3.3101 \(\int \frac{(a+b x)^m (c+d x)^{-4-m}}{e+f x} \, dx\)

Optimal. Leaf size=330 \[ \frac{d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )+a b d f (m+3) (d e-c f (2 m+5))+b^2 \left (c^2 f^2 \left (m^2+6 m+11\right )-c d e f (m+7)+2 d^2 e^2\right )\right )}{(m+1) (m+2) (m+3) (b c-a d)^3 (d e-c f)^3}+\frac{f^3 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (d e-c f)^4}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-3}}{(m+3) (b c-a d) (d e-c f)}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-2} (a d f (m+3)+b (2 d e-c f (m+5)))}{(m+2) (m+3) (b c-a d)^2 (d e-c f)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.49992, antiderivative size = 344, 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, 155, 12, 131} \[ \frac{d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )+a b d f (m+3) (d e-c f (2 m+5))+b^2 \left (c^2 f^2 \left (m^2+6 m+11\right )-c d e f (m+7)+2 d^2 e^2\right )\right )}{(m+1) (m+2) (m+3) (b c-a d)^3 (d e-c f)^3}-\frac{f^3 (a+b x)^{m+1} (c+d x)^{-m-1} \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f) (d e-c f)^3}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-3}}{(m+3) (b c-a d) (d e-c f)}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-2} (a d f (m+3)-b c f (m+5)+2 b d e)}{(m+2) (m+3) (b c-a d)^2 (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.862909, size = 301, normalized size = 0.91 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m-3} \left (-\frac{(c+d x)^2 \left (f^3 \left (m^2+5 m+6\right ) (b c-a d)^3 \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )-d (b e-a f) \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )+a b d f (m+3) (d e-c f (2 m+5))+b^2 \left (c^2 f^2 \left (m^2+6 m+11\right )-c d e f (m+7)+2 d^2 e^2\right )\right )\right )}{(m+1) (m+2) (b c-a d)^2 (b e-a f) (d e-c f)^2}+\frac{d (c+d x) (-a d f (m+3)+b c f (m+5)-2 b d e)}{(m+2) (b c-a d) (c f-d e)}+d\right )}{(m+3) (b c-a d) (c f-d e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-4-m}}{fx+e}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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 - 4}}{f x + e}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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 - 4}}{f x + e}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)**(-4-m)/(f*x+e),x)

[Out]

Timed out

________________________________________________________________________________________

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 - 4}}{f x + e}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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