Integrand size = 24, antiderivative size = 94 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\frac {519}{32} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {173}{88} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {7 (3+5 x)^{5/2}}{11 \sqrt {1-2 x}}-\frac {5709 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{32 \sqrt {10}} \]
-5709/320*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(3+5*x)^(5/2)/ (1-2*x)^(1/2)+173/88*(3+5*x)^(3/2)*(1-2*x)^(1/2)+519/32*(1-2*x)^(1/2)*(3+5 *x)^(1/2)
Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\frac {1}{160} \left (-\frac {5 \sqrt {3+5 x} \left (-891+490 x+120 x^2\right )}{\sqrt {1-2 x}}+5709 \sqrt {10} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )\right ) \]
((-5*Sqrt[3 + 5*x]*(-891 + 490*x + 120*x^2))/Sqrt[1 - 2*x] + 5709*Sqrt[10] *ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/160
Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {87, 60, 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.
\(\displaystyle \int \frac {(3 x+2) (5 x+3)^{3/2}}{(1-2 x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}-\frac {173}{22} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}-\frac {173}{22} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}-\frac {173}{22} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}-\frac {173}{22} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}-\frac {173}{22} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
(7*(3 + 5*x)^(5/2))/(11*Sqrt[1 - 2*x]) - (173*(-1/4*(Sqrt[1 - 2*x]*(3 + 5* x)^(3/2)) + (33*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11 ]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/8))/22
3.26.32.3.1 Defintions of rubi rules used
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 106, normalized size of antiderivative = 1.13
method | result | size |
default | \(-\frac {\left (11418 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -2400 x^{2} \sqrt {-10 x^{2}-x +3}-5709 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-9800 x \sqrt {-10 x^{2}-x +3}+17820 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{640 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(106\) |
-1/640*(11418*10^(1/2)*arcsin(20/11*x+1/11)*x-2400*x^2*(-10*x^2-x+3)^(1/2) -5709*10^(1/2)*arcsin(20/11*x+1/11)-9800*x*(-10*x^2-x+3)^(1/2)+17820*(-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 = 81, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\frac {5709 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (120 \, x^{2} + 490 \, x - 891\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{640 \, {\left (2 \, x - 1\right )}} \]
1/640*(5709*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*(120*x^2 + 490*x - 891)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=-\frac {5709}{640} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {99}{32} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {7 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{4 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{8 \, {\left (2 \, x - 1\right )}} - \frac {231 \, \sqrt {-10 \, x^{2} - x + 3}}{8 \, {\left (2 \, x - 1\right )}} \]
-5709/640*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 99/32*sqrt(-10*x^2 - x + 3) - 7/4*(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) - 3/8*(-10*x^2 - x + 3)^(3/2)/(2*x - 1) - 231/8*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=-\frac {5709}{320} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 173 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 5709 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{800 \, {\left (2 \, x - 1\right )}} \]
-5709/320*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/800*(2*(12*sqrt (5)*(5*x + 3) + 173*sqrt(5))*(5*x + 3) - 5709*sqrt(5))*sqrt(5*x + 3)*sqrt( -10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx=\int \frac {\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{3/2}} \,d x \]