Integrand size = 26, antiderivative size = 113 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {243}{220} \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}+\frac {7 (2+3 x)^3 \sqrt {3+5 x}}{11 \sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} \sqrt {3+5 x} (27269+11316 x)}{7040}-\frac {184641 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{640 \sqrt {10}} \]
-184641/6400*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(2+3*x)^3*( 3+5*x)^(1/2)/(1-2*x)^(1/2)+243/220*(2+3*x)^2*(1-2*x)^(1/2)*(3+5*x)^(1/2)+9 /7040*(27269+11316*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {-10 \sqrt {3+5 x} \left (-312365+196614 x+78408 x^2+19008 x^3\right )+2031051 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{70400 \sqrt {1-2 x}} \]
(-10*Sqrt[3 + 5*x]*(-312365 + 196614*x + 78408*x^2 + 19008*x^3) + 2031051* Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(70400*Sqrt[1 - 2*x ])
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, 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.
\(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^{3/2} \sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {1}{11} \int \frac {3 (3 x+2)^2 (243 x+148)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \int \frac {(3 x+2)^2 (243 x+148)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 170 |
\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (-\frac {1}{30} \int -\frac {3 (3 x+2) (14145 x+8674)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {81}{10} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \int \frac {(3 x+2) (14145 x+8674)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \left (\frac {677017}{32} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {3}{16} \sqrt {1-2 x} \sqrt {5 x+3} (11316 x+27269)\right )-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \left (\frac {677017}{80} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {3}{16} \sqrt {1-2 x} \sqrt {5 x+3} (11316 x+27269)\right )-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \left (\frac {677017 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{16 \sqrt {10}}-\frac {3}{16} \sqrt {1-2 x} \sqrt {5 x+3} (11316 x+27269)\right )-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\) |
(7*(2 + 3*x)^3*Sqrt[3 + 5*x])/(11*Sqrt[1 - 2*x]) - (3*((-81*Sqrt[1 - 2*x]* (2 + 3*x)^2*Sqrt[3 + 5*x])/10 + ((-3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(27269 + 11316*x))/16 + (677017*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(16*Sqrt[10]))/20 ))/22
3.26.51.3.1 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[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 = 1.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09
method | result | size |
default | \(-\frac {\left (-380160 x^{3} \sqrt {-10 x^{2}-x +3}+4062102 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -1568160 x^{2} \sqrt {-10 x^{2}-x +3}-2031051 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-3932280 x \sqrt {-10 x^{2}-x +3}+6247300 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}\, \sqrt {1-2 x}}{140800 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(123\) |
-1/140800*(-380160*x^3*(-10*x^2-x+3)^(1/2)+4062102*10^(1/2)*arcsin(20/11*x +1/11)*x-1568160*x^2*(-10*x^2-x+3)^(1/2)-2031051*10^(1/2)*arcsin(20/11*x+1 /11)-3932280*x*(-10*x^2-x+3)^(1/2)+6247300*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1 /2)*(1-2*x)^(1/2)/(-1+2*x)/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {2031051 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (19008 \, x^{3} + 78408 \, x^{2} + 196614 \, x - 312365\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{140800 \, {\left (2 \, x - 1\right )}} \]
1/140800*(2031051*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt( 5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(19008*x^3 + 78408*x^2 + 19 6614*x - 312365)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\int \frac {\left (3 x + 2\right )^{4}}{\left (1 - 2 x\right )^{\frac {3}{2}} \sqrt {5 x + 3}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {27}{20} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} - \frac {184641}{12800} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {999}{160} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {2187}{128} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2401 \, \sqrt {-10 \, x^{2} - x + 3}}{88 \, {\left (2 \, x - 1\right )}} \]
27/20*sqrt(-10*x^2 - x + 3)*x^2 - 184641/12800*sqrt(5)*sqrt(2)*arcsin(20/1 1*x + 1/11) + 999/160*sqrt(-10*x^2 - x + 3)*x + 2187/128*sqrt(-10*x^2 - x + 3) - 2401/88*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.35 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=-\frac {184641}{6400} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (594 \, {\left (4 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} + 93 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 5179 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 50776531 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{4400000 \, {\left (2 \, x - 1\right )}} \]
-184641/6400*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/4400000*(594 *(4*(8*sqrt(5)*(5*x + 3) + 93*sqrt(5))*(5*x + 3) + 5179*sqrt(5))*(5*x + 3) - 50776531*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\int \frac {{\left (3\,x+2\right )}^4}{{\left (1-2\,x\right )}^{3/2}\,\sqrt {5\,x+3}} \,d x \]