\(\int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx\) [1882]

Optimal result

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

[Out]

-(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/(m^2+3*m+2)

Rubi [A] (verified)

Time = 0.03 (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 = {47, 37} \[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-1}}{a^2 b c^2 (m+1) (m+2)}-\frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-2}}{a b c (m+2)} \]

[In]

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

[Out]

-(((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))

Rule 37

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] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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] - Dist[d*(Simplify[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] && NeQ[b*c - a*d, 0] && 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] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{1+m} (a c (1+m)+b c (2+m) x)^{-2-m}}{a b c (2+m)}-\frac {\int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-2-m} \, dx}{a c (2+m)} \\ & = -\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)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(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)

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.60

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\frac {{\left ({\left (b^{2} m + 2 \, b^{2}\right )} x^{3} + {\left (2 \, a b m + 3 \, a b\right )} x^{2} + {\left (a^{2} m + a^{2}\right )} x\right )} {\left (a c m + a c + {\left (b c m + 2 \, b c\right )} x\right )}^{-m - 3} {\left (b x + a\right )}^{m}}{a^{2} m + a^{2}} \]

[In]

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

[Out]

((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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx=\frac {x\,{\left (a+b\,x\right )}^m+\frac {b\,x^2\,\left (2\,m+3\right )\,{\left (a+b\,x\right )}^m}{a\,\left (m+1\right )}+\frac {b^2\,x^3\,\left (m+2\right )\,{\left (a+b\,x\right )}^m}{a^2\,\left (m+1\right )}}{{\left (a\,c\,\left (m+1\right )+b\,c\,x\,\left (m+2\right )\right )}^{m+3}} \]

[In]

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

[Out]

(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)