Integrand size = 24, antiderivative size = 118 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {13365}{128} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {405}{32} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {81}{44} \sqrt {1-2 x} (3+5 x)^{5/2}+\frac {7 (3+5 x)^{7/2}}{11 \sqrt {1-2 x}}-\frac {29403}{128} \sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]
-29403/256*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(3+5*x)^(7/2) /(1-2*x)^(1/2)+405/32*(3+5*x)^(3/2)*(1-2*x)^(1/2)+81/44*(3+5*x)^(5/2)*(1-2 *x)^(1/2)+13365/128*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {-2 \sqrt {3+5 x} \left (-22545+14526 x+6120 x^2+1600 x^3\right )+29403 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{256 \sqrt {1-2 x}} \]
(-2*Sqrt[3 + 5*x]*(-22545 + 14526*x + 6120*x^2 + 1600*x^3) + 29403*Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(256*Sqrt[1 - 2*x])
Time = 0.19 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 60, 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.
\(\displaystyle \int \frac {(3 x+2) (5 x+3)^{5/2}}{(1-2 x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {7 (5 x+3)^{7/2}}{11 \sqrt {1-2 x}}-\frac {243}{22} \int \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7 (5 x+3)^{7/2}}{11 \sqrt {1-2 x}}-\frac {243}{22} \left (\frac {55}{12} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7 (5 x+3)^{7/2}}{11 \sqrt {1-2 x}}-\frac {243}{22} \left (\frac {55}{12} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7 (5 x+3)^{7/2}}{11 \sqrt {1-2 x}}-\frac {243}{22} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {7 (5 x+3)^{7/2}}{11 \sqrt {1-2 x}}-\frac {243}{22} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7 (5 x+3)^{7/2}}{11 \sqrt {1-2 x}}-\frac {243}{22} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )\) |
(7*(3 + 5*x)^(7/2))/(11*Sqrt[1 - 2*x]) - (243*(-1/6*(Sqrt[1 - 2*x]*(3 + 5* x)^(5/2)) + (55*(-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))/12))/22
3.26.42.3.1 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 = 1.18 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.04
method | result | size |
default | \(-\frac {\left (-6400 x^{3} \sqrt {-10 x^{2}-x +3}+58806 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -24480 x^{2} \sqrt {-10 x^{2}-x +3}-29403 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-58104 x \sqrt {-10 x^{2}-x +3}+90180 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{512 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(123\) |
-1/512*(-6400*x^3*(-10*x^2-x+3)^(1/2)+58806*10^(1/2)*arcsin(20/11*x+1/11)* x-24480*x^2*(-10*x^2-x+3)^(1/2)-29403*10^(1/2)*arcsin(20/11*x+1/11)-58104* x*(-10*x^2-x+3)^(1/2)+90180*(-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 = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {29403 \, \sqrt {5} \sqrt {2} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 4 \, {\left (1600 \, x^{3} + 6120 \, x^{2} + 14526 \, x - 22545\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{512 \, {\left (2 \, x - 1\right )}} \]
1/512*(29403*sqrt(5)*sqrt(2)*(2*x - 1)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 4*(1600*x^3 + 6120*x^ 2 + 14526*x - 22545)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {5}{2}}}{\left (1 - 2 x\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=-\frac {125 \, x^{4}}{2 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {4425 \, x^{3}}{16 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {45495 \, x^{2}}{64 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {29403}{512} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {69147 \, x}{128 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {67635}{128 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-125/2*x^4/sqrt(-10*x^2 - x + 3) - 4425/16*x^3/sqrt(-10*x^2 - x + 3) - 454 95/64*x^2/sqrt(-10*x^2 - x + 3) + 29403/512*sqrt(10)*arcsin(-20/11*x - 1/1 1) + 69147/128*x/sqrt(-10*x^2 - x + 3) + 67635/128/sqrt(-10*x^2 - x + 3)
Time = 0.31 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=-\frac {29403}{256} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, {\left (4 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} + 81 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 4455 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 147015 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{3200 \, {\left (2 \, x - 1\right )}} \]
-29403/256*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/3200*(2*(4*(8* sqrt(5)*(5*x + 3) + 81*sqrt(5))*(5*x + 3) + 4455*sqrt(5))*(5*x + 3) - 1470 15*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\int \frac {\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{5/2}}{{\left (1-2\,x\right )}^{3/2}} \,d x \]