Integrand size = 26, antiderivative size = 115 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {1815 \sqrt {1-2 x}}{4 \sqrt {3+5 x}}+\frac {(1-2 x)^{5/2}}{2 (2+3 x)^2 \sqrt {3+5 x}}+\frac {55 (1-2 x)^{3/2}}{4 (2+3 x) \sqrt {3+5 x}}+\frac {1815}{4} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]
1815/4*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+1/2*(1-2*x) ^(5/2)/(2+3*x)^2/(3+5*x)^(1/2)+55/4*(1-2*x)^(3/2)/(2+3*x)/(3+5*x)^(1/2)-18 15/4*(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {\sqrt {1-2 x} \left (7148+21843 x+16657 x^2\right )}{4 (2+3 x)^2 \sqrt {3+5 x}}+\frac {1815}{4} \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]
-1/4*(Sqrt[1 - 2*x]*(7148 + 21843*x + 16657*x^2))/((2 + 3*x)^2*Sqrt[3 + 5* x]) + (1815*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/4
Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {105, 105, 105, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2)^3 (5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {55}{4} \int \frac {(1-2 x)^{3/2}}{(3 x+2)^2 (5 x+3)^{3/2}}dx+\frac {(1-2 x)^{5/2}}{2 (3 x+2)^2 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {55}{4} \left (\frac {33}{2} \int \frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}dx+\frac {(1-2 x)^{3/2}}{(3 x+2) \sqrt {5 x+3}}\right )+\frac {(1-2 x)^{5/2}}{2 (3 x+2)^2 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {55}{4} \left (\frac {33}{2} \left (-7 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {(1-2 x)^{3/2}}{(3 x+2) \sqrt {5 x+3}}\right )+\frac {(1-2 x)^{5/2}}{2 (3 x+2)^2 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {55}{4} \left (\frac {33}{2} \left (-14 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {(1-2 x)^{3/2}}{(3 x+2) \sqrt {5 x+3}}\right )+\frac {(1-2 x)^{5/2}}{2 (3 x+2)^2 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {55}{4} \left (\frac {33}{2} \left (2 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {(1-2 x)^{3/2}}{(3 x+2) \sqrt {5 x+3}}\right )+\frac {(1-2 x)^{5/2}}{2 (3 x+2)^2 \sqrt {5 x+3}}\) |
(1 - 2*x)^(5/2)/(2*(2 + 3*x)^2*Sqrt[3 + 5*x]) + (55*((1 - 2*x)^(3/2)/((2 + 3*x)*Sqrt[3 + 5*x]) + (33*((-2*Sqrt[1 - 2*x])/Sqrt[3 + 5*x] + 2*Sqrt[7]*A rcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])]))/2))/4
3.25.43.3.1 Defintions of rubi rules used
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[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
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. \(2(88)=176\).
Time = 3.87 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.76
method | result | size |
default | \(-\frac {\left (81675 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}+157905 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+101640 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +33314 x^{2} \sqrt {-10 x^{2}-x +3}+21780 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+43686 x \sqrt {-10 x^{2}-x +3}+14296 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{8 \left (2+3 x \right )^{2} \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(202\) |
-1/8*(81675*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3 +157905*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+101 640*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+33314*x^2 *(-10*x^2-x+3)^(1/2)+21780*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2- x+3)^(1/2))+43686*x*(-10*x^2-x+3)^(1/2)+14296*(-10*x^2-x+3)^(1/2))*(1-2*x) ^(1/2)/(2+3*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 = 0.88 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx=\frac {1815 \, \sqrt {7} {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 2 \, {\left (16657 \, x^{2} + 21843 \, x + 7148\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{8 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} \]
1/8*(1815*sqrt(7)*(45*x^3 + 87*x^2 + 56*x + 12)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 2*(16657*x^2 + 2184 3*x + 7148)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(45*x^3 + 87*x^2 + 56*x + 12)
\[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}}}{\left (3 x + 2\right )^{3} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {1815}{8} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {16657 \, x}{18 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {52169}{108 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {343}{54 \, {\left (9 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt {-10 \, x^{2} - x + 3} x + 4 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} + \frac {833}{12 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]
-1815/8*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 16657/ 18*x/sqrt(-10*x^2 - x + 3) - 52169/108/sqrt(-10*x^2 - x + 3) + 343/54/(9*s qrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sqrt(-10*x^2 - x + 3)) + 833/12/(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. 311 vs. \(2 (88) = 176\).
Time = 0.43 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.70 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {363}{16} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {121}{10} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {847 \, \sqrt {10} {\left (9 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {1960 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {7840 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{2 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]
-363/16*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((s qrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 121/10*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/s qrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) - 847 /2*sqrt(10)*(9*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqr t(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 1960*(sqrt(2)*sqrt(-1 0*x + 5) - sqrt(22))/sqrt(5*x + 3) - 7840*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)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2
Timed out. \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}}{{\left (3\,x+2\right )}^3\,{\left (5\,x+3\right )}^{3/2}} \,d x \]