Integrand size = 22, antiderivative size = 86 \[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 (3 b B-4 A c)}{3 b^2 \sqrt {b x+c x^2}}-\frac {2 A}{3 b x \sqrt {b x+c x^2}}-\frac {4 (3 b B-4 A c) \sqrt {b x+c x^2}}{3 b^3 x} \] Output:
2/3*(-4*A*c+3*B*b)/b^2/(c*x^2+b*x)^(1/2)-2/3*A/b/x/(c*x^2+b*x)^(1/2)-4/3*( -4*A*c+3*B*b)*(c*x^2+b*x)^(1/2)/b^3/x
Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (3 b B x (b+2 c x)+A \left (b^2-4 b c x-8 c^2 x^2\right )\right )}{3 b^3 x \sqrt {x (b+c x)}} \] Input:
Integrate[(A + B*x)/(x*(b*x + c*x^2)^(3/2)),x]
Output:
(-2*(3*b*B*x*(b + 2*c*x) + A*(b^2 - 4*b*c*x - 8*c^2*x^2)))/(3*b^3*x*Sqrt[x *(b + c*x)])
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1220, 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.
\(\displaystyle \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(3 b B-4 A c) \int \frac {1}{\left (c x^2+b x\right )^{3/2}}dx}{3 b}-\frac {2 A}{3 b x \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle -\frac {2 (b+2 c x) (3 b B-4 A c)}{3 b^3 \sqrt {b x+c x^2}}-\frac {2 A}{3 b x \sqrt {b x+c x^2}}\) |
Input:
Int[(A + B*x)/(x*(b*x + c*x^2)^(3/2)),x]
Output:
(-2*A)/(3*b*x*Sqrt[b*x + c*x^2]) - (2*(3*b*B - 4*A*c)*(b + 2*c*x))/(3*b^3* 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[(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 = 0.94 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.57
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (3 B x +A \right ) b^{2}-4 c x \left (-\frac {3 B x}{2}+A \right ) b -8 A \,c^{2} x^{2}\right )}{3 \sqrt {x \left (c x +b \right )}\, x \,b^{3}}\) | \(49\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-8 A \,c^{2} x^{2}+6 x^{2} B b c -4 A b c x +3 x B \,b^{2}+b^{2} A \right )}{3 b^{3} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(58\) |
orering | \(-\frac {2 \left (c x +b \right ) \left (-8 A \,c^{2} x^{2}+6 x^{2} B b c -4 A b c x +3 x B \,b^{2}+b^{2} A \right )}{3 b^{3} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(58\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (-5 A c x +3 B b x +A b \right )}{3 b^{3} x \sqrt {x \left (c x +b \right )}}+\frac {2 c \left (A c -B b \right ) x}{\sqrt {x \left (c x +b \right )}\, b^{3}}\) | \(62\) |
trager | \(-\frac {2 \left (-8 A \,c^{2} x^{2}+6 x^{2} B b c -4 A b c x +3 x B \,b^{2}+b^{2} A \right ) \sqrt {c \,x^{2}+b x}}{3 \left (c x +b \right ) b^{3} x^{2}}\) | \(63\) |
default | \(-\frac {2 B \left (2 c x +b \right )}{b^{2} \sqrt {c \,x^{2}+b x}}+A \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )\) | \(70\) |
Input:
int((B*x+A)/x/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/3/(x*(c*x+b))^(1/2)*((3*B*x+A)*b^2-4*c*x*(-3/2*B*x+A)*b-8*A*c^2*x^2)/x/ b^3
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (A b^{2} + 2 \, {\left (3 \, B b c - 4 \, A c^{2}\right )} x^{2} + {\left (3 \, B b^{2} - 4 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{3} c x^{3} + b^{4} x^{2}\right )}} \] Input:
integrate((B*x+A)/x/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
Output:
-2/3*(A*b^2 + 2*(3*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*x^3 + b^4*x^2)
\[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{x \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((B*x+A)/x/(c*x**2+b*x)**(3/2),x)
Output:
Integral((A + B*x)/(x*(x*(b + c*x))**(3/2)), x)
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {4 \, B c x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {16 \, A c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {2 \, B}{\sqrt {c x^{2} + b x} b} + \frac {8 \, A c}{3 \, \sqrt {c x^{2} + b x} b^{2}} - \frac {2 \, A}{3 \, \sqrt {c x^{2} + b x} b x} \] Input:
integrate((B*x+A)/x/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
Output:
-4*B*c*x/(sqrt(c*x^2 + b*x)*b^2) + 16/3*A*c^2*x/(sqrt(c*x^2 + b*x)*b^3) - 2*B/(sqrt(c*x^2 + b*x)*b) + 8/3*A*c/(sqrt(c*x^2 + b*x)*b^2) - 2/3*A/(sqrt( c*x^2 + b*x)*b*x)
\[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x} \,d x } \] Input:
integrate((B*x+A)/x/(c*x^2+b*x)^(3/2),x, algorithm="giac")
Output:
integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x), x)
Time = 5.35 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (3\,B\,b^2\,x+A\,b^2+6\,B\,b\,c\,x^2-4\,A\,b\,c\,x-8\,A\,c^2\,x^2\right )}{3\,b^3\,x^2\,\left (b+c\,x\right )} \] Input:
int((A + B*x)/(x*(b*x + c*x^2)^(3/2)),x)
Output:
-(2*(b*x + c*x^2)^(1/2)*(A*b^2 - 8*A*c^2*x^2 + 3*B*b^2*x + 6*B*b*c*x^2 - 4 *A*b*c*x))/(3*b^3*x^2*(b + c*x))
Time = 0.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx=\frac {-\frac {16 \sqrt {c}\, \sqrt {c x +b}\, a c \,x^{2}}{3}+4 \sqrt {c}\, \sqrt {c x +b}\, b^{2} x^{2}-\frac {2 \sqrt {x}\, a \,b^{2}}{3}+\frac {8 \sqrt {x}\, a b c x}{3}+\frac {16 \sqrt {x}\, a \,c^{2} x^{2}}{3}-2 \sqrt {x}\, b^{3} x -4 \sqrt {x}\, b^{2} c \,x^{2}}{\sqrt {c x +b}\, b^{3} x^{2}} \] Input:
int((B*x+A)/x/(c*x^2+b*x)^(3/2),x)
Output:
(2*( - 8*sqrt(c)*sqrt(b + c*x)*a*c*x**2 + 6*sqrt(c)*sqrt(b + c*x)*b**2*x** 2 - sqrt(x)*a*b**2 + 4*sqrt(x)*a*b*c*x + 8*sqrt(x)*a*c**2*x**2 - 3*sqrt(x) *b**3*x - 6*sqrt(x)*b**2*c*x**2))/(3*sqrt(b + c*x)*b**3*x**2)