2)^4}{\sqrt {1-2 x}}\) |
((2 + 3*x)^4*Sqrt[3 + 5*x])/Sqrt[1 - 2*x] + ((27*Sqrt[1 - 2*x]*(2 + 3*x)^3 *Sqrt[3 + 5*x])/8 + ((2203*Sqrt[1 - 2*x]*(2 + 3*x)^2*Sqrt[3 + 5*x])/10 + ( (Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(11129753 + 4618500*x))/80 - (92108287*ArcSin [Sqrt[2/11]*Sqrt[3 + 5*x]])/(80*Sqrt[10]))/20)/16)/2
3.26.19.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp [2/b Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] || PosQ[b])
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[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Time = 1.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(-\frac {\left (-10368000 x^{4} \sqrt {-10 x^{2}-x +3}-45676800 x^{3} \sqrt {-10 x^{2}-x +3}+184216574 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -100404000 x^{2} \sqrt {-10 x^{2}-x +3}-92108287 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-191757720 x \sqrt {-10 x^{2}-x +3}+281001460 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{1024000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(140\) |
-1/1024000*(-10368000*x^4*(-10*x^2-x+3)^(1/2)-45676800*x^3*(-10*x^2-x+3)^( 1/2)+184216574*10^(1/2)*arcsin(20/11*x+1/11)*x-100404000*x^2*(-10*x^2-x+3) ^(1/2)-92108287*10^(1/2)*arcsin(20/11*x+1/11)-191757720*x*(-10*x^2-x+3)^(1 /2)+281001460*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(-1+2*x)/(- 10*x^2-x+3)^(1/2)
Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\frac {92108287 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (518400 \, x^{4} + 2283840 \, x^{3} + 5020200 \, x^{2} + 9587886 \, x - 14050073\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1024000 \, {\left (2 \, x - 1\right )}} \]
1/1024000*(92108287*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqr t(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(518400*x^4 + 2283840*x^3 + 5020200*x^2 + 9587886*x - 14050073)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4} \sqrt {5 x + 3}}{\left (1 - 2 x\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=-\frac {81}{160} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x - \frac {92108287}{1024000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {1557}{640} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {154953}{2560} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {6740553}{51200} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2401 \, \sqrt {-10 \, x^{2} - x + 3}}{16 \, {\left (2 \, x - 1\right )}} \]
-81/160*(-10*x^2 - x + 3)^(3/2)*x - 92108287/1024000*sqrt(5)*sqrt(2)*arcsi n(20/11*x + 1/11) - 1557/640*(-10*x^2 - x + 3)^(3/2) + 154953/2560*sqrt(-1 0*x^2 - x + 3)*x + 6740553/51200*sqrt(-10*x^2 - x + 3) - 2401/16*sqrt(-10* x^2 - x + 3)/(2*x - 1)
Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.70 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=-\frac {92108287}{512000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (6 \, {\left (12 \, {\left (8 \, {\left (36 \, \sqrt {5} {\left (5 \, x + 3\right )} + 361 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 28181 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 4651913 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 460541435 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{6400000 \, {\left (2 \, x - 1\right )}} \]
-92108287/512000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/6400000* (6*(12*(8*(36*sqrt(5)*(5*x + 3) + 361*sqrt(5))*(5*x + 3) + 28181*sqrt(5))* (5*x + 3) + 4651913*sqrt(5))*(5*x + 3) - 460541435*sqrt(5))*sqrt(5*x + 3)* sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4\,\sqrt {5\,x+3}}{{\left (1-2\,x\right )}^{3/2}} \,d x \]