Integrand size = 24, antiderivative size = 133 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=-\frac {16 b^2 (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{3465 c^4 x^{5/2}}+\frac {8 b (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{693 c^3 x^{3/2}}-\frac {2 (6 b B-11 A c) \left (b x+c x^2\right )^{5/2}}{99 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c} \]
-16/3465*b^2*(-11*A*c+6*B*b)*(c*x^2+b*x)^(5/2)/c^4/x^(5/2)+8/693*b*(-11*A* c+6*B*b)*(c*x^2+b*x)^(5/2)/c^3/x^(3/2)-2/99*(-11*A*c+6*B*b)*(c*x^2+b*x)^(5 /2)/c^2/x^(1/2)+2/11*B*(c*x^2+b*x)^(5/2)*x^(1/2)/c
Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.56 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 (x (b+c x))^{5/2} \left (-48 b^3 B+35 c^3 x^2 (11 A+9 B x)+8 b^2 c (11 A+15 B x)-10 b c^2 x (22 A+21 B x)\right )}{3465 c^4 x^{5/2}} \]
(2*(x*(b + c*x))^(5/2)*(-48*b^3*B + 35*c^3*x^2*(11*A + 9*B*x) + 8*b^2*c*(1 1*A + 15*B*x) - 10*b*c^2*x*(22*A + 21*B*x)))/(3465*c^4*x^(5/2))
Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1221, 1128, 1128, 1122}
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.
\(\displaystyle \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {(6 b B-11 A c) \int \sqrt {x} \left (c x^2+b x\right )^{3/2}dx}{11 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {(6 b B-11 A c) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {4 b \int \frac {\left (c x^2+b x\right )^{3/2}}{\sqrt {x}}dx}{9 c}\right )}{11 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {(6 b B-11 A c) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {2 b \int \frac {\left (c x^2+b x\right )^{3/2}}{x^{3/2}}dx}{7 c}\right )}{9 c}\right )}{11 c}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {\left (\frac {2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}}\right )}{9 c}\right ) (6 b B-11 A c)}{11 c}\) |
(2*B*Sqrt[x]*(b*x + c*x^2)^(5/2))/(11*c) - ((6*b*B - 11*A*c)*((2*(b*x + c* x^2)^(5/2))/(9*c*Sqrt[x]) - (4*b*((-4*b*(b*x + c*x^2)^(5/2))/(35*c^2*x^(5/ 2)) + (2*(b*x + c*x^2)^(5/2))/(7*c*x^(3/2))))/(9*c)))/(11*c)
3.3.5.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[Simplify[m + p], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x ] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(\frac {2 \left (c x +b \right ) \left (315 B \,c^{3} x^{3}+385 A \,c^{3} x^{2}-210 B b \,c^{2} x^{2}-220 A b \,c^{2} x +120 B \,b^{2} c x +88 A \,b^{2} c -48 B \,b^{3}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3465 c^{4} x^{\frac {3}{2}}}\) | \(83\) |
default | \(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{2} \left (315 B \,c^{3} x^{3}+385 A \,c^{3} x^{2}-210 B b \,c^{2} x^{2}-220 A b \,c^{2} x +120 B \,b^{2} c x +88 A \,b^{2} c -48 B \,b^{3}\right )}{3465 \sqrt {x}\, c^{4}}\) | \(83\) |
risch | \(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (315 B \,c^{5} x^{5}+385 A \,c^{5} x^{4}+420 B b \,c^{4} x^{4}+550 A b \,c^{4} x^{3}+15 B \,b^{2} c^{3} x^{3}+33 A \,b^{2} c^{3} x^{2}-18 B \,b^{3} c^{2} x^{2}-44 A \,b^{3} c^{2} x +24 B \,b^{4} c x +88 A \,b^{4} c -48 B \,b^{5}\right )}{3465 \sqrt {x \left (c x +b \right )}\, c^{4}}\) | \(129\) |
2/3465*(c*x+b)*(315*B*c^3*x^3+385*A*c^3*x^2-210*B*b*c^2*x^2-220*A*b*c^2*x+ 120*B*b^2*c*x+88*A*b^2*c-48*B*b^3)*(c*x^2+b*x)^(3/2)/c^4/x^(3/2)
Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left (315 \, B c^{5} x^{5} - 48 \, B b^{5} + 88 \, A b^{4} c + 35 \, {\left (12 \, B b c^{4} + 11 \, A c^{5}\right )} x^{4} + 5 \, {\left (3 \, B b^{2} c^{3} + 110 \, A b c^{4}\right )} x^{3} - 3 \, {\left (6 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{2} + 4 \, {\left (6 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{3465 \, c^{4} \sqrt {x}} \]
2/3465*(315*B*c^5*x^5 - 48*B*b^5 + 88*A*b^4*c + 35*(12*B*b*c^4 + 11*A*c^5) *x^4 + 5*(3*B*b^2*c^3 + 110*A*b*c^4)*x^3 - 3*(6*B*b^3*c^2 - 11*A*b^2*c^3)* x^2 + 4*(6*B*b^4*c - 11*A*b^3*c^2)*x)*sqrt(c*x^2 + b*x)/(c^4*sqrt(x))
\[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\int \sqrt {x} \left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (109) = 218\).
Time = 0.21 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.72 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left ({\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} x^{3} + 3 \, {\left (15 \, b c^{3} x^{4} + 3 \, b^{2} c^{2} x^{3} - 4 \, b^{3} c x^{2} + 8 \, b^{4} x\right )} x^{2}\right )} \sqrt {c x + b} A}{315 \, c^{3} x^{3}} + \frac {2 \, {\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 11 \, {\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3}\right )} \sqrt {c x + b} B}{3465 \, c^{4} x^{4}} \]
2/315*((35*c^4*x^4 + 5*b*c^3*x^3 - 6*b^2*c^2*x^2 + 8*b^3*c*x - 16*b^4)*x^3 + 3*(15*b*c^3*x^4 + 3*b^2*c^2*x^3 - 4*b^3*c*x^2 + 8*b^4*x)*x^2)*sqrt(c*x + b)*A/(c^3*x^3) + 2/3465*((315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*x^4 + 11*(35*b*c^4*x^5 + 5*b^2*c^3* x^4 - 6*b^3*c^2*x^3 + 8*b^4*c*x^2 - 16*b^5*x)*x^3)*sqrt(c*x + b)*B/(c^4*x^ 4)
Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (109) = 218\).
Time = 0.28 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.86 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=-\frac {2}{3465} \, B c {\left (\frac {128 \, b^{\frac {11}{2}}}{c^{5}} - \frac {315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}}{c^{5}}\right )} + \frac {2}{315} \, B b {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} + \frac {2}{315} \, A c {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} - \frac {2}{105} \, A b {\left (\frac {8 \, b^{\frac {7}{2}}}{c^{3}} - \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} \]
-2/3465*B*c*(128*b^(11/2)/c^5 - (315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/ 2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b )^(3/2)*b^4)/c^5) + 2/315*B*b*(16*b^(9/2)/c^4 + (35*(c*x + b)^(9/2) - 135* (c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4 ) + 2/315*A*c*(16*b^(9/2)/c^4 + (35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)* b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4) - 2/105*A*b*(8 *b^(7/2)/c^3 - (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^( 3/2)*b^2)/c^3)
Timed out. \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\int \sqrt {x}\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]