\(\int \frac {(c+d x)^2}{x^2 (a x+b x^2)^{3/2}} \, dx\) [174]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {(c+d x)^2}{x^2 \left (a x+b x^2\right )^{3/2}} \, dx=-\frac {8 (b c-a d) (6 b c-a d)}{15 a^3 \sqrt {a x+b x^2}}-\frac {8 (b c-a d) (2 b c-a d) (6 b c-a d) x}{15 a^4 c \sqrt {a x+b x^2}}+\frac {2 (6 b c-a d) (c+d x)^2}{15 a^2 c x \sqrt {a x+b x^2}}-\frac {2 (c+d x)^3}{5 a c x^2 \sqrt {a x+b x^2}} \] Output:

-8/15*(-a*d+b*c)*(-a*d+6*b*c)/a^3/(b*x^2+a*x)^(1/2)-8/15*(-a*d+b*c)*(-a*d+ 
2*b*c)*(-a*d+6*b*c)*x/a^4/c/(b*x^2+a*x)^(1/2)+2/15*(-a*d+6*b*c)*(d*x+c)^2/ 
a^2/c/x/(b*x^2+a*x)^(1/2)-2/5*(d*x+c)^3/a/c/x^2/(b*x^2+a*x)^(1/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.61 \[ \int \frac {(c+d x)^2}{x^2 \left (a x+b x^2\right )^{3/2}} \, dx=-\frac {2 \left (48 b^3 c^2 x^3+8 a b^2 c x^2 (3 c-10 d x)+2 a^2 b x \left (-3 c^2-20 c d x+15 d^2 x^2\right )+a^3 \left (3 c^2+10 c d x+15 d^2 x^2\right )\right )}{15 a^4 x^2 \sqrt {x (a+b x)}} \] Input:

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

Output:

(-2*(48*b^3*c^2*x^3 + 8*a*b^2*c*x^2*(3*c - 10*d*x) + 2*a^2*b*x*(-3*c^2 - 2 
0*c*d*x + 15*d^2*x^2) + a^3*(3*c^2 + 10*c*d*x + 15*d^2*x^2)))/(15*a^4*x^2* 
Sqrt[x*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1262, 27, 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.

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

\(\Big \downarrow \) 1262

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1220

\(\displaystyle \frac {-\frac {\left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \int \frac {1}{x \left (b x^2+a x\right )^{3/2}}dx}{5 a}-\frac {8 b c^2}{5 a x^2 \sqrt {a x+b x^2}}}{4 b}-\frac {d^2}{2 b x \sqrt {a x+b x^2}}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {-\frac {\left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \left (-\frac {4 b \int \frac {1}{\left (b x^2+a x\right )^{3/2}}dx}{3 a}-\frac {2}{3 a x \sqrt {a x+b x^2}}\right )}{5 a}-\frac {8 b c^2}{5 a x^2 \sqrt {a x+b x^2}}}{4 b}-\frac {d^2}{2 b x \sqrt {a x+b x^2}}\)

\(\Big \downarrow \) 1088

\(\displaystyle \frac {-\frac {\left (\frac {8 b (a+2 b x)}{3 a^3 \sqrt {a x+b x^2}}-\frac {2}{3 a x \sqrt {a x+b x^2}}\right ) \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right )}{5 a}-\frac {8 b c^2}{5 a x^2 \sqrt {a x+b x^2}}}{4 b}-\frac {d^2}{2 b x \sqrt {a x+b x^2}}\)

Input:

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

Output:

-1/2*d^2/(b*x*Sqrt[a*x + b*x^2]) + ((-8*b*c^2)/(5*a*x^2*Sqrt[a*x + b*x^2]) 
 - ((24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d^2)*(-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))/(4*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 
]
 

rule 1262

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b 
*x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m 
 + n + 2*p + 1))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m 
 + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n + e*g^n*( 
m + p + n)*(d + e*x)^(n - 2)*(b*d - 2*a*e + (2*c*d - b*e)*x), x], x], x] /; 
 FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && 
IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.56

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (5 d^{2} x^{2}+\frac {10}{3} c d x +c^{2}\right ) a^{3}-2 x b \left (-5 d^{2} x^{2}+\frac {20}{3} c d x +c^{2}\right ) a^{2}+8 x^{2} \left (-\frac {10 d x}{3}+c \right ) b^{2} c a +16 b^{3} c^{2} x^{3}\right )}{5 \sqrt {x \left (b x +a \right )}\, x^{2} a^{4}}\)	\(91\)
risch	\(-\frac {2 \left (b x +a \right ) \left (15 a^{2} d^{2} x^{2}-50 a b c d \,x^{2}+33 b^{2} c^{2} x^{2}+10 a^{2} c d x -9 a b \,c^{2} x +3 a^{2} c^{2}\right )}{15 a^{4} x^{2} \sqrt {x \left (b x +a \right )}}-\frac {2 b \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x}{\sqrt {x \left (b x +a \right )}\, a^{4}}\)	\(117\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (30 d^{2} x^{3} a^{2} b -80 a \,b^{2} c d \,x^{3}+48 b^{3} c^{2} x^{3}+15 a^{3} d^{2} x^{2}-40 x^{2} a^{2} b c d +24 a \,b^{2} c^{2} x^{2}+10 a^{3} c d x -6 a^{2} b \,c^{2} x +3 c^{2} a^{3}\right )}{15 x \,a^{4} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}\)	\(120\)
orering	\(-\frac {2 \left (b x +a \right ) \left (30 d^{2} x^{3} a^{2} b -80 a \,b^{2} c d \,x^{3}+48 b^{3} c^{2} x^{3}+15 a^{3} d^{2} x^{2}-40 x^{2} a^{2} b c d +24 a \,b^{2} c^{2} x^{2}+10 a^{3} c d x -6 a^{2} b \,c^{2} x +3 c^{2} a^{3}\right )}{15 x \,a^{4} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}\)	\(120\)
trager	\(-\frac {2 \left (30 d^{2} x^{3} a^{2} b -80 a \,b^{2} c d \,x^{3}+48 b^{3} c^{2} x^{3}+15 a^{3} d^{2} x^{2}-40 x^{2} a^{2} b c d +24 a \,b^{2} c^{2} x^{2}+10 a^{3} c d x -6 a^{2} b \,c^{2} x +3 c^{2} a^{3}\right ) \sqrt {b \,x^{2}+a x}}{15 \left (b x +a \right ) a^{4} x^{3}}\)	\(122\)
default	\(-\frac {2 d^{2} \left (2 b x +a \right )}{a^{2} \sqrt {b \,x^{2}+a x}}+c^{2} \left (-\frac {2}{5 a \,x^{2} \sqrt {b \,x^{2}+a x}}-\frac {6 b \left (-\frac {2}{3 a x \sqrt {b \,x^{2}+a x}}+\frac {8 b \left (2 b x +a \right )}{3 a^{3} \sqrt {b \,x^{2}+a x}}\right )}{5 a}\right )+2 c d \left (-\frac {2}{3 a x \sqrt {b \,x^{2}+a x}}+\frac {8 b \left (2 b x +a \right )}{3 a^{3} \sqrt {b \,x^{2}+a x}}\right )\)	\(147\)

Input:

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

Output:

-2/5/(x*(b*x+a))^(1/2)*((5*d^2*x^2+10/3*c*d*x+c^2)*a^3-2*x*b*(-5*d^2*x^2+2 
0/3*c*d*x+c^2)*a^2+8*x^2*(-10/3*d*x+c)*b^2*c*a+16*b^3*c^2*x^3)/x^2/a^4
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.74 \[ \int \frac {(c+d x)^2}{x^2 \left (a x+b x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (3 \, a^{3} c^{2} + 2 \, {\left (24 \, b^{3} c^{2} - 40 \, a b^{2} c d + 15 \, a^{2} b d^{2}\right )} x^{3} + {\left (24 \, a b^{2} c^{2} - 40 \, a^{2} b c d + 15 \, a^{3} d^{2}\right )} x^{2} - 2 \, {\left (3 \, a^{2} b c^{2} - 5 \, a^{3} c d\right )} x\right )} \sqrt {b x^{2} + a x}}{15 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \] Input:

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

Output:

-2/15*(3*a^3*c^2 + 2*(24*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24* 
a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2 - 2*(3*a^2*b*c^2 - 5*a^3*c*d)*x 
)*sqrt(b*x^2 + a*x)/(a^4*b*x^4 + a^5*x^3)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.18 \[ \int \frac {(c+d x)^2}{x^2 \left (a x+b x^2\right )^{3/2}} \, dx=-\frac {32 \, b^{3} c^{2} x}{5 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {32 \, b^{2} c d x}{3 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {4 \, b d^{2} x}{\sqrt {b x^{2} + a x} a^{2}} - \frac {16 \, b^{2} c^{2}}{5 \, \sqrt {b x^{2} + a x} a^{3}} + \frac {16 \, b c d}{3 \, \sqrt {b x^{2} + a x} a^{2}} - \frac {2 \, d^{2}}{\sqrt {b x^{2} + a x} a} + \frac {4 \, b c^{2}}{5 \, \sqrt {b x^{2} + a x} a^{2} x} - \frac {4 \, c d}{3 \, \sqrt {b x^{2} + a x} a x} - \frac {2 \, c^{2}}{5 \, \sqrt {b x^{2} + a x} a x^{2}} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.52 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.74 \[ \int \frac {(c+d x)^2}{x^2 \left (a x+b x^2\right )^{3/2}} \, dx=-\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (3\,a^3\,c^2+10\,a^3\,c\,d\,x+15\,a^3\,d^2\,x^2-6\,a^2\,b\,c^2\,x-40\,a^2\,b\,c\,d\,x^2+30\,a^2\,b\,d^2\,x^3+24\,a\,b^2\,c^2\,x^2-80\,a\,b^2\,c\,d\,x^3+48\,b^3\,c^2\,x^3\right )}{15\,a^4\,x^3\,\left (a+b\,x\right )} \] Input:

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

Output:

-(2*(a*x + b*x^2)^(1/2)*(3*a^3*c^2 + 15*a^3*d^2*x^2 + 48*b^3*c^2*x^3 + 24* 
a*b^2*c^2*x^2 + 30*a^2*b*d^2*x^3 + 10*a^3*c*d*x - 6*a^2*b*c^2*x - 40*a^2*b 
*c*d*x^2 - 80*a*b^2*c*d*x^3))/(15*a^4*x^3*(a + b*x))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x)^2}{x^2 \left (a x+b x^2\right )^{3/2}} \, dx=\frac {4 \sqrt {b}\, \sqrt {b x +a}\, a^{2} d^{2} x^{3}-\frac {32 \sqrt {b}\, \sqrt {b x +a}\, a b c d \,x^{3}}{3}+\frac {32 \sqrt {b}\, \sqrt {b x +a}\, b^{2} c^{2} x^{3}}{5}-\frac {2 \sqrt {x}\, a^{3} c^{2}}{5}-\frac {4 \sqrt {x}\, a^{3} c d x}{3}-2 \sqrt {x}\, a^{3} d^{2} x^{2}+\frac {4 \sqrt {x}\, a^{2} b \,c^{2} x}{5}+\frac {16 \sqrt {x}\, a^{2} b c d \,x^{2}}{3}-4 \sqrt {x}\, a^{2} b \,d^{2} x^{3}-\frac {16 \sqrt {x}\, a \,b^{2} c^{2} x^{2}}{5}+\frac {32 \sqrt {x}\, a \,b^{2} c d \,x^{3}}{3}-\frac {32 \sqrt {x}\, b^{3} c^{2} x^{3}}{5}}{\sqrt {b x +a}\, a^{4} x^{3}} \] Input:

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

Output:

(2*(30*sqrt(b)*sqrt(a + b*x)*a**2*d**2*x**3 - 80*sqrt(b)*sqrt(a + b*x)*a*b 
*c*d*x**3 + 48*sqrt(b)*sqrt(a + b*x)*b**2*c**2*x**3 - 3*sqrt(x)*a**3*c**2 
- 10*sqrt(x)*a**3*c*d*x - 15*sqrt(x)*a**3*d**2*x**2 + 6*sqrt(x)*a**2*b*c** 
2*x + 40*sqrt(x)*a**2*b*c*d*x**2 - 30*sqrt(x)*a**2*b*d**2*x**3 - 24*sqrt(x 
)*a*b**2*c**2*x**2 + 80*sqrt(x)*a*b**2*c*d*x**3 - 48*sqrt(x)*b**3*c**2*x** 
3))/(15*sqrt(a + b*x)*a**4*x**3)