Integrand size = 26, antiderivative size = 108 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\frac {505}{84} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x}}-\frac {475}{36} \sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )+\frac {2 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{63 \sqrt {7}} \]
-475/72*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+2/441*arctan(1/7*(1-2 *x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+11/7*(3+5*x)^(3/2)/(1-2*x)^(1/2)+ 505/84*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\frac {42 (901-350 x) \sqrt {3+5 x}+23275 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )+16 \sqrt {7-14 x} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{3528 \sqrt {1-2 x}} \]
(42*(901 - 350*x)*Sqrt[3 + 5*x] + 23275*Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] + 16*Sqrt[7 - 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt [3 + 5*x])])/(3528*Sqrt[1 - 2*x])
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {109, 27, 171, 27, 175, 64, 104, 217, 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.
| \(\displaystyle \int \frac {(5 x+3)^{5/2}}{(1-2 x)^{3/2} (3 x+2)} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}-\frac {1}{7} \int \frac {\sqrt {5 x+3} (505 x+336)}{2 \sqrt {1-2 x} (3 x+2)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}-\frac {1}{14} \int \frac {\sqrt {5 x+3} (505 x+336)}{\sqrt {1-2 x} (3 x+2)}dx\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{14} \left (\frac {1}{6} \int -\frac {16625 x+11086}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {505}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{14} \left (\frac {505}{6} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {1}{12} \int \frac {16625 x+11086}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{14} \left (\frac {1}{12} \left (-\frac {16625}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {8}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {505}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{14} \left (\frac {1}{12} \left (-\frac {8}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {6650}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {505}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{14} \left (\frac {1}{12} \left (-\frac {16}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {6650}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {505}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{14} \left (\frac {1}{12} \left (\frac {16 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}-\frac {6650}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {505}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{14} \left (\frac {1}{12} \left (\frac {16 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}-\frac {3325}{3} \sqrt {10} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )\right )+\frac {505}{6} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}\) |
(11*(3 + 5*x)^(3/2))/(7*Sqrt[1 - 2*x]) + ((505*Sqrt[1 - 2*x]*Sqrt[3 + 5*x] )/6 + ((-3325*Sqrt[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 + (16*ArcTan[Sq rt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3*Sqrt[7]))/12)/14
3.26.44.3.1 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_))^(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[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
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])
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 = 1.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.35
| method | result | size |
| default | \(-\frac {\left (46550 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +32 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -23275 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-16 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-29400 x \sqrt {-10 x^{2}-x +3}+75684 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{7056 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(146\) |
-1/7056*(46550*10^(1/2)*arcsin(20/11*x+1/11)*x+32*7^(1/2)*arctan(1/14*(37* x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x-23275*10^(1/2)*arcsin(20/11*x+1/11)-1 6*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-29400*x*(-10* x^2-x+3)^(1/2)+75684*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(-1+ 2*x)/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.18 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\frac {23275 \, \sqrt {5} \sqrt {2} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 16 \, \sqrt {7} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 84 \, {\left (350 \, x - 901\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{7056 \, {\left (2 \, x - 1\right )}} \]
1/7056*(23275*sqrt(5)*sqrt(2)*(2*x - 1)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 16*sqrt(7)*(2*x - 1) *arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 84*(350*x - 901)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\int \frac {\left (5 x + 3\right )^{\frac {5}{2}}}{\left (1 - 2 x\right )^{\frac {3}{2}} \cdot \left (3 x + 2\right )}\, dx \]
Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=-\frac {125 \, x^{2}}{6 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {475}{144} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {1}{441} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {3455 \, x}{84 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {901}{28 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-125/6*x^2/sqrt(-10*x^2 - x + 3) - 475/144*sqrt(10)*arcsin(20/11*x + 1/11) - 1/441*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 3455/ 84*x/sqrt(-10*x^2 - x + 3) + 901/28/sqrt(-10*x^2 - x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (76) = 152\).
Time = 0.36 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.67 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=-\frac {1}{4410} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {475}{144} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {{\left (70 \, \sqrt {5} {\left (5 \, x + 3\right )} - 1111 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{420 \, {\left (2 \, x - 1\right )}} \]
-1/4410*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)))) - 475/144*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sq rt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5 ) - sqrt(22)))) + 1/420*(70*sqrt(5)*(5*x + 3) - 1111*sqrt(5))*sqrt(5*x + 3 )*sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\int \frac {{\left (5\,x+3\right )}^{5/2}}{{\left (1-2\,x\right )}^{3/2}\,\left (3\,x+2\right )} \,d x \]