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

3.32.29.1 Optimal result

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

output

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)
 

3.32.29.2 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 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(e+f x)^3} \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {AppellF1}\left (1+m,-2+m,3,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{(b e-a f)^3 (1+m)} \]

input

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

output

((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)
 

3.32.29.3 Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 358, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {140, 80, 79, 2116, 27, 27, 168, 27, 141}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 140

\(\displaystyle \int \frac {(a+b x)^{m-2} (c+d x)^{1-m} \left (-\frac {b^2 d e^3}{f^3}+b \left (2 a d+b \left (c-\frac {3 d e}{f}\right )\right ) x^2+a^2 c+\left (d a^2+2 b c a-\frac {3 b^2 d e^2}{f^2}\right ) x\right )}{(e+f x)^3}dx+\frac {b^2 d \int (a+b x)^{m-2} (c+d x)^{1-m}dx}{f^3}\)

\(\Big \downarrow \) 80

\(\displaystyle \int \frac {(a+b x)^{m-2} (c+d x)^{1-m} \left (-\frac {b^2 d e^3}{f^3}+b \left (2 a d+b \left (c-\frac {3 d e}{f}\right )\right ) x^2+a^2 c+\left (d a^2+2 b c a-\frac {3 b^2 d e^2}{f^2}\right ) x\right )}{(e+f x)^3}dx+\frac {b d (b c-a d) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \int (a+b x)^{m-2} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m}dx}{f^3}\)

\(\Big \downarrow \) 79

\(\displaystyle \int \frac {(a+b x)^{m-2} (c+d x)^{1-m} \left (-\frac {b^2 d e^3}{f^3}+b \left (2 a d+b \left (c-\frac {3 d e}{f}\right )\right ) x^2+a^2 c+\left (d a^2+2 b c a-\frac {3 b^2 d e^2}{f^2}\right ) x\right )}{(e+f x)^3}dx-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}\)

\(\Big \downarrow \) 2116

\(\displaystyle -\frac {\int \frac {(d e-c f) (a+b x)^{m-2} (c+d x)^{1-m} \left ((b e-a f) \left (e (2 d e-c f (1-m)) b^2+a f (d e (2-m)-c f (m+1)) b-a^2 d f^2 (2-m)\right )+b f (b e-a f) (5 b d e-2 b c f-3 a d f) x\right )}{f^3 (e+f x)^2}dx}{2 (b e-a f) (d e-c f)}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {(b e-a f) (a+b x)^{m-2} (c+d x)^{1-m} \left (e (2 d e-c f (1-m)) b^2+a f (d e (2-m)-c f (m+1)) b+f (5 b d e-2 b c f-3 a d f) x b-a^2 d f^2 (2-m)\right )}{(e+f x)^2}dx}{2 f^3 (b e-a f)}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {(a+b x)^{m-2} (c+d x)^{1-m} \left (e (2 d e-c f (1-m)) b^2+a f (d e (2-m)-c f (m+1)) b+f (5 b d e-2 b c f-3 a d f) x b-a^2 d f^2 (2-m)\right )}{(e+f x)^2}dx}{2 f^3}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {\frac {f (a+b x)^{m-1} (c+d x)^{2-m} (-a d f (2-m)-b c f (m+1)+3 b d e)}{(e+f x) (d e-c f)}-\frac {\int \frac {(b e-a f) \left (-\left (\left (2 d^2 e^2-2 c d f m e-c^2 f^2 (1-m) m\right ) b^2\right )+2 a d f (2-m) (d e-c f m) b-a^2 d^2 f^2 \left (m^2-3 m+2\right )\right ) (a+b x)^{m-2} (c+d x)^{1-m}}{e+f x}dx}{(b e-a f) (d e-c f)}}{2 f^3}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {f (a+b x)^{m-1} (c+d x)^{2-m} (-a d f (2-m)-b c f (m+1)+3 b d e)}{(e+f x) (d e-c f)}-\frac {\left (-a^2 d^2 f^2 \left (m^2-3 m+2\right )+2 a b d f (2-m) (d e-c f m)-\left (b^2 \left (-c^2 f^2 (1-m) m-2 c d e f m+2 d^2 e^2\right )\right )\right ) \int \frac {(a+b x)^{m-2} (c+d x)^{1-m}}{e+f x}dx}{d e-c f}}{2 f^3}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}\)

\(\Big \downarrow \) 141

\(\displaystyle -\frac {\frac {(a+b x)^{m-1} (c+d x)^{1-m} \left (-a^2 d^2 f^2 \left (m^2-3 m+2\right )+2 a b d f (2-m) (d e-c f m)-\left (b^2 \left (-c^2 f^2 (1-m) m-2 c d e f m+2 d^2 e^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m-1,m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(1-m) (b e-a f) (d e-c f)}+\frac {f (a+b x)^{m-1} (c+d x)^{2-m} (-a d f (2-m)-b c f (m+1)+3 b d e)}{(e+f x) (d e-c f)}}{2 f^3}+\frac {(b e-a f) (a+b x)^{m-1} (c+d x)^{2-m}}{2 f^2 (e+f x)^2}-\frac {d (b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m-1,m,-\frac {d (a+b x)}{b c-a d}\right )}{f^3 (1-m)}\)

input

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

output

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

3.32.29.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

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] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(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] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 140

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[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)*ExpandToSum[(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 141

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> 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] ||  !Su 
mSimplerQ[p, 1]) &&  !ILtQ[m, 0]
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] + S 
imp[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 2116

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] + Si 
mp[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]
 

3.32.29.4 Maple [F]

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

input

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

output

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

3.32.29.5 Fricas [F]

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

input

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

output

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)
 

3.32.29.6 Sympy [F(-2)]

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

input

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

output

Exception raised: HeuristicGCDFailed >> no luck
 

3.32.29.7 Maxima [F]

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

input

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

output

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

3.32.29.8 Giac [F]

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

input

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

output

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

3.32.29.9 Mupad [F(-1)]

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

input

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

output

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