Integrand size = 25, antiderivative size = 95 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=-\frac {2 a (a+b x)^{3/2}}{3 b^2 (b-c)}+\frac {2 (a+b x)^{5/2}}{5 b^2 (b-c)}+\frac {2 a (a+c x)^{3/2}}{3 (b-c) c^2}-\frac {2 (a+c x)^{5/2}}{5 (b-c) c^2} \] Output:
-2/3*a*(b*x+a)^(3/2)/b^2/(b-c)+2/5*(b*x+a)^(5/2)/b^2/(b-c)+2/3*a*(c*x+a)^( 3/2)/(b-c)/c^2-2/5*(c*x+a)^(5/2)/(b-c)/c^2
Time = 1.48 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.19 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\frac {6 b^2 c^2 x^2 \left (\sqrt {a+b x}-\sqrt {a+c x}\right )+2 a b c x \left (c \sqrt {a+b x}-b \sqrt {a+c x}\right )+a^2 \left (-4 c^2 \sqrt {a+b x}+4 b^2 \sqrt {a+c x}\right )}{15 b^2 (b-c) c^2} \] Input:
Integrate[x^2/(Sqrt[a + b*x] + Sqrt[a + c*x]),x]
Output:
(6*b^2*c^2*x^2*(Sqrt[a + b*x] - Sqrt[a + c*x]) + 2*a*b*c*x*(c*Sqrt[a + b*x ] - b*Sqrt[a + c*x]) + a^2*(-4*c^2*Sqrt[a + b*x] + 4*b^2*Sqrt[a + c*x]))/( 15*b^2*(b - c)*c^2)
Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2528, 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.
\(\displaystyle \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx\) |
\(\Big \downarrow \) 2528 |
\(\displaystyle \frac {\int x \sqrt {a+b x}dx}{b-c}-\frac {\int x \sqrt {a+c x}dx}{b-c}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (\frac {(a+b x)^{3/2}}{b}-\frac {a \sqrt {a+b x}}{b}\right )dx}{b-c}-\frac {\int \left (\frac {(a+c x)^{3/2}}{c}-\frac {a \sqrt {a+c x}}{c}\right )dx}{b-c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 (a+b x)^{5/2}}{5 b^2}-\frac {2 a (a+b x)^{3/2}}{3 b^2}}{b-c}-\frac {\frac {2 (a+c x)^{5/2}}{5 c^2}-\frac {2 a (a+c x)^{3/2}}{3 c^2}}{b-c}\) |
Input:
Int[x^2/(Sqrt[a + b*x] + Sqrt[a + c*x]),x]
Output:
((-2*a*(a + b*x)^(3/2))/(3*b^2) + (2*(a + b*x)^(5/2))/(5*b^2))/(b - c) - ( (-2*a*(a + c*x)^(3/2))/(3*c^2) + (2*(a + c*x)^(5/2))/(5*c^2))/(b - c)
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[c/(e*(b*c - a*d)) Int[(u*Sqrt[a + b*x])/x, x], x] - Si mp[a/(f*(b*c - a*d)) Int[(u*Sqrt[c + d*x])/x, x], x] /; FreeQ[{a, b, c, d , e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a*e^2 - c*f^2, 0]
Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69
method | result | size |
default | \(\frac {\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}-\frac {2 a \left (b x +a \right )^{\frac {3}{2}}}{3}}{\left (b -c \right ) b^{2}}-\frac {2 \left (\frac {\left (c x +a \right )^{\frac {5}{2}}}{5}-\frac {a \left (c x +a \right )^{\frac {3}{2}}}{3}\right )}{\left (b -c \right ) c^{2}}\) | \(66\) |
Input:
int(x^2/((b*x+a)^(1/2)+(c*x+a)^(1/2)),x,method=_RETURNVERBOSE)
Output:
2/(b-c)/b^2*(1/5*(b*x+a)^(5/2)-1/3*a*(b*x+a)^(3/2))-2/(b-c)/c^2*(1/5*(c*x+ a)^(5/2)-1/3*a*(c*x+a)^(3/2))
Time = 0.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\frac {2 \, {\left ({\left (3 \, b^{2} c^{2} x^{2} + a b c^{2} x - 2 \, a^{2} c^{2}\right )} \sqrt {b x + a} - {\left (3 \, b^{2} c^{2} x^{2} + a b^{2} c x - 2 \, a^{2} b^{2}\right )} \sqrt {c x + a}\right )}}{15 \, {\left (b^{3} c^{2} - b^{2} c^{3}\right )}} \] Input:
integrate(x^2/((b*x+a)^(1/2)+(c*x+a)^(1/2)),x, algorithm="fricas")
Output:
2/15*((3*b^2*c^2*x^2 + a*b*c^2*x - 2*a^2*c^2)*sqrt(b*x + a) - (3*b^2*c^2*x ^2 + a*b^2*c*x - 2*a^2*b^2)*sqrt(c*x + a))/(b^3*c^2 - b^2*c^3)
\[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\int \frac {x^{2}}{\sqrt {a + b x} + \sqrt {a + c x}}\, dx \] Input:
integrate(x**2/((b*x+a)**(1/2)+(c*x+a)**(1/2)),x)
Output:
Integral(x**2/(sqrt(a + b*x) + sqrt(a + c*x)), x)
\[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\int { \frac {x^{2}}{\sqrt {b x + a} + \sqrt {c x + a}} \,d x } \] Input:
integrate(x^2/((b*x+a)^(1/2)+(c*x+a)^(1/2)),x, algorithm="maxima")
Output:
integrate(x^2/(sqrt(b*x + a) + sqrt(c*x + a)), x)
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (79) = 158\).
Time = 0.15 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.67 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=-\frac {2 \, {\left (\sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c} {\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (b^{5} c^{3} {\left | b \right |} - b^{4} c^{4} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{8} c^{3} - 2 \, b^{7} c^{4} + b^{6} c^{5}} + \frac {a b^{6} c^{2} {\left | b \right |} - 7 \, a b^{5} c^{3} {\left | b \right |} + 6 \, a b^{4} c^{4} {\left | b \right |}}{b^{8} c^{3} - 2 \, b^{7} c^{4} + b^{6} c^{5}}\right )} - \frac {2 \, a^{2} b^{7} c {\left | b \right |} - a^{2} b^{6} c^{2} {\left | b \right |} - 4 \, a^{2} b^{5} c^{3} {\left | b \right |} + 3 \, a^{2} b^{4} c^{4} {\left | b \right |}}{b^{8} c^{3} - 2 \, b^{7} c^{4} + b^{6} c^{5}}\right )} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a}{b - c}\right )}}{15 \, b^{2}} \] Input:
integrate(x^2/((b*x+a)^(1/2)+(c*x+a)^(1/2)),x, algorithm="giac")
Output:
-2/15*(sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)*((b*x + a)*(3*(b^5*c^3*abs(b) - b^4*c^4*abs(b))*(b*x + a)/(b^8*c^3 - 2*b^7*c^4 + b^6*c^5) + (a*b^6*c^2*ab s(b) - 7*a*b^5*c^3*abs(b) + 6*a*b^4*c^4*abs(b))/(b^8*c^3 - 2*b^7*c^4 + b^6 *c^5)) - (2*a^2*b^7*c*abs(b) - a^2*b^6*c^2*abs(b) - 4*a^2*b^5*c^3*abs(b) + 3*a^2*b^4*c^4*abs(b))/(b^8*c^3 - 2*b^7*c^4 + b^6*c^5)) - (3*(b*x + a)^(5/ 2) - 5*(b*x + a)^(3/2)*a)/(b - c))/b^2
Time = 22.80 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.36 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\frac {2\,x^2\,\sqrt {a+b\,x}}{5\,\left (b-c\right )}-\frac {2\,x^2\,\sqrt {a+c\,x}}{5\,\left (b-c\right )}-\frac {4\,a^2\,\sqrt {a+b\,x}}{15\,b^2\,\left (b-c\right )}+\frac {4\,a^2\,\sqrt {a+c\,x}}{15\,c^2\,\left (b-c\right )}+\frac {2\,a\,x\,\sqrt {a+b\,x}}{15\,b\,\left (b-c\right )}-\frac {2\,a\,x\,\sqrt {a+c\,x}}{15\,c\,\left (b-c\right )} \] Input:
int(x^2/((a + b*x)^(1/2) + (a + c*x)^(1/2)),x)
Output:
(2*x^2*(a + b*x)^(1/2))/(5*(b - c)) - (2*x^2*(a + c*x)^(1/2))/(5*(b - c)) - (4*a^2*(a + b*x)^(1/2))/(15*b^2*(b - c)) + (4*a^2*(a + c*x)^(1/2))/(15*c ^2*(b - c)) + (2*a*x*(a + b*x)^(1/2))/(15*b*(b - c)) - (2*a*x*(a + c*x)^(1 /2))/(15*c*(b - c))
Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.11 \[ \int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\frac {-\frac {4 \sqrt {b x +a}\, a^{2} c^{2}}{15}+\frac {2 \sqrt {b x +a}\, a b \,c^{2} x}{15}+\frac {2 \sqrt {b x +a}\, b^{2} c^{2} x^{2}}{5}+\frac {4 \sqrt {c x +a}\, a^{2} b^{2}}{15}-\frac {2 \sqrt {c x +a}\, a \,b^{2} c x}{15}-\frac {2 \sqrt {c x +a}\, b^{2} c^{2} x^{2}}{5}}{b^{2} c^{2} \left (b -c \right )} \] Input:
int(x^2/((b*x+a)^(1/2)+(c*x+a)^(1/2)),x)
Output:
(2*( - 2*sqrt(a + b*x)*a**2*c**2 + sqrt(a + b*x)*a*b*c**2*x + 3*sqrt(a + b *x)*b**2*c**2*x**2 + 2*sqrt(a + c*x)*a**2*b**2 - sqrt(a + c*x)*a*b**2*c*x - 3*sqrt(a + c*x)*b**2*c**2*x**2))/(15*b**2*c**2*(b - c))