Integrand size = 26, antiderivative size = 113 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx=-\frac {1099 (2+3 x)^2}{726 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {\sqrt {1-2 x} (4898747+8200665 x)}{798600 \sqrt {3+5 x}}+\frac {4887 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{200 \sqrt {10}} \]
4887/2000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/33*(2+3*x)^3/(1-2 *x)^(3/2)/(3+5*x)^(1/2)-1099/726*(2+3*x)^2/(1-2*x)^(1/2)/(3+5*x)^(1/2)-1/7 98600*(4898747+8200665*x)*(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx=-\frac {10 \sqrt {3+5 x} \left (8379147-12657123 x-40488772 x^2+6468660 x^3\right )-19513791 \sqrt {10-20 x} \left (-3+x+10 x^2\right ) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{7986000 (1-2 x)^{3/2} (3+5 x)} \]
-1/7986000*(10*Sqrt[3 + 5*x]*(8379147 - 12657123*x - 40488772*x^2 + 646866 0*x^3) - 19513791*Sqrt[10 - 20*x]*(-3 + x + 10*x^2)*ArcTan[Sqrt[5/2 - 5*x] /Sqrt[3 + 5*x]])/((1 - 2*x)^(3/2)*(3 + 5*x))
Time = 0.20 (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.
\(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^{5/2} (5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}-\frac {1}{33} \int \frac {(3 x+2)^2 (507 x+296)}{2 (1-2 x)^{3/2} (5 x+3)^{3/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}-\frac {1}{66} \int \frac {(3 x+2)^2 (507 x+296)}{(1-2 x)^{3/2} (5 x+3)^{3/2}}dx\) |
\(\Big \downarrow \) 167 |
\(\displaystyle \frac {1}{66} \left (-\frac {1}{11} \int -\frac {(3 x+2) (49701 x+28738)}{2 \sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {1099 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{66} \left (\frac {1}{22} \int \frac {(3 x+2) (49701 x+28738)}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {1099 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 160 |
\(\displaystyle \frac {1}{66} \left (\frac {1}{22} \left (\frac {1773981}{100} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} (8200665 x+4898747)}{550 \sqrt {5 x+3}}\right )-\frac {1099 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{66} \left (\frac {1}{22} \left (\frac {1773981}{250} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {\sqrt {1-2 x} (8200665 x+4898747)}{550 \sqrt {5 x+3}}\right )-\frac {1099 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{66} \left (\frac {1}{22} \left (\frac {1773981 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{50 \sqrt {10}}-\frac {\sqrt {1-2 x} (8200665 x+4898747)}{550 \sqrt {5 x+3}}\right )-\frac {1099 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
(7*(2 + 3*x)^3)/(33*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) + ((-1099*(2 + 3*x)^2)/ (11*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (-1/550*(Sqrt[1 - 2*x]*(4898747 + 82006 65*x))/Sqrt[3 + 5*x] + (1773981*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(50*Sqrt [10]))/22)/66
3.27.19.3.1 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 = 4.12 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.34
method | result | size |
default | \(\frac {\sqrt {1-2 x}\, \left (390275820 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{3}-156110328 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-129373200 x^{3} \sqrt {-10 x^{2}-x +3}-136596537 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +809775440 x^{2} \sqrt {-10 x^{2}-x +3}+58541373 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+253142460 x \sqrt {-10 x^{2}-x +3}-167582940 \sqrt {-10 x^{2}-x +3}\right )}{15972000 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(151\) |
1/15972000*(1-2*x)^(1/2)*(390275820*10^(1/2)*arcsin(20/11*x+1/11)*x^3-1561 10328*10^(1/2)*arcsin(20/11*x+1/11)*x^2-129373200*x^3*(-10*x^2-x+3)^(1/2)- 136596537*10^(1/2)*arcsin(20/11*x+1/11)*x+809775440*x^2*(-10*x^2-x+3)^(1/2 )+58541373*10^(1/2)*arcsin(20/11*x+1/11)+253142460*x*(-10*x^2-x+3)^(1/2)-1 67582940*(-10*x^2-x+3)^(1/2))/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx=-\frac {19513791 \, \sqrt {10} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (6468660 \, x^{3} - 40488772 \, x^{2} - 12657123 \, x + 8379147\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{15972000 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]
-1/15972000*(19513791*sqrt(10)*(20*x^3 - 8*x^2 - 7*x + 3)*arctan(1/20*sqrt (10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(64686 60*x^3 - 40488772*x^2 - 12657123*x + 8379147)*sqrt(5*x + 3)*sqrt(-2*x + 1) )/(20*x^3 - 8*x^2 - 7*x + 3)
\[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4}}{\left (1 - 2 x\right )^{\frac {5}{2}} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx=\frac {4887}{4000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {81 \, x^{2}}{20 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {18627221 \, x}{798600 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {3910543}{199650 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {2401}{264 \, {\left (2 \, \sqrt {-10 \, x^{2} - x + 3} x - \sqrt {-10 \, x^{2} - x + 3}\right )}} \]
4887/4000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 81/20*x^2/sqrt(-10*x^2 - x + 3) - 18627221/798600*x/sqrt(-10*x^2 - x + 3) - 3910543/199650/sqrt(- 10*x^2 - x + 3) - 2401/264/(2*sqrt(-10*x^2 - x + 3)*x - sqrt(-10*x^2 - x + 3))
Time = 0.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.16 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx=\frac {4887}{2000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{332750 \, \sqrt {5 \, x + 3}} - \frac {{\left (4 \, {\left (323433 \, \sqrt {5} {\left (5 \, x + 3\right )} - 13033138 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 214579893 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{99825000 \, {\left (2 \, x - 1\right )}^{2}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{166375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
4887/2000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/332750*sqrt(10) *(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 1/99825000*(4*(32343 3*sqrt(5)*(5*x + 3) - 13033138*sqrt(5))*(5*x + 3) + 214579893*sqrt(5))*sqr t(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2 + 2/166375*sqrt(10)*sqrt(5*x + 3)/( sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4}{{\left (1-2\,x\right )}^{5/2}\,{\left (5\,x+3\right )}^{3/2}} \,d x \]