\(\int \frac {A+B x}{x^3 (b x+c x^2)^{3/2}} \, dx\) [160]

Optimal result

Integrand size = 22, antiderivative size = 157 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 A}{7 b x^3 \sqrt {b x+c x^2}}+\frac {2 (7 b B-8 A c)}{7 b^2 x^2 \sqrt {b x+c x^2}}-\frac {12 (7 b B-8 A c) \sqrt {b x+c x^2}}{35 b^3 x^3}+\frac {16 c (7 b B-8 A c) \sqrt {b x+c x^2}}{35 b^4 x^2}-\frac {32 c^2 (7 b B-8 A c) \sqrt {b x+c x^2}}{35 b^5 x} \] Output:

-2/7*A/b/x^3/(c*x^2+b*x)^(1/2)+2/7*(-8*A*c+7*B*b)/b^2/x^2/(c*x^2+b*x)^(1/2 
)-12/35*(-8*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/b^3/x^3+16/35*c*(-8*A*c+7*B*b)*(c 
*x^2+b*x)^(1/2)/b^4/x^2-32/35*c^2*(-8*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/b^5/x
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.62 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (7 b B x \left (b^3-2 b^2 c x+8 b c^2 x^2+16 c^3 x^3\right )+A \left (5 b^4-8 b^3 c x+16 b^2 c^2 x^2-64 b c^3 x^3-128 c^4 x^4\right )\right )}{35 b^5 x^3 \sqrt {x (b+c x)}} \] Input:

Integrate[(A + B*x)/(x^3*(b*x + c*x^2)^(3/2)),x]
 

Output:

(-2*(7*b*B*x*(b^3 - 2*b^2*c*x + 8*b*c^2*x^2 + 16*c^3*x^3) + A*(5*b^4 - 8*b 
^3*c*x + 16*b^2*c^2*x^2 - 64*b*c^3*x^3 - 128*c^4*x^4)))/(35*b^5*x^3*Sqrt[x 
*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1220, 1129, 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.

\(\displaystyle \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(7 b B-8 A c) \int \frac {1}{x^2 \left (c x^2+b x\right )^{3/2}}dx}{7 b}-\frac {2 A}{7 b x^3 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(7 b B-8 A c) \left (-\frac {6 c \int \frac {1}{x \left (c x^2+b x\right )^{3/2}}dx}{5 b}-\frac {2}{5 b x^2 \sqrt {b x+c x^2}}\right )}{7 b}-\frac {2 A}{7 b x^3 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(7 b B-8 A c) \left (-\frac {6 c \left (-\frac {4 c \int \frac {1}{\left (c x^2+b x\right )^{3/2}}dx}{3 b}-\frac {2}{3 b x \sqrt {b x+c x^2}}\right )}{5 b}-\frac {2}{5 b x^2 \sqrt {b x+c x^2}}\right )}{7 b}-\frac {2 A}{7 b x^3 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1088

\(\displaystyle \frac {\left (-\frac {6 c \left (\frac {8 c (b+2 c x)}{3 b^3 \sqrt {b x+c x^2}}-\frac {2}{3 b x \sqrt {b x+c x^2}}\right )}{5 b}-\frac {2}{5 b x^2 \sqrt {b x+c x^2}}\right ) (7 b B-8 A c)}{7 b}-\frac {2 A}{7 b x^3 \sqrt {b x+c x^2}}\)

Input:

Int[(A + B*x)/(x^3*(b*x + c*x^2)^(3/2)),x]
 

Output:

(-2*A)/(7*b*x^3*Sqrt[b*x + c*x^2]) + ((7*b*B - 8*A*c)*(-2/(5*b*x^2*Sqrt[b* 
x + c*x^2]) - (6*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)))/(7*b)
 

Defintions of rubi rules used

rule 1088

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]
 

rule 1129

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]
 

rule 1220

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 
]
 

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (\frac {7 B x}{5}+A \right ) b^{4}-\frac {8 \left (\frac {7 B x}{4}+A \right ) c x \,b^{3}}{5}+\frac {16 c^{2} x^{2} \left (\frac {7 B x}{2}+A \right ) b^{2}}{5}-\frac {64 c^{3} \left (-\frac {7 B x}{4}+A \right ) x^{3} b}{5}-\frac {128 A \,c^{4} x^{4}}{5}\right )}{7 \sqrt {x \left (c x +b \right )}\, x^{3} b^{5}}\)	\(83\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (-128 A \,c^{4} x^{4}+112 B b \,c^{3} x^{4}-64 A b \,c^{3} x^{3}+56 B \,b^{2} c^{2} x^{3}+16 A \,b^{2} c^{2} x^{2}-14 B \,b^{3} c \,x^{2}-8 A \,b^{3} c x +7 B \,b^{4} x +5 A \,b^{4}\right )}{35 x^{2} b^{5} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(110\)
orering	\(-\frac {2 \left (c x +b \right ) \left (-128 A \,c^{4} x^{4}+112 B b \,c^{3} x^{4}-64 A b \,c^{3} x^{3}+56 B \,b^{2} c^{2} x^{3}+16 A \,b^{2} c^{2} x^{2}-14 B \,b^{3} c \,x^{2}-8 A \,b^{3} c x +7 B \,b^{4} x +5 A \,b^{4}\right )}{35 x^{2} b^{5} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(110\)
risch	\(-\frac {2 \left (c x +b \right ) \left (-93 A \,c^{3} x^{3}+77 x^{3} B b \,c^{2}+29 A b \,c^{2} x^{2}-21 x^{2} B \,b^{2} c -13 A \,b^{2} c x +7 x B \,b^{3}+5 A \,b^{3}\right )}{35 b^{5} x^{3} \sqrt {x \left (c x +b \right )}}+\frac {2 c^{3} \left (A c -B b \right ) x}{\sqrt {x \left (c x +b \right )}\, b^{5}}\)	\(111\)
trager	\(-\frac {2 \left (-128 A \,c^{4} x^{4}+112 B b \,c^{3} x^{4}-64 A b \,c^{3} x^{3}+56 B \,b^{2} c^{2} x^{3}+16 A \,b^{2} c^{2} x^{2}-14 B \,b^{3} c \,x^{2}-8 A \,b^{3} c x +7 B \,b^{4} x +5 A \,b^{4}\right ) \sqrt {c \,x^{2}+b x}}{35 \left (c x +b \right ) b^{5} x^{4}}\)	\(112\)
default	\(A \left (-\frac {2}{7 b \,x^{3} \sqrt {c \,x^{2}+b x}}-\frac {8 c \left (-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )}{7 b}\right )+B \left (-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )\)	\(170\)

Input:

int((B*x+A)/x^3/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/7*((7/5*B*x+A)*b^4-8/5*(7/4*B*x+A)*c*x*b^3+16/5*c^2*x^2*(7/2*B*x+A)*b^2 
-64/5*c^3*(-7/4*B*x+A)*x^3*b-128/5*A*c^4*x^4)/(x*(c*x+b))^(1/2)/x^3/b^5
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (5 \, A b^{4} + 16 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{4} + 8 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{3} - 2 \, {\left (7 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{2} + {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{35 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}} \] Input:

integrate((B*x+A)/x^3/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
 

Output:

-2/35*(5*A*b^4 + 16*(7*B*b*c^3 - 8*A*c^4)*x^4 + 8*(7*B*b^2*c^2 - 8*A*b*c^3 
)*x^3 - 2*(7*B*b^3*c - 8*A*b^2*c^2)*x^2 + (7*B*b^4 - 8*A*b^3*c)*x)*sqrt(c* 
x^2 + b*x)/(b^5*c*x^5 + b^6*x^4)
 

Sympy [F]

\[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{x^{3} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((B*x+A)/x**3/(c*x**2+b*x)**(3/2),x)
 

Output:

Integral((A + B*x)/(x**3*(x*(b + c*x))**(3/2)), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {32 \, B c^{3} x}{5 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {256 \, A c^{4} x}{35 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {16 \, B c^{2}}{5 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {128 \, A c^{3}}{35 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {4 \, B c}{5 \, \sqrt {c x^{2} + b x} b^{2} x} - \frac {32 \, A c^{2}}{35 \, \sqrt {c x^{2} + b x} b^{3} x} - \frac {2 \, B}{5 \, \sqrt {c x^{2} + b x} b x^{2}} + \frac {16 \, A c}{35 \, \sqrt {c x^{2} + b x} b^{2} x^{2}} - \frac {2 \, A}{7 \, \sqrt {c x^{2} + b x} b x^{3}} \] Input:

integrate((B*x+A)/x^3/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
 

Output:

-32/5*B*c^3*x/(sqrt(c*x^2 + b*x)*b^4) + 256/35*A*c^4*x/(sqrt(c*x^2 + b*x)* 
b^5) - 16/5*B*c^2/(sqrt(c*x^2 + b*x)*b^3) + 128/35*A*c^3/(sqrt(c*x^2 + b*x 
)*b^4) + 4/5*B*c/(sqrt(c*x^2 + b*x)*b^2*x) - 32/35*A*c^2/(sqrt(c*x^2 + b*x 
)*b^3*x) - 2/5*B/(sqrt(c*x^2 + b*x)*b*x^2) + 16/35*A*c/(sqrt(c*x^2 + b*x)* 
b^2*x^2) - 2/7*A/(sqrt(c*x^2 + b*x)*b*x^3)
 

Giac [F]

\[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{3}} \,d x } \] Input:

integrate((B*x+A)/x^3/(c*x^2+b*x)^(3/2),x, algorithm="giac")
 

Output:

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^3), x)
 

Mupad [B] (verification not implemented)

Time = 5.72 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {\left (14\,B\,b^2-26\,A\,b\,c\right )\,\sqrt {c\,x^2+b\,x}}{35\,b^4\,x^3}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{7\,b^2\,x^4}-\frac {\sqrt {c\,x^2+b\,x}\,\left (x\,\left (\frac {116\,A\,c^4-84\,B\,b\,c^3}{35\,b^5}-\frac {4\,c^3\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^5}\right )-\frac {2\,c^2\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^4}\right )}{x\,\left (b+c\,x\right )}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}\,\left (29\,A\,c-21\,B\,b\right )}{35\,b^4\,x^2} \] Input:

int((A + B*x)/(x^3*(b*x + c*x^2)^(3/2)),x)
 

Output:

- ((14*B*b^2 - 26*A*b*c)*(b*x + c*x^2)^(1/2))/(35*b^4*x^3) - (2*A*(b*x + c 
*x^2)^(1/2))/(7*b^2*x^4) - ((b*x + c*x^2)^(1/2)*(x*((116*A*c^4 - 84*B*b*c^ 
3)/(35*b^5) - (4*c^3*(93*A*c - 77*B*b))/(35*b^5)) - (2*c^2*(93*A*c - 77*B* 
b))/(35*b^4)))/(x*(b + c*x)) - (2*c*(b*x + c*x^2)^(1/2)*(29*A*c - 21*B*b)) 
/(35*b^4*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {-\frac {256 \sqrt {c}\, \sqrt {c x +b}\, a \,c^{3} x^{4}}{35}+\frac {32 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{2} x^{4}}{5}-\frac {2 \sqrt {x}\, a \,b^{4}}{7}+\frac {16 \sqrt {x}\, a \,b^{3} c x}{35}-\frac {32 \sqrt {x}\, a \,b^{2} c^{2} x^{2}}{35}+\frac {128 \sqrt {x}\, a b \,c^{3} x^{3}}{35}+\frac {256 \sqrt {x}\, a \,c^{4} x^{4}}{35}-\frac {2 \sqrt {x}\, b^{5} x}{5}+\frac {4 \sqrt {x}\, b^{4} c \,x^{2}}{5}-\frac {16 \sqrt {x}\, b^{3} c^{2} x^{3}}{5}-\frac {32 \sqrt {x}\, b^{2} c^{3} x^{4}}{5}}{\sqrt {c x +b}\, b^{5} x^{4}} \] Input:

int((B*x+A)/x^3/(c*x^2+b*x)^(3/2),x)
 

Output:

(2*( - 128*sqrt(c)*sqrt(b + c*x)*a*c**3*x**4 + 112*sqrt(c)*sqrt(b + c*x)*b 
**2*c**2*x**4 - 5*sqrt(x)*a*b**4 + 8*sqrt(x)*a*b**3*c*x - 16*sqrt(x)*a*b** 
2*c**2*x**2 + 64*sqrt(x)*a*b*c**3*x**3 + 128*sqrt(x)*a*c**4*x**4 - 7*sqrt( 
x)*b**5*x + 14*sqrt(x)*b**4*c*x**2 - 56*sqrt(x)*b**3*c**2*x**3 - 112*sqrt( 
x)*b**2*c**3*x**4))/(35*sqrt(b + c*x)*b**5*x**4)