3.3078 \(\int \frac{(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx\)

Optimal. Leaf size=498 \[ \frac{(a+b x)^m (c+d x)^{-m} \left (-3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (c f m+3 d e)+a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )+3 a b^2 d f (m+1) \left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )+b^3 \left (-\left (-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (m^2-3 m+2\right )+18 c d^2 e^2 f m+6 d^3 e^3\right )\right )\right ) \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{6 m (b e-a f)^3 (d e-c f)^4}-\frac{f (a+b x)^{m+1} (c+d x)^{-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (d e (8 m+15)-c f \left (-2 m^2-2 m+3\right )\right )+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-c d e f (7-8 m)+11 d^2 e^2\right )\right )}{6 (e+f x) (b e-a f)^3 (d e-c f)^3}-\frac{f (a+b x)^{m+1} (c+d x)^{-m} (b (5 d e-c f (2-m))-a d f (m+3))}{6 (e+f x)^2 (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \]

[Out]

-(f*(a + b*x)^(1 + m))/(3*(b*e - a*f)*(d*e - c*f)*(c + d*x)^m*(e + f*x)^3) - (f*(b*(5*d*e - c*f*(2 - m)) - a*d
*f*(3 + m))*(a + b*x)^(1 + m))/(6*(b*e - a*f)^2*(d*e - c*f)^2*(c + d*x)^m*(e + f*x)^2) - (f*(a^2*d^2*f^2*(6 +
5*m + m^2) - a*b*d*f*(d*e*(15 + 8*m) - c*f*(3 - 2*m - 2*m^2)) + b^2*(11*d^2*e^2 - c*d*e*f*(7 - 8*m) + c^2*f^2*
(2 - 3*m + m^2)))*(a + b*x)^(1 + m))/(6*(b*e - a*f)^3*(d*e - c*f)^3*(c + d*x)^m*(e + f*x)) + ((3*a*b^2*d*f*(1
+ m)*(6*d^2*e^2 + 6*c*d*e*f*m - c^2*f^2*(1 - m)*m) - 3*a^2*b*d^2*f^2*(3*d*e + c*f*m)*(2 + 3*m + m^2) + a^3*d^3
*f^3*(6 + 11*m + 6*m^2 + m^3) - b^3*(6*d^3*e^3 + 18*c*d^2*e^2*f*m - 9*c^2*d*e*f^2*(1 - m)*m + c^3*f^3*m*(2 - 3
*m + m^2)))*(a + b*x)^m*Hypergeometric2F1[1, -m, 1 - m, ((b*e - a*f)*(c + d*x))/((d*e - c*f)*(a + b*x))])/(6*(
b*e - a*f)^3*(d*e - c*f)^4*m*(c + d*x)^m)

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Rubi [A]  time = 0.877868, antiderivative size = 520, normalized size of antiderivative = 1.04, 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 = {129, 151, 12, 131} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (-3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (c f m+3 d e)+a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )+3 a b^2 d f (m+1) \left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )+b^3 \left (-\left (-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (m^2-3 m+2\right )+18 c d^2 e^2 f m+6 d^3 e^3\right )\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 m (m+1) (b e-a f)^4 (d e-c f)^3}+\frac{f (a+b x)^{m+1} (c+d x)^{1-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f (c f m (2 m+3)+d e (7 m+12))+b^2 \left (-c^2 f^2 (2-m) m+7 c d e f m+6 d^2 e^2\right )\right )}{6 m (e+f x)^2 (b c-a d) (b e-a f)^2 (d e-c f)^3}-\frac{f (a+b x)^{m+1} (c+d x)^{1-m} (a d f (m+3)-b (c f m+3 d e))}{3 m (e+f x)^3 (b c-a d) (b e-a f) (d e-c f)^2}+\frac{d (a+b x)^{m+1} (c+d x)^{-m}}{m (e+f x)^3 (b c-a d) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^m*(c + d*x)^(-1 - m))/(e + f*x)^4,x]

[Out]

-(f*(a*d*f*(3 + m) - b*(3*d*e + c*f*m))*(a + b*x)^(1 + m)*(c + d*x)^(1 - m))/(3*(b*c - a*d)*(b*e - a*f)*(d*e -
c*f)^2*m*(e + f*x)^3) + (d*(a + b*x)^(1 + m))/((b*c - a*d)*(d*e - c*f)*m*(c + d*x)^m*(e + f*x)^3) + (f*(b^2*(
6*d^2*e^2 + 7*c*d*e*f*m - c^2*f^2*(2 - m)*m) + a^2*d^2*f^2*(6 + 5*m + m^2) - a*b*d*f*(c*f*m*(3 + 2*m) + d*e*(1
2 + 7*m)))*(a + b*x)^(1 + m)*(c + d*x)^(1 - m))/(6*(b*c - a*d)*(b*e - a*f)^2*(d*e - c*f)^3*m*(e + f*x)^2) + ((
3*a*b^2*d*f*(1 + m)*(6*d^2*e^2 + 6*c*d*e*f*m - c^2*f^2*(1 - m)*m) - 3*a^2*b*d^2*f^2*(3*d*e + c*f*m)*(2 + 3*m +
m^2) + a^3*d^3*f^3*(6 + 11*m + 6*m^2 + m^3) - b^3*(6*d^3*e^3 + 18*c*d^2*e^2*f*m - 9*c^2*d*e*f^2*(1 - m)*m + c
^3*f^3*m*(2 - 3*m + m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e - c*f
)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(6*(b*e - a*f)^4*(d*e - c*f)^3*m*(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 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)^{-1-m}}{(e+f x)^4} \, dx &=\frac{d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac{\int \frac{(a+b x)^m (c+d x)^{-m} (a d f (3+m)-b (d e+c f m)+2 b d f x)}{(e+f x)^4} \, dx}{(b c-a d) (d e-c f) m}\\ &=-\frac{f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac{d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}-\frac{\int \frac{(a+b x)^m (c+d x)^{-m} \left (-a b d f (3+2 m) (3 d e+c f m)+b^2 \left (3 d^2 e^2+6 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )+b d f (a d f (3+m)-b (3 d e+c f m)) x\right )}{(e+f x)^3} \, dx}{3 (b c-a d) (b e-a f) (d e-c f)^2 m}\\ &=-\frac{f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac{d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac{f \left (b^2 \left (6 d^2 e^2+7 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m (e+f x)^2}+\frac{\int \frac{\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m}\\ &=-\frac{f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac{d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac{f \left (b^2 \left (6 d^2 e^2+7 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m (e+f x)^2}+\frac{\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) \int \frac{(a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m}\\ &=-\frac{f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac{d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac{f \left (b^2 \left (6 d^2 e^2+7 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m (e+f x)^2}+\frac{\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (2,1+m;2+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (b e-a f)^4 (d e-c f)^3 m (1+m)}\\ \end{align*}

Mathematica [A]  time = 1.81316, size = 466, normalized size = 0.94 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m} \left (-\frac{(e+f x) \left ((e+f x)^2 (b c-a d) \left (3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (c f m+3 d e)-a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )-3 a b^2 d f (m+1) \left (c^2 f^2 (m-1) m+6 c d e f m+6 d^2 e^2\right )+b^3 \left (9 c^2 d e f^2 (m-1) m+c^3 f^3 m \left (m^2-3 m+2\right )+18 c d^2 e^2 f m+6 d^3 e^3\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )+f (m+1) (c+d x)^2 (b e-a f)^2 \left (-a^2 d^2 f^2 \left (m^2+5 m+6\right )+a b d f (c f m (2 m+3)+d e (7 m+12))+b^2 \left (-\left (c^2 f^2 (m-2) m+7 c d e f m+6 d^2 e^2\right )\right )\right )\right )}{c+d x}+2 f (m+1) (c+d x) (a f-b e)^3 (c f-d e) (b (c f m+3 d e)-a d f (m+3))+6 d (m+1) (b e-a f)^4 (d e-c f)^2\right )}{6 m (m+1) (e+f x)^3 (b c-a d) (b e-a f)^4 (d e-c f)^2 (c f-d e)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^m*(c + d*x)^(-1 - m))/(e + f*x)^4,x]

[Out]

-((a + b*x)^(1 + m)*(6*d*(b*e - a*f)^4*(d*e - c*f)^2*(1 + m) + 2*f*(-(b*e) + a*f)^3*(-(d*e) + c*f)*(1 + m)*(-(
a*d*f*(3 + m)) + b*(3*d*e + c*f*m))*(c + d*x) - ((e + f*x)*(f*(b*e - a*f)^2*(1 + m)*(-(b^2*(6*d^2*e^2 + 7*c*d*
e*f*m + c^2*f^2*(-2 + m)*m)) - a^2*d^2*f^2*(6 + 5*m + m^2) + a*b*d*f*(c*f*m*(3 + 2*m) + d*e*(12 + 7*m)))*(c +
d*x)^2 + (b*c - a*d)*(-3*a*b^2*d*f*(1 + m)*(6*d^2*e^2 + 6*c*d*e*f*m + c^2*f^2*(-1 + m)*m) + 3*a^2*b*d^2*f^2*(3
*d*e + c*f*m)*(2 + 3*m + m^2) - a^3*d^3*f^3*(6 + 11*m + 6*m^2 + m^3) + b^3*(6*d^3*e^3 + 18*c*d^2*e^2*f*m + 9*c
^2*d*e*f^2*(-1 + m)*m + c^3*f^3*m*(2 - 3*m + m^2)))*(e + f*x)^2*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e - c*f
)*(a + b*x))/((b*e - a*f)*(c + d*x))]))/(c + d*x)))/(6*(b*c - a*d)*(b*e - a*f)^4*(d*e - c*f)^2*(-(d*e) + c*f)*
m*(1 + m)*(c + d*x)^m*(e + f*x)^3)

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Maple [F]  time = 0.132, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-1-m}}{ \left ( fx+e \right ) ^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x)

[Out]

int((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,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 - 1}}{{\left (f x + e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^4, 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 - 1}}{f^{4} x^{4} + 4 \, e f^{3} x^{3} + 6 \, e^{2} f^{2} x^{2} + 4 \, e^{3} f x + e^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m - 1)/(f^4*x^4 + 4*e*f^3*x^3 + 6*e^2*f^2*x^2 + 4*e^3*f*x + e^4), 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)**(-1-m)/(f*x+e)**4,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 - 1}}{{\left (f x + e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^4, x)