Integrand size = 24, antiderivative size = 133 \[ \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx=-\frac {16 b^2 (2 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{315 c^4 x^{3/2}}+\frac {8 b (2 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{105 c^3 \sqrt {x}}-\frac {2 (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c} \]
-16/315*b^2*(-3*A*c+2*B*b)*(c*x^2+b*x)^(3/2)/c^4/x^(3/2)+2/9*B*x^(3/2)*(c* x^2+b*x)^(3/2)/c+8/105*b*(-3*A*c+2*B*b)*(c*x^2+b*x)^(3/2)/c^3/x^(1/2)-2/21 *(-3*A*c+2*B*b)*(c*x^2+b*x)^(3/2)*x^(1/2)/c^2
Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.54 \[ \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 (x (b+c x))^{3/2} \left (-16 b^3 B+24 b^2 c (A+B x)-6 b c^2 x (6 A+5 B x)+5 c^3 x^2 (9 A+7 B x)\right )}{315 c^4 x^{3/2}} \]
(2*(x*(b + c*x))^(3/2)*(-16*b^3*B + 24*b^2*c*(A + B*x) - 6*b*c^2*x*(6*A + 5*B*x) + 5*c^3*x^2*(9*A + 7*B*x)))/(315*c^4*x^(3/2))
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.
\(\displaystyle \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {(2 b B-3 A c) \int x^{3/2} \sqrt {c x^2+b x}dx}{3 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {(2 b B-3 A c) \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {4 b \int \sqrt {x} \sqrt {c x^2+b x}dx}{7 c}\right )}{3 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {(2 b B-3 A c) \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {2 b \int \frac {\sqrt {c x^2+b x}}{\sqrt {x}}dx}{5 c}\right )}{7 c}\right )}{3 c}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 B x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {\left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}}\right )}{7 c}\right ) (2 b B-3 A c)}{3 c}\) |
(2*B*x^(3/2)*(b*x + c*x^2)^(3/2))/(9*c) - ((2*b*B - 3*A*c)*((2*Sqrt[x]*(b* x + c*x^2)^(3/2))/(7*c) - (4*b*((-4*b*(b*x + c*x^2)^(3/2))/(15*c^2*x^(3/2) ) + (2*(b*x + c*x^2)^(3/2))/(5*c*Sqrt[x])))/(7*c)))/(3*c)
3.2.96.3.1 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 = 81, normalized size of antiderivative = 0.61
method | result | size |
default | \(\frac {2 \left (c x +b \right ) \left (35 B \,c^{3} x^{3}+45 A \,c^{3} x^{2}-30 B b \,c^{2} x^{2}-36 A b \,c^{2} x +24 B \,b^{2} c x +24 A \,b^{2} c -16 B \,b^{3}\right ) \sqrt {x \left (c x +b \right )}}{315 c^{4} \sqrt {x}}\) | \(81\) |
gosper | \(\frac {2 \left (c x +b \right ) \left (35 B \,c^{3} x^{3}+45 A \,c^{3} x^{2}-30 B b \,c^{2} x^{2}-36 A b \,c^{2} x +24 B \,b^{2} c x +24 A \,b^{2} c -16 B \,b^{3}\right ) \sqrt {c \,x^{2}+b x}}{315 c^{4} \sqrt {x}}\) | \(83\) |
risch | \(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (35 B \,x^{4} c^{4}+45 A \,c^{4} x^{3}+5 B b \,c^{3} x^{3}+9 A b \,c^{3} x^{2}-6 B \,b^{2} c^{2} x^{2}-12 A \,b^{2} c^{2} x +8 B \,b^{3} c x +24 A \,b^{3} c -16 b^{4} B \right )}{315 \sqrt {x \left (c x +b \right )}\, c^{4}}\) | \(105\) |
2/315*(c*x+b)*(35*B*c^3*x^3+45*A*c^3*x^2-30*B*b*c^2*x^2-36*A*b*c^2*x+24*B* b^2*c*x+24*A*b^2*c-16*B*b^3)*(x*(c*x+b))^(1/2)/c^4/x^(1/2)
Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.77 \[ \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (35 \, B c^{4} x^{4} - 16 \, B b^{4} + 24 \, A b^{3} c + 5 \, {\left (B b c^{3} + 9 \, A c^{4}\right )} x^{3} - 3 \, {\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{2} + 4 \, {\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, c^{4} \sqrt {x}} \]
2/315*(35*B*c^4*x^4 - 16*B*b^4 + 24*A*b^3*c + 5*(B*b*c^3 + 9*A*c^4)*x^3 - 3*(2*B*b^2*c^2 - 3*A*b*c^3)*x^2 + 4*(2*B*b^3*c - 3*A*b^2*c^2)*x)*sqrt(c*x^ 2 + b*x)/(c^4*sqrt(x))
\[ \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx=\int x^{\frac {3}{2}} \sqrt {x \left (b + c x\right )} \left (A + B x\right )\, dx \]
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.74 \[ \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x + b} A}{105 \, c^{3}} + \frac {2 \, {\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x + b} B}{315 \, c^{4}} \]
2/105*(15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*sqrt(c*x + b)*A/c^3 + 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)*sqr t(c*x + b)*B/c^4
Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.83 \[ \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2}{315} \, B {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} - \frac {2}{105} \, A {\left (\frac {8 \, b^{\frac {7}{2}}}{c^{3}} - \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} \]
2/315*B*(16*b^(9/2)/c^4 + (35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)*b + 18 9*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4) - 2/105*A*(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. \[ \int x^{3/2} (A+B x) \sqrt {b x+c x^2} \, dx=\int x^{3/2}\,\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right ) \,d x \]