Integrand size = 20, antiderivative size = 81 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 A x}{b \left (b x+c x^2\right )^{3/2}}+\frac {2 (b B-4 A c) x^2}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 (b B-4 A c) x}{3 b^3 \sqrt {b x+c x^2}} \] Output:
-2*A*x/b/(c*x^2+b*x)^(3/2)+2/3*(-4*A*c+B*b)*x^2/b^2/(c*x^2+b*x)^(3/2)+4/3* (-4*A*c+B*b)*x/b^3/(c*x^2+b*x)^(1/2)
Time = 0.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {x \left (2 b B x (3 b+2 c x)-2 A \left (3 b^2+12 b c x+8 c^2 x^2\right )\right )}{3 b^3 (x (b+c x))^{3/2}} \] Input:
Integrate[(x*(A + B*x))/(b*x + c*x^2)^(5/2),x]
Output:
(x*(2*b*B*x*(3*b + 2*c*x) - 2*A*(3*b^2 + 12*b*c*x + 8*c^2*x^2)))/(3*b^3*(x *(b + c*x))^(3/2))
Time = 0.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1218, 1088}
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.
\(\displaystyle \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1218 |
\(\displaystyle -\frac {(b B-4 A c) \int \frac {1}{\left (c x^2+b x\right )^{3/2}}dx}{3 b c}-\frac {2 x (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle \frac {2 (b+2 c x) (b B-4 A c)}{3 b^3 c \sqrt {b x+c x^2}}-\frac {2 x (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
Input:
Int[(x*(A + B*x))/(b*x + c*x^2)^(5/2),x]
Output:
(-2*(b*B - A*c)*x)/(3*b*c*(b*x + c*x^2)^(3/2)) + (2*(b*B - 4*A*c)*(b + 2*c *x))/(3*b^3*c*Sqrt[b*x + c*x^2])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + ( c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(c*d - b*e) + c*e*f)*(d + e*x)^m*(( a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(2*c*d - b*e))), x] - Simp[e*((m*(g*(c* d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g))/(c*(p + 1)*(2*c*d - b*e))) I nt[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d , e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
Time = 0.96 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (-B x +A \right ) b^{2}+4 c x \left (-\frac {B x}{6}+A \right ) b +\frac {8 A \,c^{2} x^{2}}{3}\right )}{\sqrt {x \left (c x +b \right )}\, \left (c x +b \right ) b^{3}}\) | \(53\) |
gosper | \(-\frac {2 x^{2} \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-2 x^{2} B b c +12 A b c x -3 x B \,b^{2}+3 b^{2} A \right )}{3 b^{3} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\) | \(62\) |
orering | \(-\frac {2 x^{2} \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-2 x^{2} B b c +12 A b c x -3 x B \,b^{2}+3 b^{2} A \right )}{3 b^{3} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\) | \(62\) |
trager | \(-\frac {2 \left (8 A \,c^{2} x^{2}-2 x^{2} B b c +12 A b c x -3 x B \,b^{2}+3 b^{2} A \right ) \sqrt {c \,x^{2}+b x}}{3 b^{3} \left (c x +b \right )^{2} x}\) | \(64\) |
risch | \(-\frac {2 A \left (c x +b \right )}{b^{3} \sqrt {x \left (c x +b \right )}}-\frac {2 \left (5 A \,c^{2} x -2 B b c x +6 A b c -3 B \,b^{2}\right ) x}{3 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right ) b^{3}}\) | \(69\) |
default | \(A \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )+B \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )\) | \(168\) |
Input:
int(x*(B*x+A)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2*((-B*x+A)*b^2+4*c*x*(-1/6*B*x+A)*b+8/3*A*c^2*x^2)/(x*(c*x+b))^(1/2)/(c* x+b)/b^3
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, A b^{2} - 2 \, {\left (B b c - 4 \, A c^{2}\right )} x^{2} - 3 \, {\left (B b^{2} - 4 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{3} c^{2} x^{3} + 2 \, b^{4} c x^{2} + b^{5} x\right )}} \] Input:
integrate(x*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")
Output:
-2/3*(3*A*b^2 - 2*(B*b*c - 4*A*c^2)*x^2 - 3*(B*b^2 - 4*A*b*c)*x)*sqrt(c*x^ 2 + b*x)/(b^3*c^2*x^3 + 2*b^4*c*x^2 + b^5*x)
\[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {x \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x*(B*x+A)/(c*x**2+b*x)**(5/2),x)
Output:
Integral(x*(A + B*x)/(x*(b + c*x))**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.37 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {4 \, B x}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, A x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {2 \, B x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {16 \, A c x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {8 \, A}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, B}{3 \, \sqrt {c x^{2} + b x} b c} \] Input:
integrate(x*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")
Output:
4/3*B*x/(sqrt(c*x^2 + b*x)*b^2) + 2/3*A*x/((c*x^2 + b*x)^(3/2)*b) - 2/3*B* x/((c*x^2 + b*x)^(3/2)*c) - 16/3*A*c*x/(sqrt(c*x^2 + b*x)*b^3) - 8/3*A/(sq rt(c*x^2 + b*x)*b^2) + 2/3*B/(sqrt(c*x^2 + b*x)*b*c)
\[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (B x + A\right )} x}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(x*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")
Output:
integrate((B*x + A)*x/(c*x^2 + b*x)^(5/2), x)
Time = 5.51 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (-3\,B\,b^2\,x+3\,A\,b^2-2\,B\,b\,c\,x^2+12\,A\,b\,c\,x+8\,A\,c^2\,x^2\right )}{3\,b^3\,x\,{\left (b+c\,x\right )}^2} \] Input:
int((x*(A + B*x))/(b*x + c*x^2)^(5/2),x)
Output:
-(2*(b*x + c*x^2)^(1/2)*(3*A*b^2 + 8*A*c^2*x^2 - 3*B*b^2*x - 2*B*b*c*x^2 + 12*A*b*c*x))/(3*b^3*x*(b + c*x)^2)
Time = 0.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.74 \[ \int \frac {x (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {\frac {16 \sqrt {c}\, \sqrt {c x +b}\, a b c x}{3}+\frac {16 \sqrt {c}\, \sqrt {c x +b}\, a \,c^{2} x^{2}}{3}-\frac {4 \sqrt {c}\, \sqrt {c x +b}\, b^{3} x}{3}-\frac {4 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c \,x^{2}}{3}-2 \sqrt {x}\, a \,b^{2} c -8 \sqrt {x}\, a b \,c^{2} x -\frac {16 \sqrt {x}\, a \,c^{3} x^{2}}{3}+2 \sqrt {x}\, b^{3} c x +\frac {4 \sqrt {x}\, b^{2} c^{2} x^{2}}{3}}{\sqrt {c x +b}\, b^{3} c x \left (c x +b \right )} \] Input:
int(x*(B*x+A)/(c*x^2+b*x)^(5/2),x)
Output:
(2*(8*sqrt(c)*sqrt(b + c*x)*a*b*c*x + 8*sqrt(c)*sqrt(b + c*x)*a*c**2*x**2 - 2*sqrt(c)*sqrt(b + c*x)*b**3*x - 2*sqrt(c)*sqrt(b + c*x)*b**2*c*x**2 - 3 *sqrt(x)*a*b**2*c - 12*sqrt(x)*a*b*c**2*x - 8*sqrt(x)*a*c**3*x**2 + 3*sqrt (x)*b**3*c*x + 2*sqrt(x)*b**2*c**2*x**2))/(3*sqrt(b + c*x)*b**3*c*x*(b + c *x))