Integrand size = 26, antiderivative size = 103 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=-\frac {25 \sqrt {1-2 x}}{3 (3+5 x)^{3/2}}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^{3/2}}+\frac {2495 \sqrt {1-2 x}}{33 \sqrt {3+5 x}}-\frac {519 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{\sqrt {7}} \]
-519/7*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-25/3*(1-2*x )^(1/2)/(3+5*x)^(3/2)+(1-2*x)^(1/2)/(2+3*x)/(3+5*x)^(3/2)+2495/33*(1-2*x)^ (1/2)/(3+5*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=\frac {7 \sqrt {1-2 x} \left (14453+46580 x+37425 x^2\right )-17127 \sqrt {7} \sqrt {3+5 x} \left (6+19 x+15 x^2\right ) \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{231 (2+3 x) (3+5 x)^{3/2}} \]
(7*Sqrt[1 - 2*x]*(14453 + 46580*x + 37425*x^2) - 17127*Sqrt[7]*Sqrt[3 + 5* x]*(6 + 19*x + 15*x^2)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(231 *(2 + 3*x)*(3 + 5*x)^(3/2))
Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {110, 27, 169, 27, 169, 27, 104, 217}
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.
\(\displaystyle \int \frac {\sqrt {1-2 x}}{(3 x+2)^2 (5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 110 |
\(\displaystyle \frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}-\int -\frac {31-40 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)^{5/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {31-40 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{5/2}}dx+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{2} \left (-\frac {2}{33} \int \frac {11 (319-300 x)}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {50 \sqrt {1-2 x}}{3 (5 x+3)^{3/2}}\right )+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{3} \int \frac {319-300 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {50 \sqrt {1-2 x}}{3 (5 x+3)^{3/2}}\right )+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {2}{11} \int \frac {17127}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {4990 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {50 \sqrt {1-2 x}}{3 (5 x+3)^{3/2}}\right )+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (1557 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {4990 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {50 \sqrt {1-2 x}}{3 (5 x+3)^{3/2}}\right )+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3114 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}+\frac {4990 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {50 \sqrt {1-2 x}}{3 (5 x+3)^{3/2}}\right )+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {4990 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}-\frac {3114 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}\right )-\frac {50 \sqrt {1-2 x}}{3 (5 x+3)^{3/2}}\right )+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}\) |
Sqrt[1 - 2*x]/((2 + 3*x)*(3 + 5*x)^(3/2)) + ((-50*Sqrt[1 - 2*x])/(3*(3 + 5 *x)^(3/2)) + ((4990*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) - (3114*ArcTan[Sqrt[ 1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7])/3)/2
3.24.24.3.1 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[(((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[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f)) Int[(a + b*x)^(m + 1) *(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt Q[n, 0] && (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_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n, 2*p]
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])
Leaf count of result is larger than twice the leaf count of optimal. \(201\) vs. \(2(80)=160\).
Time = 1.13 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.96
method | result | size |
default | \(\frac {\left (1284525 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}+2397780 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+1490049 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +523950 x^{2} \sqrt {-10 x^{2}-x +3}+308286 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+652120 x \sqrt {-10 x^{2}-x +3}+202342 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{462 \left (2+3 x \right ) \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(202\) |
1/462*(1284525*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))* x^3+2397780*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2 +1490049*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+5239 50*x^2*(-10*x^2-x+3)^(1/2)+308286*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(- 10*x^2-x+3)^(1/2))+652120*x*(-10*x^2-x+3)^(1/2)+202342*(-10*x^2-x+3)^(1/2) )*(1-2*x)^(1/2)/(2+3*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=-\frac {17127 \, \sqrt {7} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (37425 \, x^{2} + 46580 \, x + 14453\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{462 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \]
-1/462*(17127*sqrt(7)*(75*x^3 + 140*x^2 + 87*x + 18)*arctan(1/14*sqrt(7)*( 37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(37425*x^2 + 46580*x + 14453)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(75*x^3 + 140*x^2 + 87*x + 18)
\[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=\int \frac {\sqrt {1 - 2 x}}{\left (3 x + 2\right )^{2} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=\frac {519}{14} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {4990 \, x}{33 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {2605}{33 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {38 \, x}{{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {49}{9 \, {\left (3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 2 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} - \frac {185}{9 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
519/14*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 4990/33 *x/sqrt(-10*x^2 - x + 3) + 2605/33/sqrt(-10*x^2 - x + 3) + 38*x/(-10*x^2 - x + 3)^(3/2) + 49/9/(3*(-10*x^2 - x + 3)^(3/2)*x + 2*(-10*x^2 - x + 3)^(3 /2)) - 185/9/(-10*x^2 - x + 3)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (80) = 160\).
Time = 0.40 (sec) , antiderivative size = 318, normalized size of antiderivative = 3.09 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=-\frac {1}{18480} \, \sqrt {5} {\left (35 \, \sqrt {2} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - 68508 \, \sqrt {70} \sqrt {2} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - 55440 \, \sqrt {2} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {3659040 \, \sqrt {2} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{{\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280}\right )} \]
-1/18480*sqrt(5)*(35*sqrt(2)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5* x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 - 68508*s qrt(70)*sqrt(2)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqr t(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt( 22)))) - 55440*sqrt(2)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) - 3659040*sqrt(2) *((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sq rt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/ sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280))
Timed out. \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^{5/2}} \, dx=\int \frac {\sqrt {1-2\,x}}{{\left (3\,x+2\right )}^2\,{\left (5\,x+3\right )}^{5/2}} \,d x \]