\(\int \frac {c+d x}{x^3 (a x+b x^2)^{3/2}} \, dx\) [167]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.60 \[ \int \frac {c+d x}{x^3 \left (a x+b x^2\right )^{3/2}} \, dx=-\frac {2 \left (-128 b^4 c x^4+16 a b^3 x^3 (-4 c+7 d x)+8 a^2 b^2 x^2 (2 c+7 d x)-2 a^3 b x (4 c+7 d x)+a^4 (5 c+7 d x)\right )}{35 a^5 x^3 \sqrt {x (a+b x)}} \] Input:

Integrate[(c + d*x)/(x^3*(a*x + b*x^2)^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (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.

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1088

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

Input:

Int[(c + d*x)/(x^3*(a*x + b*x^2)^(3/2)),x]
 

Output:

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

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 = 0.52 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.75 \[ \int \frac {c+d x}{x^3 \left (a x+b x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (5 \, a^{4} c - 16 \, {\left (8 \, b^{4} c - 7 \, a b^{3} d\right )} x^{4} - 8 \, {\left (8 \, a b^{3} c - 7 \, a^{2} b^{2} d\right )} x^{3} + 2 \, {\left (8 \, a^{2} b^{2} c - 7 \, a^{3} b d\right )} x^{2} - {\left (8 \, a^{3} b c - 7 \, a^{4} d\right )} x\right )} \sqrt {b x^{2} + a x}}{35 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

integrate((d*x+c)/x**3/(b*x**2+a*x)**(3/2),x)
 

Output:

Integral((c + d*x)/(x**3*(x*(a + b*x))**(3/2)), x)
 

Maxima [A] (verification not implemented)

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

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

Output:

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

Giac [F]

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

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

Output:

integrate((d*x + c)/((b*x^2 + a*x)^(3/2)*x^3), x)
 

Mupad [B] (verification not implemented)

Time = 9.71 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x}{x^3 \left (a x+b x^2\right )^{3/2}} \, dx=\frac {2\,b\,\sqrt {b\,x^2+a\,x}\,\left (21\,a\,d-29\,b\,c\right )}{35\,a^4\,x^2}-\frac {2\,c\,\sqrt {b\,x^2+a\,x}}{7\,a^2\,x^4}-\frac {\sqrt {b\,x^2+a\,x}\,\left (14\,a^2\,d-26\,a\,b\,c\right )}{35\,a^4\,x^3}-\frac {\sqrt {b\,x^2+a\,x}\,\left (x\,\left (\frac {116\,b^4\,c-84\,a\,b^3\,d}{35\,a^5}+\frac {4\,b^3\,\left (77\,a\,d-93\,b\,c\right )}{35\,a^5}\right )+\frac {2\,b^2\,\left (77\,a\,d-93\,b\,c\right )}{35\,a^4}\right )}{x\,\left (a+b\,x\right )} \] Input:

int((c + d*x)/(x^3*(a*x + b*x^2)^(3/2)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x}{x^3 \left (a x+b x^2\right )^{3/2}} \, dx=\frac {\frac {32 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} d \,x^{4}}{5}-\frac {256 \sqrt {b}\, \sqrt {b x +a}\, b^{3} c \,x^{4}}{35}-\frac {2 \sqrt {x}\, a^{4} c}{7}-\frac {2 \sqrt {x}\, a^{4} d x}{5}+\frac {16 \sqrt {x}\, a^{3} b c x}{35}+\frac {4 \sqrt {x}\, a^{3} b d \,x^{2}}{5}-\frac {32 \sqrt {x}\, a^{2} b^{2} c \,x^{2}}{35}-\frac {16 \sqrt {x}\, a^{2} b^{2} d \,x^{3}}{5}+\frac {128 \sqrt {x}\, a \,b^{3} c \,x^{3}}{35}-\frac {32 \sqrt {x}\, a \,b^{3} d \,x^{4}}{5}+\frac {256 \sqrt {x}\, b^{4} c \,x^{4}}{35}}{\sqrt {b x +a}\, a^{5} x^{4}} \] Input:

int((d*x+c)/x^3/(b*x^2+a*x)^(3/2),x)
 

Output:

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