Integrand size = 26, antiderivative size = 84 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\frac {7 (2+3 x)^2}{11 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {\sqrt {1-2 x} (30443+50985 x)}{12100 \sqrt {3+5 x}}-\frac {999 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{100 \sqrt {10}} \]
-999/1000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(2+3*x)^2/(1-2 *x)^(1/2)/(3+5*x)^(1/2)+1/12100*(30443+50985*x)*(1-2*x)^(1/2)/(3+5*x)^(1/2 )
Time = 0.13 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\frac {612430+824990 x-326700 x^2+120879 \sqrt {10-20 x} \sqrt {3+5 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{121000 \sqrt {1-2 x} \sqrt {3+5 x}} \]
(612430 + 824990*x - 326700*x^2 + 120879*Sqrt[10 - 20*x]*Sqrt[3 + 5*x]*Arc Tan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(121000*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {109, 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.
\(\displaystyle \int \frac {(3 x+2)^3}{(1-2 x)^{3/2} (5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {1}{11} \int \frac {(3 x+2) (309 x+178)}{2 \sqrt {1-2 x} (5 x+3)^{3/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {1}{22} \int \frac {(3 x+2) (309 x+178)}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx\) |
\(\Big \downarrow \) 160 |
\(\displaystyle \frac {1}{22} \left (\frac {\sqrt {1-2 x} (50985 x+30443)}{550 \sqrt {5 x+3}}-\frac {10989}{100} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{22} \left (\frac {\sqrt {1-2 x} (50985 x+30443)}{550 \sqrt {5 x+3}}-\frac {10989}{250} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{22} \left (\frac {\sqrt {1-2 x} (50985 x+30443)}{550 \sqrt {5 x+3}}-\frac {10989 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{50 \sqrt {10}}\right )+\frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + ((Sqrt[1 - 2*x]*(30443 + 50985*x))/(550*Sqrt[3 + 5*x]) - (10989*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]]) /(50*Sqrt[10]))/22
3.26.63.3.1 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[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 = 120, normalized size of antiderivative = 1.43
method | result | size |
default | \(-\frac {\sqrt {1-2 x}\, \left (1208790 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+120879 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -653400 x^{2} \sqrt {-10 x^{2}-x +3}-362637 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1649980 x \sqrt {-10 x^{2}-x +3}+1224860 \sqrt {-10 x^{2}-x +3}\right )}{242000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(120\) |
-1/242000*(1-2*x)^(1/2)*(1208790*10^(1/2)*arcsin(20/11*x+1/11)*x^2+120879* 10^(1/2)*arcsin(20/11*x+1/11)*x-653400*x^2*(-10*x^2-x+3)^(1/2)-362637*10^( 1/2)*arcsin(20/11*x+1/11)+1649980*x*(-10*x^2-x+3)^(1/2)+1224860*(-10*x^2-x +3)^(1/2))/(-1+2*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\frac {120879 \, \sqrt {10} {\left (10 \, x^{2} + x - 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (32670 \, x^{2} - 82499 \, x - 61243\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{242000 \, {\left (10 \, x^{2} + x - 3\right )}} \]
1/242000*(120879*sqrt(10)*(10*x^2 + x - 3)*arctan(1/20*sqrt(10)*(20*x + 1) *sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(32670*x^2 - 82499*x - 61243)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(10*x^2 + x - 3)
\[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{3}}{\left (1 - 2 x\right )^{\frac {3}{2}} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=-\frac {27 \, x^{2}}{10 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {999}{2000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {82499 \, x}{12100 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {61243}{12100 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-27/10*x^2/sqrt(-10*x^2 - x + 3) + 999/2000*sqrt(10)*arcsin(-20/11*x - 1/1 1) + 82499/12100*x/sqrt(-10*x^2 - x + 3) + 61243/12100/sqrt(-10*x^2 - x + 3)
Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.40 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=-\frac {999}{1000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (6534 \, \sqrt {5} {\left (5 \, x + 3\right )} - 121687 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{302500 \, {\left (2 \, x - 1\right )}} - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{30250 \, \sqrt {5 \, x + 3}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{15125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
-999/1000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/302500*(6534*sq rt(5)*(5*x + 3) - 121687*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) - 1/30250*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 2/ 15125*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^3}{{\left (1-2\,x\right )}^{3/2}\,{\left (5\,x+3\right )}^{3/2}} \,d x \]