Integrand size = 26, antiderivative size = 64 \[ \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx=-\frac {\sqrt {1-2 x} \sqrt {3+5 x}}{7 (2+3 x)}-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}} \]
-11/49*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-1/7*(1-2*x) ^(1/2)*(3+5*x)^(1/2)/(2+3*x)
Leaf count is larger than twice the leaf count of optimal. \(130\) vs. \(2(64)=128\).
Time = 1.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.03 \[ \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {1}{49} \left (-\frac {7 \sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x}+11 \sqrt {7} \arctan \left (\frac {\sqrt {2 \left (34+\sqrt {1155}\right )} \sqrt {3+5 x}}{-\sqrt {11}+\sqrt {5-10 x}}\right )+11 \sqrt {7} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {34+\sqrt {1155}} \left (-\sqrt {11}+\sqrt {5-10 x}\right )}\right )\right ) \]
((-7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x) + 11*Sqrt[7]*ArcTan[(Sqrt[2*(3 4 + Sqrt[1155])]*Sqrt[3 + 5*x])/(-Sqrt[11] + Sqrt[5 - 10*x])] + 11*Sqrt[7] *ArcTan[Sqrt[6 + 10*x]/(Sqrt[34 + Sqrt[1155]]*(-Sqrt[11] + Sqrt[5 - 10*x]) )])/49
Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {11}{14} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {11}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\) |
-1/7*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2 + 3*x) - (11*ArcTan[Sqrt[1 - 2*x]/(S qrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])
3.25.64.3.1 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. \(107\) vs. \(2(49)=98\).
Time = 1.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.69
method | result | size |
default | \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (33 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +22 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-14 \sqrt {-10 x^{2}-x +3}\right )}{98 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )}\) | \(108\) |
risch | \(\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{7 \left (2+3 x \right ) \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {11 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{98 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(114\) |
1/98*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(33*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2) /(-10*x^2-x+3)^(1/2))*x+22*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2- x+3)^(1/2))-14*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(2+3*x)
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx=-\frac {11 \, \sqrt {7} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 14 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{98 \, {\left (3 \, x + 2\right )}} \]
-1/98*(11*sqrt(7)*(3*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)* sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 14*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)
\[ \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\int \frac {\sqrt {5 x + 3}}{\sqrt {1 - 2 x} \left (3 x + 2\right )^{2}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {11}{98} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {\sqrt {-10 \, x^{2} - x + 3}}{7 \, {\left (3 \, x + 2\right )}} \]
11/98*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 1/7*sqrt (-10*x^2 - x + 3)/(3*x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (49) = 98\).
Time = 0.35 (sec) , antiderivative size = 193, normalized size of antiderivative = 3.02 \[ \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {11}{980} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {22 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{7 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \]
11/980*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)))) - 22/7*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)))/(((sqrt( 2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sq rt(-10*x + 5) - sqrt(22)))^2 + 280)
Time = 7.60 (sec) , antiderivative size = 716, normalized size of antiderivative = 11.19 \[ \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx=\frac {4\,\left (\sqrt {1-2\,x}-1\right )}{35\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )}-\frac {2\,{\left (\sqrt {1-2\,x}-1\right )}^3}{7\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )}-\frac {11\,\sqrt {7}\,\mathrm {atan}\left (\frac {2904\,\sqrt {3}\,\sqrt {7}}{42875\,\left (\frac {2904\,{\left (\sqrt {1-2\,x}-1\right )}^2}{1715\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {53724\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {5808}{8575}\right )}+\frac {1452\,\sqrt {7}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )\,\left (\frac {2904\,{\left (\sqrt {1-2\,x}-1\right )}^2}{1715\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {53724\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {5808}{8575}\right )}-\frac {1452\,\sqrt {3}\,\sqrt {7}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{8575\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2\,\left (\frac {2904\,{\left (\sqrt {1-2\,x}-1\right )}^2}{1715\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {53724\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{42875\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {5808}{8575}\right )}\right )}{49}-\frac {37\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{175\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2\,\left (\frac {14\,{\left (\sqrt {1-2\,x}-1\right )}^2}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {6\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {12\,\sqrt {3}\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {4}{25}\right )} \]
(4*((1 - 2*x)^(1/2) - 1))/(35*(3^(1/2) - (5*x + 3)^(1/2))*((14*((1 - 2*x)^ (1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4 /(3^(1/2) - (5*x + 3)^(1/2))^4 + (6*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(5*(3 ^(1/2) - (5*x + 3)^(1/2))^3) - (12*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(25*(3^( 1/2) - (5*x + 3)^(1/2))) + 4/25)) - (2*((1 - 2*x)^(1/2) - 1)^3)/(7*(3^(1/2 ) - (5*x + 3)^(1/2))^3*((14*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + (6 *3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (12* 3^(1/2)*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + 4/25)) - (11*7^(1/2)*atan((2904*3^(1/2)*7^(1/2))/(42875*((2904*((1 - 2*x)^(1/2) - 1)^2)/(1715*(3^(1/2) - (5*x + 3)^(1/2))^2) + (53724*3^(1/2)*((1 - 2*x)^(1/ 2) - 1))/(42875*(3^(1/2) - (5*x + 3)^(1/2))) - 5808/8575)) + (1452*7^(1/2) *((1 - 2*x)^(1/2) - 1))/(42875*(3^(1/2) - (5*x + 3)^(1/2))*((2904*((1 - 2* x)^(1/2) - 1)^2)/(1715*(3^(1/2) - (5*x + 3)^(1/2))^2) + (53724*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(42875*(3^(1/2) - (5*x + 3)^(1/2))) - 5808/8575)) - (1 452*3^(1/2)*7^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(8575*(3^(1/2) - (5*x + 3)^(1 /2))^2*((2904*((1 - 2*x)^(1/2) - 1)^2)/(1715*(3^(1/2) - (5*x + 3)^(1/2))^2 ) + (53724*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(42875*(3^(1/2) - (5*x + 3)^(1/2 ))) - 5808/8575))))/49 - (37*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(175*(3^(1/2 ) - (5*x + 3)^(1/2))^2*((14*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*...