Integrand size = 26, antiderivative size = 74 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {2 \sqrt {1-2 x}}{825 (3+5 x)^{3/2}}-\frac {404 \sqrt {1-2 x}}{9075 \sqrt {3+5 x}}+\frac {9}{25} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]
9/125*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/825*(1-2*x)^(1/2)/(3+ 5*x)^(3/2)-404/9075*(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {2 \sqrt {1-2 x} (617+1010 x)}{9075 (3+5 x)^{3/2}}-\frac {9}{25} \sqrt {\frac {2}{5}} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right ) \]
(-2*Sqrt[1 - 2*x]*(617 + 1010*x))/(9075*(3 + 5*x)^(3/2)) - (9*Sqrt[2/5]*Ar cTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/25
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {100, 27, 87, 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.
\(\displaystyle \int \frac {(3 x+2)^2}{\sqrt {1-2 x} (5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2}{825} \int \frac {1485 x+1093}{2 \sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{825} \int \frac {1485 x+1093}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{825} \left (297 \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {404 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{825} \left (\frac {594}{5} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {404 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{825} \left (297 \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {404 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x}}{825 (5 x+3)^{3/2}}\) |
(-2*Sqrt[1 - 2*x])/(825*(3 + 5*x)^(3/2)) + ((-404*Sqrt[1 - 2*x])/(11*Sqrt[ 3 + 5*x]) + 297*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/825
3.26.10.3.1 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_))*((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[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 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 = 96, normalized size of antiderivative = 1.30
method | result | size |
default | \(\frac {\left (81675 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+98010 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +29403 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-20200 x \sqrt {-10 x^{2}-x +3}-12340 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{90750 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(96\) |
1/90750*(81675*10^(1/2)*arcsin(20/11*x+1/11)*x^2+98010*10^(1/2)*arcsin(20/ 11*x+1/11)*x+29403*10^(1/2)*arcsin(20/11*x+1/11)-20200*x*(-10*x^2-x+3)^(1/ 2)-12340*(-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 = 92, normalized size of antiderivative = 1.24 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {3267 \, \sqrt {5} \sqrt {2} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (1010 \, x + 617\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{90750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
-1/90750*(3267*sqrt(5)*sqrt(2)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(5)*sqr t(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(1010* x + 617)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{2}}{\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\frac {9}{250} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{825 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {404 \, \sqrt {-10 \, x^{2} - x + 3}}{9075 \, {\left (5 \, x + 3\right )}} \]
9/250*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 2/825*sqrt(-10*x^2 - x + 3) /(25*x^2 + 30*x + 9) - 404/9075*sqrt(-10*x^2 - x + 3)/(5*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (51) = 102\).
Time = 0.32 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.89 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{726000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {9}{125} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {27 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{12100 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {405 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{45375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
-1/726000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 9/125*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 27/12100*sqrt(10)*( sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 1/45375*sqrt(10)*(5*x + 3)^(3/2)*(405*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqr t(2)*sqrt(-10*x + 5) - sqrt(22))^3
Timed out. \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^2}{\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{5/2}} \,d x \]