Integrand size = 26, antiderivative size = 122 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^4} \, dx=-\frac {121 \sqrt {1-2 x} \sqrt {3+5 x}}{392 (2+3 x)}-\frac {11 \sqrt {1-2 x} (3+5 x)^{3/2}}{84 (2+3 x)^2}+\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{3 (2+3 x)^3}-\frac {1331 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{392 \sqrt {7}} \]
-1331/2744*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-11/84*( 3+5*x)^(3/2)*(1-2*x)^(1/2)/(2+3*x)^2+1/3*(3+5*x)^(5/2)*(1-2*x)^(1/2)/(2+3* x)^3-121/392*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^4} \, dx=\frac {\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} \left (1152+4478 x+4223 x^2\right )}{(2+3 x)^3}-3993 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{8232} \]
((7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(1152 + 4478*x + 4223*x^2))/(2 + 3*x)^3 - 3993*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/8232
Time = 0.21 (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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{(3 x+2)^4} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {11}{6} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x} (3 x+2)^3}dx+\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{3 (3 x+2)^3}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {11}{6} \left (\frac {33}{28} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2}dx-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )+\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{3 (3 x+2)^3}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {11}{6} \left (\frac {33}{28} \left (\frac {11}{14} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )+\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{3 (3 x+2)^3}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {11}{6} \left (\frac {33}{28} \left (\frac {11}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )+\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{3 (3 x+2)^3}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {11}{6} \left (\frac {33}{28} \left (-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )+\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{3 (3 x+2)^3}\) |
(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(3*(2 + 3*x)^3) + (11*(-1/14*(Sqrt[1 - 2*x ]*(3 + 5*x)^(3/2))/(2 + 3*x)^2 + (33*(-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])) )/28))/6
3.23.82.3.1 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])
Time = 3.63 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02
method | result | size |
risch | \(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (4223 x^{2}+4478 x +1152\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{1176 \left (2+3 x \right )^{3} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {1331 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{5488 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(124\) |
default | \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (107811 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}+215622 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+143748 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +59122 x^{2} \sqrt {-10 x^{2}-x +3}+31944 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+62692 x \sqrt {-10 x^{2}-x +3}+16128 \sqrt {-10 x^{2}-x +3}\right )}{16464 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{3}}\) | \(202\) |
-1/1176*(-1+2*x)*(3+5*x)^(1/2)*(4223*x^2+4478*x+1152)/(2+3*x)^3/(-(-1+2*x) *(3+5*x))^(1/2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)+1331/5488*7^(1/2)*ar ctan(9/14*(20/3+37/3*x)*7^(1/2)/(-90*(2/3+x)^2+67+111*x)^(1/2))*((1-2*x)*( 3+5*x))^(1/2)/(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^4} \, dx=-\frac {3993 \, \sqrt {7} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (4223 \, x^{2} + 4478 \, x + 1152\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{16464 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]
-1/16464*(3993*sqrt(7)*(27*x^3 + 54*x^2 + 36*x + 8)*arctan(1/14*sqrt(7)*(3 7*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(4223*x^2 + 4478*x + 1152)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)
\[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^4} \, dx=\int \frac {\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {3}{2}}}{\left (3 x + 2\right )^{4}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^4} \, dx=\frac {1331}{5488} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {55}{294} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{21 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + \frac {33 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{196 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {407 \, \sqrt {-10 \, x^{2} - x + 3}}{1176 \, {\left (3 \, x + 2\right )}} \]
1331/5488*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 55/2 94*sqrt(-10*x^2 - x + 3) - 1/21*(-10*x^2 - x + 3)^(3/2)/(27*x^3 + 54*x^2 + 36*x + 8) + 33/196*(-10*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 407/1176* sqrt(-10*x^2 - x + 3)/(3*x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (95) = 190\).
Time = 0.42 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.54 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^4} \, dx=\frac {1331}{54880} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {1331 \, \sqrt {10} {\left (3 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{5} + 2240 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - \frac {235200 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {940800 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{588 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{3}} \]
1331/54880*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)))) - 1331/588*sqrt(10)*(3*((sqrt(2)*sqrt(-10*x + 5) - sqr t(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22) ))^5 + 2240*((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)))^3 - 235200*(sqrt(2)*sqrt(-10 *x + 5) - sqrt(22))/sqrt(5*x + 3) + 940800*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10 *x + 5) - sqrt(22)))/(((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)))^2 + 280)^3
Timed out. \[ \int \frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^4} \, dx=\int \frac {\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{3/2}}{{\left (3\,x+2\right )}^4} \,d x \]