Integrand size = 26, antiderivative size = 101 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\frac {4}{231 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {42230 \sqrt {1-2 x}}{195657 \sqrt {3+5 x}}+\frac {54 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{49 \sqrt {7}} \]
54/343*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+4/231/(1-2* x)^(3/2)/(3+5*x)^(1/2)+956/17787/(1-2*x)^(1/2)/(3+5*x)^(1/2)-42230/195657* (1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {2 \left (14163-73944 x+84460 x^2\right )}{195657 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {54 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{49 \sqrt {7}} \]
(-2*(14163 - 73944*x + 84460*x^2))/(195657*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) + (54*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(49*Sqrt[7])
Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {115, 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.
\(\displaystyle \int \frac {1}{(1-2 x)^{5/2} (3 x+2) (5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 115 |
\(\displaystyle \frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}-\frac {2}{231} \int -\frac {120 x+179}{2 (1-2 x)^{3/2} (3 x+2) (5 x+3)^{3/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{231} \int \frac {120 x+179}{(1-2 x)^{3/2} (3 x+2) (5 x+3)^{3/2}}dx+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{231} \left (\frac {956}{77 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {2}{77} \int -\frac {14340 x+12827}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx\right )+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{231} \left (\frac {1}{77} \int \frac {14340 x+12827}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx+\frac {956}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{231} \left (\frac {1}{77} \left (-\frac {2}{11} \int \frac {107811}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {42230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {956}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{231} \left (\frac {1}{77} \left (-9801 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {42230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {956}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {1}{231} \left (\frac {1}{77} \left (-19602 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {42230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {956}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{231} \left (\frac {1}{77} \left (\frac {19602 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {42230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {956}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}\) |
4/(231*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) + (956/(77*Sqrt[1 - 2*x]*Sqrt[3 + 5* x]) + ((-42230*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (19602*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7])/77)/231
3.27.24.3.1 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[b*(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] + Simp[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*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2 *n, 2*p]
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. \(2(74)=148\).
Time = 1.24 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.00
method | result | size |
default | \(-\frac {\sqrt {1-2 x}\, \left (2156220 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}-862488 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-754677 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +1182440 x^{2} \sqrt {-10 x^{2}-x +3}+323433 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-1035216 x \sqrt {-10 x^{2}-x +3}+198282 \sqrt {-10 x^{2}-x +3}\right )}{1369599 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(202\) |
-1/1369599*(1-2*x)^(1/2)*(2156220*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(- 10*x^2-x+3)^(1/2))*x^3-862488*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x ^2-x+3)^(1/2))*x^2-754677*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x +3)^(1/2))*x+1182440*x^2*(-10*x^2-x+3)^(1/2)+323433*7^(1/2)*arctan(1/14*(3 7*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-1035216*x*(-10*x^2-x+3)^(1/2)+198282* (-10*x^2-x+3)^(1/2))/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\frac {107811 \, \sqrt {7} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (84460 \, x^{2} - 73944 \, x + 14163\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1369599 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]
1/1369599*(107811*sqrt(7)*(20*x^3 - 8*x^2 - 7*x + 3)*arctan(1/14*sqrt(7)*( 37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(84460*x^2 - 73944*x + 14163)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3 )
\[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\int \frac {1}{\left (1 - 2 x\right )^{\frac {5}{2}} \cdot \left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\int { \frac {1}{{\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (3 \, x + 2\right )} {\left (-2 \, x + 1\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (74) = 148\).
Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {27}{3430} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {25}{2662} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {8 \, {\left (548 \, \sqrt {5} {\left (5 \, x + 3\right )} - 3399 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{4891425 \, {\left (2 \, x - 1\right )}^{2}} \]
-27/3430*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*(( sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 25/2662*sqrt(10)*((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))) - 8 /4891425*(548*sqrt(5)*(5*x + 3) - 3399*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\int \frac {1}{{\left (1-2\,x\right )}^{5/2}\,\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{3/2}} \,d x \]