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

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

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Rubi [A]  time = 0.12, 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} \begin {gather*} -\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} \end {gather*}

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 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]

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 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]

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*}

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Mathematica [A]  time = 0.20, size = 180, normalized size = 0.87 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x)^{-n} (c+d x) (e+f x)^{-4+n} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 1.48, size = 884, normalized size = 4.27 \begin {gather*} \frac {{\left (2 \, a^{3} c e f^{2} + {\left (b^{3} d e f^{2} + {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} f^{3} - {\left (b^{3} d e f^{2} - a b^{2} d f^{3}\right )} n\right )} x^{4} + 3 \, {\left (2 \, a b^{2} c - a^{2} b d\right )} e^{3} - {\left (6 \, a^{2} b c - a^{3} d\right )} e^{2} f + {\left (4 \, b^{3} d e^{2} f + 4 \, {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} e f^{2} + {\left (b^{3} d e^{2} f - 2 \, a b^{2} d e f^{2} + a^{2} b d f^{3}\right )} n^{2} - {\left (5 \, b^{3} d e^{2} f + 2 \, {\left (b^{3} c - 4 \, a b^{2} d\right )} e f^{2} - {\left (2 \, a b^{2} c - 3 \, a^{2} b d\right )} f^{3}\right )} n\right )} x^{3} + {\left (a b^{2} c e^{3} - 2 \, a^{2} b c e^{2} f + a^{3} c e f^{2}\right )} n^{2} + {\left (3 \, b^{3} d e^{3} - 9 \, a^{2} b d e f^{2} + 3 \, a^{3} d f^{3} + 3 \, {\left (4 \, b^{3} c - 3 \, a b^{2} d\right )} e^{2} f + {\left (b^{3} d e^{3} + {\left (b^{3} c - a b^{2} d\right )} e^{2} f - {\left (2 \, a b^{2} c + a^{2} b d\right )} e f^{2} + {\left (a^{2} b c + a^{3} d\right )} f^{3}\right )} n^{2} - {\left (4 \, b^{3} d e^{3} + {\left (7 \, b^{3} c - 4 \, a b^{2} d\right )} e^{2} f - 4 \, {\left (2 \, a b^{2} c + a^{2} b d\right )} e f^{2} + {\left (a^{2} b c + 4 \, a^{3} d\right )} f^{3}\right )} n\right )} x^{2} - {\left (3 \, a^{3} c e f^{2} + {\left (5 \, a b^{2} c - a^{2} b d\right )} e^{3} - {\left (8 \, a^{2} b c - a^{3} d\right )} e^{2} f\right )} n + {\left (6 \, b^{3} c e^{3} + 2 \, a^{3} c f^{3} + 6 \, {\left (a b^{2} c - 2 \, a^{2} b d\right )} e^{2} f - 2 \, {\left (3 \, a^{2} b c - 2 \, a^{3} d\right )} e f^{2} + {\left (a^{3} c f^{3} + {\left (b^{3} c + a b^{2} d\right )} e^{3} - {\left (a b^{2} c + 2 \, a^{2} b d\right )} e^{2} f - {\left (a^{2} b c - a^{3} d\right )} e f^{2}\right )} n^{2} - {\left (3 \, a^{3} c f^{3} + {\left (5 \, b^{3} c + 3 \, a b^{2} d\right )} e^{3} - {\left (a b^{2} c + 8 \, a^{2} b d\right )} e^{2} f - {\left (7 \, a^{2} b c - 5 \, a^{3} d\right )} e f^{2}\right )} n\right )} x\right )} {\left (f x + e\right )}^{n - 4}}{{\left (6 \, b^{3} e^{3} - 18 \, a b^{2} e^{2} f + 18 \, a^{2} b e f^{2} - 6 \, a^{3} f^{3} - {\left (b^{3} e^{3} - 3 \, a b^{2} e^{2} f + 3 \, a^{2} b e f^{2} - a^{3} f^{3}\right )} n^{3} + 6 \, {\left (b^{3} e^{3} - 3 \, a b^{2} e^{2} f + 3 \, a^{2} b e f^{2} - a^{3} f^{3}\right )} n^{2} - 11 \, {\left (b^{3} e^{3} - 3 \, a b^{2} e^{2} f + 3 \, a^{2} b e f^{2} - a^{3} f^{3}\right )} n\right )} {\left (b x + a\right )}^{n}} \end {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x + c\right )} {\left (f x + e\right )}^{n - 4}}{{\left (b x + a\right )}^{n}}\,{d x} \end {gather*}

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)

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maple [B]  time = 0.01, size = 505, normalized size = 2.44 \begin {gather*} \frac {\left (b x +a \right ) \left (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}\right ) \left (b x +a \right )^{-n} \left (f x +e \right )^{n -3}}{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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)*(f*x+e)^(n-3)*(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x + c\right )} {\left (f x + e\right )}^{n - 4}}{{\left (b x + a\right )}^{n}}\,{d x} \end {gather*}

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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mupad [B]  time = 3.48, size = 869, normalized size = 4.20 \begin {gather*} \frac {x\,{\left (e+f\,x\right )}^{n-4}\,\left (d\,a^3\,e\,f^2\,n^2-5\,d\,a^3\,e\,f^2\,n+4\,d\,a^3\,e\,f^2+c\,a^3\,f^3\,n^2-3\,c\,a^3\,f^3\,n+2\,c\,a^3\,f^3-2\,d\,a^2\,b\,e^2\,f\,n^2+8\,d\,a^2\,b\,e^2\,f\,n-12\,d\,a^2\,b\,e^2\,f-c\,a^2\,b\,e\,f^2\,n^2+7\,c\,a^2\,b\,e\,f^2\,n-6\,c\,a^2\,b\,e\,f^2+d\,a\,b^2\,e^3\,n^2-3\,d\,a\,b^2\,e^3\,n-c\,a\,b^2\,e^2\,f\,n^2+c\,a\,b^2\,e^2\,f\,n+6\,c\,a\,b^2\,e^2\,f+c\,b^3\,e^3\,n^2-5\,c\,b^3\,e^3\,n+6\,c\,b^3\,e^3\right )}{{\left (a\,f-b\,e\right )}^3\,{\left (a+b\,x\right )}^n\,\left (n^3-6\,n^2+11\,n-6\right )}+\frac {x^2\,{\left (e+f\,x\right )}^{n-4}\,\left (d\,a^3\,f^3\,n^2-4\,d\,a^3\,f^3\,n+3\,d\,a^3\,f^3-d\,a^2\,b\,e\,f^2\,n^2+4\,d\,a^2\,b\,e\,f^2\,n-9\,d\,a^2\,b\,e\,f^2+c\,a^2\,b\,f^3\,n^2-c\,a^2\,b\,f^3\,n-d\,a\,b^2\,e^2\,f\,n^2+4\,d\,a\,b^2\,e^2\,f\,n-9\,d\,a\,b^2\,e^2\,f-2\,c\,a\,b^2\,e\,f^2\,n^2+8\,c\,a\,b^2\,e\,f^2\,n+d\,b^3\,e^3\,n^2-4\,d\,b^3\,e^3\,n+3\,d\,b^3\,e^3+c\,b^3\,e^2\,f\,n^2-7\,c\,b^3\,e^2\,f\,n+12\,c\,b^3\,e^2\,f\right )}{{\left (a\,f-b\,e\right )}^3\,{\left (a+b\,x\right )}^n\,\left (n^3-6\,n^2+11\,n-6\right )}+\frac {a\,e\,{\left (e+f\,x\right )}^{n-4}\,\left (-d\,a^2\,e\,f\,n+d\,a^2\,e\,f+c\,a^2\,f^2\,n^2-3\,c\,a^2\,f^2\,n+2\,c\,a^2\,f^2+d\,a\,b\,e^2\,n-3\,d\,a\,b\,e^2-2\,c\,a\,b\,e\,f\,n^2+8\,c\,a\,b\,e\,f\,n-6\,c\,a\,b\,e\,f+c\,b^2\,e^2\,n^2-5\,c\,b^2\,e^2\,n+6\,c\,b^2\,e^2\right )}{{\left (a\,f-b\,e\right )}^3\,{\left (a+b\,x\right )}^n\,\left (n^3-6\,n^2+11\,n-6\right )}+\frac {b^2\,f^2\,x^4\,{\left (e+f\,x\right )}^{n-4}\,\left (2\,b\,c\,f-3\,a\,d\,f+b\,d\,e+a\,d\,f\,n-b\,d\,e\,n\right )}{{\left (a\,f-b\,e\right )}^3\,{\left (a+b\,x\right )}^n\,\left (n^3-6\,n^2+11\,n-6\right )}+\frac {b\,f\,x^3\,{\left (e+f\,x\right )}^{n-4}\,\left (4\,b\,e+a\,f\,n-b\,e\,n\right )\,\left (2\,b\,c\,f-3\,a\,d\,f+b\,d\,e+a\,d\,f\,n-b\,d\,e\,n\right )}{{\left (a\,f-b\,e\right )}^3\,{\left (a+b\,x\right )}^n\,\left (n^3-6\,n^2+11\,n-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(e + f*x)^(n - 4)*(2*a^3*c*f^3 + 6*b^3*c*e^3 + a^3*c*f^3*n^2 + b^3*c*e^3*n^2 + 4*a^3*d*e*f^2 - 3*a^3*c*f^3*
n - 5*b^3*c*e^3*n + 6*a*b^2*c*e^2*f - 6*a^2*b*c*e*f^2 - 12*a^2*b*d*e^2*f - 3*a*b^2*d*e^3*n - 5*a^3*d*e*f^2*n +
a*b^2*d*e^3*n^2 + a^3*d*e*f^2*n^2 + a*b^2*c*e^2*f*n + 7*a^2*b*c*e*f^2*n + 8*a^2*b*d*e^2*f*n - a*b^2*c*e^2*f*n
^2 - a^2*b*c*e*f^2*n^2 - 2*a^2*b*d*e^2*f*n^2))/((a*f - b*e)^3*(a + b*x)^n*(11*n - 6*n^2 + n^3 - 6)) + (x^2*(e
+ f*x)^(n - 4)*(3*a^3*d*f^3 + 3*b^3*d*e^3 + a^3*d*f^3*n^2 + b^3*d*e^3*n^2 + 12*b^3*c*e^2*f - 4*a^3*d*f^3*n - 4
*b^3*d*e^3*n - 9*a*b^2*d*e^2*f - 9*a^2*b*d*e*f^2 - a^2*b*c*f^3*n - 7*b^3*c*e^2*f*n + a^2*b*c*f^3*n^2 + b^3*c*e
^2*f*n^2 + 8*a*b^2*c*e*f^2*n + 4*a*b^2*d*e^2*f*n + 4*a^2*b*d*e*f^2*n - 2*a*b^2*c*e*f^2*n^2 - a*b^2*d*e^2*f*n^2
- a^2*b*d*e*f^2*n^2))/((a*f - b*e)^3*(a + b*x)^n*(11*n - 6*n^2 + n^3 - 6)) + (a*e*(e + f*x)^(n - 4)*(2*a^2*c*
f^2 + 6*b^2*c*e^2 + a^2*c*f^2*n^2 + b^2*c*e^2*n^2 - 3*a*b*d*e^2 + a^2*d*e*f - 3*a^2*c*f^2*n - 5*b^2*c*e^2*n -
6*a*b*c*e*f + a*b*d*e^2*n - a^2*d*e*f*n - 2*a*b*c*e*f*n^2 + 8*a*b*c*e*f*n))/((a*f - b*e)^3*(a + b*x)^n*(11*n -
6*n^2 + n^3 - 6)) + (b^2*f^2*x^4*(e + f*x)^(n - 4)*(2*b*c*f - 3*a*d*f + b*d*e + a*d*f*n - b*d*e*n))/((a*f - b
*e)^3*(a + b*x)^n*(11*n - 6*n^2 + n^3 - 6)) + (b*f*x^3*(e + f*x)^(n - 4)*(4*b*e + a*f*n - b*e*n)*(2*b*c*f - 3*
a*d*f + b*d*e + a*d*f*n - b*d*e*n))/((a*f - b*e)^3*(a + b*x)^n*(11*n - 6*n^2 + n^3 - 6))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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