Integrand size = 25, antiderivative size = 147 \[ \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\frac {2 a^2 (a+b x)^{3/2}}{3 b^3 (b-c)}-\frac {4 a (a+b x)^{5/2}}{5 b^3 (b-c)}+\frac {2 (a+b x)^{7/2}}{7 b^3 (b-c)}-\frac {2 a^2 (a+c x)^{3/2}}{3 (b-c) c^3}+\frac {4 a (a+c x)^{5/2}}{5 (b-c) c^3}-\frac {2 (a+c x)^{7/2}}{7 (b-c) c^3} \]
2/3*a^2*(b*x+a)^(3/2)/b^3/(b-c)-4/5*a*(b*x+a)^(5/2)/b^3/(b-c)+2/7*(b*x+a)^ (7/2)/b^3/(b-c)-2/3*a^2*(c*x+a)^(3/2)/(b-c)/c^3+4/5*a*(c*x+a)^(5/2)/(b-c)/ c^3-2/7*(c*x+a)^(7/2)/(b-c)/c^3
Leaf count is larger than twice the leaf count of optimal. \(1832\) vs. \(2(147)=294\).
Time = 5.98 (sec) , antiderivative size = 1832, normalized size of antiderivative = 12.46 \[ \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx =\text {Too large to display} \]
(-2*a^3*(b - c)^2*(a + c*x)*(15*b^3*Sqrt[a - (a*b)/c]*c^6*x^5*(b*x - Sqrt[ a + b*x]*Sqrt[a + c*x]) + 3*a*b^2*c^5*x^4*(109*b*Sqrt[a - (a*b)/c]*Sqrt[a + b*x]*Sqrt[a + c*x] - 120*Sqrt[a - (a*b)/c]*c*Sqrt[a + b*x]*Sqrt[a + c*x] + 7*b*c*x*(18*Sqrt[a - (a*b)/c] - 5*Sqrt[a + b*x] + 5*Sqrt[a + c*x]) - 5* b^2*x*(22*Sqrt[a - (a*b)/c] - 7*Sqrt[a + b*x] + 7*Sqrt[a + c*x])) + a^6*(b ^3*c*(-378*Sqrt[a - (a*b)/c] + 966*Sqrt[a + b*x] - 200*Sqrt[a + c*x]) + b^ 4*(15*Sqrt[a - (a*b)/c] - 105*Sqrt[a + b*x] + 8*Sqrt[a + c*x]) + 32*c^4*(3 5*Sqrt[a - (a*b)/c] - 35*Sqrt[a + b*x] + 16*Sqrt[a + c*x]) - 4*b*c^3*(595* Sqrt[a - (a*b)/c] - 735*Sqrt[a + b*x] + 288*Sqrt[a + c*x]) + b^2*c^2*(1631 *Sqrt[a - (a*b)/c] - 2681*Sqrt[a + b*x] + 832*Sqrt[a + c*x])) + a^3*c^3*x^ 2*(-960*Sqrt[a - (a*b)/c]*c^3*Sqrt[a + b*x]*Sqrt[a + c*x] + b^4*x*(-960*Sq rt[a - (a*b)/c] + 945*Sqrt[a + b*x] - 791*Sqrt[a + c*x]) + 20*b*c^2*(132*S qrt[a - (a*b)/c]*Sqrt[a + b*x]*Sqrt[a + c*x] + c*x*(119*Sqrt[a - (a*b)/c] - 91*Sqrt[a + b*x] + 84*Sqrt[a + c*x])) + b^2*(-2376*Sqrt[a - (a*b)/c]*c*S qrt[a + b*x]*Sqrt[a + c*x] - 14*c^2*x*(379*Sqrt[a - (a*b)/c] - 319*Sqrt[a + b*x] + 288*Sqrt[a + c*x])) + b^3*(693*Sqrt[a - (a*b)/c]*Sqrt[a + b*x]*Sq rt[a + c*x] + 7*c*x*(558*Sqrt[a - (a*b)/c] - 513*Sqrt[a + b*x] + 449*Sqrt[ a + c*x]))) + a^2*b*c^4*x^3*(-1200*Sqrt[a - (a*b)/c]*c^2*Sqrt[a + b*x]*Sqr t[a + c*x] + b^3*x*(855*Sqrt[a - (a*b)/c] - 630*Sqrt[a + b*x] + 609*Sqrt[a + c*x]) + b^2*(-785*Sqrt[a - (a*b)/c]*Sqrt[a + b*x]*Sqrt[a + c*x] - 21...
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 = {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.
\(\displaystyle \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx\) |
\(\Big \downarrow \) 2528 |
\(\displaystyle \frac {\int x^2 \sqrt {a+b x}dx}{b-c}-\frac {\int x^2 \sqrt {a+c x}dx}{b-c}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (\frac {(a+b x)^{5/2}}{b^2}-\frac {2 a (a+b x)^{3/2}}{b^2}+\frac {a^2 \sqrt {a+b x}}{b^2}\right )dx}{b-c}-\frac {\int \left (\frac {(a+c x)^{5/2}}{c^2}-\frac {2 a (a+c x)^{3/2}}{c^2}+\frac {a^2 \sqrt {a+c x}}{c^2}\right )dx}{b-c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 a^2 (a+b x)^{3/2}}{3 b^3}+\frac {2 (a+b x)^{7/2}}{7 b^3}-\frac {4 a (a+b x)^{5/2}}{5 b^3}}{b-c}-\frac {\frac {2 a^2 (a+c x)^{3/2}}{3 c^3}+\frac {2 (a+c x)^{7/2}}{7 c^3}-\frac {4 a (a+c x)^{5/2}}{5 c^3}}{b-c}\) |
((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))/(b - c) - ((2*a^2*(a + c*x)^(3/2))/(3*c^3) - (4*a*(a + c*x)^(5/2))/(5*c^3) + (2*(a + c*x)^(7/2))/(7*c^3))/(b - c)
3.5.26.3.1 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[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 = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.61
method | result | size |
default | \(\frac {\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {4 \left (b x +a \right )^{\frac {5}{2}} a}{5}+\frac {2 \left (b x +a \right )^{\frac {3}{2}} a^{2}}{3}}{\left (b -c \right ) b^{3}}-\frac {2 \left (\frac {\left (c x +a \right )^{\frac {7}{2}}}{7}-\frac {2 \left (c x +a \right )^{\frac {5}{2}} a}{5}+\frac {\left (c x +a \right )^{\frac {3}{2}} a^{2}}{3}\right )}{\left (b -c \right ) c^{3}}\) | \(90\) |
2/(b-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/(b-c)/c^3*(1/7*(c*x+a)^(7/2)-2/5*(c*x+a)^(5/2)*a+1/3*(c*x+a)^(3/2)*a^2)
Time = 0.31 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\frac {2 \, {\left ({\left (15 \, b^{3} c^{3} x^{3} + 3 \, a b^{2} c^{3} x^{2} - 4 \, a^{2} b c^{3} x + 8 \, a^{3} c^{3}\right )} \sqrt {b x + a} - {\left (15 \, b^{3} c^{3} x^{3} + 3 \, a b^{3} c^{2} x^{2} - 4 \, a^{2} b^{3} c x + 8 \, a^{3} b^{3}\right )} \sqrt {c x + a}\right )}}{105 \, {\left (b^{4} c^{3} - b^{3} c^{4}\right )}} \]
2/105*((15*b^3*c^3*x^3 + 3*a*b^2*c^3*x^2 - 4*a^2*b*c^3*x + 8*a^3*c^3)*sqrt (b*x + a) - (15*b^3*c^3*x^3 + 3*a*b^3*c^2*x^2 - 4*a^2*b^3*c*x + 8*a^3*b^3) *sqrt(c*x + a))/(b^4*c^3 - b^3*c^4)
\[ \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\int \frac {x^{3}}{\sqrt {a + b x} + \sqrt {a + c x}}\, dx \]
\[ \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\int { \frac {x^{3}}{\sqrt {b x + a} + \sqrt {c x + a}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (123) = 246\).
Time = 0.39 (sec) , antiderivative size = 451, normalized size of antiderivative = 3.07 \[ \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=-\frac {2}{105} \, \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c} {\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {5 \, {\left (b^{17} c^{5} {\left | b \right |} - 2 \, b^{16} c^{6} {\left | b \right |} + b^{15} c^{7} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{23} c^{5} - 3 \, b^{22} c^{6} + 3 \, b^{21} c^{7} - b^{20} c^{8}} + \frac {a b^{18} c^{4} {\left | b \right |} - 17 \, a b^{17} c^{5} {\left | b \right |} + 31 \, a b^{16} c^{6} {\left | b \right |} - 15 \, a b^{15} c^{7} {\left | b \right |}}{b^{23} c^{5} - 3 \, b^{22} c^{6} + 3 \, b^{21} c^{7} - b^{20} c^{8}}\right )} - \frac {4 \, a^{2} b^{19} c^{3} {\left | b \right |} - 2 \, a^{2} b^{18} c^{4} {\left | b \right |} - 53 \, a^{2} b^{17} c^{5} {\left | b \right |} + 96 \, a^{2} b^{16} c^{6} {\left | b \right |} - 45 \, a^{2} b^{15} c^{7} {\left | b \right |}}{b^{23} c^{5} - 3 \, b^{22} c^{6} + 3 \, b^{21} c^{7} - b^{20} c^{8}}\right )} {\left (b x + a\right )} + \frac {8 \, a^{3} b^{20} c^{2} {\left | b \right |} - 12 \, a^{3} b^{19} c^{3} {\left | b \right |} + 3 \, a^{3} b^{18} c^{4} {\left | b \right |} - 17 \, a^{3} b^{17} c^{5} {\left | b \right |} + 33 \, a^{3} b^{16} c^{6} {\left | b \right |} - 15 \, a^{3} b^{15} c^{7} {\left | b \right |}}{b^{23} c^{5} - 3 \, b^{22} c^{6} + 3 \, b^{21} c^{7} - b^{20} c^{8}}\right )} + \frac {2 \, {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} - 42 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}\right )}}{105 \, {\left (b^{4} - b^{3} c\right )}} \]
-2/105*sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)*((3*(b*x + a)*(5*(b^17*c^5*abs( b) - 2*b^16*c^6*abs(b) + b^15*c^7*abs(b))*(b*x + a)/(b^23*c^5 - 3*b^22*c^6 + 3*b^21*c^7 - b^20*c^8) + (a*b^18*c^4*abs(b) - 17*a*b^17*c^5*abs(b) + 31 *a*b^16*c^6*abs(b) - 15*a*b^15*c^7*abs(b))/(b^23*c^5 - 3*b^22*c^6 + 3*b^21 *c^7 - b^20*c^8)) - (4*a^2*b^19*c^3*abs(b) - 2*a^2*b^18*c^4*abs(b) - 53*a^ 2*b^17*c^5*abs(b) + 96*a^2*b^16*c^6*abs(b) - 45*a^2*b^15*c^7*abs(b))/(b^23 *c^5 - 3*b^22*c^6 + 3*b^21*c^7 - b^20*c^8))*(b*x + a) + (8*a^3*b^20*c^2*ab s(b) - 12*a^3*b^19*c^3*abs(b) + 3*a^3*b^18*c^4*abs(b) - 17*a^3*b^17*c^5*ab s(b) + 33*a^3*b^16*c^6*abs(b) - 15*a^3*b^15*c^7*abs(b))/(b^23*c^5 - 3*b^22 *c^6 + 3*b^21*c^7 - b^20*c^8)) + 2/105*(15*(b*x + a)^(7/2) - 42*(b*x + a)^ (5/2)*a + 35*(b*x + a)^(3/2)*a^2)/(b^4 - b^3*c)
Time = 17.61 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.22 \[ \int \frac {x^3}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx=\frac {2\,x^3\,\sqrt {a+b\,x}}{7\,\left (b-c\right )}-\frac {2\,x^3\,\sqrt {a+c\,x}}{7\,\left (b-c\right )}+\frac {16\,a^3\,\sqrt {a+b\,x}}{105\,b^3\,\left (b-c\right )}-\frac {16\,a^3\,\sqrt {a+c\,x}}{105\,c^3\,\left (b-c\right )}+\frac {2\,a\,x^2\,\sqrt {a+b\,x}}{35\,b\,\left (b-c\right )}-\frac {8\,a^2\,x\,\sqrt {a+b\,x}}{105\,b^2\,\left (b-c\right )}-\frac {2\,a\,x^2\,\sqrt {a+c\,x}}{35\,c\,\left (b-c\right )}+\frac {8\,a^2\,x\,\sqrt {a+c\,x}}{105\,c^2\,\left (b-c\right )} \]
(2*x^3*(a + b*x)^(1/2))/(7*(b - c)) - (2*x^3*(a + c*x)^(1/2))/(7*(b - c)) + (16*a^3*(a + b*x)^(1/2))/(105*b^3*(b - c)) - (16*a^3*(a + c*x)^(1/2))/(1 05*c^3*(b - c)) + (2*a*x^2*(a + b*x)^(1/2))/(35*b*(b - c)) - (8*a^2*x*(a + b*x)^(1/2))/(105*b^2*(b - c)) - (2*a*x^2*(a + c*x)^(1/2))/(35*c*(b - c)) + (8*a^2*x*(a + c*x)^(1/2))/(105*c^2*(b - c))