Integrand size = 26, antiderivative size = 84 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {2 \sqrt {1-2 x} (2+3 x)^2}{165 (3+5 x)^{3/2}}-\frac {\sqrt {1-2 x} (5831+9405 x)}{18150 \sqrt {3+5 x}}+\frac {81 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{50 \sqrt {10}} \]
81/500*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/165*(2+3*x)^2*(1-2*x )^(1/2)/(3+5*x)^(3/2)-1/18150*(5831+9405*x)*(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {1-2 x} \left (18373+60010 x+49005 x^2\right )}{18150 (3+5 x)^{3/2}}-\frac {81 \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{50 \sqrt {10}} \]
-1/18150*(Sqrt[1 - 2*x]*(18373 + 60010*x + 49005*x^2))/(3 + 5*x)^(3/2) - ( 81*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(50*Sqrt[10])
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {109, 27, 160, 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.
\(\displaystyle \int \frac {(3 x+2)^3}{\sqrt {1-2 x} (5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle -\frac {2}{165} \int -\frac {(3 x+2) (285 x+218)}{2 \sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{165} \int \frac {(3 x+2) (285 x+218)}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 160 |
\(\displaystyle \frac {1}{165} \left (\frac {2673}{20} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} (9405 x+5831)}{110 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{165} \left (\frac {2673}{50} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {\sqrt {1-2 x} (9405 x+5831)}{110 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{165} \left (\frac {2673 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{10 \sqrt {10}}-\frac {\sqrt {1-2 x} (9405 x+5831)}{110 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{165 (5 x+3)^{3/2}}\) |
(-2*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(165*(3 + 5*x)^(3/2)) + (-1/110*(Sqrt[1 - 2 *x]*(5831 + 9405*x))/Sqrt[3 + 5*x] + (2673*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x] ])/(10*Sqrt[10]))/165
3.26.9.3.1 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_) )^(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_) )*((g_.) + (h_.)*(x_)), x_] :> Simp[(b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d*(b*c - a*d)*(m + 1))), x] + Simp[(a*d*f*h*m + b*(d* (f*g + e*h) - c*f*h*(m + 2)))/(b^2*d) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
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 = 113, normalized size of antiderivative = 1.35
method | result | size |
default | \(\frac {\left (735075 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+882090 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -980100 x^{2} \sqrt {-10 x^{2}-x +3}+264627 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-1200200 x \sqrt {-10 x^{2}-x +3}-367460 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{363000 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(113\) |
1/363000*(735075*10^(1/2)*arcsin(20/11*x+1/11)*x^2+882090*10^(1/2)*arcsin( 20/11*x+1/11)*x-980100*x^2*(-10*x^2-x+3)^(1/2)+264627*10^(1/2)*arcsin(20/1 1*x+1/11)-1200200*x*(-10*x^2-x+3)^(1/2)-367460*(-10*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 = 91, normalized size of antiderivative = 1.08 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {29403 \, \sqrt {10} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (49005 \, x^{2} + 60010 \, x + 18373\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{363000 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
-1/363000*(29403*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*(49005*x^2 + 60010 *x + 18373)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{3}}{\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\frac {81}{1000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {27}{250} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{4125 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {602 \, \sqrt {-10 \, x^{2} - x + 3}}{45375 \, {\left (5 \, x + 3\right )}} \]
81/1000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 27/250*sqrt(-10*x^2 - x + 3) - 2/4125*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) - 602/45375*sqrt(-1 0*x^2 - x + 3)/(5*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (63) = 126\).
Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.89 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{3630000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {27}{1250} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {81}{500} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {201 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{302500 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {603 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{226875 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
-1/3630000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 27/1250*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 81/500*sqrt(10)*arcsin( 1/11*sqrt(22)*sqrt(5*x + 3)) - 201/302500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5 ) - sqrt(22))/sqrt(5*x + 3) + 1/226875*sqrt(10)*(5*x + 3)^(3/2)*(603*(sqrt (2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3
Timed out. \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^3}{\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{5/2}} \,d x \]