Integrand size = 26, antiderivative size = 77 \[ \int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {10 \sqrt {1-2 x}}{33 (3+5 x)^{3/2}}+\frac {950 \sqrt {1-2 x}}{363 \sqrt {3+5 x}}-\frac {18 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{\sqrt {7}} \]
-18/7*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-10/33*(1-2*x )^(1/2)/(3+5*x)^(3/2)+950/363*(1-2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx=\frac {10 \sqrt {1-2 x} (274+475 x)}{363 (3+5 x)^{3/2}}-\frac {18 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{\sqrt {7}} \]
(10*Sqrt[1 - 2*x]*(274 + 475*x))/(363*(3 + 5*x)^(3/2)) - (18*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int \frac {1}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 115 |
\(\displaystyle -\frac {2}{33} \int \frac {59-60 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {10 \sqrt {1-2 x}}{33 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{33} \int \frac {59-60 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {10 \sqrt {1-2 x}}{33 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {1}{33} \left (\frac {2}{11} \int \frac {3267}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {950 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {10 \sqrt {1-2 x}}{33 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{33} \left (297 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {950 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {10 \sqrt {1-2 x}}{33 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {1}{33} \left (594 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}+\frac {950 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {10 \sqrt {1-2 x}}{33 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{33} \left (\frac {950 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}-\frac {594 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}\right )-\frac {10 \sqrt {1-2 x}}{33 (5 x+3)^{3/2}}\) |
(-10*Sqrt[1 - 2*x])/(33*(3 + 5*x)^(3/2)) + ((950*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) - (594*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7])/33
3.26.13.3.1 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. \(146\) vs. \(2(58)=116\).
Time = 3.97 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.91
method | result | size |
default | \(\frac {\left (81675 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+98010 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +29403 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+33250 x \sqrt {-10 x^{2}-x +3}+19180 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{2541 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(147\) |
1/2541*(81675*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x ^2+98010*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+2940 3*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+33250*x*(-10* x^2-x+3)^(1/2)+19180*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2 )/(3+5*x)^(3/2)
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {3267 \, \sqrt {7} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 70 \, {\left (475 \, x + 274\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{2541 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
-1/2541*(3267*sqrt(7)*(25*x^2 + 30*x + 9)*arctan(1/14*sqrt(7)*(37*x + 20)* sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 70*(475*x + 274)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx=\int \frac {1}{\sqrt {1 - 2 x} \left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx=\int { \frac {1}{{\left (5 \, x + 3\right )}^{\frac {5}{2}} {\left (3 \, x + 2\right )} \sqrt {-2 \, x + 1}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (58) = 116\).
Time = 0.31 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.52 \[ \int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {1}{5808} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {9}{70} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {31}{242} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} \]
-1/5808*sqrt(10)*((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)))^3 + 9/70*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)))) + 31/ 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)))
Timed out. \[ \int \frac {1}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx=\int \frac {1}{\sqrt {1-2\,x}\,\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{5/2}} \,d x \]