Integrand size = 26, antiderivative size = 122 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\frac {95 \sqrt {1-2 x} \sqrt {3+5 x}}{3773 (2+3 x)}+\frac {190 (3+5 x)^{3/2}}{1617 \sqrt {1-2 x} (2+3 x)}+\frac {4 (3+5 x)^{5/2}}{231 (1-2 x)^{3/2} (2+3 x)}+\frac {95 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{343 \sqrt {7}} \]
4/231*(3+5*x)^(5/2)/(1-2*x)^(3/2)/(2+3*x)+95/2401*arctan(1/7*(1-2*x)^(1/2) *7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+190/1617*(3+5*x)^(3/2)/(2+3*x)/(1-2*x)^(1/ 2)+95/3773*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)
Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.70 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=-\frac {7 \sqrt {3+5 x} \left (-549-310 x+660 x^2\right )+285 \sqrt {7-14 x} \left (-2+x+6 x^2\right ) \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7203 (1-2 x)^{3/2} (2+3 x)} \]
-1/7203*(7*Sqrt[3 + 5*x]*(-549 - 310*x + 660*x^2) + 285*Sqrt[7 - 14*x]*(-2 + x + 6*x^2)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/((1 - 2*x)^(3 /2)*(2 + 3*x))
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 = {107, 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.
\(\displaystyle \int \frac {(5 x+3)^{3/2}}{(1-2 x)^{5/2} (3 x+2)^2} \, dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle \frac {95}{231} \int \frac {(5 x+3)^{3/2}}{(1-2 x)^{3/2} (3 x+2)^2}dx+\frac {4 (5 x+3)^{5/2}}{231 (1-2 x)^{3/2} (3 x+2)}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {95}{231} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2}dx\right )+\frac {4 (5 x+3)^{5/2}}{231 (1-2 x)^{3/2} (3 x+2)}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {95}{231} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \left (\frac {11}{14} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )\right )+\frac {4 (5 x+3)^{5/2}}{231 (1-2 x)^{3/2} (3 x+2)}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {95}{231} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \left (\frac {11}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )\right )+\frac {4 (5 x+3)^{5/2}}{231 (1-2 x)^{3/2} (3 x+2)}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {95}{231} \left (\frac {2 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}-\frac {3}{7} \left (-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )\right )+\frac {4 (5 x+3)^{5/2}}{231 (1-2 x)^{3/2} (3 x+2)}\) |
(4*(3 + 5*x)^(5/2))/(231*(1 - 2*x)^(3/2)*(2 + 3*x)) + (95*((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))/231
3.26.95.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[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 | \(-\frac {\left (3420 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}-1140 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-1425 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +9240 x^{2} \sqrt {-10 x^{2}-x +3}+570 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-4340 x \sqrt {-10 x^{2}-x +3}-7686 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{14406 \left (2+3 x \right ) \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(209\) |
-1/14406*(3420*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))* x^3-1140*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-14 25*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+9240*x^2*( -10*x^2-x+3)^(1/2)+570*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3) ^(1/2))-4340*x*(-10*x^2-x+3)^(1/2)-7686*(-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 = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\frac {285 \, \sqrt {7} {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (660 \, x^{2} - 310 \, x - 549\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14406 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \]
1/14406*(285*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*(660*x^2 - 310*x - 549)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(12*x^3 - 4*x^2 - 5*x + 2)
\[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\int \frac {\left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {5}{2}} \left (3 x + 2\right )^{2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=-\frac {95}{4802} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {550 \, x}{1029 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {20}{1029 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1825 \, x}{441 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {1}{189 \, {\left (3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 2 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} + \frac {3250}{1323 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
-95/4802*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 550/1 029*x/sqrt(-10*x^2 - x + 3) - 20/1029/sqrt(-10*x^2 - x + 3) + 1825/441*x/( -10*x^2 - x + 3)^(3/2) + 1/189/(3*(-10*x^2 - x + 3)^(3/2)*x + 2*(-10*x^2 - x + 3)^(3/2)) + 3250/1323/(-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 = 0.42 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.90 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=-\frac {19}{9604} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {66 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{343 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} - \frac {2 \, {\left (116 \, \sqrt {5} {\left (5 \, x + 3\right )} - 1023 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{25725 \, {\left (2 \, x - 1\right )}^{2}} \]
-19/9604*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*(( sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 66/343*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/ sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((s qrt(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) - 2/25725*(116*sqrt(5)*(5*x + 3) - 1023*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)^2} \, dx=\int \frac {{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{5/2}\,{\left (3\,x+2\right )}^2} \,d x \]