\(\int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx\) [796]

Optimal result

Integrand size = 28, antiderivative size = 95 \[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=-\frac {(a+b x)^{1+m} (a c (1+m)+b c (2+m) x)^{-2-m}}{a b c (2+m)}+\frac {(a+b x)^{1+m} (a c (1+m)+b c (2+m) x)^{-1-m}}{a^2 b c^2 (1+m) (2+m)} \] Output:

-(b*x+a)^(1+m)*(a*c*(1+m)+b*c*(2+m)*x)^(-2-m)/a/b/c/(2+m)+(b*x+a)^(1+m)*(a 
*c*(1+m)+b*c*(2+m)*x)^(-1-m)/a^2/b/c^2/(1+m)/(2+m)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.57 \[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\frac {x (a+b x)^{1+m} (a c (1+m)+b c (2+m) x)^{-m}}{a^2 c^3 (1+m) (a (1+m)+b (2+m) x)^2} \] Input:

Integrate[(a + b*x)^m*(a*c*(1 + m) + b*c*(2 + m)*x)^(-3 - m),x]
 

Output:

(x*(a + b*x)^(1 + m))/(a^2*c^3*(1 + m)*(a*(1 + m) + b*(2 + m)*x)^2*(a*c*(1 
 + m) + b*c*(2 + m)*x)^m)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {55, 48}

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.

\(\displaystyle \int (a+b x)^m (a c (m+1)+b c (m+2) x)^{-m-3} \, dx\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {\int (a+b x)^m (a c (m+1)+b c (m+2) x)^{-m-2}dx}{a c (m+2)}-\frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-2}}{a b c (m+2)}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-1}}{a^2 b c^2 (m+1) (m+2)}-\frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-2}}{a b c (m+2)}\)

Input:

Int[(a + b*x)^m*(a*c*(1 + m) + b*c*(2 + m)*x)^(-3 - m),x]
 

Output:

-(((a + b*x)^(1 + m)*(a*c*(1 + m) + b*c*(2 + m)*x)^(-2 - m))/(a*b*c*(2 + m 
))) + ((a + b*x)^(1 + m)*(a*c*(1 + m) + b*c*(2 + m)*x)^(-1 - m))/(a^2*b*c^ 
2*(1 + m)*(2 + m))
 

Defintions of rubi rules used

rule 48

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

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.59

method	result	size
orering	\(\frac {\left (b x +a \right ) \left (b m x +a m +2 b x +a \right ) x \left (b x +a \right )^{m} \left (a c \left (1+m \right )+b c \left (2+m \right ) x \right )^{-3-m}}{a^{2} \left (1+m \right )}\)	\(56\)
gosper	\(\frac {x \left (b x +a \right )^{1+m} \left (b m x +a m +2 b x +a \right ) \left (b c x m +a c m +2 b c x +a c \right )^{-3-m}}{a^{2} \left (1+m \right )}\)	\(57\)
parallelrisch	\(\frac {x^{3} \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} b^{4} m^{2}+4 x^{3} \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} b^{4} m +2 x^{2} \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} a \,b^{3} m^{2}+4 x^{3} \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} b^{4}+7 x^{2} \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} a \,b^{3} m +x \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} a^{2} b^{2} m^{2}+6 x^{2} \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} a \,b^{3}+3 x \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} a^{2} b^{2} m +2 x \left (b x +a \right )^{m} \left (c \left (b m x +a m +2 b x +a \right )\right )^{-3-m} a^{2} b^{2}}{a^{2} \left (1+m \right ) b^{2} \left (2+m \right )}\)	\(359\)

Input:

int((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x,method=_RETURNVERBOSE)
 

Output:

(b*x+a)*(b*m*x+a*m+2*b*x+a)/a^2/(1+m)*x*(b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^ 
(-3-m)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\frac {{\left ({\left (b^{2} m + 2 \, b^{2}\right )} x^{3} + {\left (2 \, a b m + 3 \, a b\right )} x^{2} + {\left (a^{2} m + a^{2}\right )} x\right )} {\left (a c m + a c + {\left (b c m + 2 \, b c\right )} x\right )}^{-m - 3} {\left (b x + a\right )}^{m}}{a^{2} m + a^{2}} \] Input:

integrate((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x, algorithm="fricas")
 

Output:

((b^2*m + 2*b^2)*x^3 + (2*a*b*m + 3*a*b)*x^2 + (a^2*m + a^2)*x)*(a*c*m + a 
*c + (b*c*m + 2*b*c)*x)^(-m - 3)*(b*x + a)^m/(a^2*m + a^2)
 

Sympy [F]

\[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\int \left (c \left (a m + a + b m x + 2 b x\right )\right )^{- m - 3} \left (a + b x\right )^{m}\, dx \] Input:

integrate((b*x+a)**m*(a*c*(1+m)+b*c*(2+m)*x)**(-3-m),x)
 

Output:

Integral((c*(a*m + a + b*m*x + 2*b*x))**(-m - 3)*(a + b*x)**m, x)
 

Maxima [F]

\[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\int { {\left (b c {\left (m + 2\right )} x + a c {\left (m + 1\right )}\right )}^{-m - 3} {\left (b x + a\right )}^{m} \,d x } \] Input:

integrate((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x, algorithm="maxima")
 

Output:

integrate((b*c*(m + 2)*x + a*c*(m + 1))^(-m - 3)*(b*x + a)^m, x)
 

Giac [F]

\[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\int { {\left (b c {\left (m + 2\right )} x + a c {\left (m + 1\right )}\right )}^{-m - 3} {\left (b x + a\right )}^{m} \,d x } \] Input:

integrate((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x, algorithm="giac")
 

Output:

integrate((b*c*(m + 2)*x + a*c*(m + 1))^(-m - 3)*(b*x + a)^m, x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\frac {x\,{\left (a+b\,x\right )}^m+\frac {b\,x^2\,\left (2\,m+3\right )\,{\left (a+b\,x\right )}^m}{a\,\left (m+1\right )}+\frac {b^2\,x^3\,\left (m+2\right )\,{\left (a+b\,x\right )}^m}{a^2\,\left (m+1\right )}}{{\left (a\,c\,\left (m+1\right )+b\,c\,x\,\left (m+2\right )\right )}^{m+3}} \] Input:

int((a + b*x)^m/(a*c*(m + 1) + b*c*x*(m + 2))^(m + 3),x)
 

Output:

(x*(a + b*x)^m + (b*x^2*(2*m + 3)*(a + b*x)^m)/(a*(m + 1)) + (b^2*x^3*(m + 
 2)*(a + b*x)^m)/(a^2*(m + 1)))/(a*c*(m + 1) + b*c*x*(m + 2))^(m + 3)
 

Reduce [F]

\[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\frac {\int \frac {\left (b x +a \right )^{m}}{\left (b c m x +a c m +2 b c x +a c \right )^{m} a^{3} m^{3}+3 \left (b c m x +a c m +2 b c x +a c \right )^{m} a^{3} m^{2}+3 \left (b c m x +a c m +2 b c x +a c \right )^{m} a^{3} m +\left (b c m x +a c m +2 b c x +a c \right )^{m} a^{3}+3 \left (b c m x +a c m +2 b c x +a c \right )^{m} a^{2} b \,m^{3} x +12 \left (b c m x +a c m +2 b c x +a c \right )^{m} a^{2} b \,m^{2} x +15 \left (b c m x +a c m +2 b c x +a c \right )^{m} a^{2} b m x +6 \left (b c m x +a c m +2 b c x +a c \right )^{m} a^{2} b x +3 \left (b c m x +a c m +2 b c x +a c \right )^{m} a \,b^{2} m^{3} x^{2}+15 \left (b c m x +a c m +2 b c x +a c \right )^{m} a \,b^{2} m^{2} x^{2}+24 \left (b c m x +a c m +2 b c x +a c \right )^{m} a \,b^{2} m \,x^{2}+12 \left (b c m x +a c m +2 b c x +a c \right )^{m} a \,b^{2} x^{2}+\left (b c m x +a c m +2 b c x +a c \right )^{m} b^{3} m^{3} x^{3}+6 \left (b c m x +a c m +2 b c x +a c \right )^{m} b^{3} m^{2} x^{3}+12 \left (b c m x +a c m +2 b c x +a c \right )^{m} b^{3} m \,x^{3}+8 \left (b c m x +a c m +2 b c x +a c \right )^{m} b^{3} x^{3}}d x}{c^{3}} \] Input:

int((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x)
 

Output:

int((a + b*x)**m/((a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*a**3*m**3 + 3*(a*c* 
m + a*c + b*c*m*x + 2*b*c*x)**m*a**3*m**2 + 3*(a*c*m + a*c + b*c*m*x + 2*b 
*c*x)**m*a**3*m + (a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*a**3 + 3*(a*c*m + a 
*c + b*c*m*x + 2*b*c*x)**m*a**2*b*m**3*x + 12*(a*c*m + a*c + b*c*m*x + 2*b 
*c*x)**m*a**2*b*m**2*x + 15*(a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*a**2*b*m* 
x + 6*(a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*a**2*b*x + 3*(a*c*m + a*c + b*c 
*m*x + 2*b*c*x)**m*a*b**2*m**3*x**2 + 15*(a*c*m + a*c + b*c*m*x + 2*b*c*x) 
**m*a*b**2*m**2*x**2 + 24*(a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*a*b**2*m*x* 
*2 + 12*(a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*a*b**2*x**2 + (a*c*m + a*c + 
b*c*m*x + 2*b*c*x)**m*b**3*m**3*x**3 + 6*(a*c*m + a*c + b*c*m*x + 2*b*c*x) 
**m*b**3*m**2*x**3 + 12*(a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*b**3*m*x**3 + 
 8*(a*c*m + a*c + b*c*m*x + 2*b*c*x)**m*b**3*x**3),x)/c**3