Integrand size = 26, antiderivative size = 113 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=-\frac {203 (2+3 x)^2}{242 \sqrt {1-2 x} (3+5 x)^{3/2}}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac {\sqrt {1-2 x} (627287+991010 x)}{2196150 (3+5 x)^{3/2}}+\frac {81 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{50 \sqrt {10}} \]
7/33*(2+3*x)^3/(1-2*x)^(3/2)/(3+5*x)^(3/2)+81/500*arcsin(1/11*22^(1/2)*(3+ 5*x)^(1/2))*10^(1/2)-203/242*(2+3*x)^2/(3+5*x)^(3/2)/(1-2*x)^(1/2)+1/21961 50*(627287+991010*x)*(1-2*x)^(1/2)/(3+5*x)^(3/2)
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {-3014813+7883562 x+51334383 x^2+49702040 x^3}{2196150 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac {81 \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{50 \sqrt {10}} \]
(-3014813 + 7883562*x + 51334383*x^2 + 49702040*x^3)/(2196150*(1 - 2*x)^(3 /2)*(3 + 5*x)^(3/2)) - (81*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(50*Sqrt [10])
Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.11, 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, 162, 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.
| \(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^{5/2} (5 x+3)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac {1}{33} \int \frac {3 (3 x+2)^2 (99 x+52)}{2 (1-2 x)^{3/2} (5 x+3)^{5/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac {1}{22} \int \frac {(3 x+2)^2 (99 x+52)}{(1-2 x)^{3/2} (5 x+3)^{5/2}}dx\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {1}{22} \left (-\frac {1}{11} \int -\frac {(3 x+2) (3267 x+1366)}{2 \sqrt {1-2 x} (5 x+3)^{5/2}}dx-\frac {203 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{22} \left (\frac {1}{22} \int \frac {(3 x+2) (3267 x+1366)}{\sqrt {1-2 x} (5 x+3)^{5/2}}dx-\frac {203 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 162 |
| \(\displaystyle \frac {1}{22} \left (\frac {1}{22} \left (\frac {9801}{25} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {2 \sqrt {1-2 x} (991010 x+627287)}{9075 (5 x+3)^{3/2}}\right )-\frac {203 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{22} \left (\frac {1}{22} \left (\frac {19602}{125} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {2 \sqrt {1-2 x} (991010 x+627287)}{9075 (5 x+3)^{3/2}}\right )-\frac {203 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{22} \left (\frac {1}{22} \left (\frac {9801}{25} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {2 \sqrt {1-2 x} (991010 x+627287)}{9075 (5 x+3)^{3/2}}\right )-\frac {203 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
(7*(2 + 3*x)^3)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2)) + ((-203*(2 + 3*x)^2) /(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + ((2*Sqrt[1 - 2*x]*(627287 + 991010*x ))/(9075*(3 + 5*x)^(3/2)) + (9801*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x ]])/25)/22)/22
3.27.29.3.1 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^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e *g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b *c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d *(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( b^2*(b*c - a*d)^2*(m + 1)*(m + 2))) Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] , x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] && !LtQ[n, -2]))
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 = 4.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.46
| method | result | size |
| default | \(\frac {\sqrt {1-2 x}\, \left (355776300 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{4}+71155260 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{3}-209908017 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+994040800 x^{3} \sqrt {-10 x^{2}-x +3}-21346578 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +1026687660 x^{2} \sqrt {-10 x^{2}-x +3}+32019867 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+157671240 x \sqrt {-10 x^{2}-x +3}-60296260 \sqrt {-10 x^{2}-x +3}\right )}{43923000 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(165\) |
1/43923000*(1-2*x)^(1/2)*(355776300*10^(1/2)*arcsin(20/11*x+1/11)*x^4+7115 5260*10^(1/2)*arcsin(20/11*x+1/11)*x^3-209908017*10^(1/2)*arcsin(20/11*x+1 /11)*x^2+994040800*x^3*(-10*x^2-x+3)^(1/2)-21346578*10^(1/2)*arcsin(20/11* x+1/11)*x+1026687660*x^2*(-10*x^2-x+3)^(1/2)+32019867*10^(1/2)*arcsin(20/1 1*x+1/11)+157671240*x*(-10*x^2-x+3)^(1/2)-60296260*(-10*x^2-x+3)^(1/2))/(- 1+2*x)^2/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)
Time = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.03 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=-\frac {3557763 \, \sqrt {10} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (49702040 \, x^{3} + 51334383 \, x^{2} + 7883562 \, x - 3014813\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{43923000 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \]
-1/43923000*(3557763*sqrt(10)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*arctan (1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 20*(49702040*x^3 + 51334383*x^2 + 7883562*x - 3014813)*sqrt(5*x + 3)*sqrt (-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)
\[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4}}{\left (1 - 2 x\right )^{\frac {5}{2}} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (86) = 172\).
Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.59 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {27}{1464100} \, x {\left (\frac {7220 \, x}{\sqrt {-10 \, x^{2} - x + 3}} + \frac {439230 \, x^{2}}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {361}{\sqrt {-10 \, x^{2} - x + 3}} + \frac {21901 \, x}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {87483}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}\right )} - \frac {81}{1000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {9747}{732050} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {1588351 \, x}{1098075 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {108 \, x^{2}}{5 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {34823}{1098075 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {86854 \, x}{9075 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {12682}{9075 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
27/1464100*x*(7220*x/sqrt(-10*x^2 - x + 3) + 439230*x^2/(-10*x^2 - x + 3)^ (3/2) + 361/sqrt(-10*x^2 - x + 3) + 21901*x/(-10*x^2 - x + 3)^(3/2) - 8748 3/(-10*x^2 - x + 3)^(3/2)) - 81/1000*sqrt(10)*arcsin(-20/11*x - 1/11) + 97 47/732050*sqrt(-10*x^2 - x + 3) - 1588351/1098075*x/sqrt(-10*x^2 - x + 3) + 108/5*x^2/(-10*x^2 - x + 3)^(3/2) - 34823/1098075/sqrt(-10*x^2 - x + 3) + 86854/9075*x/(-10*x^2 - x + 3)^(3/2) - 12682/9075/(-10*x^2 - x + 3)^(3/2 )
Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (86) = 172\).
Time = 0.37 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.58 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{87846000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {81}{500} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{26620 \, \sqrt {5 \, x + 3}} + \frac {343 \, {\left (232 \, \sqrt {5} {\left (5 \, x + 3\right )} - 891 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{2196150 \, {\left (2 \, x - 1\right )}^{2}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {825 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{5490375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
-1/87846000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2 ) + 81/500*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/26620*sqrt(10) *(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 343/2196150*(232*sqr t(5)*(5*x + 3) - 891*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2 + 1/5490375*sqrt(10)*(5*x + 3)^(3/2)*(825*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22 ))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3
Timed out. \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4}{{\left (1-2\,x\right )}^{5/2}\,{\left (5\,x+3\right )}^{5/2}} \,d x \]