3.32.58 \(\int (a+b x)^m (c+d x)^n (e+f x)^{-4-m-n} \, dx\) [3158]

Optimal. Leaf size=402 \[ -\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {f (a d f (2+m)+b (c f (2+n)-d e (4+m+n))) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}+\frac {\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )+2 a b d f (1+m) (c f (1+n)-d e (3+m+n))-b^2 \left (2 c d e f (1+n) (3+m+n)-c^2 f^2 \left (2+3 n+n^2\right )-d^2 e^2 \left (6+m^2+5 n+n^2+m (5+2 n)\right )\right )\right ) (a+b x)^{1+m} (c+d x)^n \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{-n} (e+f x)^{-1-m-n} \, _2F_1\left (1+m,-n;2+m;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^3 (d e-c f)^2 (1+m) (2+m+n) (3+m+n)} \]

[Out]

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

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Rubi [A]
time = 0.39, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {136, 160, 12, 134} \begin {gather*} \frac {(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (c f (n+1)-d e (m+n+3))-\left (b^2 \left (-c^2 f^2 \left (n^2+3 n+2\right )+2 c d e f (n+1) (m+n+3)-d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right )\right ) \left (\frac {(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac {f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a d f (m+2)+b c f (n+2)-b d e (m+n+4))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((f*(a + b*x)^(1 + m)*(c + d*x)^(1 + n)*(e + f*x)^(-3 - m - n))/((b*e - a*f)*(d*e - c*f)*(3 + m + n))) + (f*(
a*d*f*(2 + m) + b*c*f*(2 + n) - b*d*e*(4 + m + n))*(a + b*x)^(1 + m)*(c + d*x)^(1 + n)*(e + f*x)^(-2 - m - n))
/((b*e - a*f)^2*(d*e - c*f)^2*(2 + m + n)*(3 + m + n)) + ((a^2*d^2*f^2*(2 + 3*m + m^2) + 2*a*b*d*f*(1 + m)*(c*
f*(1 + n) - d*e*(3 + m + n)) - b^2*(2*c*d*e*f*(1 + n)*(3 + m + n) - c^2*f^2*(2 + 3*n + n^2) - d^2*e^2*(6 + m^2
+ 5*n + n^2 + m*(5 + 2*n))))*(a + b*x)^(1 + m)*(c + d*x)^n*(e + f*x)^(-1 - m - n)*Hypergeometric2F1[1 + m, -n
, 2 + m, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)^3*(d*e - c*f)^2*(1 + m)*(2 + m + n)
*(3 + m + n)*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 134

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c
*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f*x))))^n, x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rule 136

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !SumSimplerQ[n, 1] &&  !SumSimplerQ[p, 1]))

Rule 160

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

Rubi steps

\begin {align*} \int (a+b x)^m (c+d x)^n (e+f x)^{-4-m-n} \, dx &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}-\frac {\int (a+b x)^m (c+d x)^n (e+f x)^{-3-m-n} (a d f (2+m)+b c f (2+n)-b d e (3+m+n)+b d f x) \, dx}{(b e-a f) (d e-c f) (3+m+n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}+\frac {\int \left (-f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))-(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) (a+b x)^m (c+d x)^n (e+f x)^{-2-m-n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}-\frac {\left (f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))+(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) \int (a+b x)^m (c+d x)^n (e+f x)^{-2-m-n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}-\frac {\left (f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))+(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) (a+b x)^{1+m} (c+d x)^n \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{-n} (e+f x)^{-1-m-n} \, _2F_1\left (1+m,-n;2+m;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^3 (d e-c f)^2 (1+m) (2+m+n) (3+m+n)}\\ \end {align*}

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Mathematica [A]
time = 0.80, size = 346, normalized size = 0.86 \begin {gather*} -\frac {(a+b x)^{1+m} (c+d x)^n (e+f x)^{-3-m-n} \left (f (c+d x)+\frac {f (-a d f (2+m)-b c f (2+n)+b d e (4+m+n)) (c+d x) (e+f x)}{(b e-a f) (d e-c f) (2+m+n)}-\frac {\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )-2 a b d f (1+m) (-c f (1+n)+d e (3+m+n))+b^2 \left (-2 c d e f (1+n) (3+m+n)+c^2 f^2 \left (2+3 n+n^2\right )+d^2 e^2 \left (6+m^2+5 n+n^2+m (5+2 n)\right )\right )\right ) \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{-n} (e+f x)^2 \, _2F_1\left (1+m,-n;2+m;\frac {(-d e+c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^2 (d e-c f) (1+m) (2+m+n)}\right )}{(b e-a f) (d e-c f) (3+m+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*x)^(1 + m)*(c + d*x)^n*(e + f*x)^(-3 - m - n)*(f*(c + d*x) + (f*(-(a*d*f*(2 + m)) - b*c*f*(2 + n) +
b*d*e*(4 + m + n))*(c + d*x)*(e + f*x))/((b*e - a*f)*(d*e - c*f)*(2 + m + n)) - ((a^2*d^2*f^2*(2 + 3*m + m^2)
- 2*a*b*d*f*(1 + m)*(-(c*f*(1 + n)) + d*e*(3 + m + n)) + b^2*(-2*c*d*e*f*(1 + n)*(3 + m + n) + c^2*f^2*(2 + 3*
n + n^2) + d^2*e^2*(6 + m^2 + 5*n + n^2 + m*(5 + 2*n))))*(e + f*x)^2*Hypergeometric2F1[1 + m, -n, 2 + m, ((-(d
*e) + c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))])/((b*e - a*f)^2*(d*e - c*f)*(1 + m)*(2 + m + n)*(((b*e - a*f)*(
c + d*x))/((b*c - a*d)*(e + f*x)))^n)))/((b*e - a*f)*(d*e - c*f)*(3 + m + n)))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{m} \left (d x +c \right )^{n} \left (f x +e \right )^{-4-m -n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
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((b*x+a)**m*(d*x+c)**n*(f*x+e)**(-4-m-n),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n}{{\left (e+f\,x\right )}^{m+n+4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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