Integrand size = 25, antiderivative size = 375 \[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=-\frac {8 a^3 (a+b x)^{3/2}}{3 b^3 (a-c)^3}+\frac {2 a^2 (a+3 c) (a+b x)^{3/2}}{3 b^3 (a-c)^3}+\frac {24 a^2 (a+b x)^{5/2}}{5 b^3 (a-c)^3}-\frac {4 a (a+3 c) (a+b x)^{5/2}}{5 b^3 (a-c)^3}-\frac {24 a (a+b x)^{7/2}}{7 b^3 (a-c)^3}+\frac {2 (a+3 c) (a+b x)^{7/2}}{7 b^3 (a-c)^3}+\frac {8 (a+b x)^{9/2}}{9 b^3 (a-c)^3}+\frac {8 c^3 (c+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac {2 c^2 (3 a+c) (c+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac {24 c^2 (c+b x)^{5/2}}{5 b^3 (a-c)^3}+\frac {4 c (3 a+c) (c+b x)^{5/2}}{5 b^3 (a-c)^3}+\frac {24 c (c+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac {2 (3 a+c) (c+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac {8 (c+b x)^{9/2}}{9 b^3 (a-c)^3} \] Output:
-8/3*a^3*(b*x+a)^(3/2)/b^3/(a-c)^3+2/3*a^2*(a+3*c)*(b*x+a)^(3/2)/b^3/(a-c) ^3+24/5*a^2*(b*x+a)^(5/2)/b^3/(a-c)^3-4/5*a*(a+3*c)*(b*x+a)^(5/2)/b^3/(a-c )^3-24/7*a*(b*x+a)^(7/2)/b^3/(a-c)^3+2/7*(a+3*c)*(b*x+a)^(7/2)/b^3/(a-c)^3 +8/9*(b*x+a)^(9/2)/b^3/(a-c)^3+8/3*c^3*(b*x+c)^(3/2)/b^3/(a-c)^3-2/3*c^2*( 3*a+c)*(b*x+c)^(3/2)/b^3/(a-c)^3-24/5*c^2*(b*x+c)^(5/2)/b^3/(a-c)^3+4/5*c* (3*a+c)*(b*x+c)^(5/2)/b^3/(a-c)^3+24/7*c*(b*x+c)^(7/2)/b^3/(a-c)^3-2/7*(3* a+c)*(b*x+c)^(7/2)/b^3/(a-c)^3-8/9*(b*x+c)^(9/2)/b^3/(a-c)^3
Time = 1.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.37 \[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=\frac {2 \left ((a+b x)^{3/2} \left (-40 a^3+12 a^2 (6 c+5 b x)-3 a b x (36 c+25 b x)+5 b^2 x^2 (27 c+28 b x)\right )+(c+b x)^{3/2} \left (-9 a \left (8 c^2-12 b c x+15 b^2 x^2\right )+5 \left (8 c^3-12 b c^2 x+15 b^2 c x^2-28 b^3 x^3\right )\right )\right )}{315 b^3 (a-c)^3} \] Input:
Integrate[x^2/(Sqrt[a + b*x] + Sqrt[c + b*x])^3,x]
Output:
(2*((a + b*x)^(3/2)*(-40*a^3 + 12*a^2*(6*c + 5*b*x) - 3*a*b*x*(36*c + 25*b *x) + 5*b^2*x^2*(27*c + 28*b*x)) + (c + b*x)^(3/2)*(-9*a*(8*c^2 - 12*b*c*x + 15*b^2*x^2) + 5*(8*c^3 - 12*b*c^2*x + 15*b^2*c*x^2 - 28*b^3*x^3))))/(31 5*b^3*(a - c)^3)
Time = 0.90 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.76, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {7240, 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.
| \(\displaystyle \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {b x+c}\right )^3} \, dx\) |
| \(\Big \downarrow \) 7240 |
| \(\displaystyle \frac {\int \left (4 b \sqrt {a+b x} x^3-4 b \sqrt {c+b x} x^3+(a+3 c) \sqrt {a+b x} x^2-(3 a+c) \sqrt {c+b x} x^2\right )dx}{(a-c)^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {8 a^3 (a+b x)^{3/2}}{3 b^3}+\frac {2 a^2 (a+3 c) (a+b x)^{3/2}}{3 b^3}+\frac {24 a^2 (a+b x)^{5/2}}{5 b^3}-\frac {2 c^2 (3 a+c) (b x+c)^{3/2}}{3 b^3}+\frac {2 (a+3 c) (a+b x)^{7/2}}{7 b^3}-\frac {4 a (a+3 c) (a+b x)^{5/2}}{5 b^3}-\frac {2 (3 a+c) (b x+c)^{7/2}}{7 b^3}+\frac {4 c (3 a+c) (b x+c)^{5/2}}{5 b^3}+\frac {8 (a+b x)^{9/2}}{9 b^3}-\frac {24 a (a+b x)^{7/2}}{7 b^3}+\frac {8 c^3 (b x+c)^{3/2}}{3 b^3}-\frac {24 c^2 (b x+c)^{5/2}}{5 b^3}-\frac {8 (b x+c)^{9/2}}{9 b^3}+\frac {24 c (b x+c)^{7/2}}{7 b^3}}{(a-c)^3}\) |
Input:
Int[x^2/(Sqrt[a + b*x] + Sqrt[c + b*x])^3,x]
Output:
((-8*a^3*(a + b*x)^(3/2))/(3*b^3) + (2*a^2*(a + 3*c)*(a + b*x)^(3/2))/(3*b ^3) + (24*a^2*(a + b*x)^(5/2))/(5*b^3) - (4*a*(a + 3*c)*(a + b*x)^(5/2))/( 5*b^3) - (24*a*(a + b*x)^(7/2))/(7*b^3) + (2*(a + 3*c)*(a + b*x)^(7/2))/(7 *b^3) + (8*(a + b*x)^(9/2))/(9*b^3) + (8*c^3*(c + b*x)^(3/2))/(3*b^3) - (2 *c^2*(3*a + c)*(c + b*x)^(3/2))/(3*b^3) - (24*c^2*(c + b*x)^(5/2))/(5*b^3) + (4*c*(3*a + c)*(c + b*x)^(5/2))/(5*b^3) + (24*c*(c + b*x)^(7/2))/(7*b^3 ) - (2*(3*a + c)*(c + b*x)^(7/2))/(7*b^3) - (8*(c + b*x)^(9/2))/(9*b^3))/( a - c)^3
Int[(u_.)*((e_.)*Sqrt[(a_.) + (b_.)*(x_)^(n_.)] + (f_.)*Sqrt[(c_.) + (d_.)* (x_)^(n_.)])^(m_), x_Symbol] :> Simp[(a*e^2 - c*f^2)^m Int[ExpandIntegran d[u/(e*Sqrt[a + b*x^n] - f*Sqrt[c + d*x^n])^m, x], x], x] /; FreeQ[{a, b, c , d, e, f, n}, x] && ILtQ[m, 0] && EqQ[b*e^2 - d*f^2, 0]
Time = 0.02 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.78
| method | result | size |
| default | \(\frac {2 a \left (\frac {\left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {a^{2} \left (b x +a \right )^{\frac {3}{2}}}{3}\right )}{\left (a -c \right )^{3} b^{3}}+\frac {6 c \left (\frac {\left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {a^{2} \left (b x +a \right )^{\frac {3}{2}}}{3}\right )}{\left (a -c \right )^{3} b^{3}}-\frac {6 a \left (\frac {\left (b x +c \right )^{\frac {7}{2}}}{7}-\frac {2 c \left (b x +c \right )^{\frac {5}{2}}}{5}+\frac {c^{2} \left (b x +c \right )^{\frac {3}{2}}}{3}\right )}{\left (a -c \right )^{3} b^{3}}-\frac {2 c \left (\frac {\left (b x +c \right )^{\frac {7}{2}}}{7}-\frac {2 c \left (b x +c \right )^{\frac {5}{2}}}{5}+\frac {c^{2} \left (b x +c \right )^{\frac {3}{2}}}{3}\right )}{\left (a -c \right )^{3} b^{3}}+\frac {\frac {8 \left (b x +a \right )^{\frac {9}{2}}}{9}-\frac {24 a \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {24 a^{2} \left (b x +a \right )^{\frac {5}{2}}}{5}-\frac {8 a^{3} \left (b x +a \right )^{\frac {3}{2}}}{3}}{\left (a -c \right )^{3} b^{3}}-\frac {8 \left (\frac {\left (b x +c \right )^{\frac {9}{2}}}{9}-\frac {3 c \left (b x +c \right )^{\frac {7}{2}}}{7}+\frac {3 c^{2} \left (b x +c \right )^{\frac {5}{2}}}{5}-\frac {c^{3} \left (b x +c \right )^{\frac {3}{2}}}{3}\right )}{\left (a -c \right )^{3} b^{3}}\) | \(294\) |
Input:
int(x^2/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x,method=_RETURNVERBOSE)
Output:
2/(a-c)^3*a/b^3*(1/7*(b*x+a)^(7/2)-2/5*a*(b*x+a)^(5/2)+1/3*a^2*(b*x+a)^(3/ 2))+6/(a-c)^3*c/b^3*(1/7*(b*x+a)^(7/2)-2/5*a*(b*x+a)^(5/2)+1/3*a^2*(b*x+a) ^(3/2))-6/(a-c)^3*a/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))-2/(a-c)^3*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))+8/(a-c)^3/b^3*(1/9*(b*x+a)^(9/2)-3/7*a*(b*x+a)^(7/2)+3/5*a ^2*(b*x+a)^(5/2)-1/3*a^3*(b*x+a)^(3/2))-8/(a-c)^3/b^3*(1/9*(b*x+c)^(9/2)-3 /7*c*(b*x+c)^(7/2)+3/5*c^2*(b*x+c)^(5/2)-1/3*c^3*(b*x+c)^(3/2))
Time = 0.07 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.55 \[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=\frac {2 \, {\left ({\left (140 \, b^{4} x^{4} - 40 \, a^{4} + 72 \, a^{3} c + 5 \, {\left (13 \, a b^{3} + 27 \, b^{3} c\right )} x^{3} - 3 \, {\left (5 \, a^{2} b^{2} - 9 \, a b^{2} c\right )} x^{2} + 4 \, {\left (5 \, a^{3} b - 9 \, a^{2} b c\right )} x\right )} \sqrt {b x + a} - {\left (140 \, b^{4} x^{4} + 72 \, a c^{3} - 40 \, c^{4} + 5 \, {\left (27 \, a b^{3} + 13 \, b^{3} c\right )} x^{3} + 3 \, {\left (9 \, a b^{2} c - 5 \, b^{2} c^{2}\right )} x^{2} - 4 \, {\left (9 \, a b c^{2} - 5 \, b c^{3}\right )} x\right )} \sqrt {b x + c}\right )}}{315 \, {\left (a^{3} b^{3} - 3 \, a^{2} b^{3} c + 3 \, a b^{3} c^{2} - b^{3} c^{3}\right )}} \] Input:
integrate(x^2/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x, algorithm="fricas")
Output:
2/315*((140*b^4*x^4 - 40*a^4 + 72*a^3*c + 5*(13*a*b^3 + 27*b^3*c)*x^3 - 3* (5*a^2*b^2 - 9*a*b^2*c)*x^2 + 4*(5*a^3*b - 9*a^2*b*c)*x)*sqrt(b*x + a) - ( 140*b^4*x^4 + 72*a*c^3 - 40*c^4 + 5*(27*a*b^3 + 13*b^3*c)*x^3 + 3*(9*a*b^2 *c - 5*b^2*c^2)*x^2 - 4*(9*a*b*c^2 - 5*b*c^3)*x)*sqrt(b*x + c))/(a^3*b^3 - 3*a^2*b^3*c + 3*a*b^3*c^2 - b^3*c^3)
\[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=\int \frac {x^{2}}{\left (\sqrt {a + b x} + \sqrt {b x + c}\right )^{3}}\, dx \] Input:
integrate(x**2/((b*x+a)**(1/2)+(b*x+c)**(1/2))**3,x)
Output:
Integral(x**2/(sqrt(a + b*x) + sqrt(b*x + c))**3, x)
\[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=\int { \frac {x^{2}}{{\left (\sqrt {b x + a} + \sqrt {b x + c}\right )}^{3}} \,d x } \] Input:
integrate(x^2/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x, algorithm="maxima")
Output:
integrate(x^2/(sqrt(b*x + a) + sqrt(b*x + c))^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 1443 vs. \(2 (319) = 638\).
Time = 0.25 (sec) , antiderivative size = 1443, normalized size of antiderivative = 3.85 \[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(x^2/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x, algorithm="giac")
Output:
-2/315*((((5*(b*x + a)*(28*(a^9*b^4 - 9*a^8*b^4*c + 36*a^7*b^4*c^2 - 84*a^ 6*b^4*c^3 + 126*a^5*b^4*c^4 - 126*a^4*b^4*c^5 + 84*a^3*b^4*c^6 - 36*a^2*b^ 4*c^7 + 9*a*b^4*c^8 - b^4*c^9)*(b*x + a)/(a^12*b^5 - 12*a^11*b^5*c + 66*a^ 10*b^5*c^2 - 220*a^9*b^5*c^3 + 495*a^8*b^5*c^4 - 792*a^7*b^5*c^5 + 924*a^6 *b^5*c^6 - 792*a^5*b^5*c^7 + 495*a^4*b^5*c^8 - 220*a^3*b^5*c^9 + 66*a^2*b^ 5*c^10 - 12*a*b^5*c^11 + b^5*c^12) - (85*a^10*b^4 - 778*a^9*b^4*c + 3177*a ^8*b^4*c^2 - 7608*a^7*b^4*c^3 + 11802*a^6*b^4*c^4 - 12348*a^5*b^4*c^5 + 87 78*a^4*b^4*c^6 - 4152*a^3*b^4*c^7 + 1233*a^2*b^4*c^8 - 202*a*b^4*c^9 + 13* b^4*c^10)/(a^12*b^5 - 12*a^11*b^5*c + 66*a^10*b^5*c^2 - 220*a^9*b^5*c^3 + 495*a^8*b^5*c^4 - 792*a^7*b^5*c^5 + 924*a^6*b^5*c^6 - 792*a^5*b^5*c^7 + 49 5*a^4*b^5*c^8 - 220*a^3*b^5*c^9 + 66*a^2*b^5*c^10 - 12*a*b^5*c^11 + b^5*c^ 12)) + 3*(145*a^11*b^4 - 1361*a^10*b^4*c + 5719*a^9*b^4*c^2 - 14151*a^8*b^ 4*c^3 + 22794*a^7*b^4*c^4 - 24906*a^6*b^4*c^5 + 18606*a^5*b^4*c^6 - 9294*a ^4*b^4*c^7 + 2901*a^3*b^4*c^8 - 469*a^2*b^4*c^9 + 11*a*b^4*c^10 + 5*b^4*c^ 11)/(a^12*b^5 - 12*a^11*b^5*c + 66*a^10*b^5*c^2 - 220*a^9*b^5*c^3 + 495*a^ 8*b^5*c^4 - 792*a^7*b^5*c^5 + 924*a^6*b^5*c^6 - 792*a^5*b^5*c^7 + 495*a^4* b^5*c^8 - 220*a^3*b^5*c^9 + 66*a^2*b^5*c^10 - 12*a*b^5*c^11 + b^5*c^12))*( b*x + a) - (155*a^12*b^4 - 1536*a^11*b^4*c + 6855*a^10*b^4*c^2 - 18170*a^9 *b^4*c^3 + 31770*a^8*b^4*c^4 - 38520*a^7*b^4*c^5 + 33222*a^6*b^4*c^6 - 207 00*a^5*b^4*c^7 + 9495*a^4*b^4*c^8 - 3320*a^3*b^4*c^9 + 915*a^2*b^4*c^10...
Time = 22.72 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.41 \[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=\frac {x^3\,\left (\frac {64\,b\,c}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (3\,a+5\,c\right )}{{\left (a-c\right )}^3}\right )\,\sqrt {c+b\,x}}{7\,b}-\frac {x^3\,\left (\frac {64\,a\,b}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (5\,a+3\,c\right )}{{\left (a-c\right )}^3}\right )\,\sqrt {a+b\,x}}{7\,b}-\frac {8\,c^2\,\left (\frac {2\,c\,\left (3\,a+c\right )}{{\left (a-c\right )}^3}+\frac {6\,c\,\left (\frac {64\,b\,c}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (3\,a+5\,c\right )}{{\left (a-c\right )}^3}\right )}{7\,b}\right )\,\sqrt {c+b\,x}}{15\,b^3}-\frac {x^2\,\left (\frac {2\,c\,\left (3\,a+c\right )}{{\left (a-c\right )}^3}+\frac {6\,c\,\left (\frac {64\,b\,c}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (3\,a+5\,c\right )}{{\left (a-c\right )}^3}\right )}{7\,b}\right )\,\sqrt {c+b\,x}}{5\,b}+\frac {8\,a^2\,\left (\frac {2\,\left (a^2+3\,c\,a\right )}{{\left (a-c\right )}^3}+\frac {6\,a\,\left (\frac {64\,a\,b}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (5\,a+3\,c\right )}{{\left (a-c\right )}^3}\right )}{7\,b}\right )\,\sqrt {a+b\,x}}{15\,b^3}+\frac {x^2\,\left (\frac {2\,\left (a^2+3\,c\,a\right )}{{\left (a-c\right )}^3}+\frac {6\,a\,\left (\frac {64\,a\,b}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (5\,a+3\,c\right )}{{\left (a-c\right )}^3}\right )}{7\,b}\right )\,\sqrt {a+b\,x}}{5\,b}+\frac {8\,b\,x^4\,\sqrt {a+b\,x}}{9\,{\left (a-c\right )}^3}-\frac {8\,b\,x^4\,\sqrt {c+b\,x}}{9\,{\left (a-c\right )}^3}+\frac {4\,c\,x\,\left (\frac {2\,c\,\left (3\,a+c\right )}{{\left (a-c\right )}^3}+\frac {6\,c\,\left (\frac {64\,b\,c}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (3\,a+5\,c\right )}{{\left (a-c\right )}^3}\right )}{7\,b}\right )\,\sqrt {c+b\,x}}{15\,b^2}-\frac {4\,a\,x\,\left (\frac {2\,\left (a^2+3\,c\,a\right )}{{\left (a-c\right )}^3}+\frac {6\,a\,\left (\frac {64\,a\,b}{9\,{\left (a-c\right )}^3}-\frac {2\,b\,\left (5\,a+3\,c\right )}{{\left (a-c\right )}^3}\right )}{7\,b}\right )\,\sqrt {a+b\,x}}{15\,b^2} \] Input:
int(x^2/((a + b*x)^(1/2) + (c + b*x)^(1/2))^3,x)
Output:
(x^3*((64*b*c)/(9*(a - c)^3) - (2*b*(3*a + 5*c))/(a - c)^3)*(c + b*x)^(1/2 ))/(7*b) - (x^3*((64*a*b)/(9*(a - c)^3) - (2*b*(5*a + 3*c))/(a - c)^3)*(a + b*x)^(1/2))/(7*b) - (8*c^2*((2*c*(3*a + c))/(a - c)^3 + (6*c*((64*b*c)/( 9*(a - c)^3) - (2*b*(3*a + 5*c))/(a - c)^3))/(7*b))*(c + b*x)^(1/2))/(15*b ^3) - (x^2*((2*c*(3*a + c))/(a - c)^3 + (6*c*((64*b*c)/(9*(a - c)^3) - (2* b*(3*a + 5*c))/(a - c)^3))/(7*b))*(c + b*x)^(1/2))/(5*b) + (8*a^2*((2*(3*a *c + a^2))/(a - c)^3 + (6*a*((64*a*b)/(9*(a - c)^3) - (2*b*(5*a + 3*c))/(a - c)^3))/(7*b))*(a + b*x)^(1/2))/(15*b^3) + (x^2*((2*(3*a*c + a^2))/(a - c)^3 + (6*a*((64*a*b)/(9*(a - c)^3) - (2*b*(5*a + 3*c))/(a - c)^3))/(7*b)) *(a + b*x)^(1/2))/(5*b) + (8*b*x^4*(a + b*x)^(1/2))/(9*(a - c)^3) - (8*b*x ^4*(c + b*x)^(1/2))/(9*(a - c)^3) + (4*c*x*((2*c*(3*a + c))/(a - c)^3 + (6 *c*((64*b*c)/(9*(a - c)^3) - (2*b*(3*a + 5*c))/(a - c)^3))/(7*b))*(c + b*x )^(1/2))/(15*b^2) - (4*a*x*((2*(3*a*c + a^2))/(a - c)^3 + (6*a*((64*a*b)/( 9*(a - c)^3) - (2*b*(5*a + 3*c))/(a - c)^3))/(7*b))*(a + b*x)^(1/2))/(15*b ^2)
Time = 0.17 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^3} \, dx=\frac {-\frac {6 \sqrt {b x +c}\, a \,b^{3} x^{3}}{7}-\frac {6 \sqrt {b x +c}\, a \,b^{2} c \,x^{2}}{35}+\frac {8 \sqrt {b x +c}\, a b \,c^{2} x}{35}-\frac {16 \sqrt {b x +c}\, a \,c^{3}}{35}-\frac {8 \sqrt {b x +c}\, b^{4} x^{4}}{9}-\frac {26 \sqrt {b x +c}\, b^{3} c \,x^{3}}{63}+\frac {2 \sqrt {b x +c}\, b^{2} c^{2} x^{2}}{21}-\frac {8 \sqrt {b x +c}\, b \,c^{3} x}{63}+\frac {16 \sqrt {b x +c}\, c^{4}}{63}-\frac {16 \sqrt {b x +a}\, a^{4}}{63}+\frac {8 \sqrt {b x +a}\, a^{3} b x}{63}+\frac {16 \sqrt {b x +a}\, a^{3} c}{35}-\frac {2 \sqrt {b x +a}\, a^{2} b^{2} x^{2}}{21}-\frac {8 \sqrt {b x +a}\, a^{2} b c x}{35}+\frac {26 \sqrt {b x +a}\, a \,b^{3} x^{3}}{63}+\frac {6 \sqrt {b x +a}\, a \,b^{2} c \,x^{2}}{35}+\frac {8 \sqrt {b x +a}\, b^{4} x^{4}}{9}+\frac {6 \sqrt {b x +a}\, b^{3} c \,x^{3}}{7}}{b^{3} \left (a^{3}-3 a^{2} c +3 a \,c^{2}-c^{3}\right )} \] Input:
int(x^2/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x)
Output:
(2*( - 135*sqrt(b*x + c)*a*b**3*x**3 - 27*sqrt(b*x + c)*a*b**2*c*x**2 + 36 *sqrt(b*x + c)*a*b*c**2*x - 72*sqrt(b*x + c)*a*c**3 - 140*sqrt(b*x + c)*b* *4*x**4 - 65*sqrt(b*x + c)*b**3*c*x**3 + 15*sqrt(b*x + c)*b**2*c**2*x**2 - 20*sqrt(b*x + c)*b*c**3*x + 40*sqrt(b*x + c)*c**4 - 40*sqrt(a + b*x)*a**4 + 20*sqrt(a + b*x)*a**3*b*x + 72*sqrt(a + b*x)*a**3*c - 15*sqrt(a + b*x)* a**2*b**2*x**2 - 36*sqrt(a + b*x)*a**2*b*c*x + 65*sqrt(a + b*x)*a*b**3*x** 3 + 27*sqrt(a + b*x)*a*b**2*c*x**2 + 140*sqrt(a + b*x)*b**4*x**4 + 135*sqr t(a + b*x)*b**3*c*x**3))/(315*b**3*(a**3 - 3*a**2*c + 3*a*c**2 - c**3))