3.1859 \(\int (a+b x)^{-4+n} (c+d x)^{-n} \, dx\)

Optimal. Leaf size=143 \[ -\frac{2 d^2 (a+b x)^{n-1} (c+d x)^{1-n}}{(1-n) (2-n) (3-n) (b c-a d)^3}-\frac{(a+b x)^{n-3} (c+d x)^{1-n}}{(3-n) (b c-a d)}+\frac{2 d (a+b x)^{n-2} (c+d x)^{1-n}}{(2-n) (3-n) (b c-a d)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0627202, antiderivative size = 143, 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} \[ -\frac{2 d^2 (a+b x)^{n-1} (c+d x)^{1-n}}{(1-n) (2-n) (3-n) (b c-a d)^3}-\frac{(a+b x)^{n-3} (c+d x)^{1-n}}{(3-n) (b c-a d)}+\frac{2 d (a+b x)^{n-2} (c+d x)^{1-n}}{(2-n) (3-n) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(-4 + n)/(c + d*x)^n,x]

[Out]

-(((a + b*x)^(-3 + n)*(c + d*x)^(1 - n))/((b*c - a*d)*(3 - n))) + (2*d*(a + b*x)^(-2 + n)*(c + d*x)^(1 - n))/(
(b*c - a*d)^2*(2 - n)*(3 - n)) - (2*d^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)^{-4+n} (c+d x)^{-n} \, dx &=-\frac{(a+b x)^{-3+n} (c+d x)^{1-n}}{(b c-a d) (3-n)}-\frac{(2 d) \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx}{(b c-a d) (3-n)}\\ &=-\frac{(a+b x)^{-3+n} (c+d x)^{1-n}}{(b c-a d) (3-n)}+\frac{2 d (a+b x)^{-2+n} (c+d x)^{1-n}}{(b c-a d)^2 (2-n) (3-n)}+\frac{\left (2 d^2\right ) \int (a+b x)^{-2+n} (c+d x)^{-n} \, dx}{(b c-a d)^2 (2-n) (3-n)}\\ &=-\frac{(a+b x)^{-3+n} (c+d x)^{1-n}}{(b c-a d) (3-n)}+\frac{2 d (a+b x)^{-2+n} (c+d x)^{1-n}}{(b c-a d)^2 (2-n) (3-n)}-\frac{2 d^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 = 0.0647185, size = 112, normalized size = 0.78 \[ \frac{(a+b x)^{n-3} (c+d x)^{1-n} \left (a^2 d^2 \left (n^2-5 n+6\right )-2 a b d (n-3) (c (n-1)+d x)+b^2 \left (c^2 \left (n^2-3 n+2\right )+2 c d (n-1) x+2 d^2 x^2\right )\right )}{(n-3) (n-2) (n-1) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.006, size = 322, normalized size = 2.3 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{-3+n} \left ( dx+c \right ) \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}-5\,{a}^{2}{d}^{2}n+8\,abcdn+6\,ab{d}^{2}x-3\,{b}^{2}{c}^{2}n-2\,{b}^{2}cdx+6\,{a}^{2}{d}^{2}-6\,abcd+2\,{b}^{2}{c}^{2} \right ) }{ \left ({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} \right ) \left ( dx+c \right ) ^{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.04388, size = 1023, normalized size = 7.15 \begin{align*} -\frac{{\left (2 \, b^{3} d^{3} x^{4} + 2 \, a b^{2} c^{3} - 6 \, a^{2} b c^{2} d + 6 \, a^{3} c d^{2} + 2 \,{\left (4 \, a b^{2} d^{3} +{\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 \, a^{2} b d^{3} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} n^{2} -{\left (b^{3} c^{2} d - 8 \, a b^{2} c d^{2} + 7 \, a^{2} b d^{3}\right )} n\right )} x^{2} -{\left (3 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d + 5 \, a^{3} c d^{2}\right )} n +{\left (2 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d + 6 \, a^{2} b c d^{2} + 6 \, 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 (3 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d - a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} n\right )} x\right )}{\left (b x + a\right )}^{n - 4}}{{\left (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\right )}{\left (d x + c\right )}^{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)**(-4+n)/((d*x+c)**n),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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