Integrand size = 26, antiderivative size = 122 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx=\frac {165 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}+\frac {5 \sqrt {1-2 x} (3+5 x)^{3/2}}{98 (2+3 x)^2}+\frac {2 (3+5 x)^{5/2}}{7 \sqrt {1-2 x} (2+3 x)^2}+\frac {1815 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1372 \sqrt {7}} \]
1815/9604*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+2/7*(3+5 *x)^(5/2)/(2+3*x)^2/(1-2*x)^(1/2)+5/98*(3+5*x)^(3/2)*(1-2*x)^(1/2)/(2+3*x) ^2+165/1372*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)
Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.70 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx=\frac {7 \sqrt {3+5 x} \left (4068+11525 x+8110 x^2\right )+1815 \sqrt {7-14 x} (2+3 x)^2 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{9604 \sqrt {1-2 x} (2+3 x)^2} \]
(7*Sqrt[3 + 5*x]*(4068 + 11525*x + 8110*x^2) + 1815*Sqrt[7 - 14*x]*(2 + 3* x)^2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(9604*Sqrt[1 - 2*x]*(2 + 3*x)^2)
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.
\(\displaystyle \int \frac {(5 x+3)^{5/2}}{(1-2 x)^{3/2} (3 x+2)^3} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{7 \sqrt {1-2 x} (3 x+2)^2}-\frac {5}{7} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x} (3 x+2)^3}dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{7 \sqrt {1-2 x} (3 x+2)^2}-\frac {5}{7} \left (\frac {33}{28} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2}dx-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{7 \sqrt {1-2 x} (3 x+2)^2}-\frac {5}{7} \left (\frac {33}{28} \left (\frac {11}{14} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{7 \sqrt {1-2 x} (3 x+2)^2}-\frac {5}{7} \left (\frac {33}{28} \left (\frac {11}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 (5 x+3)^{5/2}}{7 \sqrt {1-2 x} (3 x+2)^2}-\frac {5}{7} \left (\frac {33}{28} \left (-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )-\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{14 (3 x+2)^2}\right )\) |
(2*(3 + 5*x)^(5/2))/(7*Sqrt[1 - 2*x]*(2 + 3*x)^2) - (5*(-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))/7
3.26.46.3.1 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 = 4.02 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.71
method | result | size |
default | \(-\frac {\left (32670 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}+27225 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-7260 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +113540 x^{2} \sqrt {-10 x^{2}-x +3}-7260 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+161350 x \sqrt {-10 x^{2}-x +3}+56952 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{19208 \left (2+3 x \right )^{2} \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(209\) |
-1/19208*(32670*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2)) *x^3+27225*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2- 7260*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+113540*x ^2*(-10*x^2-x+3)^(1/2)-7260*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2 -x+3)^(1/2))+161350*x*(-10*x^2-x+3)^(1/2)+56952*(-10*x^2-x+3)^(1/2))*(1-2* x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)^2/(-1+2*x)/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx=\frac {1815 \, \sqrt {7} {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (8110 \, x^{2} + 11525 \, x + 4068\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{19208 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} \]
1/19208*(1815*sqrt(7)*(18*x^3 + 15*x^2 - 4*x - 4)*arctan(1/14*sqrt(7)*(37* x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(8110*x^2 + 11 525*x + 4068)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(18*x^3 + 15*x^2 - 4*x - 4)
\[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx=\int \frac {\left (5 x + 3\right )^{\frac {5}{2}}}{\left (1 - 2 x\right )^{\frac {3}{2}} \left (3 x + 2\right )^{3}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.17 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx=-\frac {1815}{19208} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {20275 \, x}{6174 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {83665}{37044 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1}{378 \, {\left (9 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt {-10 \, x^{2} - x + 3} x + 4 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} - \frac {125}{1764 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]
-1815/19208*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 20 275/6174*x/sqrt(-10*x^2 - x + 3) + 83665/37044/sqrt(-10*x^2 - x + 3) + 1/3 78/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sqrt(-10* x^2 - x + 3)) - 125/1764/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-10*x^2 - x + 3))
Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (95) = 190\).
Time = 0.48 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.26 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx=-\frac {363}{38416} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {242 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1715 \, {\left (2 \, x - 1\right )}} + \frac {121 \, \sqrt {10} {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {360 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {1440 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{98 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]
-363/38416*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)))) - 242/1715*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) + 121/98*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)))^3 + 360*(sqrt(2) *sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 1440*sqrt(5*x + 3)/(sqrt(2)*s qrt(-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)^2
Timed out. \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx=\int \frac {{\left (5\,x+3\right )}^{5/2}}{{\left (1-2\,x\right )}^{3/2}\,{\left (3\,x+2\right )}^3} \,d x \]