Integrand size = 26, antiderivative size = 72 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\frac {49 \sqrt {3+5 x}}{66 (1-2 x)^{3/2}}-\frac {448 \sqrt {3+5 x}}{363 \sqrt {1-2 x}}+\frac {9 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2 \sqrt {10}} \]
9/20*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+49/66*(3+5*x)^(1/2)/(1-2 *x)^(3/2)-448/363*(3+5*x)^(1/2)/(1-2*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\frac {70 \sqrt {3+5 x} (-51+256 x)+3267 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{7260 (1-2 x)^{3/2}} \]
(70*Sqrt[3 + 5*x]*(-51 + 256*x) + 3267*Sqrt[10 - 20*x]*(-1 + 2*x)*ArcTan[S qrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(7260*(1 - 2*x)^(3/2))
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {100, 27, 87, 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.
\(\displaystyle \int \frac {(3 x+2)^2}{(1-2 x)^{5/2} \sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {49 \sqrt {5 x+3}}{66 (1-2 x)^{3/2}}-\frac {1}{66} \int \frac {594 x+599}{2 (1-2 x)^{3/2} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {49 \sqrt {5 x+3}}{66 (1-2 x)^{3/2}}-\frac {1}{132} \int \frac {594 x+599}{(1-2 x)^{3/2} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{132} \left (297 \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1792 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}\right )+\frac {49 \sqrt {5 x+3}}{66 (1-2 x)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{132} \left (\frac {594}{5} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1792 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}\right )+\frac {49 \sqrt {5 x+3}}{66 (1-2 x)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{132} \left (297 \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {1792 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}\right )+\frac {49 \sqrt {5 x+3}}{66 (1-2 x)^{3/2}}\) |
(49*Sqrt[3 + 5*x])/(66*(1 - 2*x)^(3/2)) + ((-1792*Sqrt[3 + 5*x])/(11*Sqrt[ 1 - 2*x]) + 297*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/132
3.27.11.3.1 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_))*((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[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 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 = 4.15 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.43
method | result | size |
default | \(\frac {\left (13068 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-13068 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +3267 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+35840 x \sqrt {-10 x^{2}-x +3}-7140 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}\, \sqrt {1-2 x}}{14520 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(103\) |
1/14520*(13068*10^(1/2)*arcsin(20/11*x+1/11)*x^2-13068*10^(1/2)*arcsin(20/ 11*x+1/11)*x+3267*10^(1/2)*arcsin(20/11*x+1/11)+35840*x*(-10*x^2-x+3)^(1/2 )-7140*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)*(1-2*x)^(1/2)/(-1+2*x)^2/(-10*x^ 2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=-\frac {3267 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 140 \, {\left (256 \, x - 51\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14520 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]
-1/14520*(3267*sqrt(10)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(10)*(20*x + 1)* sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 140*(256*x - 51)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
\[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\int \frac {\left (3 x + 2\right )^{2}}{\left (1 - 2 x\right )^{\frac {5}{2}} \sqrt {5 x + 3}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\frac {9}{40} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{66 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {448 \, \sqrt {-10 \, x^{2} - x + 3}}{363 \, {\left (2 \, x - 1\right )}} \]
9/40*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 49/66*sqrt(-10*x^2 - x + 3)/ (4*x^2 - 4*x + 1) + 448/363*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\frac {9}{20} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {7 \, {\left (256 \, \sqrt {5} {\left (5 \, x + 3\right )} - 1023 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{18150 \, {\left (2 \, x - 1\right )}^{2}} \]
9/20*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 7/18150*(256*sqrt(5)*( 5*x + 3) - 1023*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\int \frac {{\left (3\,x+2\right )}^2}{{\left (1-2\,x\right )}^{5/2}\,\sqrt {5\,x+3}} \,d x \]