Integrand size = 26, antiderivative size = 113 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=-\frac {37 \sqrt {1-2 x} (2+3 x)^2}{605 \sqrt {3+5 x}}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x} \sqrt {3+5 x} (173063+72060 x)}{96800}-\frac {35451 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{800 \sqrt {10}} \]
-35451/8000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(2+3*x)^3/(1 -2*x)^(1/2)/(3+5*x)^(1/2)-37/605*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)^(1/2)+3/9 6800*(173063+72060*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\frac {10 \left (2026687+2323271 x-1992870 x^2-392040 x^3\right )+4289571 \sqrt {10-20 x} \sqrt {3+5 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{968000 \sqrt {1-2 x} \sqrt {3+5 x}} \]
(10*(2026687 + 2323271*x - 1992870*x^2 - 392040*x^3) + 4289571*Sqrt[10 - 2 0*x]*Sqrt[3 + 5*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(968000*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])
Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {109, 27, 167, 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.
| \(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^{3/2} (5 x+3)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {1}{11} \int \frac {(3 x+2)^2 (519 x+304)}{2 \sqrt {1-2 x} (5 x+3)^{3/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {1}{22} \int \frac {(3 x+2)^2 (519 x+304)}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {1}{22} \left (-\frac {2}{55} \int \frac {3 (3 x+2) (6005 x+3658)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {74 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{22} \left (-\frac {3}{55} \int \frac {(3 x+2) (6005 x+3658)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {74 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{22} \left (-\frac {3}{55} \left (\frac {1429857}{160} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (72060 x+173063)\right )-\frac {74 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{22} \left (-\frac {3}{55} \left (\frac {1429857}{400} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (72060 x+173063)\right )-\frac {74 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{22} \left (-\frac {3}{55} \left (\frac {1429857 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{80 \sqrt {10}}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (72060 x+173063)\right )-\frac {74 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
(7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + ((-74*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(55*Sqrt[3 + 5*x]) - (3*(-1/80*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(173 063 + 72060*x)) + (1429857*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(80*Sqrt[10]) ))/55)/22
3.26.62.3.1 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[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
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 = 1.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21
| method | result | size |
| default | \(-\frac {\sqrt {1-2 x}\, \left (42895710 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-7840800 x^{3} \sqrt {-10 x^{2}-x +3}+4289571 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -39857400 x^{2} \sqrt {-10 x^{2}-x +3}-12868713 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+46465420 x \sqrt {-10 x^{2}-x +3}+40533740 \sqrt {-10 x^{2}-x +3}\right )}{1936000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(137\) |
-1/1936000*(1-2*x)^(1/2)*(42895710*10^(1/2)*arcsin(20/11*x+1/11)*x^2-78408 00*x^3*(-10*x^2-x+3)^(1/2)+4289571*10^(1/2)*arcsin(20/11*x+1/11)*x-3985740 0*x^2*(-10*x^2-x+3)^(1/2)-12868713*10^(1/2)*arcsin(20/11*x+1/11)+46465420* x*(-10*x^2-x+3)^(1/2)+40533740*(-10*x^2-x+3)^(1/2))/(-1+2*x)/(-10*x^2-x+3) ^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\frac {4289571 \, \sqrt {10} {\left (10 \, x^{2} + x - 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (392040 \, x^{3} + 1992870 \, x^{2} - 2323271 \, x - 2026687\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1936000 \, {\left (10 \, x^{2} + x - 3\right )}} \]
1/1936000*(4289571*sqrt(10)*(10*x^2 + x - 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(392040*x^3 + 19928 70*x^2 - 2323271*x - 2026687)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(10*x^2 + x - 3)
\[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4}}{\left (1 - 2 x\right )^{\frac {3}{2}} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=-\frac {81 \, x^{3}}{20 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {1647 \, x^{2}}{80 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {35451}{16000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {2323271 \, x}{96800 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {2026687}{96800 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-81/20*x^3/sqrt(-10*x^2 - x + 3) - 1647/80*x^2/sqrt(-10*x^2 - x + 3) + 354 51/16000*sqrt(10)*arcsin(-20/11*x - 1/11) + 2323271/96800*x/sqrt(-10*x^2 - x + 3) + 2026687/96800/sqrt(-10*x^2 - x + 3)
Time = 0.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.16 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=-\frac {35451}{8000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (6534 \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 197 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 21456431 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{12100000 \, {\left (2 \, x - 1\right )}} - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{151250 \, \sqrt {5 \, x + 3}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{75625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
-35451/8000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/12100000*(653 4*(12*sqrt(5)*(5*x + 3) + 197*sqrt(5))*(5*x + 3) - 21456431*sqrt(5))*sqrt( 5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) - 1/151250*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 2/75625*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)* sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4}{{\left (1-2\,x\right )}^{3/2}\,{\left (5\,x+3\right )}^{3/2}} \,d x \]