3.3092 \(\int (a+b x)^m (c+d x)^{-3-m} \, dx\)

Optimal. Leaf size=79 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d)}+\frac{b (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (m+2) (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.0176636, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d)}+\frac{b (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (m+2) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(-3 - m),x]

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/((b*c - a*d)*(2 + m)) + (b*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/((b*c
- a*d)^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 (c+d x)^{-3-m} \, dx &=\frac{(a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (2+m)}+\frac{b \int (a+b x)^m (c+d x)^{-2-m} \, dx}{(b c-a d) (2+m)}\\ &=\frac{(a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (2+m)}+\frac{b (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (1+m) (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.0244873, size = 59, normalized size = 0.75 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-2} (-a d (m+1)+b c (m+2)+b d x)}{(m+1) (m+2) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 124, normalized size = 1.6 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+m} \left ( dx+c \right ) ^{-2-m} \left ( adm-bcm-bdx+ad-2\,bc \right ) }{{a}^{2}{d}^{2}{m}^{2}-2\,abcd{m}^{2}+{b}^{2}{c}^{2}{m}^{2}+3\,{a}^{2}{d}^{2}m-6\,abcdm+3\,{b}^{2}{c}^{2}m+2\,{a}^{2}{d}^{2}-4\,abcd+2\,{b}^{2}{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.62617, size = 416, normalized size = 5.27 \begin{align*} \frac{{\left (b^{2} d^{2} x^{3} + 2 \, a b c^{2} - a^{2} c d +{\left (3 \, b^{2} c d +{\left (b^{2} c d - a b d^{2}\right )} m\right )} x^{2} +{\left (a b c^{2} - a^{2} c d\right )} m +{\left (2 \, b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} +{\left (b^{2} c^{2} - a^{2} d^{2}\right )} m\right )} x\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}}{2 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2} +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m^{2} + 3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*d^2*x^3 + 2*a*b*c^2 - a^2*c*d + (3*b^2*c*d + (b^2*c*d - a*b*d^2)*m)*x^2 + (a*b*c^2 - a^2*c*d)*m + (2*b^2*
c^2 + 2*a*b*c*d - a^2*d^2 + (b^2*c^2 - a^2*d^2)*m)*x)*(b*x + a)^m*(d*x + c)^(-m - 3)/(2*b^2*c^2 - 4*a*b*c*d +
2*a^2*d^2 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*m^2 + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*m)

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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*(d*x+c)**(-3-m),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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