3.3052 \(\int (a+b x) (c+d x)^{-5+n} (e+f x)^{-n} \, dx\)

Optimal. Leaf size=299 \[ \frac{2 f^2 (c+d x)^{n-1} (e+f x)^{1-n} (3 a d f+b (c f (1-n)-d e (4-n)))}{d (1-n) (2-n) (3-n) (4-n) (d e-c f)^4}+\frac{(b c-a d) (c+d x)^{n-4} (e+f x)^{1-n}}{d (4-n) (d e-c f)}+\frac{(c+d x)^{n-3} (e+f x)^{1-n} (3 a d f+b (c f (1-n)-d e (4-n)))}{d (3-n) (4-n) (d e-c f)^2}-\frac{2 f (c+d x)^{n-2} (e+f x)^{1-n} (3 a d f+b (c f (1-n)-d e (4-n)))}{d (2-n) (3-n) (4-n) (d e-c f)^3} \]

[Out]

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

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Rubi [A]  time = 0.195902, antiderivative size = 296, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {79, 45, 37} \[ \frac{2 f^2 (c+d x)^{n-1} (e+f x)^{1-n} (3 a d f+b c f (1-n)-b d e (4-n))}{d (1-n) (2-n) (3-n) (4-n) (d e-c f)^4}+\frac{(b c-a d) (c+d x)^{n-4} (e+f x)^{1-n}}{d (4-n) (d e-c f)}+\frac{(c+d x)^{n-3} (e+f x)^{1-n} (3 a d f+b c f (1-n)-b d e (4-n))}{d (3-n) (4-n) (d e-c f)^2}-\frac{2 f (c+d x)^{n-2} (e+f x)^{1-n} (3 a d f+b c f (1-n)-b d e (4-n))}{d (2-n) (3-n) (4-n) (d e-c f)^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(c + d*x)^(-5 + n))/(e + f*x)^n,x]

[Out]

((b*c - a*d)*(c + d*x)^(-4 + n)*(e + f*x)^(1 - n))/(d*(d*e - c*f)*(4 - n)) + ((3*a*d*f + b*c*f*(1 - n) - b*d*e
*(4 - n))*(c + d*x)^(-3 + n)*(e + f*x)^(1 - n))/(d*(d*e - c*f)^2*(3 - n)*(4 - n)) - (2*f*(3*a*d*f + b*c*f*(1 -
n) - b*d*e*(4 - n))*(c + d*x)^(-2 + n)*(e + f*x)^(1 - n))/(d*(d*e - c*f)^3*(2 - n)*(3 - n)*(4 - n)) + (2*f^2*
(3*a*d*f + b*c*f*(1 - n) - b*d*e*(4 - n))*(c + d*x)^(-1 + n)*(e + f*x)^(1 - n))/(d*(d*e - c*f)^4*(1 - n)*(2 -
n)*(3 - n)*(4 - 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) (c+d x)^{-5+n} (e+f x)^{-n} \, dx &=\frac{(b c-a d) (c+d x)^{-4+n} (e+f x)^{1-n}}{d (d e-c f) (4-n)}-\frac{(3 a d f+b c f (1-n)-b d e (4-n)) \int (c+d x)^{-4+n} (e+f x)^{-n} \, dx}{d (d e-c f) (4-n)}\\ &=\frac{(b c-a d) (c+d x)^{-4+n} (e+f x)^{1-n}}{d (d e-c f) (4-n)}+\frac{(3 a d f+b c f (1-n)-b d e (4-n)) (c+d x)^{-3+n} (e+f x)^{1-n}}{d (d e-c f)^2 (3-n) (4-n)}+\frac{(2 f (3 a d f+b c f (1-n)-b d e (4-n))) \int (c+d x)^{-3+n} (e+f x)^{-n} \, dx}{d (d e-c f)^2 (3-n) (4-n)}\\ &=\frac{(b c-a d) (c+d x)^{-4+n} (e+f x)^{1-n}}{d (d e-c f) (4-n)}+\frac{(3 a d f+b c f (1-n)-b d e (4-n)) (c+d x)^{-3+n} (e+f x)^{1-n}}{d (d e-c f)^2 (3-n) (4-n)}-\frac{2 f (3 a d f+b c f (1-n)-b d e (4-n)) (c+d x)^{-2+n} (e+f x)^{1-n}}{d (d e-c f)^3 (2-n) (3-n) (4-n)}-\frac{\left (2 f^2 (3 a d f+b c f (1-n)-b d e (4-n))\right ) \int (c+d x)^{-2+n} (e+f x)^{-n} \, dx}{d (d e-c f)^3 (2-n) (3-n) (4-n)}\\ &=\frac{(b c-a d) (c+d x)^{-4+n} (e+f x)^{1-n}}{d (d e-c f) (4-n)}+\frac{(3 a d f+b c f (1-n)-b d e (4-n)) (c+d x)^{-3+n} (e+f x)^{1-n}}{d (d e-c f)^2 (3-n) (4-n)}-\frac{2 f (3 a d f+b c f (1-n)-b d e (4-n)) (c+d x)^{-2+n} (e+f x)^{1-n}}{d (d e-c f)^3 (2-n) (3-n) (4-n)}+\frac{2 f^2 (3 a d f+b c f (1-n)-b d e (4-n)) (c+d x)^{-1+n} (e+f x)^{1-n}}{d (d e-c f)^4 (1-n) (2-n) (3-n) (4-n)}\\ \end{align*}

Mathematica [A]  time = 0.28042, size = 165, normalized size = 0.55 \[ \frac{(c+d x)^{n-4} (e+f x)^{1-n} \left (\frac{(c+d x) \left (c^2 f^2 \left (n^2-5 n+6\right )-2 c d f (n-3) (e (n-1)+f x)+d^2 \left (e^2 \left (n^2-3 n+2\right )+2 e f (n-1) x+2 f^2 x^2\right )\right ) (3 a d f-b c f (n-1)+b d e (n-4))}{(n-3) (n-2) (n-1) (d e-c f)^3}+a d-b c\right )}{d (n-4) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(c + d*x)^(-5 + n))/(e + f*x)^n,x]

[Out]

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

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Maple [B]  time = 0.01, size = 1187, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(d*x+c)^(-5+n)/((f*x+e)^n),x)

[Out]

-(d*x+c)^(-4+n)*(f*x+e)*(b*c^3*f^3*n^3*x-3*b*c^2*d*e*f^2*n^3*x-2*b*c^2*d*f^3*n^2*x^2+3*b*c*d^2*e^2*f*n^3*x+4*b
*c*d^2*e*f^2*n^2*x^2+2*b*c*d^2*f^3*n*x^3-b*d^3*e^3*n^3*x-2*b*d^3*e^2*f*n^2*x^2-2*b*d^3*e*f^2*n*x^3+a*c^3*f^3*n
^3-3*a*c^2*d*e*f^2*n^3-3*a*c^2*d*f^3*n^2*x+3*a*c*d^2*e^2*f*n^3+6*a*c*d^2*e*f^2*n^2*x+6*a*c*d^2*f^3*n*x^2-a*d^3
*e^3*n^3-3*a*d^3*e^2*f*n^2*x-6*a*d^3*e*f^2*n*x^2-6*a*d^3*f^3*x^3-8*b*c^3*f^3*n^2*x+23*b*c^2*d*e*f^2*n^2*x+10*b
*c^2*d*f^3*n*x^2-22*b*c*d^2*e^2*f*n^2*x-20*b*c*d^2*e*f^2*n*x^2-2*b*c*d^2*f^3*x^3+7*b*d^3*e^3*n^2*x+10*b*d^3*e^
2*f*n*x^2+8*b*d^3*e*f^2*x^3-9*a*c^3*f^3*n^2+24*a*c^2*d*e*f^2*n^2+21*a*c^2*d*f^3*n*x-21*a*c*d^2*e^2*f*n^2-30*a*
c*d^2*e*f^2*n*x-24*a*c*d^2*f^3*x^2+6*a*d^3*e^3*n^2+9*a*d^3*e^2*f*n*x+6*a*d^3*e*f^2*x^2+b*c^3*e*f^2*n^2+19*b*c^
3*f^3*n*x-2*b*c^2*d*e^2*f*n^2-58*b*c^2*d*e*f^2*n*x-8*b*c^2*d*f^3*x^2+b*c*d^2*e^3*n^2+53*b*c*d^2*e^2*f*n*x+34*b
*c*d^2*e*f^2*x^2-14*b*d^3*e^3*n*x-8*b*d^3*e^2*f*x^2+26*a*c^3*f^3*n-57*a*c^2*d*e*f^2*n-36*a*c^2*d*f^3*x+42*a*c*
d^2*e^2*f*n+24*a*c*d^2*e*f^2*x-11*a*d^3*e^3*n-6*a*d^3*e^2*f*x-7*b*c^3*e*f^2*n-12*b*c^3*f^3*x+10*b*c^2*d*e^2*f*
n+56*b*c^2*d*e*f^2*x-3*b*c*d^2*e^3*n-34*b*c*d^2*e^2*f*x+8*b*d^3*e^3*x-24*a*c^3*f^3+36*a*c^2*d*e*f^2-24*a*c*d^2
*e^2*f+6*a*d^3*e^3+12*b*c^3*e*f^2-8*b*c^2*d*e^2*f+2*b*c*d^2*e^3)/(c^4*f^4*n^4-4*c^3*d*e*f^3*n^4+6*c^2*d^2*e^2*
f^2*n^4-4*c*d^3*e^3*f*n^4+d^4*e^4*n^4-10*c^4*f^4*n^3+40*c^3*d*e*f^3*n^3-60*c^2*d^2*e^2*f^2*n^3+40*c*d^3*e^3*f*
n^3-10*d^4*e^4*n^3+35*c^4*f^4*n^2-140*c^3*d*e*f^3*n^2+210*c^2*d^2*e^2*f^2*n^2-140*c*d^3*e^3*f*n^2+35*d^4*e^4*n
^2-50*c^4*f^4*n+200*c^3*d*e*f^3*n-300*c^2*d^2*e^2*f^2*n+200*c*d^3*e^3*f*n-50*d^4*e^4*n+24*c^4*f^4-96*c^3*d*e*f
^3+144*c^2*d^2*e^2*f^2-96*c*d^3*e^3*f+24*d^4*e^4)/((f*x+e)^n)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}{\left (d x + c\right )}^{n - 5}}{{\left (f x + e\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-5+n)/((f*x+e)^n),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(d*x + c)^(n - 5)/(f*x + e)^n, x)

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Fricas [B]  time = 1.87983, size = 3572, normalized size = 11.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-5+n)/((f*x+e)^n),x, algorithm="fricas")

[Out]

(24*a*c^4*e*f^3 - 2*(4*b*d^4*e*f^3 - (b*c*d^3 + 3*a*d^4)*f^4 - (b*d^4*e*f^3 - b*c*d^3*f^4)*n)*x^5 - 2*(b*c^2*d
^2 + 3*a*c*d^3)*e^4 + 8*(b*c^3*d + 3*a*c^2*d^2)*e^3*f - 12*(b*c^4 + 3*a*c^3*d)*e^2*f^2 - 2*(20*b*c*d^3*e*f^3 -
5*(b*c^2*d^2 + 3*a*c*d^3)*f^4 - (b*d^4*e^2*f^2 - 2*b*c*d^3*e*f^3 + b*c^2*d^2*f^4)*n^2 + (4*b*d^4*e^2*f^2 - (1
0*b*c*d^3 + 3*a*d^4)*e*f^3 + 3*(2*b*c^2*d^2 + a*c*d^3)*f^4)*n)*x^4 + (a*c*d^3*e^4 - 3*a*c^2*d^2*e^3*f + 3*a*c^
3*d*e^2*f^2 - a*c^4*e*f^3)*n^3 - (80*b*c^2*d^2*e*f^3 - 20*(b*c^3*d + 3*a*c^2*d^2)*f^4 - (b*d^4*e^3*f - 3*b*c*d
^3*e^2*f^2 + 3*b*c^2*d^2*e*f^3 - b*c^3*d*f^4)*n^3 + (5*b*d^4*e^3*f - (20*b*c*d^3 + 3*a*d^4)*e^2*f^2 + (25*b*c^
2*d^2 + 6*a*c*d^3)*e*f^3 - (10*b*c^3*d + 3*a*c^2*d^2)*f^4)*n^2 - (4*b*d^4*e^3*f - (41*b*c*d^3 + 3*a*d^4)*e^2*f
^2 + 6*(11*b*c^2*d^2 + 5*a*c*d^3)*e*f^3 - (29*b*c^3*d + 27*a*c^2*d^2)*f^4)*n)*x^3 + (9*a*c^4*e*f^3 - (b*c^2*d^
2 + 6*a*c*d^3)*e^4 + (2*b*c^3*d + 21*a*c^2*d^2)*e^3*f - (b*c^4 + 24*a*c^3*d)*e^2*f^2)*n^2 - (8*b*d^4*e^4 - 32*
b*c*d^3*e^3*f + 48*b*c^2*d^2*e^2*f^2 + 48*b*c^3*d*e*f^3 - 12*(b*c^4 + 5*a*c^3*d)*f^4 - (b*d^4*e^4 - 3*a*c*d^3*
e^2*f^2 - (2*b*c*d^3 - a*d^4)*e^3*f + (2*b*c^3*d + 3*a*c^2*d^2)*e*f^3 - (b*c^4 + a*c^3*d)*f^4)*n^3 + (7*b*d^4*
e^4 - (16*b*c*d^3 - 3*a*d^4)*e^3*f + 3*(b*c^2*d^2 - 6*a*c*d^3)*e^2*f^2 + (14*b*c^3*d + 27*a*c^2*d^2)*e*f^3 - 4
*(2*b*c^4 + 3*a*c^3*d)*f^4)*n^2 - (14*b*d^4*e^4 - 2*(23*b*c*d^3 - a*d^4)*e^3*f + 15*(b*c^2*d^2 - a*c*d^3)*e^2*
f^2 + 12*(3*b*c^3*d + 5*a*c^2*d^2)*e*f^3 - (19*b*c^4 + 47*a*c^3*d)*f^4)*n)*x^2 - (26*a*c^4*e*f^3 - (3*b*c^2*d^
2 + 11*a*c*d^3)*e^4 + 2*(5*b*c^3*d + 21*a*c^2*d^2)*e^3*f - (7*b*c^4 + 57*a*c^3*d)*e^2*f^2)*n + (24*a*c^3*d*e*f
^3 + 24*a*c^4*f^4 - 2*(5*b*c*d^3 + 3*a*d^4)*e^4 + 8*(5*b*c^2*d^2 + 3*a*c*d^3)*e^3*f - 12*(5*b*c^3*d + 3*a*c^2*
d^2)*e^2*f^2 + (3*b*c^3*d*e^2*f^2 - a*c^4*f^4 + (b*c*d^3 + a*d^4)*e^4 - (3*b*c^2*d^2 + 2*a*c*d^3)*e^3*f - (b*c
^4 - 2*a*c^3*d)*e*f^3)*n^3 + (9*a*c^4*f^4 - 2*(4*b*c*d^3 + 3*a*d^4)*e^4 + (23*b*c^2*d^2 + 18*a*c*d^3)*e^3*f -
(22*b*c^3*d + 9*a*c^2*d^2)*e^2*f^2 + (7*b*c^4 - 12*a*c^3*d)*e*f^3)*n^2 - (26*a*c^4*f^4 - (17*b*c*d^3 + 11*a*d^
4)*e^4 + 20*(3*b*c^2*d^2 + 2*a*c*d^3)*e^3*f - 5*(11*b*c^3*d + 9*a*c^2*d^2)*e^2*f^2 + 2*(6*b*c^4 - 5*a*c^3*d)*e
*f^3)*n)*x)*(d*x + c)^(n - 5)/((24*d^4*e^4 - 96*c*d^3*e^3*f + 144*c^2*d^2*e^2*f^2 - 96*c^3*d*e*f^3 + 24*c^4*f^
4 + (d^4*e^4 - 4*c*d^3*e^3*f + 6*c^2*d^2*e^2*f^2 - 4*c^3*d*e*f^3 + c^4*f^4)*n^4 - 10*(d^4*e^4 - 4*c*d^3*e^3*f
+ 6*c^2*d^2*e^2*f^2 - 4*c^3*d*e*f^3 + c^4*f^4)*n^3 + 35*(d^4*e^4 - 4*c*d^3*e^3*f + 6*c^2*d^2*e^2*f^2 - 4*c^3*d
*e*f^3 + c^4*f^4)*n^2 - 50*(d^4*e^4 - 4*c*d^3*e^3*f + 6*c^2*d^2*e^2*f^2 - 4*c^3*d*e*f^3 + c^4*f^4)*n)*(f*x + e
)^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)*(d*x+c)**(-5+n)/((f*x+e)**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 )}{\left (d x + c\right )}^{n - 5}}{{\left (f x + e\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-5+n)/((f*x+e)^n),x, algorithm="giac")

[Out]

integrate((b*x + a)*(d*x + c)^(n - 5)/(f*x + e)^n, x)