Integrand size = 26, antiderivative size = 113 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{165 (3+5 x)^{3/2}}-\frac {602 \sqrt {1-2 x} (2+3 x)^2}{9075 \sqrt {3+5 x}}-\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (12199+1020 x)}{242000}+\frac {8127 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2000 \sqrt {10}} \]
8127/20000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/165*(2+3*x)^3*(1 -2*x)^(1/2)/(3+5*x)^(3/2)-602/9075*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)^(1/2)-7 /242000*(12199+1020*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {1-2 x} \left (2953931+10891910 x+11712195 x^2+2940300 x^3\right )}{726000 (3+5 x)^{3/2}}-\frac {8127 \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{2000 \sqrt {10}} \]
-1/726000*(Sqrt[1 - 2*x]*(2953931 + 10891910*x + 11712195*x^2 + 2940300*x^ 3))/(3 + 5*x)^(3/2) - (8127*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(2000*S qrt[10])
Time = 0.21 (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.
| \(\displaystyle \int \frac {(3 x+2)^4}{\sqrt {1-2 x} (5 x+3)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {2}{165} \int -\frac {7 (3 x+2)^2 (39 x+32)}{2 \sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7}{165} \int \frac {(3 x+2)^2 (39 x+32)}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {7}{165} \left (\frac {2}{55} \int \frac {3 (3 x+2) (85 x+458)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {86 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7}{165} \left (\frac {3}{55} \int \frac {(3 x+2) (85 x+458)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {86 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {7}{165} \left (\frac {3}{55} \left (\frac {140481}{160} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (1020 x+12199)\right )-\frac {86 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {7}{165} \left (\frac {3}{55} \left (\frac {140481}{400} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (1020 x+12199)\right )-\frac {86 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {7}{165} \left (\frac {3}{55} \left (\frac {140481 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{80 \sqrt {10}}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (1020 x+12199)\right )-\frac {86 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}\) |
(-2*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(165*(3 + 5*x)^(3/2)) + (7*((-86*Sqrt[1 - 2 *x]*(2 + 3*x)^2)/(55*Sqrt[3 + 5*x]) + (3*(-1/80*(Sqrt[1 - 2*x]*Sqrt[3 + 5* x]*(12199 + 1020*x)) + (140481*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(80*Sqrt[ 10])))/55))/165
3.26.8.3.1 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 = 5.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(\frac {\left (73752525 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-58806000 x^{3} \sqrt {-10 x^{2}-x +3}+88503030 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -234243900 x^{2} \sqrt {-10 x^{2}-x +3}+26550909 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-217838200 x \sqrt {-10 x^{2}-x +3}-59078620 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{14520000 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(130\) |
1/14520000*(73752525*10^(1/2)*arcsin(20/11*x+1/11)*x^2-58806000*x^3*(-10*x ^2-x+3)^(1/2)+88503030*10^(1/2)*arcsin(20/11*x+1/11)*x-234243900*x^2*(-10* x^2-x+3)^(1/2)+26550909*10^(1/2)*arcsin(20/11*x+1/11)-217838200*x*(-10*x^2 -x+3)^(1/2)-59078620*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2 )/(3+5*x)^(3/2)
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {2950101 \, \sqrt {10} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (2940300 \, x^{3} + 11712195 \, x^{2} + 10891910 \, x + 2953931\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14520000 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
-1/14520000*(2950101*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20 *x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(2940300*x^3 + 11712195*x^2 + 10891910*x + 2953931)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^ 2 + 30*x + 9)
\[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4}}{\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\frac {8127}{40000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {81}{500} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {4509}{10000} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{20625 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {32 \, \sqrt {-10 \, x^{2} - x + 3}}{9075 \, {\left (5 \, x + 3\right )}} \]
8127/40000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 81/500*sqrt(-10*x^2 - x + 3)*x - 4509/10000*sqrt(-10*x^2 - x + 3) - 2/20625*sqrt(-10*x^2 - x + 3 )/(25*x^2 + 30*x + 9) - 32/9075*sqrt(-10*x^2 - x + 3)/(5*x + 3)
Time = 0.37 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.52 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {27}{50000} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 131 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{18150000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {8127}{20000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {267 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{1512500 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {801 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{1134375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
-27/50000*(12*sqrt(5)*(5*x + 3) + 131*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 1/18150000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^ (3/2) + 8127/20000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 267/1512 500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 1/113437 5*sqrt(10)*(5*x + 3)^(3/2)*(801*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5* x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3
Timed out. \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4}{\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{5/2}} \,d x \]