Integrand size = 26, antiderivative size = 93 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=-\frac {11 \sqrt {1-2 x} \sqrt {3+5 x}}{28 (2+3 x)}+\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{2 (2+3 x)^2}-\frac {121 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{28 \sqrt {7}} \]
-121/196*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+1/2*(3+5* x)^(3/2)*(1-2*x)^(1/2)/(2+3*x)^2-11/28*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)
Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\frac {1}{196} \left (\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (20+37 x)}{(2+3 x)^2}-121 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\right ) \]
((7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(20 + 37*x))/(2 + 3*x)^2 - 121*Sqrt[7]*Arc Tan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/196
Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {1-2 x} \sqrt {5 x+3}}{(3 x+2)^3} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {11}{4} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x} (3 x+2)^2}dx+\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{2 (3 x+2)^2}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {11}{4} \left (\frac {11}{14} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )+\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{2 (3 x+2)^2}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {11}{4} \left (\frac {11}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )+\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{2 (3 x+2)^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {11}{4} \left (-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{7 (3 x+2)}\right )+\frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{2 (3 x+2)^2}\) |
(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(2*(2 + 3*x)^2) + (11*(-1/7*(Sqrt[1 - 2*x] *Sqrt[3 + 5*x])/(2 + 3*x) - (11*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x ])])/(7*Sqrt[7])))/4
3.23.69.3.1 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])
Time = 1.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28
method | result | size |
risch | \(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (37 x +20\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{28 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {121 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{392 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(119\) |
default | \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (1089 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+1452 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +484 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+518 x \sqrt {-10 x^{2}-x +3}+280 \sqrt {-10 x^{2}-x +3}\right )}{392 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\) | \(154\) |
-1/28*(-1+2*x)*(3+5*x)^(1/2)*(37*x+20)/(2+3*x)^2/(-(-1+2*x)*(3+5*x))^(1/2) *((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)+121/392*7^(1/2)*arctan(9/14*(20/3+3 7/3*x)*7^(1/2)/(-90*(2/3+x)^2+67+111*x)^(1/2))*((1-2*x)*(3+5*x))^(1/2)/(1- 2*x)^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=-\frac {121 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{392 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]
-1/392*(121*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqr t(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(37*x + 20)*sqrt(5*x + 3) *sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)
\[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\int \frac {\sqrt {1 - 2 x} \sqrt {5 x + 3}}{\left (3 x + 2\right )^{3}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\frac {121}{392} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {5}{21} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{14 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {37 \, \sqrt {-10 \, x^{2} - x + 3}}{84 \, {\left (3 \, x + 2\right )}} \]
121/392*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 5/21*s qrt(-10*x^2 - x + 3) + 3/14*(-10*x^2 - x + 3)^(3/2)/(9*x^2 + 12*x + 4) - 3 7/84*sqrt(-10*x^2 - x + 3)/(3*x + 2)
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (72) = 144\).
Time = 0.39 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.69 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\frac {121}{3920} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {121 \, \sqrt {10} {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - \frac {280 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {1120 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{14 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]
121/3920*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)))) - 121/14*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)))^3 - 280*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 1120*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
Time = 13.70 (sec) , antiderivative size = 1037, normalized size of antiderivative = 11.15 \[ \int \frac {\sqrt {1-2 x} \sqrt {3+5 x}}{(2+3 x)^3} \, dx=\text {Too large to display} \]
((2129*((1 - 2*x)^(1/2) - 1)^5)/(875*(3^(1/2) - (5*x + 3)^(1/2))^5) - (425 8*((1 - 2*x)^(1/2) - 1)^3)/(4375*(3^(1/2) - (5*x + 3)^(1/2))^3) - (158*((1 - 2*x)^(1/2) - 1))/(4375*(3^(1/2) - (5*x + 3)^(1/2))) + (79*((1 - 2*x)^(1 /2) - 1)^7)/(140*(3^(1/2) - (5*x + 3)^(1/2))^7) + (991*3^(1/2)*((1 - 2*x)^ (1/2) - 1)^2)/(4375*(3^(1/2) - (5*x + 3)^(1/2))^2) - (376*3^(1/2)*((1 - 2* x)^(1/2) - 1)^4)/(875*(3^(1/2) - (5*x + 3)^(1/2))^4) + (991*3^(1/2)*((1 - 2*x)^(1/2) - 1)^6)/(700*(3^(1/2) - (5*x + 3)^(1/2))^6))/((544*((1 - 2*x)^( 1/2) - 1)^2)/(625*(3^(1/2) - (5*x + 3)^(1/2))^2) - (1764*((1 - 2*x)^(1/2) - 1)^4)/(625*(3^(1/2) - (5*x + 3)^(1/2))^4) + (136*((1 - 2*x)^(1/2) - 1)^6 )/(25*(3^(1/2) - (5*x + 3)^(1/2))^6) + ((1 - 2*x)^(1/2) - 1)^8/(3^(1/2) - (5*x + 3)^(1/2))^8 - (96*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(625*(3^(1/2) - (5*x + 3)^(1/2))^3) + (48*3^(1/2)*((1 - 2*x)^(1/2) - 1)^5)/(125*(3^(1/2) - (5*x + 3)^(1/2))^5) + (12*3^(1/2)*((1 - 2*x)^(1/2) - 1)^7)/(5*(3^(1/2) - (5*x + 3)^(1/2))^7) - (96*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(625*(3^(1/2) - ( 5*x + 3)^(1/2))) + 16/625) - (121*7^(1/2)*atan(((121*7^(1/2)*((726*3^(1/2) )/875 + (363*((1 - 2*x)^(1/2) - 1))/(875*(3^(1/2) - (5*x + 3)^(1/2))) - (7 ^(1/2)*((212*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 3)^(1/2))^2) + (888*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(125*(3^(1/2) - (5*x + 3)^(1/2))) - 5 36/125)*121i)/392 - (363*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(175*(3^(1/2) - (5*x + 3)^(1/2))^2)))/392 + (121*7^(1/2)*((726*3^(1/2))/875 + (363*((1 ...