Integrand size = 24, antiderivative size = 116 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {2 (1-2 x)^{7/2}}{165 (3+5 x)^{3/2}}-\frac {182 (1-2 x)^{5/2}}{825 \sqrt {3+5 x}}-\frac {91}{250} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {91}{825} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {1001 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{250 \sqrt {10}} \]
-2/165*(1-2*x)^(7/2)/(3+5*x)^(3/2)-1001/2500*arcsin(1/11*22^(1/2)*(3+5*x)^ (1/2))*10^(1/2)-182/825*(1-2*x)^(5/2)/(3+5*x)^(1/2)-91/825*(1-2*x)^(3/2)*( 3+5*x)^(1/2)-91/250*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (-3707-7970 x-2715 x^2+900 x^3\right )}{(3+5 x)^{3/2}}+3003 \sqrt {10} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{3750} \]
((5*Sqrt[1 - 2*x]*(-3707 - 7970*x - 2715*x^2 + 900*x^3))/(3 + 5*x)^(3/2) + 3003*Sqrt[10]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/3750
Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 57, 60, 60, 64, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)}{(5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {91}{165} \int \frac {(1-2 x)^{5/2}}{(5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{7/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {91}{165} \left (-2 \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{7/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {91}{165} \left (-2 \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{7/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {91}{165} \left (-2 \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{7/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {91}{165} \left (-2 \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{7/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {91}{165} \left (-2 \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{7/2}}{165 (5 x+3)^{3/2}}\) |
(-2*(1 - 2*x)^(7/2))/(165*(3 + 5*x)^(3/2)) + (91*((-2*(1 - 2*x)^(5/2))/(5* Sqrt[3 + 5*x]) - 2*(((1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/10 + (33*((Sqrt[1 - 2* x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(5*Sqrt[10]))) /20)))/165
3.25.50.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 1.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.12
method | result | size |
default | \(-\frac {\left (75075 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-18000 x^{3} \sqrt {-10 x^{2}-x +3}+90090 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +54300 x^{2} \sqrt {-10 x^{2}-x +3}+27027 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+159400 x \sqrt {-10 x^{2}-x +3}+74140 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{15000 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(130\) |
-1/15000*(75075*10^(1/2)*arcsin(20/11*x+1/11)*x^2-18000*x^3*(-10*x^2-x+3)^ (1/2)+90090*10^(1/2)*arcsin(20/11*x+1/11)*x+54300*x^2*(-10*x^2-x+3)^(1/2)+ 27027*10^(1/2)*arcsin(20/11*x+1/11)+159400*x*(-10*x^2-x+3)^(1/2)+74140*(-1 0*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\frac {3003 \, \sqrt {10} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (900 \, x^{3} - 2715 \, x^{2} - 7970 \, x - 3707\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{15000 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/15000*(3003*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1) *sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(900*x^3 - 2715*x^2 - 7970*x - 3707)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}} \cdot \left (3 x + 2\right )}{\left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (83) = 166\).
Time = 0.30 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.60 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {1001}{5000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{25 \, {\left (625 \, x^{4} + 1500 \, x^{3} + 1350 \, x^{2} + 540 \, x + 81\right )}} + \frac {3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{50 \, {\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} - \frac {11 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{150 \, {\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} + \frac {33 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{100 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {121 \, \sqrt {-10 \, x^{2} - x + 3}}{750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {2959 \, \sqrt {-10 \, x^{2} - x + 3}}{1500 \, {\left (5 \, x + 3\right )}} \]
-1001/5000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 1/25*(-10*x^2 - x + 3) ^(5/2)/(625*x^4 + 1500*x^3 + 1350*x^2 + 540*x + 81) + 3/50*(-10*x^2 - x + 3)^(5/2)/(125*x^3 + 225*x^2 + 135*x + 27) - 11/150*(-10*x^2 - x + 3)^(3/2) /(125*x^3 + 225*x^2 + 135*x + 27) + 33/100*(-10*x^2 - x + 3)^(3/2)/(25*x^2 + 30*x + 9) - 121/750*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) - 2959/15 00*sqrt(-10*x^2 - x + 3)/(5*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (83) = 166\).
Time = 0.38 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.48 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\frac {1}{6250} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} - 289 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {11 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{150000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {1001}{2500} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {627 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{12500 \, \sqrt {5 \, x + 3}} + \frac {11 \, \sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {171 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{9375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
1/6250*(12*sqrt(5)*(5*x + 3) - 289*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 11/150000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2 ) - 1001/2500*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 627/12500*sqr t(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 11/9375*sqrt(10 )*(5*x + 3)^(3/2)*(171*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3
Timed out. \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}\,\left (3\,x+2\right )}{{\left (5\,x+3\right )}^{5/2}} \,d x \]