Integrand size = 24, antiderivative size = 96 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=-\frac {845}{88} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {169 (3+5 x)^{3/2}}{66 \sqrt {1-2 x}}+\frac {7 (3+5 x)^{5/2}}{33 (1-2 x)^{3/2}}+\frac {169}{8} \sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]
7/33*(3+5*x)^(5/2)/(1-2*x)^(3/2)+169/16*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2) )*10^(1/2)-169/66*(3+5*x)^(3/2)/(1-2*x)^(1/2)-845/88*(1-2*x)^(1/2)*(3+5*x) ^(1/2)
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {-2 \sqrt {3+5 x} \left (369-1136 x+180 x^2\right )+507 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{48 (1-2 x)^{3/2}} \]
(-2*Sqrt[3 + 5*x]*(369 - 1136*x + 180*x^2) + 507*Sqrt[10 - 20*x]*(-1 + 2*x )*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(48*(1 - 2*x)^(3/2))
Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {87, 57, 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.
\(\displaystyle \int \frac {(3 x+2) (5 x+3)^{3/2}}{(1-2 x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{33 (1-2 x)^{3/2}}-\frac {169}{66} \int \frac {(5 x+3)^{3/2}}{(1-2 x)^{3/2}}dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{33 (1-2 x)^{3/2}}-\frac {169}{66} \left (\frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {15}{2} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{33 (1-2 x)^{3/2}}-\frac {169}{66} \left (\frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {15}{2} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{33 (1-2 x)^{3/2}}-\frac {169}{66} \left (\frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {15}{2} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7 (5 x+3)^{5/2}}{33 (1-2 x)^{3/2}}-\frac {169}{66} \left (\frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}-\frac {15}{2} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )\right )\) |
(7*(3 + 5*x)^(5/2))/(33*(1 - 2*x)^(3/2)) - (169*((3 + 5*x)^(3/2)/Sqrt[1 - 2*x] - (15*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqr t[3 + 5*x]])/(2*Sqrt[10])))/2))/66
3.26.92.3.1 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 = 5.44 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.25
method | result | size |
default | \(\frac {\left (2028 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-2028 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -720 x^{2} \sqrt {-10 x^{2}-x +3}+507 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+4544 x \sqrt {-10 x^{2}-x +3}-1476 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{96 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(120\) |
1/96*(2028*10^(1/2)*arcsin(20/11*x+1/11)*x^2-2028*10^(1/2)*arcsin(20/11*x+ 1/11)*x-720*x^2*(-10*x^2-x+3)^(1/2)+507*10^(1/2)*arcsin(20/11*x+1/11)+4544 *x*(-10*x^2-x+3)^(1/2)-1476*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/ 2)/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)
Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=-\frac {507 \, \sqrt {5} \sqrt {2} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 4 \, {\left (180 \, x^{2} - 1136 \, x + 369\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{96 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]
-1/96*(507*sqrt(5)*sqrt(2)*(4*x^2 - 4*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)) + 4*(180*x^2 - 11 36*x + 369)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
\[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {5}{2}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.24 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {169}{32} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {7 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{12 \, {\left (8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1\right )}} + \frac {3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{4 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {77 \, \sqrt {-10 \, x^{2} - x + 3}}{24 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {271 \, \sqrt {-10 \, x^{2} - x + 3}}{12 \, {\left (2 \, x - 1\right )}} \]
169/32*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 7/12*(-10*x^2 - x + 3)^(3/ 2)/(8*x^3 - 12*x^2 + 6*x - 1) + 3/4*(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) + 77/24*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1) + 271/12*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.74 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {169}{16} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (9 \, \sqrt {5} {\left (5 \, x + 3\right )} - 338 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 5577 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{600 \, {\left (2 \, x - 1\right )}^{2}} \]
169/16*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/600*(4*(9*sqrt(5)* (5*x + 3) - 338*sqrt(5))*(5*x + 3) + 5577*sqrt(5))*sqrt(5*x + 3)*sqrt(-10* x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\int \frac {\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{5/2}} \,d x \]