Integrand size = 26, antiderivative size = 157 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{3/2}} \, dx=-\frac {2 (1-2 x)^{5/2} (2+3 x)^3}{5 \sqrt {3+5 x}}+\frac {2210901 \sqrt {1-2 x} \sqrt {3+5 x}}{8000000}+\frac {66997 (1-2 x)^{3/2} \sqrt {3+5 x}}{800000}-\frac {9 (2127-460 x) (1-2 x)^{5/2} \sqrt {3+5 x}}{200000}+\frac {33}{125} (1-2 x)^{5/2} (2+3 x)^2 \sqrt {3+5 x}+\frac {24319911 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{8000000 \sqrt {10}} \]
24319911/80000000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/5*(1-2*x) ^(5/2)*(2+3*x)^3/(3+5*x)^(1/2)+66997/800000*(1-2*x)^(3/2)*(3+5*x)^(1/2)-9/ 200000*(2127-460*x)*(1-2*x)^(5/2)*(3+5*x)^(1/2)+33/125*(1-2*x)^(5/2)*(2+3* x)^2*(3+5*x)^(1/2)+2210901/8000000*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{3/2}} \, dx=\frac {10 \sqrt {1-2 x} \left (6089453+20337375 x-4101140 x^2-39487200 x^3+12528000 x^4+34560000 x^5\right )-24319911 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{80000000 \sqrt {3+5 x}} \]
(10*Sqrt[1 - 2*x]*(6089453 + 20337375*x - 4101140*x^2 - 39487200*x^3 + 125 28000*x^4 + 34560000*x^5) - 24319911*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x ]/Sqrt[3 + 5*x]])/(80000000*Sqrt[3 + 5*x])
Time = 0.23 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {108, 25, 170, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)^3}{(5 x+3)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {2}{5} \int -\frac {(1-2 x)^{3/2} (3 x+2)^2 (33 x+1)}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2}{5} \int \frac {(1-2 x)^{3/2} (3 x+2)^2 (33 x+1)}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle -\frac {2}{5} \left (-\frac {1}{50} \int \frac {(262-69 x) (1-2 x)^{3/2} (3 x+2)}{2 \sqrt {5 x+3}}dx-\frac {33}{50} (3 x+2)^2 \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2}{5} \left (-\frac {1}{100} \int \frac {(262-69 x) (1-2 x)^{3/2} (3 x+2)}{\sqrt {5 x+3}}dx-\frac {33}{50} (3 x+2)^2 \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{100} \left (\frac {9}{800} (2127-460 x) (1-2 x)^{5/2} \sqrt {5 x+3}-\frac {66997}{320} \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx\right )-\frac {33}{50} (1-2 x)^{5/2} (3 x+2)^2 \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{100} \left (\frac {9}{800} (2127-460 x) (1-2 x)^{5/2} \sqrt {5 x+3}-\frac {66997}{320} \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )\right )-\frac {33}{50} (1-2 x)^{5/2} (3 x+2)^2 \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{100} \left (\frac {9}{800} (2127-460 x) (1-2 x)^{5/2} \sqrt {5 x+3}-\frac {66997}{320} \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )\right )-\frac {33}{50} (1-2 x)^{5/2} (3 x+2)^2 \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{100} \left (\frac {9}{800} (2127-460 x) (1-2 x)^{5/2} \sqrt {5 x+3}-\frac {66997}{320} \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )\right )-\frac {33}{50} (1-2 x)^{5/2} (3 x+2)^2 \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle -\frac {2}{5} \left (\frac {1}{100} \left (\frac {9}{800} (2127-460 x) (1-2 x)^{5/2} \sqrt {5 x+3}-\frac {66997}{320} \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )\right )-\frac {33}{50} (1-2 x)^{5/2} (3 x+2)^2 \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2} (3 x+2)^3}{5 \sqrt {5 x+3}}\) |
(-2*(1 - 2*x)^(5/2)*(2 + 3*x)^3)/(5*Sqrt[3 + 5*x]) - (2*((-33*(1 - 2*x)^(5 /2)*(2 + 3*x)^2*Sqrt[3 + 5*x])/50 + ((9*(2127 - 460*x)*(1 - 2*x)^(5/2)*Sqr t[3 + 5*x])/800 - (66997*(((1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/10 + (33*((Sqrt[ 1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(5*Sqrt[ 10])))/20))/320)/100))/5
3.25.37.3.1 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[(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 = 1.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(\frac {\left (691200000 x^{5} \sqrt {-10 x^{2}-x +3}+250560000 x^{4} \sqrt {-10 x^{2}-x +3}-789744000 x^{3} \sqrt {-10 x^{2}-x +3}+121599555 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -82022800 x^{2} \sqrt {-10 x^{2}-x +3}+72959733 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+406747500 x \sqrt {-10 x^{2}-x +3}+121789060 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{160000000 \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(150\) |
1/160000000*(691200000*x^5*(-10*x^2-x+3)^(1/2)+250560000*x^4*(-10*x^2-x+3) ^(1/2)-789744000*x^3*(-10*x^2-x+3)^(1/2)+121599555*10^(1/2)*arcsin(20/11*x +1/11)*x-82022800*x^2*(-10*x^2-x+3)^(1/2)+72959733*10^(1/2)*arcsin(20/11*x +1/11)+406747500*x*(-10*x^2-x+3)^(1/2)+121789060*(-10*x^2-x+3)^(1/2))*(1-2 *x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.61 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{3/2}} \, dx=-\frac {24319911 \, \sqrt {10} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (34560000 \, x^{5} + 12528000 \, x^{4} - 39487200 \, x^{3} - 4101140 \, x^{2} + 20337375 \, x + 6089453\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{160000000 \, {\left (5 \, x + 3\right )}} \]
-1/160000000*(24319911*sqrt(10)*(5*x + 3)*arctan(1/20*sqrt(10)*(20*x + 1)* sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 20*(34560000*x^5 + 125280 00*x^4 - 39487200*x^3 - 4101140*x^2 + 20337375*x + 6089453)*sqrt(5*x + 3)* sqrt(-2*x + 1))/(5*x + 3)
\[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{3/2}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}} \left (3 x + 2\right )^{3}}{\left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{3/2}} \, dx=-\frac {216 \, x^{6}}{25 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {297 \, x^{5}}{250 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {57189 \, x^{4}}{5000 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {782123 \, x^{3}}{200000 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {4477589 \, x^{2}}{800000 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {24319911}{160000000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {8158469 \, x}{8000000 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {6089453}{8000000 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-216/25*x^6/sqrt(-10*x^2 - x + 3) + 297/250*x^5/sqrt(-10*x^2 - x + 3) + 57 189/5000*x^4/sqrt(-10*x^2 - x + 3) - 782123/200000*x^3/sqrt(-10*x^2 - x + 3) - 4477589/800000*x^2/sqrt(-10*x^2 - x + 3) - 24319911/160000000*sqrt(10 )*arcsin(-20/11*x - 1/11) + 8158469/8000000*x/sqrt(-10*x^2 - x + 3) + 6089 453/8000000/sqrt(-10*x^2 - x + 3)
Time = 0.40 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{3/2}} \, dx=\frac {1}{200000000} \, {\left (4 \, {\left (24 \, {\left (36 \, {\left (16 \, \sqrt {5} {\left (5 \, x + 3\right )} - 211 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 22859 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 969335 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 5816745 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {24319911}{80000000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {121 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{156250 \, \sqrt {5 \, x + 3}} + \frac {242 \, \sqrt {10} \sqrt {5 \, x + 3}}{78125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
1/200000000*(4*(24*(36*(16*sqrt(5)*(5*x + 3) - 211*sqrt(5))*(5*x + 3) + 22 859*sqrt(5))*(5*x + 3) + 969335*sqrt(5))*(5*x + 3) - 5816745*sqrt(5))*sqrt (5*x + 3)*sqrt(-10*x + 5) + 24319911/80000000*sqrt(10)*arcsin(1/11*sqrt(22 )*sqrt(5*x + 3)) - 121/156250*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22) )/sqrt(5*x + 3) + 242/78125*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5 ) - sqrt(22))
Timed out. \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^{3/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}\,{\left (3\,x+2\right )}^3}{{\left (5\,x+3\right )}^{3/2}} \,d x \]