3.3057 \(\int (a+b x)^{-n} (c+d x) (e+f x)^{-4+n} \, dx\)

Optimal. Leaf size=207 \[ -\frac{(a+b x)^{1-n} (d e-c f) (e+f x)^{n-3}}{f (3-n) (b e-a f)}+\frac{(a+b x)^{1-n} (e+f x)^{n-2} (b (2 c f+d e (1-n))-a d f (3-n))}{f (2-n) (3-n) (b e-a f)^2}+\frac{b (a+b x)^{1-n} (e+f x)^{n-1} (b (2 c f+d e (1-n))-a d f (3-n))}{f (1-n) (2-n) (3-n) (b e-a f)^3} \]

[Out]

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

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Rubi [A]  time = 0.117984, antiderivative size = 205, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {79, 45, 37} \[ -\frac{(a+b x)^{1-n} (d e-c f) (e+f x)^{n-3}}{f (3-n) (b e-a f)}+\frac{(a+b x)^{1-n} (e+f x)^{n-2} (-a d f (3-n)+2 b c f+b d e (1-n))}{f (2-n) (3-n) (b e-a f)^2}+\frac{b (a+b x)^{1-n} (e+f x)^{n-1} (-a d f (3-n)+2 b c f+b d e (1-n))}{f (1-n) (2-n) (3-n) (b e-a f)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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

Mathematica [A]  time = 0.149136, size = 180, normalized size = 0.87 \[ \frac{(a+b x)^{1-n} (e+f x)^{n-3} \left (a^2 f (n-1) (c f (n-2)-d e+d f (n-3) x)+a b \left (2 c f (n-1) (f x-e (n-3))+d \left (e^2 (n-3)-2 e f \left (n^2-4 n+5\right ) x+f^2 (n-3) x^2\right )\right )+b^2 \left (c \left (e^2 \left (n^2-5 n+6\right )-2 e f (n-3) x+2 f^2 x^2\right )+d e (n-1) x (e (n-3)-f x)\right )\right )}{(n-3) (n-2) (n-1) (a f-b e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 - n)*(e + f*x)^(-3 + n)*(a^2*f*(-1 + n)*(-(d*e) + c*f*(-2 + n) + d*f*(-3 + n)*x) + b^2*(d*e*(-1
+ n)*x*(e*(-3 + n) - f*x) + c*(e^2*(6 - 5*n + n^2) - 2*e*f*(-3 + n)*x + 2*f^2*x^2)) + a*b*(2*c*f*(-1 + n)*(-(e
*(-3 + n)) + f*x) + d*(e^2*(-3 + n) - 2*e*f*(5 - 4*n + n^2)*x + f^2*(-3 + n)*x^2))))/((-(b*e) + a*f)^3*(-3 + n
)*(-2 + n)*(-1 + n))

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Maple [B]  time = 0.007, size = 505, normalized size = 2.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( fx+e \right ) ^{-3+n} \left ({a}^{2}d{f}^{2}{n}^{2}x-2\,abdef{n}^{2}x+abd{f}^{2}n{x}^{2}+{b}^{2}d{e}^{2}{n}^{2}x-{b}^{2}defn{x}^{2}+{a}^{2}c{f}^{2}{n}^{2}-4\,{a}^{2}d{f}^{2}nx-2\,abcef{n}^{2}+2\,abc{f}^{2}nx+8\,abdefnx-3\,abd{f}^{2}{x}^{2}+{b}^{2}c{e}^{2}{n}^{2}-2\,{b}^{2}cefnx+2\,{b}^{2}c{f}^{2}{x}^{2}-4\,{b}^{2}d{e}^{2}nx+{b}^{2}def{x}^{2}-3\,{a}^{2}c{f}^{2}n-{a}^{2}defn+3\,{a}^{2}d{f}^{2}x+8\,abcefn-2\,abc{f}^{2}x+abd{e}^{2}n-10\,abdefx-5\,{b}^{2}c{e}^{2}n+6\,{b}^{2}cefx+3\,{b}^{2}d{e}^{2}x+2\,{a}^{2}c{f}^{2}+{a}^{2}def-6\,abcef-3\,abd{e}^{2}+6\,{b}^{2}c{e}^{2} \right ) }{ \left ({a}^{3}{f}^{3}{n}^{3}-3\,{a}^{2}be{f}^{2}{n}^{3}+3\,a{b}^{2}{e}^{2}f{n}^{3}-{b}^{3}{e}^{3}{n}^{3}-6\,{a}^{3}{f}^{3}{n}^{2}+18\,{a}^{2}be{f}^{2}{n}^{2}-18\,a{b}^{2}{e}^{2}f{n}^{2}+6\,{b}^{3}{e}^{3}{n}^{2}+11\,{a}^{3}{f}^{3}n-33\,{a}^{2}be{f}^{2}n+33\,a{b}^{2}{e}^{2}fn-11\,{b}^{3}{e}^{3}n-6\,{a}^{3}{f}^{3}+18\,{a}^{2}be{f}^{2}-18\,a{b}^{2}{e}^{2}f+6\,{b}^{3}{e}^{3} \right ) \left ( bx+a \right ) ^{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)*(f*x+e)^(-3+n)*(a^2*d*f^2*n^2*x-2*a*b*d*e*f*n^2*x+a*b*d*f^2*n*x^2+b^2*d*e^2*n^2*x-b^2*d*e*f*n*x^2+a^2*
c*f^2*n^2-4*a^2*d*f^2*n*x-2*a*b*c*e*f*n^2+2*a*b*c*f^2*n*x+8*a*b*d*e*f*n*x-3*a*b*d*f^2*x^2+b^2*c*e^2*n^2-2*b^2*
c*e*f*n*x+2*b^2*c*f^2*x^2-4*b^2*d*e^2*n*x+b^2*d*e*f*x^2-3*a^2*c*f^2*n-a^2*d*e*f*n+3*a^2*d*f^2*x+8*a*b*c*e*f*n-
2*a*b*c*f^2*x+a*b*d*e^2*n-10*a*b*d*e*f*x-5*b^2*c*e^2*n+6*b^2*c*e*f*x+3*b^2*d*e^2*x+2*a^2*c*f^2+a^2*d*e*f-6*a*b
*c*e*f-3*a*b*d*e^2+6*b^2*c*e^2)/(a^3*f^3*n^3-3*a^2*b*e*f^2*n^3+3*a*b^2*e^2*f*n^3-b^3*e^3*n^3-6*a^3*f^3*n^2+18*
a^2*b*e*f^2*n^2-18*a*b^2*e^2*f*n^2+6*b^3*e^3*n^2+11*a^3*f^3*n-33*a^2*b*e*f^2*n+33*a*b^2*e^2*f*n-11*b^3*e^3*n-6
*a^3*f^3+18*a^2*b*e*f^2-18*a*b^2*e^2*f+6*b^3*e^3)/((b*x+a)^n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.76781, size = 1787, normalized size = 8.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^3*c*e*f^2 + (b^3*d*e*f^2 + (2*b^3*c - 3*a*b^2*d)*f^3 - (b^3*d*e*f^2 - a*b^2*d*f^3)*n)*x^4 + 3*(2*a*b^2*c
- a^2*b*d)*e^3 - (6*a^2*b*c - a^3*d)*e^2*f + (4*b^3*d*e^2*f + 4*(2*b^3*c - 3*a*b^2*d)*e*f^2 + (b^3*d*e^2*f - 2
*a*b^2*d*e*f^2 + a^2*b*d*f^3)*n^2 - (5*b^3*d*e^2*f + 2*(b^3*c - 4*a*b^2*d)*e*f^2 - (2*a*b^2*c - 3*a^2*b*d)*f^3
)*n)*x^3 + (a*b^2*c*e^3 - 2*a^2*b*c*e^2*f + a^3*c*e*f^2)*n^2 + (3*b^3*d*e^3 - 9*a^2*b*d*e*f^2 + 3*a^3*d*f^3 +
3*(4*b^3*c - 3*a*b^2*d)*e^2*f + (b^3*d*e^3 + (b^3*c - a*b^2*d)*e^2*f - (2*a*b^2*c + a^2*b*d)*e*f^2 + (a^2*b*c
+ a^3*d)*f^3)*n^2 - (4*b^3*d*e^3 + (7*b^3*c - 4*a*b^2*d)*e^2*f - 4*(2*a*b^2*c + a^2*b*d)*e*f^2 + (a^2*b*c + 4*
a^3*d)*f^3)*n)*x^2 - (3*a^3*c*e*f^2 + (5*a*b^2*c - a^2*b*d)*e^3 - (8*a^2*b*c - a^3*d)*e^2*f)*n + (6*b^3*c*e^3
+ 2*a^3*c*f^3 + 6*(a*b^2*c - 2*a^2*b*d)*e^2*f - 2*(3*a^2*b*c - 2*a^3*d)*e*f^2 + (a^3*c*f^3 + (b^3*c + a*b^2*d)
*e^3 - (a*b^2*c + 2*a^2*b*d)*e^2*f - (a^2*b*c - a^3*d)*e*f^2)*n^2 - (3*a^3*c*f^3 + (5*b^3*c + 3*a*b^2*d)*e^3 -
(a*b^2*c + 8*a^2*b*d)*e^2*f - (7*a^2*b*c - 5*a^3*d)*e*f^2)*n)*x)*(f*x + e)^(n - 4)/((6*b^3*e^3 - 18*a*b^2*e^2
*f + 18*a^2*b*e*f^2 - 6*a^3*f^3 - (b^3*e^3 - 3*a*b^2*e^2*f + 3*a^2*b*e*f^2 - a^3*f^3)*n^3 + 6*(b^3*e^3 - 3*a*b
^2*e^2*f + 3*a^2*b*e*f^2 - a^3*f^3)*n^2 - 11*(b^3*e^3 - 3*a*b^2*e^2*f + 3*a^2*b*e*f^2 - a^3*f^3)*n)*(b*x + a)^
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((d*x+c)*(f*x+e)**(-4+n)/((b*x+a)**n),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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