Integrand size = 22, antiderivative size = 122 \[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}}+\frac {2 (5 b B-6 A c)}{5 b^2 x \sqrt {b x+c x^2}}-\frac {8 (5 b B-6 A c) \sqrt {b x+c x^2}}{15 b^3 x^2}+\frac {16 c (5 b B-6 A c) \sqrt {b x+c x^2}}{15 b^4 x} \] Output:
-2/5*A/b/x^2/(c*x^2+b*x)^(1/2)+2/5*(-6*A*c+5*B*b)/b^2/x/(c*x^2+b*x)^(1/2)- 8/15*(-6*A*c+5*B*b)*(c*x^2+b*x)^(1/2)/b^3/x^2+16/15*c*(-6*A*c+5*B*b)*(c*x^ 2+b*x)^(1/2)/b^4/x
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (5 b B x \left (b^2-4 b c x-8 c^2 x^2\right )+3 A \left (b^3-2 b^2 c x+8 b c^2 x^2+16 c^3 x^3\right )\right )}{15 b^4 x^2 \sqrt {x (b+c x)}} \] Input:
Integrate[(A + B*x)/(x^2*(b*x + c*x^2)^(3/2)),x]
Output:
(-2*(5*b*B*x*(b^2 - 4*b*c*x - 8*c^2*x^2) + 3*A*(b^3 - 2*b^2*c*x + 8*b*c^2* x^2 + 16*c^3*x^3)))/(15*b^4*x^2*Sqrt[x*(b + c*x)])
Time = 0.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1220, 1129, 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.
\(\displaystyle \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(5 b B-6 A c) \int \frac {1}{x \left (c x^2+b x\right )^{3/2}}dx}{5 b}-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {(5 b B-6 A c) \left (-\frac {4 c \int \frac {1}{\left (c x^2+b x\right )^{3/2}}dx}{3 b}-\frac {2}{3 b x \sqrt {b x+c x^2}}\right )}{5 b}-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle \frac {\left (\frac {8 c (b+2 c x)}{3 b^3 \sqrt {b x+c x^2}}-\frac {2}{3 b x \sqrt {b x+c x^2}}\right ) (5 b B-6 A c)}{5 b}-\frac {2 A}{5 b x^2 \sqrt {b x+c x^2}}\) |
Input:
Int[(A + B*x)/(x^2*(b*x + c*x^2)^(3/2)),x]
Output:
(-2*A)/(5*b*x^2*Sqrt[b*x + c*x^2]) + ((5*b*B - 6*A*c)*(-2/(3*b*x*Sqrt[b*x + c*x^2]) + (8*c*(b + 2*c*x))/(3*b^3*Sqrt[b*x + c*x^2])))/(5*b)
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_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) )) 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] && ILtQ[Simplify[m + 2*p + 2], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(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] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.96 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.54
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (\frac {5 B x}{3}+A \right ) b^{3}-2 c x \left (\frac {10 B x}{3}+A \right ) b^{2}+8 c^{2} x^{2} \left (-\frac {5 B x}{3}+A \right ) b +16 A \,c^{3} x^{3}\right )}{5 \sqrt {x \left (c x +b \right )}\, x^{2} b^{4}}\) | \(66\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (48 A \,c^{3} x^{3}-40 x^{3} B b \,c^{2}+24 A b \,c^{2} x^{2}-20 x^{2} B \,b^{2} c -6 A \,b^{2} c x +5 x B \,b^{3}+3 A \,b^{3}\right )}{15 x \,b^{4} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(86\) |
orering | \(-\frac {2 \left (c x +b \right ) \left (48 A \,c^{3} x^{3}-40 x^{3} B b \,c^{2}+24 A b \,c^{2} x^{2}-20 x^{2} B \,b^{2} c -6 A \,b^{2} c x +5 x B \,b^{3}+3 A \,b^{3}\right )}{15 x \,b^{4} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(86\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (33 A \,c^{2} x^{2}-25 x^{2} B b c -9 A b c x +5 x B \,b^{2}+3 b^{2} A \right )}{15 b^{4} x^{2} \sqrt {x \left (c x +b \right )}}-\frac {2 c^{2} \left (A c -B b \right ) x}{\sqrt {x \left (c x +b \right )}\, b^{4}}\) | \(87\) |
trager | \(-\frac {2 \left (48 A \,c^{3} x^{3}-40 x^{3} B b \,c^{2}+24 A b \,c^{2} x^{2}-20 x^{2} B \,b^{2} c -6 A \,b^{2} c x +5 x B \,b^{3}+3 A \,b^{3}\right ) \sqrt {c \,x^{2}+b x}}{15 \left (c x +b \right ) b^{4} x^{3}}\) | \(88\) |
default | \(A \left (-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )+B \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )\) | \(118\) |
Input:
int((B*x+A)/x^2/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/5/(x*(c*x+b))^(1/2)*((5/3*B*x+A)*b^3-2*c*x*(10/3*B*x+A)*b^2+8*c^2*x^2*( -5/3*B*x+A)*b+16*A*c^3*x^3)/x^2/b^4
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (3 \, A b^{3} - 8 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{3} - 4 \, {\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} x^{2} + {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}} \] Input:
integrate((B*x+A)/x^2/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
Output:
-2/15*(3*A*b^3 - 8*(5*B*b*c^2 - 6*A*c^3)*x^3 - 4*(5*B*b^2*c - 6*A*b*c^2)*x ^2 + (5*B*b^3 - 6*A*b^2*c)*x)*sqrt(c*x^2 + b*x)/(b^4*c*x^4 + b^5*x^3)
\[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{x^{2} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((B*x+A)/x**2/(c*x**2+b*x)**(3/2),x)
Output:
Integral((A + B*x)/(x**2*(x*(b + c*x))**(3/2)), x)
Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {16 \, B c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {32 \, A c^{3} x}{5 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {8 \, B c}{3 \, \sqrt {c x^{2} + b x} b^{2}} - \frac {16 \, A c^{2}}{5 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {2 \, B}{3 \, \sqrt {c x^{2} + b x} b x} + \frac {4 \, A c}{5 \, \sqrt {c x^{2} + b x} b^{2} x} - \frac {2 \, A}{5 \, \sqrt {c x^{2} + b x} b x^{2}} \] Input:
integrate((B*x+A)/x^2/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
Output:
16/3*B*c^2*x/(sqrt(c*x^2 + b*x)*b^3) - 32/5*A*c^3*x/(sqrt(c*x^2 + b*x)*b^4 ) + 8/3*B*c/(sqrt(c*x^2 + b*x)*b^2) - 16/5*A*c^2/(sqrt(c*x^2 + b*x)*b^3) - 2/3*B/(sqrt(c*x^2 + b*x)*b*x) + 4/5*A*c/(sqrt(c*x^2 + b*x)*b^2*x) - 2/5*A /(sqrt(c*x^2 + b*x)*b*x^2)
\[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:
integrate((B*x+A)/x^2/(c*x^2+b*x)^(3/2),x, algorithm="giac")
Output:
integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^2), x)
Time = 5.56 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71 \[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (5\,B\,b^3\,x+3\,A\,b^3-20\,B\,b^2\,c\,x^2-6\,A\,b^2\,c\,x-40\,B\,b\,c^2\,x^3+24\,A\,b\,c^2\,x^2+48\,A\,c^3\,x^3\right )}{15\,b^4\,x^3\,\left (b+c\,x\right )} \] Input:
int((A + B*x)/(x^2*(b*x + c*x^2)^(3/2)),x)
Output:
-(2*(b*x + c*x^2)^(1/2)*(3*A*b^3 + 48*A*c^3*x^3 + 5*B*b^3*x - 6*A*b^2*c*x + 24*A*b*c^2*x^2 - 20*B*b^2*c*x^2 - 40*B*b*c^2*x^3))/(15*b^4*x^3*(b + c*x) )
Time = 0.21 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\frac {32 \sqrt {c}\, \sqrt {c x +b}\, a \,c^{2} x^{3}}{5}-\frac {16 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c \,x^{3}}{3}-\frac {2 \sqrt {x}\, a \,b^{3}}{5}+\frac {4 \sqrt {x}\, a \,b^{2} c x}{5}-\frac {16 \sqrt {x}\, a b \,c^{2} x^{2}}{5}-\frac {32 \sqrt {x}\, a \,c^{3} x^{3}}{5}-\frac {2 \sqrt {x}\, b^{4} x}{3}+\frac {8 \sqrt {x}\, b^{3} c \,x^{2}}{3}+\frac {16 \sqrt {x}\, b^{2} c^{2} x^{3}}{3}}{\sqrt {c x +b}\, b^{4} x^{3}} \] Input:
int((B*x+A)/x^2/(c*x^2+b*x)^(3/2),x)
Output:
(2*(48*sqrt(c)*sqrt(b + c*x)*a*c**2*x**3 - 40*sqrt(c)*sqrt(b + c*x)*b**2*c *x**3 - 3*sqrt(x)*a*b**3 + 6*sqrt(x)*a*b**2*c*x - 24*sqrt(x)*a*b*c**2*x**2 - 48*sqrt(x)*a*c**3*x**3 - 5*sqrt(x)*b**4*x + 20*sqrt(x)*b**3*c*x**2 + 40 *sqrt(x)*b**2*c**2*x**3))/(15*sqrt(b + c*x)*b**4*x**3)