Integrand size = 26, antiderivative size = 106 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=-\frac {239}{450} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{15} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {17687 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1350 \sqrt {10}}-\frac {98}{27} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]
-98/27*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-17687/13500 *arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-1/15*(1-2*x)^(3/2)*(3+5*x)^( 1/2)-239/450*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.31 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\frac {30 \sqrt {1-2 x} \left (-807-1165 x+300 x^2\right )+17687 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )-49000 \sqrt {7} \sqrt {3+5 x} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{13500 \sqrt {3+5 x}} \]
(30*Sqrt[1 - 2*x]*(-807 - 1165*x + 300*x^2) + 17687*Sqrt[30 + 50*x]*ArcTan [Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]] - 49000*Sqrt[7]*Sqrt[3 + 5*x]*ArcTan[Sqrt[ 1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(13500*Sqrt[3 + 5*x])
Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {113, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2) \sqrt {5 x+3}} \, dx\) |
| \(\Big \downarrow \) 113 |
| \(\displaystyle \frac {1}{30} \int \frac {(4-239 x) \sqrt {1-2 x}}{(3 x+2) \sqrt {5 x+3}}dx-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{30} \left (\frac {1}{15} \int -\frac {17687 x+358}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {239}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{30} \left (-\frac {1}{30} \int \frac {17687 x+358}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {239}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {1}{30} \left (\frac {1}{30} \left (\frac {34300}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {17687}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )-\frac {239}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{30} \left (\frac {1}{30} \left (\frac {34300}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {35374}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {239}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{30} \left (\frac {1}{30} \left (\frac {68600}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {35374}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )-\frac {239}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{30} \left (\frac {1}{30} \left (-\frac {35374}{15} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {9800}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )-\frac {239}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{30} \left (\frac {1}{30} \left (-\frac {17687}{3} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {9800}{3} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )-\frac {239}{15} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{15} (1-2 x)^{3/2} \sqrt {5 x+3}\) |
-1/15*((1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) + ((-239*Sqrt[1 - 2*x]*Sqrt[3 + 5*x] )/15 + ((-17687*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/3 - (9800*Sqrt [7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/3)/30)/30
3.25.29.3.1 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*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]
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 = 3.86 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(-\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (17687 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-49000 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-3600 x \sqrt {-10 x^{2}-x +3}+16140 \sqrt {-10 x^{2}-x +3}\right )}{27000 \sqrt {-10 x^{2}-x +3}}\) | \(98\) |
| risch | \(-\frac {\left (-269+60 x \right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{450 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}-\frac {\left (\frac {17687 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{27000}-\frac {49 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{27}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(126\) |
-1/27000*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(17687*10^(1/2)*arcsin(20/11*x+1/11)- 49000*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-3600*x*(- 10*x^2-x+3)^(1/2)+16140*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)
Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\frac {1}{450} \, {\left (60 \, x - 269\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {49}{27} \, \sqrt {7} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + \frac {17687}{27000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
1/450*(60*x - 269)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 49/27*sqrt(7)*arctan(1/1 4*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 176 87/27000*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
\[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}}}{\left (3 x + 2\right ) \sqrt {5 x + 3}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.65 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\frac {2}{15} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {17687}{27000} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {49}{27} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {269}{450} \, \sqrt {-10 \, x^{2} - x + 3} \]
2/15*sqrt(-10*x^2 - x + 3)*x - 17687/27000*sqrt(10)*arcsin(20/11*x + 1/11) + 49/27*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 269/4 50*sqrt(-10*x^2 - x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (76) = 152\).
Time = 0.36 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.63 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\frac {49}{270} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {1}{2250} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} - 305 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {17687}{27000} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} \]
49/270*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)))) + 1/2250*(12*sqrt(5)*(5*x + 3) - 305*sqrt(5))*sqrt(5*x + 3 )*sqrt(-10*x + 5) - 17687/27000*sqrt(10)*(pi + 2*arctan(-1/4*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))))
Timed out. \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) \sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}}{\left (3\,x+2\right )\,\sqrt {5\,x+3}} \,d x \]