3.1860 \(\int (a+b x)^{-3+n} (c+d x)^{-n} \, dx\)

Optimal. Leaf size=86 \[ \frac{d (a+b x)^{n-1} (c+d x)^{1-n}}{(1-n) (2-n) (b c-a d)^2}-\frac{(a+b x)^{n-2} (c+d x)^{1-n}}{(2-n) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0124284, antiderivative size = 86, 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} \[ \frac{d (a+b x)^{n-1} (c+d x)^{1-n}}{(1-n) (2-n) (b c-a d)^2}-\frac{(a+b x)^{n-2} (c+d x)^{1-n}}{(2-n) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(-3 + n)/(c + d*x)^n,x]

[Out]

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

Mathematica [A]  time = 0.0311377, size = 59, normalized size = 0.69 \[ \frac{(a+b x)^{n-2} (c+d x)^{1-n} (-a d (n-2)+b c (n-1)+b d x)}{(n-2) (n-1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 127, normalized size = 1.5 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{-2+n} \left ( dx+c \right ) \left ( adn-bcn-bdx-2\,ad+bc \right ) }{ \left ({a}^{2}{d}^{2}{n}^{2}-2\,abcd{n}^{2}+{b}^{2}{c}^{2}{n}^{2}-3\,{a}^{2}{d}^{2}n+6\,abcdn-3\,{b}^{2}{c}^{2}n+2\,{a}^{2}{d}^{2}-4\,abcd+2\,{b}^{2}{c}^{2} \right ) \left ( dx+c \right ) ^{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{n - 3}}{{\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)^(-3+n)/((d*x+c)^n),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.95155, size = 417, normalized size = 4.85 \begin{align*} \frac{{\left (b^{2} d^{2} x^{3} - a b c^{2} + 2 \, a^{2} c d +{\left (3 \, a b d^{2} +{\left (b^{2} c d - a b d^{2}\right )} n\right )} x^{2} +{\left (a b c^{2} - a^{2} c d\right )} n -{\left (b^{2} c^{2} - 2 \, a b c d - 2 \, a^{2} d^{2} -{\left (b^{2} c^{2} - a^{2} d^{2}\right )} n\right )} x\right )}{\left (b x + a\right )}^{n - 3}}{{\left (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 )} n^{2} - 3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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)^(-3+n)/((d*x+c)^n),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{n - 3}}{{\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)^(-3+n)/((d*x+c)^n),x, algorithm="giac")

[Out]

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