\(\int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx\) [3129]

Optimal result

Integrand size = 26, antiderivative size = 362 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}+\frac {(a d f (2-m)-b (3 d e-c f (1+m))) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {\left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) (a+b x)^{-1+m} (c+d x)^{1-m} \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 f^3 (b e-a f) (d e-c f) (1-m)}-\frac {d (b c-a d) (a+b x)^{-1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-1+m,-1+m,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {132, 72, 71, 1627, 156, 12, 133} \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=-\frac {(a+b x)^{m-1} (c+d x)^{1-m} \left (-a^2 d^2 f^2 \left (m^2-3 m+2\right )+2 a b d f (2-m) (d e-c f m)-\left (b^2 \left (-c^2 f^2 (1-m) m-2 c d e f m+2 d^2 e^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m-1,m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 f^3 (1-m) (b e-a f) (d e-c f)}-\frac {(a+b x)^{m-1} (c+d x)^{2-m} (-a d f (2-m)-b c f (m+1)+3 b d e)}{2 f^2 (e+f x) (d e-c f)}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)} \]

[In]

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

[Out]

((b*e - a*f)*(a + b*x)^(-1 + m)*(c + d*x)^(2 - m))/(2*f^2*(e + f*x)^2) - ((3*b*d*e - a*d*f*(2 - m) - b*c*f*(1
+ m))*(a + b*x)^(-1 + m)*(c + d*x)^(2 - m))/(2*f^2*(d*e - c*f)*(e + f*x)) - ((2*a*b*d*f*(2 - m)*(d*e - c*f*m)
- b^2*(2*d^2*e^2 - 2*c*d*e*f*m - c^2*f^2*(1 - m)*m) - a^2*d^2*f^2*(2 - 3*m + m^2))*(a + b*x)^(-1 + m)*(c + d*x
)^(1 - m)*Hypergeometric2F1[1, -1 + m, m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(2*f^3*(b*e - a*f)
*(d*e - c*f)*(1 - m)) - (d*(b*c - a*d)*(a + b*x)^(-1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-1 +
m, -1 + m, m, -((d*(a + b*x))/(b*c - a*d))])/(f^3*(1 - 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 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 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[b*d^(m
+ n)*f^p, Int[(a + b*x)^(m - 1)/(c + d*x)^m, x], x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandTo
Sum[(a + b*x)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n,
-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 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 1627

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[{
Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[b*R*(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*ExpandToSum[(m + 1)*(b*c - a*d)*(b*e - a*f)*Qx + a*d*f
*R*(m + 1) - b*R*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*R*(m + n + p + 3)*x, x], x], x]] /; FreeQ[{a, b,
c, d, e, f, n, p}, x] && PolyQ[Px, x] && ILtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 d\right ) \int (a+b x)^{-2+m} (c+d x)^{1-m} \, dx}{f^3}+\int \frac {(a+b x)^{-2+m} (c+d x)^{1-m} \left (a^2 c-\frac {b^2 d e^3}{f^3}+\left (2 a b c+a^2 d-\frac {3 b^2 d e^2}{f^2}\right ) x+b \left (2 a d+b \left (c-\frac {3 d e}{f}\right )\right ) x^2\right )}{(e+f x)^3} \, dx \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {\int \frac {(a+b x)^{-2+m} (c+d x)^{1-m} \left (\frac {(b e-a f) (d e-c f) \left (b^2 e (2 d e-c f (1-m))-a^2 d f^2 (2-m)+a b f (d e (2-m)-c f (1+m))\right )}{f^3}+\frac {b (b e-a f) (d e-c f) (5 b d e-2 b c f-3 a d f) x}{f^2}\right )}{(e+f x)^2} \, dx}{2 (b e-a f) (d e-c f)}+\frac {\left (b d (b c-a d) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^{-2+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m} \, dx}{f^3} \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {(3 b d e-a d f (2-m)-b c f (1+m)) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {d (b c-a d) (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;m;-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}+\frac {\int \frac {(b e-a f)^2 (d e-c f) \left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) (a+b x)^{-2+m} (c+d x)^{1-m}}{f^3 (e+f x)} \, dx}{2 (b e-a f)^2 (d e-c f)^2} \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {(3 b d e-a d f (2-m)-b c f (1+m)) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {d (b c-a d) (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;m;-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}+\frac {\left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) \int \frac {(a+b x)^{-2+m} (c+d x)^{1-m}}{e+f x} \, dx}{2 f^3 (d e-c f)} \\ & = \frac {(b e-a f) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {(3 b d e-a d f (2-m)-b c f (1+m)) (a+b x)^{-1+m} (c+d x)^{2-m}}{2 f^2 (d e-c f) (e+f x)}-\frac {\left (2 a b d f (2-m) (d e-c f m)-b^2 \left (2 d^2 e^2-2 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2-3 m+m^2\right )\right ) (a+b x)^{-1+m} (c+d x)^{1-m} \, _2F_1\left (1,-1+m;m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 f^3 (b e-a f) (d e-c f) (1-m)}-\frac {d (b c-a d) (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;m;-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.30 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {AppellF1}\left (1+m,-2+m,3,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{(b e-a f)^3 (1+m)} \]

[In]

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

[Out]

((b*c - a*d)^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*AppellF1[1 + m, -2 + m, 3, 2 + m, (d*(a + b*x))
/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f)])/((b*e - a*f)^3*(1 + m)*(c + d*x)^m)

Maple [F]

\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{\left (f x +e \right )^{3}}d x\]

[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)

Fricas [F]

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

[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*e*f^2*x^2 + 3*e^2*f*x + e^3), x)

Sympy [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{{\left (e+f\,x\right )}^3} \,d x \]

[In]

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

[Out]

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