3.31.86 \(\int \frac {(a+b x)^m (c+d x)^{-2-m}}{(e+f x)^3} \, dx\) [3086]

Optimal. Leaf size=432 \[ \frac {d \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)-c^2 f^2 \left (1-m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m)}-\frac {f (a+b x)^{1+m} (c+d x)^{-1-m}}{2 (b e-a f) (d e-c f) (e+f x)^2}-\frac {f (b (4 d e-c f (1-m))-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-1-m}}{2 (b e-a f)^2 (d e-c f)^2 (e+f x)}-\frac {f \left (2 a b d f (2+m) (3 d e+c f m)-b^2 \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )\right ) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{2 (b e-a f)^2 (d e-c f)^4 m} \]

[Out]

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

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Rubi [A]
time = 0.38, antiderivative size = 431, normalized size of antiderivative = 1.00, 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 = {105, 156, 160, 12, 133} \begin {gather*} -\frac {f (a+b x)^m (c+d x)^{-m} \left (-a^2 d^2 f^2 \left (m^2+5 m+6\right )+2 a b d f (m+2) (c f m+3 d e)-\left (b^2 \left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\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 )}{2 m (b e-a f)^2 (d e-c f)^4}+\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+5 m+3\right )+d e (5 m+9)\right )+b^2 \left (-c^2 f^2 \left (1-m^2\right )+5 c d e f (m+1)+2 d^2 e^2\right )\right )}{2 (m+1) (b c-a d) (b e-a f)^2 (d e-c f)^3}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-1} (-a d f (m+3)-b c f (1-m)+4 b d e)}{2 (e+f x) (b e-a f)^2 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-1}}{2 (e+f x)^2 (b e-a f) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(a^2*d^2*f^2*(6 + 5*m + m^2) + b^2*(2*d^2*e^2 + 5*c*d*e*f*(1 + m) - c^2*f^2*(1 - m^2)) - a*b*d*f*(d*e*(9 +
5*m) + c*f*(3 + 5*m + 2*m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(2*(b*c - a*d)*(b*e - a*f)^2*(d*e - c*f)^
3*(1 + m)) - (f*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(2*(b*e - a*f)*(d*e - c*f)*(e + f*x)^2) - (f*(4*b*d*e -
b*c*f*(1 - m) - a*d*f*(3 + m))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(2*(b*e - a*f)^2*(d*e - c*f)^2*(e + f*x))
- (f*(2*a*b*d*f*(2 + m)*(3*d*e + c*f*m) - b^2*(6*d^2*e^2 + 6*c*d*e*f*m - c^2*f^2*(1 - m)*m) - a^2*d^2*f^2*(6
+ 5*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))])/(2
*(b*e - a*f)^2*(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 156

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, -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 \frac {(a+b x)^m (c+d x)^{-2-m}}{(e+f x)^3} \, dx &=\frac {d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)^2}+\frac {\int \frac {(a+b x)^m (c+d x)^{-1-m} (-f (b c (1+m)-a d (3+m))+2 b d f x)}{(e+f x)^3} \, dx}{(b c-a d) (d e-c f) (1+m)}\\ &=\frac {d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)^2}+\frac {f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}-\frac {\int \frac {(a+b x)^m (c+d x)^{-1-m} \left (f \left (b^2 c (4 d e-c f (1-m)) (1+m)+a^2 d^2 f \left (6+5 m+m^2\right )-a b d (3+2 m) (2 d e+c f (1+m))\right )-b d f (2 b d e+b c f (1+m)-a d f (3+m)) x\right )}{(e+f x)^2} \, dx}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m)}\\ &=\frac {d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)^2}+\frac {f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}+\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)-c^2 f^2 \left (1-m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m) (e+f x)}+\frac {\int \frac {(b c-a d) f (1+m) \left (2 a b d f (2+m) (3 d e+c f m)-b^2 \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )\right ) (a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m)}\\ &=\frac {d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)^2}+\frac {f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}+\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)-c^2 f^2 \left (1-m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m) (e+f x)}+\frac {\left (f \left (2 a b d f (2+m) (3 d e+c f m)-b^2 \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )\right )\right ) \int \frac {(a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{2 (b e-a f)^2 (d e-c f)^3}\\ &=\frac {d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)^2}+\frac {f (2 b d e+b c f (1+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)^2}+\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)-c^2 f^2 \left (1-m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b c-a d) (b e-a f)^2 (d e-c f)^3 (1+m) (e+f x)}+\frac {f \left (2 a b d f (2+m) (3 d e+c f m)-b^2 \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )\right ) (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 )}{2 (b e-a f)^3 (d e-c f)^3 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.81, size = 374, normalized size = 0.87 \begin {gather*} -\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (f (b e-a f)^2 (d e-c f) (2 b d e+b c f (1+m)-a d f (3+m))+\frac {2 d (b e-a f)^3 (d e-c f)^2}{c+d x}+(e+f x) \left (f (b e-a f) \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (2 d^2 e^2+5 c d e f (1+m)+c^2 f^2 \left (-1+m^2\right )\right )-a b d f \left (d e (9+5 m)+c f \left (3+5 m+2 m^2\right )\right )\right )-\frac {(b c-a d) f \left (-2 a b d f (2+m) (3 d e+c f m)+b^2 \left (6 d^2 e^2+6 c d e f m+c^2 f^2 (-1+m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )\right ) (e+f x) \, _2F_1\left (1,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{c+d x}\right )\right )}{2 (b c-a d) (b e-a f)^3 (d e-c f)^2 (-d e+c f) (1+m) (e+f x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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