Integrand size = 26, antiderivative size = 115 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx=\frac {74}{45} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {7 (1-2 x)^{3/2} \sqrt {3+5 x}}{3 (2+3 x)}+\frac {346}{135} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {175}{27} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]
346/675*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-175/27*arctan(1/7*(1- 2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+7/3*(1-2*x)^(3/2)*(3+5*x)^(1/2)/ (2+3*x)+74/45*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.36 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx=\frac {1}{675} \left (\frac {15 \sqrt {1-2 x} \left (759+1301 x+60 x^2\right )}{(2+3 x) \sqrt {3+5 x}}-346 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-4375 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\right ) \]
((15*Sqrt[1 - 2*x]*(759 + 1301*x + 60*x^2))/((2 + 3*x)*Sqrt[3 + 5*x]) - 34 6*Sqrt[10]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] - 4375*Sqrt[7]*ArcTan[Sqr t[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/675
Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {109, 27, 171, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2)^2 \sqrt {5 x+3}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {1}{3} \int \frac {\sqrt {1-2 x} (148 x+157)}{2 (3 x+2) \sqrt {5 x+3}}dx+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \int \frac {\sqrt {1-2 x} (148 x+157)}{(3 x+2) \sqrt {5 x+3}}dx+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{15} \int \frac {692 x+2503}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {148}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{15} \left (\frac {692}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {6125}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {148}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{15} \left (\frac {6125}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {1384}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {148}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{15} \left (\frac {12250}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}+\frac {1384}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {148}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{15} \left (\frac {1384}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1750}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {148}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{15} \left (\frac {692}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {1750}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )+\frac {148}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {7 \sqrt {5 x+3} (1-2 x)^{3/2}}{3 (3 x+2)}\) |
(7*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/(3*(2 + 3*x)) + ((148*Sqrt[1 - 2*x]*Sqrt [3 + 5*x])/15 + ((692*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 - (175 0*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/3)/15)/6
3.25.30.3.1 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 = 133, normalized size of antiderivative = 1.16
| method | result | size |
| risch | \(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (253+12 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{45 \left (2+3 x \right ) \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (-\frac {173 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{675}-\frac {175 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{54}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(133\) |
| default | \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (1038 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +13125 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +692 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+8750 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+360 x \sqrt {-10 x^{2}-x +3}+7590 \sqrt {-10 x^{2}-x +3}\right )}{1350 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )}\) | \(146\) |
-1/45*(-1+2*x)*(3+5*x)^(1/2)*(253+12*x)/(2+3*x)/(-(-1+2*x)*(3+5*x))^(1/2)* ((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)-(-173/675*10^(1/2)*arcsin(20/11*x+1/ 11)-175/54*7^(1/2)*arctan(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 = 127, normalized size of antiderivative = 1.10 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx=-\frac {346 \, \sqrt {5} \sqrt {2} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 4375 \, \sqrt {7} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 30 \, {\left (12 \, x + 253\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1350 \, {\left (3 \, x + 2\right )}} \]
-1/1350*(346*sqrt(5)*sqrt(2)*(3*x + 2)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 4375*sqrt(7)*(3*x + 2 )*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 30*(12*x + 253)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)
\[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}}}{\left (3 x + 2\right )^{2} \sqrt {5 x + 3}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx=\frac {173}{675} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {175}{54} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {4}{45} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{9 \, {\left (3 \, x + 2\right )}} \]
173/675*sqrt(10)*arcsin(20/11*x + 1/11) + 175/54*sqrt(7)*arcsin(37/11*x/ab s(3*x + 2) + 20/11/abs(3*x + 2)) + 4/45*sqrt(-10*x^2 - x + 3) + 49/9*sqrt( -10*x^2 - x + 3)/(3*x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (83) = 166\).
Time = 0.39 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.43 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx=\frac {35}{108} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {173}{675} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {4}{225} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {1078 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{9 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \]
35/108*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*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)))) + 173/675*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sqr t(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 4/225*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 1078/9*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)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(2 2))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^ 2 + 280)
Timed out. \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}}{{\left (3\,x+2\right )}^2\,\sqrt {5\,x+3}} \,d x \]