3.31.94 \(\int \frac {(a+b x)^m (c+d x)^{-3-m}}{(e+f x)^2} \, dx\) [3094]

Optimal. Leaf size=384 \[ -\frac {d (a d f (3+m)-b (d e+c f (2+m))) (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (2+m)}-\frac {d \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-b^2 \left (d^2 e^2-c d e f (5+2 m)-c^2 f^2 \left (2+3 m+m^2\right )\right )-a b d f \left (d e (3+2 m)+c f \left (9+8 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (b e-a f) (d e-c f)^3 (1+m) (2+m)}-\frac {f (a+b x)^{1+m} (c+d x)^{-2-m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {f^2 (a d f (3+m)-b (3 d e+c f m)) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{(b e-a f) (d e-c f)^4 m} \]

[Out]

-d*(a*d*f*(3+m)-b*(d*e+c*f*(2+m)))*(b*x+a)^(1+m)*(d*x+c)^(-2-m)/(-a*d+b*c)/(-a*f+b*e)/(-c*f+d*e)^2/(2+m)-d*(a^
2*d^2*f^2*(m^2+5*m+6)-b^2*(d^2*e^2-c*d*e*f*(5+2*m)-c^2*f^2*(m^2+3*m+2))-a*b*d*f*(d*e*(3+2*m)+c*f*(2*m^2+8*m+9)
))*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/(-a*d+b*c)^2/(-a*f+b*e)/(-c*f+d*e)^3/(1+m)/(2+m)-f*(b*x+a)^(1+m)*(d*x+c)^(-2-m
)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)+f^2*(a*d*f*(3+m)-b*(c*f*m+3*d*e))*(b*x+a)^m*hypergeom([1, -m],[1-m],(-a*f+b*e)
*(d*x+c)/(-c*f+d*e)/(b*x+a))/(-a*f+b*e)/(-c*f+d*e)^4/m/((d*x+c)^m)

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Rubi [A]
time = 0.39, antiderivative size = 382, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {105, 160, 12, 133} \begin {gather*} -\frac {d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+8 m+9\right )+d e (2 m+3)\right )-\left (b^2 \left (-c^2 f^2 \left (m^2+3 m+2\right )-c d e f (2 m+5)+d^2 e^2\right )\right )\right )}{(m+1) (m+2) (b c-a d)^2 (b e-a f) (d e-c f)^3}+\frac {f^2 (a+b x)^m (c+d x)^{-m} (a d f (m+3)-b (c f m+3 d e)) \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (b e-a f) (d e-c f)^4}+\frac {d (a+b x)^{m+1} (c+d x)^{-m-2} (-a d f (m+3)+b c f (m+2)+b d e)}{(m+2) (b c-a d) (b e-a f) (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-2}}{(e+f x) (b e-a f) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^m*(c + d*x)^(-3 - m))/(e + f*x)^2,x]

[Out]

(d*(b*d*e + b*c*f*(2 + m) - a*d*f*(3 + m))*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/((b*c - a*d)*(b*e - a*f)*(d*e
- c*f)^2*(2 + m)) - (d*(a^2*d^2*f^2*(6 + 5*m + m^2) - b^2*(d^2*e^2 - c*d*e*f*(5 + 2*m) - c^2*f^2*(2 + 3*m + m
^2)) - a*b*d*f*(d*e*(3 + 2*m) + c*f*(9 + 8*m + 2*m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/((b*c - a*d)^2*(
b*e - a*f)*(d*e - c*f)^3*(1 + m)*(2 + m)) - (f*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/((b*e - a*f)*(d*e - c*f)*
(e + f*x)) + (f^2*(a*d*f*(3 + m) - b*(3*d*e + c*f*m))*(a + b*x)^m*Hypergeometric2F1[1, -m, 1 - m, ((b*e - a*f)
*(c + d*x))/((d*e - c*f)*(a + b*x))])/((b*e - a*f)*(d*e - c*f)^4*m*(c + d*x)^m)

Rule 12

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

Rule 105

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, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 133

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

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 \frac {(a+b x)^m (c+d x)^{-3-m}}{(e+f x)^2} \, dx &=\frac {d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m) (e+f x)}+\frac {\int \frac {(a+b x)^m (c+d x)^{-2-m} (b d e-b c f (2+m)+a d f (3+m)+2 b d f x)}{(e+f x)^2} \, dx}{(b c-a d) (d e-c f) (2+m)}\\ &=\frac {d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m) (e+f x)}-\frac {f (a d f (3+m)-b (d e+c f (2+m))) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (2+m) (e+f x)}-\frac {\int \frac {(a+b x)^m (c+d x)^{-2-m} \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f (3+2 m) (d e+c f (2+m))-b^2 \left (d^2 e^2-2 c d e f (2+m)-c^2 f^2 m (2+m)\right )+b d f (a d f (3+m)-b (d e+c f (2+m))) x\right )}{e+f x} \, dx}{(b c-a d) (b e-a f) (d e-c f)^2 (2+m)}\\ &=-\frac {d \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-b^2 \left (d^2 e^2-c d e f (5+2 m)-c^2 f^2 \left (2+3 m+m^2\right )\right )-a b d f \left (d e (3+2 m)+c f \left (9+8 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (b e-a f) (d e-c f)^3 (1+m) (2+m)}+\frac {d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m) (e+f x)}-\frac {f (a d f (3+m)-b (d e+c f (2+m))) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (2+m) (e+f x)}-\frac {\int \frac {(b c-a d)^2 f^2 \left (2+3 m+m^2\right ) (a d f (3+m)-b (3 d e+c f m)) (a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(b c-a d)^2 (b e-a f) (d e-c f)^3 (1+m) (2+m)}\\ &=-\frac {d \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-b^2 \left (d^2 e^2-c d e f (5+2 m)-c^2 f^2 \left (2+3 m+m^2\right )\right )-a b d f \left (d e (3+2 m)+c f \left (9+8 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (b e-a f) (d e-c f)^3 (1+m) (2+m)}+\frac {d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m) (e+f x)}-\frac {f (a d f (3+m)-b (d e+c f (2+m))) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (2+m) (e+f x)}-\frac {\left (f^2 (a d f (3+m)-b (3 d e+c f m))\right ) \int \frac {(a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(b e-a f) (d e-c f)^3}\\ &=-\frac {d \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-b^2 \left (d^2 e^2-c d e f (5+2 m)-c^2 f^2 \left (2+3 m+m^2\right )\right )-a b d f \left (d e (3+2 m)+c f \left (9+8 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (b e-a f) (d e-c f)^3 (1+m) (2+m)}+\frac {d (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (d e-c f) (2+m) (e+f x)}-\frac {f (a d f (3+m)-b (d e+c f (2+m))) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (2+m) (e+f x)}-\frac {f^2 (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (1,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (d e-c f)^3 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 333, normalized size = 0.87 \begin {gather*} -\frac {(a+b x)^{1+m} (c+d x)^{-2-m} \left (\frac {d}{e+f x}+\frac {f (b d e+b c f (2+m)-a d f (3+m)) (c+d x)}{(b e-a f) (d e-c f) (e+f x)}+\frac {(c+d x) \left (d (b e-a f) (1+m) \left (-a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (d^2 e^2-c d e f (5+2 m)-c^2 f^2 \left (2+3 m+m^2\right )\right )+a b d f \left (d e (3+2 m)+c f \left (9+8 m+2 m^2\right )\right )\right )+(b c-a d)^2 f^2 \left (2+3 m+m^2\right ) (-a d f (3+m)+b (3 d e+c f m)) \, _2F_1\left (1,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{(b c-a d) (b e-a f)^2 (d e-c f)^2 (1+m)^2}\right )}{(b c-a d) (-d e+c f) (2+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^m*(c + d*x)^(-3 - m))/(e + f*x)^2,x]

[Out]

-(((a + b*x)^(1 + m)*(c + d*x)^(-2 - m)*(d/(e + f*x) + (f*(b*d*e + b*c*f*(2 + m) - a*d*f*(3 + m))*(c + d*x))/(
(b*e - a*f)*(d*e - c*f)*(e + f*x)) + ((c + d*x)*(d*(b*e - a*f)*(1 + m)*(-(a^2*d^2*f^2*(6 + 5*m + m^2)) + b^2*(
d^2*e^2 - c*d*e*f*(5 + 2*m) - c^2*f^2*(2 + 3*m + m^2)) + a*b*d*f*(d*e*(3 + 2*m) + c*f*(9 + 8*m + 2*m^2))) + (b
*c - a*d)^2*f^2*(2 + 3*m + m^2)*(-(a*d*f*(3 + m)) + b*(3*d*e + c*f*m))*Hypergeometric2F1[1, 1 + m, 2 + m, ((d*
e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))]))/((b*c - a*d)*(b*e - a*f)^2*(d*e - c*f)^2*(1 + m)^2)))/((b*c - a
*d)*(-(d*e) + c*f)*(2 + m)))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-3-m}}{\left (f x +e \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-3-m)/(f*x+e)^2,x)

[Out]

int((b*x+a)^m*(d*x+c)^(-3-m)/(f*x+e)^2,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)^(-3-m)/(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 3)/(f*x + e)^2, 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)^(-3-m)/(f*x+e)^2,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m - 3)/(f^2*x^2 + 2*f*x*e + e^2), 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)**(-3-m)/(f*x+e)**2,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)^(-3-m)/(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 3)/(f*x + e)^2, 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 (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^{m+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^m/((e + f*x)^2*(c + d*x)^(m + 3)),x)

[Out]

int((a + b*x)^m/((e + f*x)^2*(c + d*x)^(m + 3)), x)

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