3.32.51 \(\int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-4+n} \, dx\) [3151]

Optimal. Leaf size=428 \[ -\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}+\frac {f (a d f (2+m)-b (d e (4-n)-c f (2-m-n))) (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}+\frac {\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )-2 a b d f (1+m) (d e (3-n)-c f (1-m-n))-b^2 \left (2 c d e f (3-n) (1-m-n)-d^2 e^2 \left (6-5 n+n^2\right )-c^2 f^2 \left (2+m^2-m (3-2 n)-3 n+n^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x)^{-1+n} \, _2F_1\left (1+m,m+n;2+m;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^3 (d e-c f)^2 (1+m) (2-n) (3-n)} \]

[Out]

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

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Rubi [A]
time = 0.38, antiderivative size = 426, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {136, 160, 12, 134} \begin {gather*} \frac {(a+b x)^{m+1} (e+f x)^{n-1} (c+d x)^{-m-n} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (d e (3-n)-c f (-m-n+1))-\left (b^2 \left (-c^2 f^2 \left (m^2-m (3-2 n)+n^2-3 n+2\right )+2 c d e f (3-n) (-m-n+1)-d^2 e^2 \left (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 )^{m+n} \, _2F_1\left (m+1,m+n;m+2;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (2-n) (3-n) (b e-a f)^3 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (e+f x)^{n-3} (c+d x)^{-m-n+1}}{(3-n) (b e-a f) (d e-c f)}+\frac {f (a+b x)^{m+1} (e+f x)^{n-2} (c+d x)^{-m-n+1} (a d f (m+2)+b c f (-m-n+2)-b d e (4-n))}{(2-n) (3-n) (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)^(-m - n)*(e + f*x)^(-4 + n),x]

[Out]

-((f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m - n)*(e + f*x)^(-3 + n))/((b*e - a*f)*(d*e - c*f)*(3 - n))) + (f*(a*d*
f*(2 + m) - b*d*e*(4 - n) + b*c*f*(2 - m - n))*(a + b*x)^(1 + m)*(c + d*x)^(1 - m - n)*(e + f*x)^(-2 + n))/((b
*e - a*f)^2*(d*e - c*f)^2*(2 - n)*(3 - n)) + ((a^2*d^2*f^2*(2 + 3*m + m^2) - 2*a*b*d*f*(1 + m)*(d*e*(3 - n) -
c*f*(1 - m - n)) - b^2*(2*c*d*e*f*(3 - n)*(1 - m - n) - d^2*e^2*(6 - 5*n + n^2) - c^2*f^2*(2 + m^2 - m*(3 - 2*
n) - 3*n + n^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-m - n)*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^(m +
n)*(e + f*x)^(-1 + n)*Hypergeometric2F1[1 + m, 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 - n)*(3 - 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)^{-m-n} (e+f x)^{-4+n} \, dx &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}-\frac {\int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} (a d f (2+m)-b d e (3-n)+b c f (2-m-n)+b d f x) \, dx}{(b e-a f) (d e-c f) (3-n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}+\frac {f (a d f (2+m)-b d e (4-n)+b c f (2-m-n)) (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}+\frac {\int \left (-f (b c (1+m)+a d (1-m-n)) (a d f (2+m)-b d e (4-n)+b c f (2-m-n))-\left (a b c d f^2+b d e (a d f (2+m)-b d e (3-n)+b c f (2-m-n))-(b c+a d) f (a d f (2+m)-b d e (3-n)+b c f (2-m-n))\right ) (2-n)\right ) (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-2+n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}+\frac {f (a d f (2+m)-b d e (4-n)+b c f (2-m-n)) (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}+\frac {\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )-2 a b d f (1+m) (d e (3-n)-c f (1-m-n))-b^2 \left (2 c d e f (3-n) (1-m-n)-d^2 e^2 \left (6-5 n+n^2\right )-c^2 f^2 \left (2+m^2-m (3-2 n)-3 n+n^2\right )\right )\right ) \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-2+n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-3+n}}{(b e-a f) (d e-c f) (3-n)}+\frac {f (a d f (2+m)-b d e (4-n)+b c f (2-m-n)) (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f)^2 (d e-c f)^2 (2-n) (3-n)}+\frac {\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )-2 a b d f (1+m) (d e (3-n)-c f (1-m-n))-b^2 \left (2 c d e f (3-n) (1-m-n)-d^2 e^2 \left (6-5 n+n^2\right )-c^2 f^2 \left (2+m^2-m (3-2 n)-3 n+n^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x)^{-1+n} \, _2F_1\left (1+m,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-n) (3-n)}\\ \end {align*}

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Mathematica [A]
time = 0.76, size = 341, normalized size = 0.80 \begin {gather*} \frac {(a+b x)^{1+m} (c+d x)^{-m-n} (e+f x)^{-3+n} \left (f (c+d x)+\frac {f (a d f (2+m)+b d e (-4+n)-b c f (-2+m+n)) (c+d x) (e+f x)}{(b e-a f) (d e-c f) (-2+n)}+\frac {\left (a^2 d^2 f^2 \left (2+3 m+m^2\right )+2 a b d f (1+m) (d e (-3+n)-c f (-1+m+n))+b^2 \left (-2 c d e f (-3+n) (-1+m+n)+d^2 e^2 \left (6-5 n+n^2\right )+c^2 f^2 \left (2+m^2-3 n+n^2+m (-3+2 n)\right )\right )\right ) \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x)^2 \, _2F_1\left (1+m,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+n)}\right )}{(b e-a f) (d e-c f) (-3+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-m -n} \left (f x +e \right )^{-4+n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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