3.3051 \(\int (a+b x) (c+d x)^{-4+n} (e+f x)^{-n} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.124019, 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} \[ \frac{(b c-a d) (c+d x)^{n-3} (e+f x)^{1-n}}{d (3-n) (d e-c f)}+\frac{(c+d x)^{n-2} (e+f x)^{1-n} (2 a d f+b c f (1-n)-b d e (3-n))}{d (2-n) (3-n) (d e-c f)^2}-\frac{f (c+d x)^{n-1} (e+f x)^{1-n} (2 a d f+b c f (1-n)-b d e (3-n))}{d (1-n) (2-n) (3-n) (d e-c f)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.265382, size = 112, normalized size = 0.54 \[ \frac{(c+d x)^{n-3} (e+f x)^{1-n} \left (\frac{(c+d x) (-c f (n-2)+d e (n-1)+d f x) (2 a d f-b c f (n-1)+b d e (n-3))}{(n-2) (n-1) (d e-c f)^2}+a d-b c\right )}{d (n-3) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 506, normalized size = 2.4 \begin{align*} -{\frac{ \left ( dx+c \right ) ^{-3+n} \left ( fx+e \right ) \left ( b{c}^{2}{f}^{2}{n}^{2}x-2\,bcdef{n}^{2}x-bcd{f}^{2}n{x}^{2}+b{d}^{2}{e}^{2}{n}^{2}x+b{d}^{2}efn{x}^{2}+a{c}^{2}{f}^{2}{n}^{2}-2\,acdef{n}^{2}-2\,acd{f}^{2}nx+a{d}^{2}{e}^{2}{n}^{2}+2\,a{d}^{2}efnx+2\,a{d}^{2}{f}^{2}{x}^{2}-4\,b{c}^{2}{f}^{2}nx+8\,bcdefnx+bcd{f}^{2}{x}^{2}-4\,b{d}^{2}{e}^{2}nx-3\,b{d}^{2}ef{x}^{2}-5\,a{c}^{2}{f}^{2}n+8\,acdefn+6\,acd{f}^{2}x-3\,a{d}^{2}{e}^{2}n-2\,a{d}^{2}efx+b{c}^{2}efn+3\,b{c}^{2}{f}^{2}x-bcd{e}^{2}n-10\,bcdefx+3\,b{d}^{2}{e}^{2}x+6\,a{c}^{2}{f}^{2}-6\,acdef+2\,a{d}^{2}{e}^{2}-3\,b{c}^{2}ef+bcd{e}^{2} \right ) }{ \left ({c}^{3}{f}^{3}{n}^{3}-3\,{c}^{2}de{f}^{2}{n}^{3}+3\,c{d}^{2}{e}^{2}f{n}^{3}-{d}^{3}{e}^{3}{n}^{3}-6\,{c}^{3}{f}^{3}{n}^{2}+18\,{c}^{2}de{f}^{2}{n}^{2}-18\,c{d}^{2}{e}^{2}f{n}^{2}+6\,{d}^{3}{e}^{3}{n}^{2}+11\,{c}^{3}{f}^{3}n-33\,{c}^{2}de{f}^{2}n+33\,c{d}^{2}{e}^{2}fn-11\,{d}^{3}{e}^{3}n-6\,{c}^{3}{f}^{3}+18\,{c}^{2}de{f}^{2}-18\,c{d}^{2}{e}^{2}f+6\,{d}^{3}{e}^{3} \right ) \left ( fx+e \right ) ^{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x+c)^(-3+n)*(f*x+e)*(b*c^2*f^2*n^2*x-2*b*c*d*e*f*n^2*x-b*c*d*f^2*n*x^2+b*d^2*e^2*n^2*x+b*d^2*e*f*n*x^2+a*c
^2*f^2*n^2-2*a*c*d*e*f*n^2-2*a*c*d*f^2*n*x+a*d^2*e^2*n^2+2*a*d^2*e*f*n*x+2*a*d^2*f^2*x^2-4*b*c^2*f^2*n*x+8*b*c
*d*e*f*n*x+b*c*d*f^2*x^2-4*b*d^2*e^2*n*x-3*b*d^2*e*f*x^2-5*a*c^2*f^2*n+8*a*c*d*e*f*n+6*a*c*d*f^2*x-3*a*d^2*e^2
*n-2*a*d^2*e*f*x+b*c^2*e*f*n+3*b*c^2*f^2*x-b*c*d*e^2*n-10*b*c*d*e*f*x+3*b*d^2*e^2*x+6*a*c^2*f^2-6*a*c*d*e*f+2*
a*d^2*e^2-3*b*c^2*e*f+b*c*d*e^2)/(c^3*f^3*n^3-3*c^2*d*e*f^2*n^3+3*c*d^2*e^2*f*n^3-d^3*e^3*n^3-6*c^3*f^3*n^2+18
*c^2*d*e*f^2*n^2-18*c*d^2*e^2*f*n^2+6*d^3*e^3*n^2+11*c^3*f^3*n-33*c^2*d*e*f^2*n+33*c*d^2*e^2*f*n-11*d^3*e^3*n-
6*c^3*f^3+18*c^2*d*e*f^2-18*c*d^2*e^2*f+6*d^3*e^3)/((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 - 4}}{{\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)^(-4+n)/((f*x+e)^n),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.82412, size = 1790, normalized size = 8.65 \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)^(-4+n)/((f*x+e)^n),x, algorithm="fricas")

[Out]

-(6*a*c^3*e*f^2 - (3*b*d^3*e*f^2 - (b*c*d^2 + 2*a*d^3)*f^3 - (b*d^3*e*f^2 - b*c*d^2*f^3)*n)*x^4 + (b*c^2*d + 2
*a*c*d^2)*e^3 - 3*(b*c^3 + 2*a*c^2*d)*e^2*f - (12*b*c*d^2*e*f^2 - 4*(b*c^2*d + 2*a*c*d^2)*f^3 - (b*d^3*e^2*f -
2*b*c*d^2*e*f^2 + b*c^2*d*f^3)*n^2 + (3*b*d^3*e^2*f - 2*(4*b*c*d^2 + a*d^3)*e*f^2 + (5*b*c^2*d + 2*a*c*d^2)*f
^3)*n)*x^3 + (a*c*d^2*e^3 - 2*a*c^2*d*e^2*f + a*c^3*e*f^2)*n^2 + (3*b*d^3*e^3 - 9*b*c*d^2*e^2*f - 9*b*c^2*d*e*
f^2 + 3*(b*c^3 + 4*a*c^2*d)*f^3 + (b*d^3*e^3 - (b*c*d^2 - a*d^3)*e^2*f - (b*c^2*d + 2*a*c*d^2)*e*f^2 + (b*c^3
+ a*c^2*d)*f^3)*n^2 - (4*b*d^3*e^3 - (4*b*c*d^2 - a*d^3)*e^2*f - 4*(b*c^2*d + 2*a*c*d^2)*e*f^2 + (4*b*c^3 + 7*
a*c^2*d)*f^3)*n)*x^2 - (5*a*c^3*e*f^2 + (b*c^2*d + 3*a*c*d^2)*e^3 - (b*c^3 + 8*a*c^2*d)*e^2*f)*n + (6*a*c^2*d*
e*f^2 + 6*a*c^3*f^3 + 2*(2*b*c*d^2 + a*d^3)*e^3 - 6*(2*b*c^2*d + a*c*d^2)*e^2*f + (a*c^3*f^3 + (b*c*d^2 + a*d^
3)*e^3 - (2*b*c^2*d + a*c*d^2)*e^2*f + (b*c^3 - a*c^2*d)*e*f^2)*n^2 - (5*a*c^3*f^3 + (5*b*c*d^2 + 3*a*d^3)*e^3
- (8*b*c^2*d + 7*a*c*d^2)*e^2*f + (3*b*c^3 - a*c^2*d)*e*f^2)*n)*x)*(d*x + c)^(n - 4)/((6*d^3*e^3 - 18*c*d^2*e
^2*f + 18*c^2*d*e*f^2 - 6*c^3*f^3 - (d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*n^3 + 6*(d^3*e^3 - 3*c
*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*n^2 - 11*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*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)**(-4+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 - 4}}{{\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)^(-4+n)/((f*x+e)^n),x, algorithm="giac")

[Out]

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