Integrand size = 25, antiderivative size = 147 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2 a^2 (a+b x)^{3/2}}{3 b^3 (a-c)}-\frac {4 a (a+b x)^{5/2}}{5 b^3 (a-c)}+\frac {2 (a+b x)^{7/2}}{7 b^3 (a-c)}-\frac {2 c^2 (c+b x)^{3/2}}{3 b^3 (a-c)}+\frac {4 c (c+b x)^{5/2}}{5 b^3 (a-c)}-\frac {2 (c+b x)^{7/2}}{7 b^3 (a-c)} \]
2/3*a^2*(b*x+a)^(3/2)/b^3/(a-c)-4/5*a*(b*x+a)^(5/2)/b^3/(a-c)+2/7*(b*x+a)^ (7/2)/b^3/(a-c)-2/3*c^2*(b*x+c)^(3/2)/b^3/(a-c)+4/5*c*(b*x+c)^(5/2)/b^3/(a -c)-2/7*(b*x+c)^(7/2)/b^3/(a-c)
Time = 0.47 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.95 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2 \left (8 a^3 \sqrt {a+b x}-4 a^2 b x \sqrt {a+b x}+3 a b^2 x^2 \sqrt {a+b x}-8 c^3 \sqrt {c+b x}+4 b c^2 x \sqrt {c+b x}-3 b^2 c x^2 \sqrt {c+b x}+15 b^3 x^3 \left (\sqrt {a+b x}-\sqrt {c+b x}\right )\right )}{105 b^3 (a-c)} \]
(2*(8*a^3*Sqrt[a + b*x] - 4*a^2*b*x*Sqrt[a + b*x] + 3*a*b^2*x^2*Sqrt[a + b *x] - 8*c^3*Sqrt[c + b*x] + 4*b*c^2*x*Sqrt[c + b*x] - 3*b^2*c*x^2*Sqrt[c + b*x] + 15*b^3*x^3*(Sqrt[a + b*x] - Sqrt[c + b*x])))/(105*b^3*(a - c))
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2529, 53, 2009}
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.
\(\displaystyle \int \frac {x^2}{\sqrt {a+b x}+\sqrt {b x+c}} \, dx\) |
\(\Big \downarrow \) 2529 |
\(\displaystyle \frac {\int x^2 \sqrt {a+b x}dx}{a-c}-\frac {\int x^2 \sqrt {c+b x}dx}{a-c}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (\frac {(a+b x)^{5/2}}{b^2}-\frac {2 a (a+b x)^{3/2}}{b^2}+\frac {a^2 \sqrt {a+b x}}{b^2}\right )dx}{a-c}-\frac {\int \left (\frac {(c+b x)^{5/2}}{b^2}-\frac {2 c (c+b x)^{3/2}}{b^2}+\frac {c^2 \sqrt {c+b x}}{b^2}\right )dx}{a-c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 a^2 (a+b x)^{3/2}}{3 b^3}+\frac {2 (a+b x)^{7/2}}{7 b^3}-\frac {4 a (a+b x)^{5/2}}{5 b^3}}{a-c}-\frac {\frac {2 c^2 (b x+c)^{3/2}}{3 b^3}+\frac {2 (b x+c)^{7/2}}{7 b^3}-\frac {4 c (b x+c)^{5/2}}{5 b^3}}{a-c}\) |
((2*a^2*(a + b*x)^(3/2))/(3*b^3) - (4*a*(a + b*x)^(5/2))/(5*b^3) + (2*(a + b*x)^(7/2))/(7*b^3))/(a - c) - ((2*c^2*(c + b*x)^(3/2))/(3*b^3) - (4*c*(c + b*x)^(5/2))/(5*b^3) + (2*(c + b*x)^(7/2))/(7*b^3))/(a - c)
3.5.1.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(u_)/((e_.)*Sqrt[(a_.) + (b_.)*(x_)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[-d/(e*(b*c - a*d)) Int[u*Sqrt[a + b*x], x], x] + Simp[ b/(f*(b*c - a*d)) Int[u*Sqrt[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f} , x] && NeQ[b*c - a*d, 0] && EqQ[b*e^2 - d*f^2, 0]
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.61
method | result | size |
default | \(\frac {\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {4 \left (b x +a \right )^{\frac {5}{2}} a}{5}+\frac {2 \left (b x +a \right )^{\frac {3}{2}} a^{2}}{3}}{\left (a -c \right ) b^{3}}-\frac {2 \left (\frac {\left (b x +c \right )^{\frac {7}{2}}}{7}-\frac {2 c \left (b x +c \right )^{\frac {5}{2}}}{5}+\frac {c^{2} \left (b x +c \right )^{\frac {3}{2}}}{3}\right )}{\left (a -c \right ) b^{3}}\) | \(90\) |
2/(a-c)/b^3*(1/7*(b*x+a)^(7/2)-2/5*(b*x+a)^(5/2)*a+1/3*(b*x+a)^(3/2)*a^2)- 2/(a-c)/b^3*(1/7*(b*x+c)^(7/2)-2/5*c*(b*x+c)^(5/2)+1/3*c^2*(b*x+c)^(3/2))
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2 \, {\left ({\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a} - {\left (15 \, b^{3} x^{3} + 3 \, b^{2} c x^{2} - 4 \, b c^{2} x + 8 \, c^{3}\right )} \sqrt {b x + c}\right )}}{105 \, {\left (a b^{3} - b^{3} c\right )}} \]
2/105*((15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x + a) - (15* b^3*x^3 + 3*b^2*c*x^2 - 4*b*c^2*x + 8*c^3)*sqrt(b*x + c))/(a*b^3 - b^3*c)
\[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\int \frac {x^{2}}{\sqrt {a + b x} + \sqrt {b x + c}}\, dx \]
\[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\int { \frac {x^{2}}{\sqrt {b x + a} + \sqrt {b x + c}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (123) = 246\).
Time = 0.36 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.65 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=-\frac {2}{105} \, {\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {5 \, {\left (a^{2} b^{9} - 2 \, a b^{9} c + b^{9} c^{2}\right )} {\left (b x + a\right )}}{a^{3} b^{12} - 3 \, a^{2} b^{12} c + 3 \, a b^{12} c^{2} - b^{12} c^{3}} - \frac {15 \, a^{3} b^{9} - 31 \, a^{2} b^{9} c + 17 \, a b^{9} c^{2} - b^{9} c^{3}}{a^{3} b^{12} - 3 \, a^{2} b^{12} c + 3 \, a b^{12} c^{2} - b^{12} c^{3}}\right )} + \frac {45 \, a^{4} b^{9} - 96 \, a^{3} b^{9} c + 53 \, a^{2} b^{9} c^{2} + 2 \, a b^{9} c^{3} - 4 \, b^{9} c^{4}}{a^{3} b^{12} - 3 \, a^{2} b^{12} c + 3 \, a b^{12} c^{2} - b^{12} c^{3}}\right )} {\left (b x + a\right )} - \frac {15 \, a^{5} b^{9} - 33 \, a^{4} b^{9} c + 17 \, a^{3} b^{9} c^{2} - 3 \, a^{2} b^{9} c^{3} + 12 \, a b^{9} c^{4} - 8 \, b^{9} c^{5}}{a^{3} b^{12} - 3 \, a^{2} b^{12} c + 3 \, a b^{12} c^{2} - b^{12} c^{3}}\right )} \sqrt {b x + c} + \frac {2 \, {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} - 42 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}\right )}}{105 \, {\left (a b^{3} - b^{3} c\right )}} \]
-2/105*((3*(b*x + a)*(5*(a^2*b^9 - 2*a*b^9*c + b^9*c^2)*(b*x + a)/(a^3*b^1 2 - 3*a^2*b^12*c + 3*a*b^12*c^2 - b^12*c^3) - (15*a^3*b^9 - 31*a^2*b^9*c + 17*a*b^9*c^2 - b^9*c^3)/(a^3*b^12 - 3*a^2*b^12*c + 3*a*b^12*c^2 - b^12*c^ 3)) + (45*a^4*b^9 - 96*a^3*b^9*c + 53*a^2*b^9*c^2 + 2*a*b^9*c^3 - 4*b^9*c^ 4)/(a^3*b^12 - 3*a^2*b^12*c + 3*a*b^12*c^2 - b^12*c^3))*(b*x + a) - (15*a^ 5*b^9 - 33*a^4*b^9*c + 17*a^3*b^9*c^2 - 3*a^2*b^9*c^3 + 12*a*b^9*c^4 - 8*b ^9*c^5)/(a^3*b^12 - 3*a^2*b^12*c + 3*a*b^12*c^2 - b^12*c^3))*sqrt(b*x + c) + 2/105*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a ^2)/(a*b^3 - b^3*c)
Time = 16.57 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.22 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2\,x^3\,\sqrt {a+b\,x}}{7\,\left (a-c\right )}-\frac {2\,x^3\,\sqrt {c+b\,x}}{7\,\left (a-c\right )}+\frac {16\,a^3\,\sqrt {a+b\,x}}{105\,b^3\,\left (a-c\right )}-\frac {16\,c^3\,\sqrt {c+b\,x}}{105\,b^3\,\left (a-c\right )}+\frac {2\,a\,x^2\,\sqrt {a+b\,x}}{35\,b\,\left (a-c\right )}-\frac {8\,a^2\,x\,\sqrt {a+b\,x}}{105\,b^2\,\left (a-c\right )}-\frac {2\,c\,x^2\,\sqrt {c+b\,x}}{35\,b\,\left (a-c\right )}+\frac {8\,c^2\,x\,\sqrt {c+b\,x}}{105\,b^2\,\left (a-c\right )} \]
(2*x^3*(a + b*x)^(1/2))/(7*(a - c)) - (2*x^3*(c + b*x)^(1/2))/(7*(a - c)) + (16*a^3*(a + b*x)^(1/2))/(105*b^3*(a - c)) - (16*c^3*(c + b*x)^(1/2))/(1 05*b^3*(a - c)) + (2*a*x^2*(a + b*x)^(1/2))/(35*b*(a - c)) - (8*a^2*x*(a + b*x)^(1/2))/(105*b^2*(a - c)) - (2*c*x^2*(c + b*x)^(1/2))/(35*b*(a - c)) + (8*c^2*x*(c + b*x)^(1/2))/(105*b^2*(a - c))