Integrand size = 26, antiderivative size = 128 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {30292449 \sqrt {1-2 x} \sqrt {3+5 x}}{512000}-\frac {917953 \sqrt {1-2 x} (3+5 x)^{3/2}}{128000}-\frac {3}{50} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}-\frac {3 \sqrt {1-2 x} (3+5 x)^{5/2} (7889+3900 x)}{16000}+\frac {333216939 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{512000 \sqrt {10}} \]
333216939/5120000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-917953/1280 00*(3+5*x)^(3/2)*(1-2*x)^(1/2)-3/50*(2+3*x)^2*(3+5*x)^(5/2)*(1-2*x)^(1/2)- 3/16000*(3+5*x)^(5/2)*(7889+3900*x)*(1-2*x)^(1/2)-30292449/512000*(1-2*x)^ (1/2)*(3+5*x)^(1/2)
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=\frac {-10 \sqrt {1-2 x} \left (147689703+400508925 x+397621060 x^2+314536800 x^3+155088000 x^4+34560000 x^5\right )-333216939 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{5120000 \sqrt {3+5 x}} \]
(-10*Sqrt[1 - 2*x]*(147689703 + 400508925*x + 397621060*x^2 + 314536800*x^ 3 + 155088000*x^4 + 34560000*x^5) - 333216939*Sqrt[30 + 50*x]*ArcTan[Sqrt[ 5/2 - 5*x]/Sqrt[3 + 5*x]])/(5120000*Sqrt[3 + 5*x])
Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {111, 27, 164, 60, 60, 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.
| \(\displaystyle \int \frac {(3 x+2)^3 (5 x+3)^{3/2}}{\sqrt {1-2 x}} \, dx\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle -\frac {1}{50} \int -\frac {(3 x+2) (5 x+3)^{3/2} (975 x+622)}{2 \sqrt {1-2 x}}dx-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{100} \int \frac {(3 x+2) (5 x+3)^{3/2} (975 x+622)}{\sqrt {1-2 x}}dx-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{5/2}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{100} \left (\frac {917953}{320} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx-\frac {3}{160} \sqrt {1-2 x} (5 x+3)^{5/2} (3900 x+7889)\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{5/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{100} \left (\frac {917953}{320} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{160} \sqrt {1-2 x} (5 x+3)^{5/2} (3900 x+7889)\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{5/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{100} \left (\frac {917953}{320} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{160} \sqrt {1-2 x} (5 x+3)^{5/2} (3900 x+7889)\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{5/2}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{100} \left (\frac {917953}{320} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{160} \sqrt {1-2 x} (5 x+3)^{5/2} (3900 x+7889)\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{5/2}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{100} \left (\frac {917953}{320} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{160} \sqrt {1-2 x} (5 x+3)^{5/2} (3900 x+7889)\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{5/2}\) |
(-3*Sqrt[1 - 2*x]*(2 + 3*x)^2*(3 + 5*x)^(5/2))/50 + ((-3*Sqrt[1 - 2*x]*(3 + 5*x)^(5/2)*(7889 + 3900*x))/160 + (917953*(-1/4*(Sqrt[1 - 2*x]*(3 + 5*x) ^(3/2)) + (33*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]* Sqrt[3 + 5*x]])/(2*Sqrt[10])))/8))/320)/100
3.25.68.3.1 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[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[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
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[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 = 108, normalized size of antiderivative = 0.84
| method | result | size |
| risch | \(\frac {\left (6912000 x^{4}+26870400 x^{3}+46785120 x^{2}+51453140 x +49229901\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{512000 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {333216939 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{10240000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(108\) |
| default | \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (-138240000 x^{4} \sqrt {-10 x^{2}-x +3}-537408000 x^{3} \sqrt {-10 x^{2}-x +3}-935702400 x^{2} \sqrt {-10 x^{2}-x +3}+333216939 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-1029062800 x \sqrt {-10 x^{2}-x +3}-984598020 \sqrt {-10 x^{2}-x +3}\right )}{10240000 \sqrt {-10 x^{2}-x +3}}\) | \(121\) |
1/512000*(6912000*x^4+26870400*x^3+46785120*x^2+51453140*x+49229901)*(-1+2 *x)*(3+5*x)^(1/2)/(-(-1+2*x)*(3+5*x))^(1/2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x )^(1/2)+333216939/10240000*10^(1/2)*arcsin(20/11*x+1/11)*((1-2*x)*(3+5*x)) ^(1/2)/(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.60 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{512000} \, {\left (6912000 \, x^{4} + 26870400 \, x^{3} + 46785120 \, x^{2} + 51453140 \, x + 49229901\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {333216939}{10240000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
-1/512000*(6912000*x^4 + 26870400*x^3 + 46785120*x^2 + 51453140*x + 492299 01)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 333216939/10240000*sqrt(10)*arctan(1/20 *sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
\[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3 x + 2\right )^{3} \left (5 x + 3\right )^{\frac {3}{2}}}{\sqrt {1 - 2 x}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {27}{2} \, \sqrt {-10 \, x^{2} - x + 3} x^{4} - \frac {8397}{160} \, \sqrt {-10 \, x^{2} - x + 3} x^{3} - \frac {292407}{3200} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} - \frac {2572657}{25600} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {333216939}{10240000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) - \frac {49229901}{512000} \, \sqrt {-10 \, x^{2} - x + 3} \]
-27/2*sqrt(-10*x^2 - x + 3)*x^4 - 8397/160*sqrt(-10*x^2 - x + 3)*x^3 - 292 407/3200*sqrt(-10*x^2 - x + 3)*x^2 - 2572657/25600*sqrt(-10*x^2 - x + 3)*x - 333216939/10240000*sqrt(10)*arcsin(-20/11*x - 1/11) - 49229901/512000*s qrt(-10*x^2 - x + 3)
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.56 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{25600000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (24 \, {\left (36 \, {\left (80 \, x + 167\right )} {\left (5 \, x + 3\right )} + 27809\right )} {\left (5 \, x + 3\right )} + 4589765\right )} {\left (5 \, x + 3\right )} + 151462245\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 1666084695 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \]
-1/25600000*sqrt(5)*(2*(4*(24*(36*(80*x + 167)*(5*x + 3) + 27809)*(5*x + 3 ) + 4589765)*(5*x + 3) + 151462245)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 166608 4695*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)))
Timed out. \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=\int \frac {{\left (3\,x+2\right )}^3\,{\left (5\,x+3\right )}^{3/2}}{\sqrt {1-2\,x}} \,d x \]