3.32.34 \(\int \frac {(a+b x)^m (c+d x)^{3-m}}{e+f x} \, dx\) [3134]

Optimal. Leaf size=488 \[ \frac {b (b e-a f)^3 (a+b x)^{-3+m} (c+d x)^{4-m}}{(b c-a d) f^4 (3-m)}-\frac {b (b (3 d e-c f (1-m))-a d f (2+m)) (a+b x)^{-2+m} (c+d x)^{4-m}}{6 d^2 f^2}+\frac {b (a+b x)^{-1+m} (c+d x)^{4-m}}{3 d f}-\frac {(b e-a f)^3 (a+b x)^{-3+m} (c+d x)^{3-m} \, _2F_1\left (1,-3+m;-2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^4 (3-m)}-\frac {(b c-a d)^2 \left (3 a^2 b d^2 f^2 (d e-c f (3-m)) (1-m) m+a^3 d^3 f^3 m \left (2-3 m+m^2\right )+3 a b^2 d f m \left (2 d^2 e^2-2 c d e f (3-m)+c^2 f^2 \left (6-5 m+m^2\right )\right )-b^3 \left (6 d^3 e^3-6 c d^2 e^2 f (3-m)+3 c^2 d e f^2 \left (6-5 m+m^2\right )-c^3 f^3 \left (6-11 m+6 m^2-m^3\right )\right )\right ) (a+b x)^{-2+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-3+m,-2+m;-1+m;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 f^4 (2-m) (3-m)} \]

[Out]

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

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Rubi [A]
time = 0.71, antiderivative size = 585, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {135, 133, 1637, 965, 80, 72, 71} \begin {gather*} -\frac {d (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^3 d^3 f^3 m \left (m^2-3 m+2\right )+3 a^2 b d^2 f^2 (1-m) m (d e-c f (3-m))+3 a b^2 d f m \left (c^2 f^2 \left (m^2-5 m+6\right )-2 c d e f (3-m)+2 d^2 e^2\right )-\left (b^3 \left (-c^3 f^3 \left (-m^3+6 m^2-11 m+6\right )+3 c^2 d e f^2 \left (m^2-5 m+6\right )-6 c d^2 e^2 f (3-m)+6 d^3 e^3\right )\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 f^4 m (m+1) (b c-a d)}-\frac {d (a+b x)^{m+1} (c+d x)^{-m} \left (a^3 d^3 f^3 (4-m) m+3 a^2 b d^2 f^2 m (d e-c f (5-m))-3 a b^2 c d f^2 m (2 d e-c f (6-m))+b^3 \left (-c^3 f^3 \left (-m^2+7 m+6\right )+3 c^2 d e f^2 (m+6)-18 c d^2 e^2 f+6 d^3 e^3\right )\right )}{6 b^3 f^4 m (b c-a d)}+\frac {d^3 (a+b x)^{m+3} (c+d x)^{-m}}{3 b^3 f}-\frac {d^2 (a+b x)^{m+2} (c+d x)^{-m} (a d f (6-m)-b c f (9-m)+3 b d e)}{6 b^3 f^2}-\frac {(a+b x)^m (d e-c f)^3 (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 )}{f^4 m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 80

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

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 135

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[(c*f - d*e)^
(m + n + 1)/f^(m + n + 1), Int[(a + b*x)^m/((c + d*x)^(m + 1)*(e + f*x)), x], x] + Dist[1/f^(m + n + 1), Int[(
(a + b*x)^m/(c + d*x)^(m + 1))*ExpandToSum[(f^(m + n + 1)*(c + d*x)^(m + n + 1) - (c*f - d*e)^(m + n + 1))/(e
+ f*x), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[m + n + 1, 0] && (LtQ[m, 0] || SumSimplerQ[m,
1] ||  !(LtQ[n, 0] || SumSimplerQ[n, 1]))

Rule 965

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 1637

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{q = Expon[Px, x], k = Coef
f[Px, x, Expon[Px, x]]}, Simp[k*(a + b*x)^(m + q)*((c + d*x)^(n + 1)/(d*b^q*(m + n + q + 1))), x] + Dist[1/(d*
b^q*(m + n + q + 1)), Int[(a + b*x)^m*(c + d*x)^n*ExpandToSum[d*b^q*(m + n + q + 1)*Px - d*k*(m + n + q + 1)*(
a + b*x)^q - k*(b*c - a*d)*(m + q)*(a + b*x)^(q - 1), x], x], x] /; NeQ[m + n + q + 1, 0]] /; FreeQ[{a, b, c,
d, m, n}, x] && PolyQ[Px, x] && GtQ[Expon[Px, x], 2]

Rubi steps

\begin {align*} \int \frac {(a+b x)^m (c+d x)^{3-m}}{e+f x} \, dx &=\frac {d \int (a+b x)^m (c+d x)^{2-m} \, dx}{f}-\frac {(d e-c f) \int \frac {(a+b x)^m (c+d x)^{2-m}}{e+f x} \, dx}{f}\\ &=-\frac {(d (d e-c f)) \int (a+b x)^m (c+d x)^{1-m} \, dx}{f^2}+\frac {(d e-c f)^2 \int \frac {(a+b x)^m (c+d x)^{1-m}}{e+f x} \, dx}{f^2}+\frac {\left (d (b c-a d)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2-m} \, dx}{b^2 f}\\ &=\frac {d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}+\frac {\left (d (d e-c f)^2\right ) \int (a+b x)^m (c+d x)^{-m} \, dx}{f^3}-\frac {(d e-c f)^3 \int \frac {(a+b x)^m (c+d x)^{-m}}{e+f x} \, dx}{f^3}-\frac {\left (d (b c-a d) (d e-c f) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m} \, dx}{b f^2}\\ &=\frac {d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}-\frac {d (b c-a d) (d e-c f) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^2 f^2 (1+m)}-\frac {\left (b (d e-c f)^3\right ) \int (a+b x)^{-1+m} (c+d x)^{-m} \, dx}{f^4}+\frac {\left ((b e-a f) (d e-c f)^3\right ) \int \frac {(a+b x)^{-1+m} (c+d x)^{-m}}{e+f x} \, dx}{f^4}+\frac {\left (d (d e-c f)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m} \, dx}{f^3}\\ &=\frac {(d e-c f)^3 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,m;1+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^4 m}+\frac {d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}-\frac {d (b c-a d) (d e-c f) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^2 f^2 (1+m)}+\frac {d (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b f^3 (1+m)}-\frac {\left (b (d e-c f)^3 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^{-1+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m} \, dx}{f^4}\\ &=\frac {(d e-c f)^3 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,m;1+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^4 m}+\frac {d (b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^3 f (1+m)}-\frac {d (b c-a d) (d e-c f) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b^2 f^2 (1+m)}-\frac {(d e-c f)^3 (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;1+m;-\frac {d (a+b x)}{b c-a d}\right )}{f^4 m}+\frac {d (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{b f^3 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 337, normalized size = 0.69 \begin {gather*} \frac {(a+b x)^m (c+d x)^{-m} \left (d (b c-a d)^2 f^3 m (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-2+m,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )-b (d e-c f) \left (-d (-b c+a d) f^2 m (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )-b (d e-c f) \left (b (d e-c f) (1+m) \left (\, _2F_1\left (1,m;1+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )-\left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;1+m;\frac {d (a+b x)}{-b c+a d}\right )\right )+d f m (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )\right )\right )\right )}{b^3 f^4 m (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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