Integrand size = 26, antiderivative size = 137 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {6205}{7546 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {3125575 \sqrt {1-2 x}}{166012 \sqrt {3+5 x}}+\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}+\frac {555}{196 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}}+\frac {177255 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1372 \sqrt {7}} \]
177255/9604*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-6205/7 546/(1-2*x)^(1/2)/(3+5*x)^(1/2)+3/14/(2+3*x)^2/(1-2*x)^(1/2)/(3+5*x)^(1/2) +555/196/(2+3*x)/(1-2*x)^(1/2)/(3+5*x)^(1/2)-3125575/166012*(1-2*x)^(1/2)/ (3+5*x)^(1/2)
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.58 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=\frac {\frac {7 \left (-12072596-12730165 x+45655035 x^2+56260350 x^3\right )}{\sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}+21447855 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1162084} \]
((7*(-12072596 - 12730165*x + 45655035*x^2 + 56260350*x^3))/(Sqrt[1 - 2*x] *(2 + 3*x)^2*Sqrt[3 + 5*x]) + 21447855*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[ 7]*Sqrt[3 + 5*x])])/1162084
Time = 0.23 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {114, 27, 168, 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.
| \(\displaystyle \int \frac {1}{(1-2 x)^{3/2} (3 x+2)^3 (5 x+3)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {1}{14} \int \frac {5 (13-36 x)}{2 (1-2 x)^{3/2} (3 x+2)^2 (5 x+3)^{3/2}}dx+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{28} \int \frac {13-36 x}{(1-2 x)^{3/2} (3 x+2)^2 (5 x+3)^{3/2}}dx+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{7} \int \frac {979-4440 x}{2 (1-2 x)^{3/2} (3 x+2) (5 x+3)^{3/2}}dx+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{14} \int \frac {979-4440 x}{(1-2 x)^{3/2} (3 x+2) (5 x+3)^{3/2}}dx+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (-\frac {2}{77} \int -\frac {80347-74460 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \int \frac {80347-74460 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (-\frac {2}{11} \int \frac {4289571}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (-389961 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (-779922 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (\frac {779922 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\) |
3/(14*Sqrt[1 - 2*x]*(2 + 3*x)^2*Sqrt[3 + 5*x]) + (5*(111/(7*Sqrt[1 - 2*x]* (2 + 3*x)*Sqrt[3 + 5*x]) + (-4964/(77*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + ((-12 50230*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (779922*ArcTan[Sqrt[1 - 2*x]/(Sq rt[7]*Sqrt[3 + 5*x])])/Sqrt[7])/77)/14))/28
3.26.69.3.1 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] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
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] && ILtQ[m, -1]
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. \(256\) vs. \(2(104)=208\).
Time = 1.22 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.88
| method | result | size |
| default | \(-\frac {\sqrt {1-2 x}\, \left (1930306950 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{4}+2766773295 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}+536196375 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+787644900 x^{3} \sqrt {-10 x^{2}-x +3}-686331360 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +639170490 x^{2} \sqrt {-10 x^{2}-x +3}-257374260 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-178222310 x \sqrt {-10 x^{2}-x +3}-169016344 \sqrt {-10 x^{2}-x +3}\right )}{2324168 \left (2+3 x \right )^{2} \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(257\) |
-1/2324168*(1-2*x)^(1/2)*(1930306950*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2) /(-10*x^2-x+3)^(1/2))*x^4+2766773295*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2) /(-10*x^2-x+3)^(1/2))*x^3+536196375*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/ (-10*x^2-x+3)^(1/2))*x^2+787644900*x^3*(-10*x^2-x+3)^(1/2)-686331360*7^(1/ 2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+639170490*x^2*(-10 *x^2-x+3)^(1/2)-257374260*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x +3)^(1/2))-178222310*x*(-10*x^2-x+3)^(1/2)-169016344*(-10*x^2-x+3)^(1/2))/ (2+3*x)^2/(-1+2*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=\frac {21447855 \, \sqrt {7} {\left (90 \, x^{4} + 129 \, x^{3} + 25 \, x^{2} - 32 \, x - 12\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (56260350 \, x^{3} + 45655035 \, x^{2} - 12730165 \, x - 12072596\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{2324168 \, {\left (90 \, x^{4} + 129 \, x^{3} + 25 \, x^{2} - 32 \, x - 12\right )}} \]
1/2324168*(21447855*sqrt(7)*(90*x^4 + 129*x^3 + 25*x^2 - 32*x - 12)*arctan (1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(56260350*x^3 + 45655035*x^2 - 12730165*x - 12072596)*sqrt(5*x + 3)*sq rt(-2*x + 1))/(90*x^4 + 129*x^3 + 25*x^2 - 32*x - 12)
\[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=\int \frac {1}{\left (1 - 2 x\right )^{\frac {3}{2}} \left (3 x + 2\right )^{3} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {177255}{19208} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {3125575 \, x}{83006 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {3262085}{166012 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {3}{14 \, {\left (9 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt {-10 \, x^{2} - x + 3} x + 4 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} + \frac {555}{196 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]
-177255/19208*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 3125575/83006*x/sqrt(-10*x^2 - x + 3) - 3262085/166012/sqrt(-10*x^2 - x + 3) + 3/14/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sq rt(-10*x^2 - x + 3)) + 555/196/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-10*x^2 - x + 3))
Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (104) = 208\).
Time = 0.49 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.46 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {35451}{38416} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {125}{242} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {32 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{207515 \, {\left (2 \, x - 1\right )}} - \frac {297 \, \sqrt {10} {\left (47 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {10520 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {42080 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{98 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]
-35451/38416*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)))) - 125/242*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))) - 32/207515*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) - 297/98*sqrt (10)*(47*((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 + 10520*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 42080*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*s qrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2
Timed out. \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=\int \frac {1}{{\left (1-2\,x\right )}^{3/2}\,{\left (3\,x+2\right )}^3\,{\left (5\,x+3\right )}^{3/2}} \,d x \]