3.1882 \(\int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx\)

Optimal. Leaf size=95 \[ \frac{(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-1}}{a^2 b c^2 (m+1) (m+2)}-\frac{(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-2}}{a b c (m+2)} \]

[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))

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Rubi [A]  time = 0.0360218, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {45, 37} \[ \frac{(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-1}}{a^2 b c^2 (m+1) (m+2)}-\frac{(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-2}}{a b c (m+2)} \]

Antiderivative was successfully verified.

[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 45

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])

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]

Rubi steps

\begin{align*} \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{\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]  time = 0.0516635, size = 54, normalized size = 0.57 \[ \frac{x (a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m}}{a^2 c^3 (m+1) (a (m+1)+b (m+2) x)^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.005, size = 57, normalized size = 0.6 \begin{align*}{\frac{ \left ( bx+a \right ) ^{1+m} \left ( bxm+am+2\,bx+a \right ) x \left ( bcxm+acm+2\,bcx+ac \right ) ^{-3-m}}{{a}^{2} \left ( 1+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.61231, size = 181, normalized size = 1.91 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)