Integrand size = 24, antiderivative size = 95 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 b (b B-A c) \left (b x+c x^2\right )^{5/2}}{5 c^3 x^{5/2}}-\frac {2 (2 b B-A c) \left (b x+c x^2\right )^{7/2}}{7 c^3 x^{7/2}}+\frac {2 B \left (b x+c x^2\right )^{9/2}}{9 c^3 x^{9/2}} \] Output:
2/5*b*(-A*c+B*b)*(c*x^2+b*x)^(5/2)/c^3/x^(5/2)-2/7*(-A*c+2*B*b)*(c*x^2+b*x )^(7/2)/c^3/x^(7/2)+2/9*B*(c*x^2+b*x)^(9/2)/c^3/x^(9/2)
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.59 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 (x (b+c x))^{5/2} \left (8 b^2 B+5 c^2 x (9 A+7 B x)-2 b c (9 A+10 B x)\right )}{315 c^3 x^{5/2}} \] Input:
Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/Sqrt[x],x]
Output:
(2*(x*(b + c*x))^(5/2)*(8*b^2*B + 5*c^2*x*(9*A + 7*B*x) - 2*b*c*(9*A + 10* B*x)))/(315*c^3*x^(5/2))
Time = 0.40 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1221, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {(4 b B-9 A c) \int \frac {\left (c x^2+b x\right )^{3/2}}{\sqrt {x}}dx}{9 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {(4 b B-9 A c) \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {2 b \int \frac {\left (c x^2+b x\right )^{3/2}}{x^{3/2}}dx}{7 c}\right )}{9 c}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 B \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {\left (\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}}\right ) (4 b B-9 A c)}{9 c}\) |
Input:
Int[((A + B*x)*(b*x + c*x^2)^(3/2))/Sqrt[x],x]
Output:
(2*B*(b*x + c*x^2)^(5/2))/(9*c*Sqrt[x]) - ((4*b*B - 9*A*c)*((-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)
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 = 0.98 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-35 B \,c^{2} x^{2}-45 A \,c^{2} x +20 B b c x +18 A b c -8 B \,b^{2}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{315 c^{3} x^{\frac {3}{2}}}\) | \(59\) |
default | \(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{2} \left (-35 B \,c^{2} x^{2}-45 A \,c^{2} x +20 B b c x +18 A b c -8 B \,b^{2}\right )}{315 \sqrt {x}\, c^{3}}\) | \(59\) |
orering | \(-\frac {2 \left (c x +b \right ) \left (-35 B \,c^{2} x^{2}-45 A \,c^{2} x +20 B b c x +18 A b c -8 B \,b^{2}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{315 c^{3} x^{\frac {3}{2}}}\) | \(59\) |
risch | \(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-35 B \,c^{4} x^{4}-45 A \,c^{4} x^{3}-50 B \,c^{3} x^{3} b -72 A b \,c^{3} x^{2}-3 c^{2} x^{2} B \,b^{2}-9 A \,b^{2} c^{2} x +4 B \,b^{3} c x +18 A \,b^{3} c -8 B \,b^{4}\right )}{315 \sqrt {x \left (c x +b \right )}\, c^{3}}\) | \(105\) |
Input:
int((B*x+A)*(c*x^2+b*x)^(3/2)/x^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/315*(c*x+b)*(-35*B*c^2*x^2-45*A*c^2*x+20*B*b*c*x+18*A*b*c-8*B*b^2)*(c*x ^2+b*x)^(3/2)/c^3/x^(3/2)
Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \, {\left (35 \, B c^{4} x^{4} + 8 \, B b^{4} - 18 \, A b^{3} c + 5 \, {\left (10 \, B b c^{3} + 9 \, A c^{4}\right )} x^{3} + 3 \, {\left (B b^{2} c^{2} + 24 \, A b c^{3}\right )} x^{2} - {\left (4 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, c^{3} \sqrt {x}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(1/2),x, algorithm="fricas")
Output:
2/315*(35*B*c^4*x^4 + 8*B*b^4 - 18*A*b^3*c + 5*(10*B*b*c^3 + 9*A*c^4)*x^3 + 3*(B*b^2*c^2 + 24*A*b*c^3)*x^2 - (4*B*b^3*c - 9*A*b^2*c^2)*x)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{\sqrt {x}}\, dx \] Input:
integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**(1/2),x)
Output:
Integral((x*(b + c*x))**(3/2)*(A + B*x)/sqrt(x), x)
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (77) = 154\).
Time = 0.04 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.92 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \, {\left ({\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} x^{2} + 7 \, {\left (3 \, b c^{2} x^{3} + b^{2} c x^{2} - 2 \, b^{3} x\right )} x\right )} \sqrt {c x + b} A}{105 \, c^{2} x^{2}} + \frac {2 \, {\left ({\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} x^{3} + 3 \, {\left (15 \, b c^{3} x^{4} + 3 \, b^{2} c^{2} x^{3} - 4 \, b^{3} c x^{2} + 8 \, b^{4} x\right )} x^{2}\right )} \sqrt {c x + b} B}{315 \, c^{3} x^{3}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(1/2),x, algorithm="maxima")
Output:
2/105*((15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*x^2 + 7*(3*b*c^2*x^3 + b^2*c*x^2 - 2*b^3*x)*x)*sqrt(c*x + b)*A/(c^2*x^2) + 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)*B/(c^3*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (77) = 154\).
Time = 0.16 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b\right )} A b}{15 \, c^{2}} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}\right )} B b}{105 \, c^{3}} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}\right )} A}{105 \, c^{2}} + \frac {2 \, {\left (35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}\right )} B}{315 \, c^{3}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(1/2),x, algorithm="giac")
Output:
2/15*(3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)*A*b/c^2 + 2/105*(15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)*B*b/c^3 + 2/105* (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)*A/c^2 + 2/315*(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)*B/c^3
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{\sqrt {x}} \,d x \] Input:
int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^(1/2),x)
Output:
int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx=\frac {2 \sqrt {c x +b}\, \left (35 b \,c^{4} x^{4}+45 a \,c^{4} x^{3}+50 b^{2} c^{3} x^{3}+72 a b \,c^{3} x^{2}+3 b^{3} c^{2} x^{2}+9 a \,b^{2} c^{2} x -4 b^{4} c x -18 a \,b^{3} c +8 b^{5}\right )}{315 c^{3}} \] Input:
int((B*x+A)*(c*x^2+b*x)^(3/2)/x^(1/2),x)
Output:
(2*sqrt(b + c*x)*( - 18*a*b**3*c + 9*a*b**2*c**2*x + 72*a*b*c**3*x**2 + 45 *a*c**4*x**3 + 8*b**5 - 4*b**4*c*x + 3*b**3*c**2*x**2 + 50*b**2*c**3*x**3 + 35*b*c**4*x**4))/(315*c**3)