3.3094 \(\int \frac{(a+b x)^m (c+d x)^{-3-m}}{(e+f x)^2} \, dx\)

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

[Out]

-((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))) - (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)

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Rubi [A]  time = 0.57697, antiderivative size = 398, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {129, 151, 155, 12, 131} \[ -\frac{d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+8 m+9\right )+d e (2 m+3)\right )+b^2 \left (-\left (-c^2 f^2 \left (m^2+3 m+2\right )-c d e f (2 m+5)+d^2 e^2\right )\right )\right )}{(m+1) (m+2) (b c-a d)^2 (b e-a f) (d e-c f)^3}-\frac{f^2 (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+3)-b (c f m+3 d e)) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f)^2 (d e-c f)^3}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (e+f x) (b c-a d) (d e-c f)}+\frac{f (a+b x)^{m+1} (c+d x)^{-m-1} (-a d f (m+3)+b c f (m+2)+b d e)}{(m+2) (e+f x) (b c-a d) (b e-a f) (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 129

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] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 1])))

Rule 151

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] && LtQ[m, -1] && IntegerQ[m]

Rule 155

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])))

Rule 12

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

Rule 131

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

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*}

Mathematica [A]  time = 0.838979, size = 333, normalized size = 0.87 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m-2} \left (\frac{(c+d x) \left (d (m+1) (b e-a f) \left (-a^2 d^2 f^2 \left (m^2+5 m+6\right )+a b d f \left (c f \left (2 m^2+8 m+9\right )+d e (2 m+3)\right )+b^2 \left (-c^2 f^2 \left (m^2+3 m+2\right )-c d e f (2 m+5)+d^2 e^2\right )\right )+f^2 \left (m^2+3 m+2\right ) (b c-a d)^2 (b (c f m+3 d e)-a d f (m+3)) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{(m+1)^2 (b c-a d) (b e-a f)^2 (d e-c f)^2}+\frac{f (c+d x) (-a d f (m+3)+b c f (m+2)+b d e)}{(e+f x) (b e-a f) (d e-c f)}+\frac{d}{e+f x}\right )}{(m+2) (b c-a d) (c f-d e)} \]

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 = 0.077, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-3-m}}{ \left ( fx+e \right ) ^{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \end{align*}

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*e*f*x + e^2), x)

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

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)