Integrand size = 26, antiderivative size = 79 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\frac {4}{77 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {370 \sqrt {1-2 x}}{847 \sqrt {3+5 x}}+\frac {18 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}} \]
18/49*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+4/77/(1-2*x) ^(1/2)/(3+5*x)^(1/2)-370/847*(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\frac {2 (-163+370 x)}{847 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {18 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}} \]
(2*(-163 + 370*x))/(847*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (18*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {115, 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.
\(\displaystyle \int \frac {1}{(1-2 x)^{3/2} (3 x+2) (5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 115 |
\(\displaystyle \frac {4}{77 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {2}{77} \int -\frac {60 x+73}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{77} \int \frac {60 x+73}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx+\frac {4}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{77} \left (-\frac {2}{11} \int \frac {1089}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {370 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {4}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{77} \left (-99 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {370 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {4}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {1}{77} \left (-198 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {370 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {4}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{77} \left (\frac {198 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {370 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )+\frac {4}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\) |
4/(77*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + ((-370*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5* x]) + (198*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7])/77
3.26.67.3.1 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. \(153\) vs. \(2(58)=116\).
Time = 1.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.95
method | result | size |
default | \(-\frac {\sqrt {1-2 x}\, \left (10890 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+1089 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -3267 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+5180 x \sqrt {-10 x^{2}-x +3}-2282 \sqrt {-10 x^{2}-x +3}\right )}{5929 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(154\) |
-1/5929*(1-2*x)^(1/2)*(10890*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^ 2-x+3)^(1/2))*x^2+1089*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3) ^(1/2))*x-3267*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+ 5180*x*(-10*x^2-x+3)^(1/2)-2282*(-10*x^2-x+3)^(1/2))/(-1+2*x)/(-10*x^2-x+3 )^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\frac {1089 \, \sqrt {7} {\left (10 \, x^{2} + x - 3\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (370 \, x - 163\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{5929 \, {\left (10 \, x^{2} + x - 3\right )}} \]
1/5929*(1089*sqrt(7)*(10*x^2 + x - 3)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt (5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(370*x - 163)*sqrt(5*x + 3 )*sqrt(-2*x + 1))/(10*x^2 + x - 3)
\[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\int \frac {1}{\left (1 - 2 x\right )^{\frac {3}{2}} \cdot \left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {9}{49} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {740 \, x}{847 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {326}{847 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-9/49*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 740/847* x/sqrt(-10*x^2 - x + 3) - 326/847/sqrt(-10*x^2 - x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (58) = 116\).
Time = 0.31 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.01 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx=-\frac {9}{490} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {5}{242} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {8 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{4235 \, {\left (2 \, x - 1\right )}} \]
-9/490*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sq rt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5 ) - sqrt(22)))) - 5/242*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqr t(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) - 8/423 5*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx=\int \frac {1}{{\left (1-2\,x\right )}^{3/2}\,\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{3/2}} \,d x \]