Integrand size = 26, antiderivative size = 161 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\frac {1018114917 \sqrt {1-2 x} \sqrt {3+5 x}}{1024000}+\frac {10377 \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{3/2}}{1600}+\frac {33}{20} \sqrt {1-2 x} (2+3 x)^3 (3+5 x)^{3/2}+\frac {(2+3 x)^4 (3+5 x)^{3/2}}{\sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x} (3+5 x)^{3/2} (4772357+2253560 x)}{256000}-\frac {11199264087 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1024000 \sqrt {10}} \]
-11199264087/10240000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+(2+3*x) ^4*(3+5*x)^(3/2)/(1-2*x)^(1/2)+10377/1600*(2+3*x)^2*(3+5*x)^(3/2)*(1-2*x)^ (1/2)+33/20*(2+3*x)^3*(3+5*x)^(3/2)*(1-2*x)^(1/2)+9/256000*(3+5*x)^(3/2)*( 4772357+2253560*x)*(1-2*x)^(1/2)+1018114917/1024000*(1-2*x)^(1/2)*(3+5*x)^ (1/2)
Time = 0.70 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.61 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\frac {5 \sqrt {1-2 x} \sqrt {3+5 x} \left (-1702927233+1206337246 x+732415080 x^2+461171520 x^3+200966400 x^4+41472000 x^5\right )+11199264087 \sqrt {10} (-1+2 x) \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{5120000 (-1+2 x)} \]
(5*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(-1702927233 + 1206337246*x + 732415080*x^2 + 461171520*x^3 + 200966400*x^4 + 41472000*x^5) + 11199264087*Sqrt[10]*(- 1 + 2*x)*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/(5120000*(-1 + 2*x))
Time = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {108, 27, 170, 27, 170, 27, 164, 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.
| \(\displaystyle \int \frac {(3 x+2)^4 (5 x+3)^{3/2}}{(1-2 x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\int \frac {3 (3 x+2)^3 \sqrt {5 x+3} (55 x+34)}{2 \sqrt {1-2 x}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \int \frac {(3 x+2)^3 \sqrt {5 x+3} (55 x+34)}{\sqrt {1-2 x}}dx\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (-\frac {1}{50} \int -\frac {5 (3 x+2)^2 \sqrt {5 x+3} (3459 x+2152)}{2 \sqrt {1-2 x}}dx-\frac {11}{10} \sqrt {1-2 x} (5 x+3)^{3/2} (3 x+2)^3\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (\frac {1}{20} \int \frac {(3 x+2)^2 \sqrt {5 x+3} (3459 x+2152)}{\sqrt {1-2 x}}dx-\frac {11}{10} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\right )\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (\frac {1}{20} \left (-\frac {1}{40} \int -\frac {(3 x+2) \sqrt {5 x+3} (845085 x+531106)}{2 \sqrt {1-2 x}}dx-\frac {3459}{40} \sqrt {1-2 x} (5 x+3)^{3/2} (3 x+2)^2\right )-\frac {11}{10} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (\frac {1}{20} \left (\frac {1}{80} \int \frac {(3 x+2) \sqrt {5 x+3} (845085 x+531106)}{\sqrt {1-2 x}}dx-\frac {3459}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {11}{10} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\right )\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (\frac {1}{20} \left (\frac {1}{80} \left (\frac {339371639}{160} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (2253560 x+4772357)\right )-\frac {3459}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {11}{10} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (\frac {1}{20} \left (\frac {1}{80} \left (\frac {339371639}{160} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (2253560 x+4772357)\right )-\frac {3459}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {11}{10} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\right )\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (\frac {1}{20} \left (\frac {1}{80} \left (\frac {339371639}{160} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (2253560 x+4772357)\right )-\frac {3459}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {11}{10} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {3}{2} \left (\frac {1}{20} \left (\frac {1}{80} \left (\frac {339371639}{160} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (2253560 x+4772357)\right )-\frac {3459}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {11}{10} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\right )\) |
((2 + 3*x)^4*(3 + 5*x)^(3/2))/Sqrt[1 - 2*x] - (3*((-11*Sqrt[1 - 2*x]*(2 + 3*x)^3*(3 + 5*x)^(3/2))/10 + ((-3459*Sqrt[1 - 2*x]*(2 + 3*x)^2*(3 + 5*x)^( 3/2))/40 + ((-3*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)*(4772357 + 2253560*x))/80 + (339371639*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqr t[3 + 5*x]])/(2*Sqrt[10])))/160)/80)/20))/2
3.26.29.3.1 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[((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 = 1.19 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(-\frac {\left (-829440000 x^{5} \sqrt {-10 x^{2}-x +3}-4019328000 x^{4} \sqrt {-10 x^{2}-x +3}-9223430400 x^{3} \sqrt {-10 x^{2}-x +3}+22398528174 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -14648301600 x^{2} \sqrt {-10 x^{2}-x +3}-11199264087 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-24126744920 x \sqrt {-10 x^{2}-x +3}+34058544660 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{20480000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(157\) |
-1/20480000*(-829440000*x^5*(-10*x^2-x+3)^(1/2)-4019328000*x^4*(-10*x^2-x+ 3)^(1/2)-9223430400*x^3*(-10*x^2-x+3)^(1/2)+22398528174*10^(1/2)*arcsin(20 /11*x+1/11)*x-14648301600*x^2*(-10*x^2-x+3)^(1/2)-11199264087*10^(1/2)*arc sin(20/11*x+1/11)-24126744920*x*(-10*x^2-x+3)^(1/2)+34058544660*(-10*x^2-x +3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(-1+2*x)/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.60 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\frac {11199264087 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (41472000 \, x^{5} + 200966400 \, x^{4} + 461171520 \, x^{3} + 732415080 \, x^{2} + 1206337246 \, x - 1702927233\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20480000 \, {\left (2 \, x - 1\right )}} \]
1/20480000*(11199264087*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*(41472000*x^5 + 20096 6400*x^4 + 461171520*x^3 + 732415080*x^2 + 1206337246*x - 1702927233)*sqrt (5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4} \left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {3}{2}}}\, dx \]
Result contains complex when optimal does not.
Time = 0.31 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.23 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\frac {81}{400} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}} - \frac {6669}{640} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x - \frac {12607994487}{20480000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {1760913}{25600} i \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x - \frac {21}{11}\right ) - \frac {359469}{12800} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {14553}{64} \, \sqrt {10 \, x^{2} - 21 \, x + 8} x - \frac {2420847}{51200} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {305613}{1280} \, \sqrt {10 \, x^{2} - 21 \, x + 8} + \frac {540891153}{1024000} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2401 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{32 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {1029 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{16 \, {\left (2 \, x - 1\right )}} - \frac {79233 \, \sqrt {-10 \, x^{2} - x + 3}}{64 \, {\left (2 \, x - 1\right )}} \]
81/400*(-10*x^2 - x + 3)^(5/2) - 6669/640*(-10*x^2 - x + 3)^(3/2)*x - 1260 7994487/20480000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 1760913/25600*I* sqrt(5)*sqrt(2)*arcsin(20/11*x - 21/11) - 359469/12800*(-10*x^2 - x + 3)^( 3/2) + 14553/64*sqrt(10*x^2 - 21*x + 8)*x - 2420847/51200*sqrt(-10*x^2 - x + 3)*x - 305613/1280*sqrt(10*x^2 - 21*x + 8) + 540891153/1024000*sqrt(-10 *x^2 - x + 3) - 2401/32*(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) - 1029/1 6*(-10*x^2 - x + 3)^(3/2)/(2*x - 1) - 79233/64*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=-\frac {11199264087}{10240000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, {\left (12 \, {\left (24 \, {\left (12 \, {\left (48 \, \sqrt {5} {\left (5 \, x + 3\right )} + 443 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 44497 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 10283927 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 1696858195 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 55996320435 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{128000000 \, {\left (2 \, x - 1\right )}} \]
-11199264087/10240000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/128 000000*(2*(12*(24*(12*(48*sqrt(5)*(5*x + 3) + 443*sqrt(5))*(5*x + 3) + 444 97*sqrt(5))*(5*x + 3) + 10283927*sqrt(5))*(5*x + 3) + 1696858195*sqrt(5))* (5*x + 3) - 55996320435*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4\,{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{3/2}} \,d x \]