Integrand size = 26, antiderivative size = 137 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {1735 \sqrt {3+5 x}}{3234 (1-2 x)^{3/2}}-\frac {57595 \sqrt {3+5 x}}{249018 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {51 \sqrt {3+5 x}}{28 (1-2 x)^{3/2} (2+3 x)}-\frac {5805 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1372 \sqrt {7}} \]
-5805/9604*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-1735/32 34*(3+5*x)^(1/2)/(1-2*x)^(3/2)+3/14*(3+5*x)^(1/2)/(1-2*x)^(3/2)/(2+3*x)^2+ 51/28*(3+5*x)^(1/2)/(1-2*x)^(3/2)/(2+3*x)-57595/249018*(3+5*x)^(1/2)/(1-2* x)^(1/2)
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {-7 \sqrt {3+5 x} \left (391476-945629 x-676860 x^2+2073420 x^3\right )-2107215 \sqrt {7-14 x} (-1+2 x) (2+3 x)^2 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{3486252 (1-2 x)^{3/2} (2+3 x)^2} \]
-1/3486252*(-7*Sqrt[3 + 5*x]*(391476 - 945629*x - 676860*x^2 + 2073420*x^3 ) - 2107215*Sqrt[7 - 14*x]*(-1 + 2*x)*(2 + 3*x)^2*ArcTan[Sqrt[1 - 2*x]/(Sq rt[7]*Sqrt[3 + 5*x])])/((1 - 2*x)^(3/2)*(2 + 3*x)^2)
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.
| \(\displaystyle \int \frac {1}{(1-2 x)^{5/2} (3 x+2)^3 \sqrt {5 x+3}} \, dx\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {1}{14} \int -\frac {180 x+1}{2 (1-2 x)^{5/2} (3 x+2)^2 \sqrt {5 x+3}}dx+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}-\frac {1}{28} \int \frac {180 x+1}{(1-2 x)^{5/2} (3 x+2)^2 \sqrt {5 x+3}}dx\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {1}{7} \int \frac {35 (408 x+143)}{2 (1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {5}{2} \int \frac {408 x+143}{(1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {5}{2} \left (\frac {1388 \sqrt {5 x+3}}{231 (1-2 x)^{3/2}}-\frac {2}{231} \int -\frac {20820 x+1109}{2 (1-2 x)^{3/2} (3 x+2) \sqrt {5 x+3}}dx\right )\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {5}{2} \left (\frac {1}{231} \int \frac {20820 x+1109}{(1-2 x)^{3/2} (3 x+2) \sqrt {5 x+3}}dx+\frac {1388 \sqrt {5 x+3}}{231 (1-2 x)^{3/2}}\right )\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {5}{2} \left (\frac {1}{231} \left (\frac {46076 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}-\frac {2}{77} \int \frac {421443}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {1388 \sqrt {5 x+3}}{231 (1-2 x)^{3/2}}\right )\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {5}{2} \left (\frac {1}{231} \left (\frac {46076 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}-\frac {38313}{7} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {1388 \sqrt {5 x+3}}{231 (1-2 x)^{3/2}}\right )\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {5}{2} \left (\frac {1}{231} \left (\frac {46076 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}-\frac {76626}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {1388 \sqrt {5 x+3}}{231 (1-2 x)^{3/2}}\right )\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{28} \left (\frac {51 \sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}-\frac {5}{2} \left (\frac {1}{231} \left (\frac {76626 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}+\frac {46076 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}\right )+\frac {1388 \sqrt {5 x+3}}{231 (1-2 x)^{3/2}}\right )\right )+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}\) |
(3*Sqrt[3 + 5*x])/(14*(1 - 2*x)^(3/2)*(2 + 3*x)^2) + ((51*Sqrt[3 + 5*x])/( (1 - 2*x)^(3/2)*(2 + 3*x)) - (5*((1388*Sqrt[3 + 5*x])/(231*(1 - 2*x)^(3/2) ) + ((46076*Sqrt[3 + 5*x])/(77*Sqrt[1 - 2*x]) + (76626*ArcTan[Sqrt[1 - 2*x ]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7]))/231))/2)/28
3.27.16.3.1 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 = 1.24 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.88
| method | result | size |
| default | \(\frac {\left (75859740 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{4}+25286580 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}-48465945 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+29027880 x^{3} \sqrt {-10 x^{2}-x +3}-8428860 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -9476040 x^{2} \sqrt {-10 x^{2}-x +3}+8428860 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-13238806 x \sqrt {-10 x^{2}-x +3}+5480664 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}\, \sqrt {1-2 x}}{6972504 \left (2+3 x \right )^{2} \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(257\) |
1/6972504*(75859740*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1 /2))*x^4+25286580*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2 ))*x^3-48465945*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2)) *x^2+29027880*x^3*(-10*x^2-x+3)^(1/2)-8428860*7^(1/2)*arctan(1/14*(37*x+20 )*7^(1/2)/(-10*x^2-x+3)^(1/2))*x-9476040*x^2*(-10*x^2-x+3)^(1/2)+8428860*7 ^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-13238806*x*(-10* x^2-x+3)^(1/2)+5480664*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)*(1-2*x)^(1/2)/(2 +3*x)^2/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {2107215 \, \sqrt {7} {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (2073420 \, x^{3} - 676860 \, x^{2} - 945629 \, x + 391476\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{6972504 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )}} \]
-1/6972504*(2107215*sqrt(7)*(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)*arctan(1/ 14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14 *(2073420*x^3 - 676860*x^2 - 945629*x + 391476)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)
\[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx=\int \frac {1}{\left (1 - 2 x\right )^{\frac {5}{2}} \left (3 x + 2\right )^{3} \sqrt {5 x + 3}}\, dx \]
\[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx=\int { \frac {1}{\sqrt {5 \, x + 3} {\left (3 \, x + 2\right )}^{3} {\left (-2 \, x + 1\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (104) = 208\).
Time = 0.45 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.12 \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {1161}{38416} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {32 \, {\left (367 \, \sqrt {5} {\left (5 \, x + 3\right )} - 2211 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{21789075 \, {\left (2 \, x - 1\right )}^{2}} + \frac {297 \, \sqrt {10} {\left (197 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {36680 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {146720 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{4802 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]
1161/38416*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)))) - 32/21789075*(367*sqrt(5)*(5*x + 3) - 2211*sqrt(5))*s qrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2 + 297/4802*sqrt(10)*(197*((sqrt(2 )*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqr t(-10*x + 5) - sqrt(22)))^3 + 36680*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/s qrt(5*x + 3) - 146720*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*sqrt(5*x + 3)/(sq rt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2
Timed out. \[ \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx=\int \frac {1}{{\left (1-2\,x\right )}^{5/2}\,{\left (3\,x+2\right )}^3\,\sqrt {5\,x+3}} \,d x \]