Integrand size = 26, antiderivative size = 113 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=-\frac {107 \sqrt {1-2 x} (2+3 x)^2}{1815 (3+5 x)^{3/2}}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)^{3/2}}+\frac {\sqrt {1-2 x} (627641+1051875 x)}{399300 \sqrt {3+5 x}}-\frac {621 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{100 \sqrt {10}} \]
-621/1000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(2+3*x)^3/(3+5 *x)^(3/2)/(1-2*x)^(1/2)-107/1815*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)^(3/2)+1/3 99300*(627641+1051875*x)*(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\frac {3821563+11581424 x+6746215 x^2-3234330 x^3}{399300 \sqrt {1-2 x} (3+5 x)^{3/2}}+\frac {621 \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{100 \sqrt {10}} \]
(3821563 + 11581424*x + 6746215*x^2 - 3234330*x^3)/(399300*Sqrt[1 - 2*x]*( 3 + 5*x)^(3/2)) + (621*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(100*Sqrt[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, 160, 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.
| \(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^{3/2} (5 x+3)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {1}{11} \int \frac {(3 x+2)^2 (309 x+164)}{2 \sqrt {1-2 x} (5 x+3)^{5/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {1}{22} \int \frac {(3 x+2)^2 (309 x+164)}{\sqrt {1-2 x} (5 x+3)^{5/2}}dx\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {1}{22} \left (-\frac {2}{165} \int \frac {(3 x+2) (31875 x+18254)}{2 \sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {214 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{22} \left (-\frac {1}{165} \int \frac {(3 x+2) (31875 x+18254)}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {214 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 160 |
| \(\displaystyle \frac {1}{22} \left (\frac {1}{165} \left (\frac {\sqrt {1-2 x} (1051875 x+627641)}{110 \sqrt {5 x+3}}-\frac {225423}{20} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )-\frac {214 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{22} \left (\frac {1}{165} \left (\frac {\sqrt {1-2 x} (1051875 x+627641)}{110 \sqrt {5 x+3}}-\frac {225423}{50} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {214 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{22} \left (\frac {1}{165} \left (\frac {\sqrt {1-2 x} (1051875 x+627641)}{110 \sqrt {5 x+3}}-\frac {225423 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{10 \sqrt {10}}\right )-\frac {214 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\) |
(7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + ((-214*Sqrt[1 - 2*x]* (2 + 3*x)^2)/(165*(3 + 5*x)^(3/2)) + ((Sqrt[1 - 2*x]*(627641 + 1051875*x)) /(110*Sqrt[3 + 5*x]) - (225423*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(10*Sqrt[ 10]))/165)/22
3.26.72.3.1 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[(b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d*(b*c - a*d)*(m + 1))), x] + Simp[(a*d*f*h*m + b*(d* (f*g + e*h) - c*f*h*(m + 2)))/(b^2*d) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
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 = 1.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.34
| method | result | size |
| default | \(-\frac {\sqrt {1-2 x}\, \left (123982650 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{3}+86787855 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-64686600 x^{3} \sqrt {-10 x^{2}-x +3}-29755836 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +134924300 x^{2} \sqrt {-10 x^{2}-x +3}-22316877 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+231628480 x \sqrt {-10 x^{2}-x +3}+76431260 \sqrt {-10 x^{2}-x +3}\right )}{7986000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(151\) |
-1/7986000*(1-2*x)^(1/2)*(123982650*10^(1/2)*arcsin(20/11*x+1/11)*x^3+8678 7855*10^(1/2)*arcsin(20/11*x+1/11)*x^2-64686600*x^3*(-10*x^2-x+3)^(1/2)-29 755836*10^(1/2)*arcsin(20/11*x+1/11)*x+134924300*x^2*(-10*x^2-x+3)^(1/2)-2 2316877*10^(1/2)*arcsin(20/11*x+1/11)+231628480*x*(-10*x^2-x+3)^(1/2)+7643 1260*(-10*x^2-x+3)^(1/2))/(-1+2*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)
Time = 0.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\frac {2479653 \, \sqrt {10} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (3234330 \, x^{3} - 6746215 \, x^{2} - 11581424 \, x - 3821563\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{7986000 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]
1/7986000*(2479653*sqrt(10)*(50*x^3 + 35*x^2 - 12*x - 9)*arctan(1/20*sqrt( 10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(323433 0*x^3 - 6746215*x^2 - 11581424*x - 3821563)*sqrt(5*x + 3)*sqrt(-2*x + 1))/ (50*x^3 + 35*x^2 - 12*x - 9)
\[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4}}{\left (1 - 2 x\right )^{\frac {3}{2}} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=-\frac {621}{2000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {81 \, x^{2}}{50 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {8686813 \, x}{1996500 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {31846681}{9982500 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {2}{20625 \, {\left (5 \, \sqrt {-10 \, x^{2} - x + 3} x + 3 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]
-621/2000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 81/50*x^2/sqrt(-10*x^2 - x + 3) + 8686813/1996500*x/sqrt(-10*x^2 - x + 3) + 31846681/9982500/sqrt (-10*x^2 - x + 3) - 2/20625/(5*sqrt(-10*x^2 - x + 3)*x + 3*sqrt(-10*x^2 - x + 3))
Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (86) = 172\).
Time = 0.35 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.58 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{39930000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {621}{1000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (215622 \, \sqrt {5} {\left (5 \, x + 3\right )} - 4187171 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{16637500 \, {\left (2 \, x - 1\right )}} - \frac {271 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{3327500 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {813 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{2495625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
-1/39930000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2 ) - 621/1000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/16637500*(21 5622*sqrt(5)*(5*x + 3) - 4187171*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2 *x - 1) - 271/3327500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5 *x + 3) + 1/2495625*sqrt(10)*(5*x + 3)^(3/2)*(813*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3
Timed out. \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4}{{\left (1-2\,x\right )}^{3/2}\,{\left (5\,x+3\right )}^{5/2}} \,d x \]