3.1876 \(\int (a+b x)^n (c+d x)^{-4-n} \, dx\)

Optimal. Leaf size=130 \[ \frac{2 b^2 (a+b x)^{n+1} (c+d x)^{-n-1}}{(n+1) (n+2) (n+3) (b c-a d)^3}+\frac{(a+b x)^{n+1} (c+d x)^{-n-3}}{(n+3) (b c-a d)}+\frac{2 b (a+b x)^{n+1} (c+d x)^{-n-2}}{(n+2) (n+3) (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.0372081, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac{2 b^2 (a+b x)^{n+1} (c+d x)^{-n-1}}{(n+1) (n+2) (n+3) (b c-a d)^3}+\frac{(a+b x)^{n+1} (c+d x)^{-n-3}}{(n+3) (b c-a d)}+\frac{2 b (a+b x)^{n+1} (c+d x)^{-n-2}}{(n+2) (n+3) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n*(c + d*x)^(-4 - n),x]

[Out]

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

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)^n (c+d x)^{-4-n} \, dx &=\frac{(a+b x)^{1+n} (c+d x)^{-3-n}}{(b c-a d) (3+n)}+\frac{(2 b) \int (a+b x)^n (c+d x)^{-3-n} \, dx}{(b c-a d) (3+n)}\\ &=\frac{(a+b x)^{1+n} (c+d x)^{-3-n}}{(b c-a d) (3+n)}+\frac{2 b (a+b x)^{1+n} (c+d x)^{-2-n}}{(b c-a d)^2 (2+n) (3+n)}+\frac{\left (2 b^2\right ) \int (a+b x)^n (c+d x)^{-2-n} \, dx}{(b c-a d)^2 (2+n) (3+n)}\\ &=\frac{(a+b x)^{1+n} (c+d x)^{-3-n}}{(b c-a d) (3+n)}+\frac{2 b (a+b x)^{1+n} (c+d x)^{-2-n}}{(b c-a d)^2 (2+n) (3+n)}+\frac{2 b^2 (a+b x)^{1+n} (c+d x)^{-1-n}}{(b c-a d)^3 (1+n) (2+n) (3+n)}\\ \end{align*}

Mathematica [A]  time = 0.0581298, size = 112, normalized size = 0.86 \[ \frac{(a+b x)^{n+1} (c+d x)^{-n-3} \left (a^2 d^2 \left (n^2+3 n+2\right )-2 a b d (n+1) (c (n+3)+d x)+b^2 \left (c^2 \left (n^2+5 n+6\right )+2 c d (n+3) x+2 d^2 x^2\right )\right )}{(n+1) (n+2) (n+3) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^n*(c + d*x)^(-4 - n),x]

[Out]

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

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Maple [B]  time = 0.006, size = 319, normalized size = 2.5 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+n} \left ( dx+c \right ) ^{-3-n} \left ({a}^{2}{d}^{2}{n}^{2}-2\,abcd{n}^{2}-2\,ab{d}^{2}nx+{b}^{2}{c}^{2}{n}^{2}+2\,{b}^{2}cdnx+2\,{b}^{2}{d}^{2}{x}^{2}+3\,{a}^{2}{d}^{2}n-8\,abcdn-2\,ab{d}^{2}x+5\,{b}^{2}{c}^{2}n+6\,{b}^{2}cdx+2\,{a}^{2}{d}^{2}-6\,abcd+6\,{b}^{2}{c}^{2} \right ) }{{a}^{3}{d}^{3}{n}^{3}-3\,{a}^{2}bc{d}^{2}{n}^{3}+3\,a{b}^{2}{c}^{2}d{n}^{3}-{b}^{3}{c}^{3}{n}^{3}+6\,{a}^{3}{d}^{3}{n}^{2}-18\,{a}^{2}bc{d}^{2}{n}^{2}+18\,a{b}^{2}{c}^{2}d{n}^{2}-6\,{b}^{3}{c}^{3}{n}^{2}+11\,{a}^{3}{d}^{3}n-33\,{a}^{2}bc{d}^{2}n+33\,a{b}^{2}{c}^{2}dn-11\,{b}^{3}{c}^{3}n+6\,{a}^{3}{d}^{3}-18\,{a}^{2}cb{d}^{2}+18\,a{b}^{2}{c}^{2}d-6\,{b}^{3}{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^n*(d*x+c)^(-4-n),x)

[Out]

-(b*x+a)^(1+n)*(d*x+c)^(-3-n)*(a^2*d^2*n^2-2*a*b*c*d*n^2-2*a*b*d^2*n*x+b^2*c^2*n^2+2*b^2*c*d*n*x+2*b^2*d^2*x^2
+3*a^2*d^2*n-8*a*b*c*d*n-2*a*b*d^2*x+5*b^2*c^2*n+6*b^2*c*d*x+2*a^2*d^2-6*a*b*c*d+6*b^2*c^2)/(a^3*d^3*n^3-3*a^2
*b*c*d^2*n^3+3*a*b^2*c^2*d*n^3-b^3*c^3*n^3+6*a^3*d^3*n^2-18*a^2*b*c*d^2*n^2+18*a*b^2*c^2*d*n^2-6*b^3*c^3*n^2+1
1*a^3*d^3*n-33*a^2*b*c*d^2*n+33*a*b^2*c^2*d*n-11*b^3*c^3*n+6*a^3*d^3-18*a^2*b*c*d^2+18*a*b^2*c^2*d-6*b^3*c^3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{n}{\left (d x + c\right )}^{-n - 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n*(d*x+c)^(-4-n),x, algorithm="maxima")

[Out]

integrate((b*x + a)^n*(d*x + c)^(-n - 4), x)

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Fricas [B]  time = 2.26154, size = 1021, normalized size = 7.85 \begin{align*} \frac{{\left (2 \, b^{3} d^{3} x^{4} + 6 \, a b^{2} c^{3} - 6 \, a^{2} b c^{2} d + 2 \, a^{3} c d^{2} + 2 \,{\left (4 \, b^{3} c d^{2} +{\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} n\right )} x^{3} +{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} n^{2} +{\left (12 \, b^{3} c^{2} d +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} n^{2} +{\left (7 \, b^{3} c^{2} d - 8 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} n\right )} x^{2} +{\left (5 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d + 3 \, a^{3} c d^{2}\right )} n +{\left (6 \, b^{3} c^{3} + 6 \, a b^{2} c^{2} d - 6 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} n^{2} +{\left (5 \, b^{3} c^{3} - a b^{2} c^{2} d - 7 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} n\right )} x\right )}{\left (b x + a\right )}^{n}{\left (d x + c\right )}^{-n - 4}}{6 \, b^{3} c^{3} - 18 \, a b^{2} c^{2} d + 18 \, a^{2} b c d^{2} - 6 \, a^{3} d^{3} +{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} n^{3} + 6 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} n^{2} + 11 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n*(d*x+c)^(-4-n),x, algorithm="fricas")

[Out]

(2*b^3*d^3*x^4 + 6*a*b^2*c^3 - 6*a^2*b*c^2*d + 2*a^3*c*d^2 + 2*(4*b^3*c*d^2 + (b^3*c*d^2 - a*b^2*d^3)*n)*x^3 +
(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*n^2 + (12*b^3*c^2*d + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*n^2 + (
7*b^3*c^2*d - 8*a*b^2*c*d^2 + a^2*b*d^3)*n)*x^2 + (5*a*b^2*c^3 - 8*a^2*b*c^2*d + 3*a^3*c*d^2)*n + (6*b^3*c^3 +
6*a*b^2*c^2*d - 6*a^2*b*c*d^2 + 2*a^3*d^3 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*n^2 + (5*b^3*c^3
- a*b^2*c^2*d - 7*a^2*b*c*d^2 + 3*a^3*d^3)*n)*x)*(b*x + a)^n*(d*x + c)^(-n - 4)/(6*b^3*c^3 - 18*a*b^2*c^2*d +
18*a^2*b*c*d^2 - 6*a^3*d^3 + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*n^3 + 6*(b^3*c^3 - 3*a*b^2*c^
2*d + 3*a^2*b*c*d^2 - a^3*d^3)*n^2 + 11*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*n)

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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)**n*(d*x+c)**(-4-n),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 )}^{n}{\left (d x + c\right )}^{-n - 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^n*(d*x+c)^(-4-n),x, algorithm="giac")

[Out]

integrate((b*x + a)^n*(d*x + c)^(-n - 4), x)