Integrand size = 26, antiderivative size = 122 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=-\frac {5 \sqrt {1-2 x} \sqrt {3+5 x}}{343 (2+3 x)}-\frac {10 (3+5 x)^{3/2}}{147 \sqrt {1-2 x} (2+3 x)}+\frac {2 (3+5 x)^{5/2}}{21 (1-2 x)^{3/2} (2+3 x)}-\frac {55 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{343 \sqrt {7}} \]
2/21*(3+5*x)^(5/2)/(1-2*x)^(3/2)/(2+3*x)-55/2401*arctan(1/7*(1-2*x)^(1/2)* 7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-10/147*(3+5*x)^(3/2)/(2+3*x)/(1-2*x)^(1/2)- 5/343*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)
Time = 1.91 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.16 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\frac {5 \left (\frac {7 \sqrt {3+5 x} \left (657+3070 x+3090 x^2\right )}{5 (1-2 x)^{3/2} (2+3 x)}+33 \sqrt {7} \arctan \left (\frac {\sqrt {2 \left (34+\sqrt {1155}\right )} \sqrt {3+5 x}}{-\sqrt {11}+\sqrt {5-10 x}}\right )+33 \sqrt {7} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {34+\sqrt {1155}} \left (-\sqrt {11}+\sqrt {5-10 x}\right )}\right )\right )}{7203} \]
(5*((7*Sqrt[3 + 5*x]*(657 + 3070*x + 3090*x^2))/(5*(1 - 2*x)^(3/2)*(2 + 3* x)) + 33*Sqrt[7]*ArcTan[(Sqrt[2*(34 + Sqrt[1155])]*Sqrt[3 + 5*x])/(-Sqrt[1 1] + Sqrt[5 - 10*x])] + 33*Sqrt[7]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[34 + Sqrt[1 155]]*(-Sqrt[11] + Sqrt[5 - 10*x]))]))/7203
Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {105, 105, 105, 104, 217}
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.
\(\displaystyle \int \frac {(5 x+3)^{5/2}}{(1-2 x)^{5/2} (3 x+2)^2} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{21 (1-2 x)^{3/2} (3 x+2)}-\frac {5}{21} \int \frac {(5 x+3)^{3/2}}{(1-2 x)^{3/2} (3 x+2)^2}dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{21 (1-2 x)^{3/2} (3 x+2)}-\frac {5}{21} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2}dx\right )\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{21 (1-2 x)^{3/2} (3 x+2)}-\frac {5}{21} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \left (\frac {11}{14} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )\right )\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{21 (1-2 x)^{3/2} (3 x+2)}-\frac {5}{21} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \left (\frac {11}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{21 (1-2 x)^{3/2} (3 x+2)}-\frac {5}{21} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \left (-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )\right )\) |
(2*(3 + 5*x)^(5/2))/(21*(1 - 2*x)^(3/2)*(2 + 3*x)) - (5*((2*(3 + 5*x)^(3/2 ))/(7*Sqrt[1 - 2*x]*(2 + 3*x)) - (3*(-1/7*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x) - (11*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7]))) /7))/21
3.27.4.3.1 Defintions of rubi rules used
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(208\) vs. \(2(95)=190\).
Time = 1.22 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.71
method | result | size |
default | \(\frac {\left (1980 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}-660 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-825 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +43260 x^{2} \sqrt {-10 x^{2}-x +3}+330 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+42980 x \sqrt {-10 x^{2}-x +3}+9198 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{14406 \left (2+3 x \right ) \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(209\) |
1/14406*(1980*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x ^3-660*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-825* 7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+43260*x^2*(-1 0*x^2-x+3)^(1/2)+330*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^( 1/2))+42980*x*(-10*x^2-x+3)^(1/2)+9198*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)* (3+5*x)^(1/2)/(2+3*x)/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=-\frac {165 \, \sqrt {7} {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (3090 \, x^{2} + 3070 \, x + 657\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14406 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \]
-1/14406*(165*sqrt(7)*(12*x^3 - 4*x^2 - 5*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(3090*x^2 + 307 0*x + 657)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(12*x^3 - 4*x^2 - 5*x + 2)
\[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\int \frac {\left (5 x + 3\right )^{\frac {5}{2}}}{\left (1 - 2 x\right )^{\frac {5}{2}} \left (3 x + 2\right )^{2}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.13 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\frac {55}{4802} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {2575 \, x}{1029 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {625 \, x^{2}}{18 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {135}{1372 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {138125 \, x}{5292 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {1}{567 \, {\left (3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 2 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} + \frac {50315}{15876 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
55/4802*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 2575/1 029*x/sqrt(-10*x^2 - x + 3) + 625/18*x^2/(-10*x^2 - x + 3)^(3/2) - 135/137 2/sqrt(-10*x^2 - x + 3) + 138125/5292*x/(-10*x^2 - x + 3)^(3/2) - 1/567/(3 *(-10*x^2 - x + 3)^(3/2)*x + 2*(-10*x^2 - x + 3)^(3/2)) + 50315/15876/(-10 *x^2 - x + 3)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (95) = 190\).
Time = 0.43 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.90 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\frac {11}{9604} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {22 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{343 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} + \frac {22 \, {\left (47 \, \sqrt {5} {\left (5 \, x + 3\right )} - 66 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{25725 \, {\left (2 \, x - 1\right )}^{2}} \]
11/9604*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((s qrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 22/343*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/s qrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sq rt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2) *sqrt(-10*x + 5) - sqrt(22)))^2 + 280) + 22/25725*(47*sqrt(5)*(5*x + 3) - 66*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\int \frac {{\left (5\,x+3\right )}^{5/2}}{{\left (1-2\,x\right )}^{5/2}\,{\left (3\,x+2\right )}^2} \,d x \]