3.31.78 \(\int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx\) [3078]

3.31.78.1 Optimal result

Integrand size = 26, antiderivative size = 498 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=-\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (b (5 d e-c f (2-m))-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (d e (15+8 m)-c f \left (3-2 m-2 m^2\right )\right )+b^2 \left (11 d^2 e^2-c d e f (7-8 m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^3 (d e-c f)^3 (e+f x)}+\frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (1,-m,1-m,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{6 (b e-a f)^3 (d e-c f)^4 m} \]

output

-1/3*f*(b*x+a)^(1+m)/(-a*f+b*e)/(-c*f+d*e)/((d*x+c)^m)/(f*x+e)^3-1/6*f*(b* 
(5*d*e-c*f*(2-m))-a*d*f*(3+m))*(b*x+a)^(1+m)/(-a*f+b*e)^2/(-c*f+d*e)^2/((d 
*x+c)^m)/(f*x+e)^2-1/6*f*(a^2*d^2*f^2*(m^2+5*m+6)-a*b*d*f*(d*e*(15+8*m)-c* 
f*(-2*m^2-2*m+3))+b^2*(11*d^2*e^2-c*d*e*f*(7-8*m)+c^2*f^2*(m^2-3*m+2)))*(b 
*x+a)^(1+m)/(-a*f+b*e)^3/(-c*f+d*e)^3/((d*x+c)^m)/(f*x+e)+1/6*(3*a*b^2*d*f 
*(1+m)*(6*d^2*e^2+6*c*d*e*f*m-c^2*f^2*(1-m)*m)-3*a^2*b*d^2*f^2*(c*f*m+3*d* 
e)*(m^2+3*m+2)+a^3*d^3*f^3*(m^3+6*m^2+11*m+6)-b^3*(6*d^3*e^3+18*c*d^2*e^2* 
f*m-9*c^2*d*e*f^2*(1-m)*m+c^3*f^3*m*(m^2-3*m+2)))*(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)^3/(-c*f+d*e)^4 
/m/((d*x+c)^m)
 

3.31.78.2 Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 466, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=-\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (6 d (b e-a f)^4 (d e-c f)^2 (1+m)+2 f (-b e+a f)^3 (-d e+c f) (1+m) (-a d f (3+m)+b (3 d e+c f m)) (c+d x)-\frac {(e+f x) \left (f (b e-a f)^2 (1+m) \left (-b^2 \left (6 d^2 e^2+7 c d e f m+c^2 f^2 (-2+m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )+a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (c+d x)^2+(b c-a d) \left (-3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m+c^2 f^2 (-1+m) m\right )+3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )-a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )+b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m+9 c^2 d e f^2 (-1+m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (e+f x)^2 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{c+d x}\right )}{6 (b c-a d) (b e-a f)^4 (d e-c f)^2 (-d e+c f) m (1+m) (e+f x)^3} \]

input

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

output

-1/6*((a + b*x)^(1 + m)*(6*d*(b*e - a*f)^4*(d*e - c*f)^2*(1 + m) + 2*f*(-( 
b*e) + a*f)^3*(-(d*e) + c*f)*(1 + m)*(-(a*d*f*(3 + m)) + b*(3*d*e + c*f*m) 
)*(c + d*x) - ((e + f*x)*(f*(b*e - a*f)^2*(1 + m)*(-(b^2*(6*d^2*e^2 + 7*c* 
d*e*f*m + c^2*f^2*(-2 + m)*m)) - a^2*d^2*f^2*(6 + 5*m + m^2) + a*b*d*f*(c* 
f*m*(3 + 2*m) + d*e*(12 + 7*m)))*(c + d*x)^2 + (b*c - a*d)*(-3*a*b^2*d*f*( 
1 + m)*(6*d^2*e^2 + 6*c*d*e*f*m + c^2*f^2*(-1 + m)*m) + 3*a^2*b*d^2*f^2*(3 
*d*e + c*f*m)*(2 + 3*m + m^2) - a^3*d^3*f^3*(6 + 11*m + 6*m^2 + m^3) + b^3 
*(6*d^3*e^3 + 18*c*d^2*e^2*f*m + 9*c^2*d*e*f^2*(-1 + m)*m + c^3*f^3*m*(2 - 
 3*m + m^2)))*(e + f*x)^2*Hypergeometric2F1[2, 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) 
^4*(d*e - c*f)^2*(-(d*e) + c*f)*m*(1 + m)*(c + d*x)^m*(e + f*x)^3)
 

3.31.78.3 Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 542, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {114, 25, 168, 25, 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.

\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{-m-1}}{(e+f x)^4} \, dx\)

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 168

\(\displaystyle \frac {-\frac {\int -\frac {(a+b x)^m (c+d x)^{-m-1} \left (\left (6 d^2 e^2-c d f (5-7 m) e+c^2 f^2 \left (m^2-3 m+2\right )\right ) b^2-a d f \left (d e (7 m+12)-c f \left (-2 m^2-2 m+3\right )\right ) b-d f (5 b d e-b c f (2-m)-a d f (m+3)) x b+a^2 d^2 f^2 \left (m^2+5 m+6\right )\right )}{(e+f x)^2}dx}{2 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)-b c f (2-m)+5 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {(a+b x)^m (c+d x)^{-m-1} \left (\left (6 d^2 e^2-c d f (5-7 m) e+c^2 f^2 \left (m^2-3 m+2\right )\right ) b^2-a d f \left (d e (7 m+12)-c f \left (-2 m^2-2 m+3\right )\right ) b-d f (5 b d e-b c f (2-m)-a d f (m+3)) x b+a^2 d^2 f^2 \left (m^2+5 m+6\right )\right )}{(e+f x)^2}dx}{2 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)-b c f (2-m)+5 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {\frac {-\frac {\int \frac {\left (-\left (\left (6 d^3 e^3+18 c d^2 f m e^2-9 c^2 d f^2 (1-m) m e+c^3 f^3 m \left (m^2-3 m+2\right )\right ) b^3\right )+3 a d f (m+1) \left (6 d^2 e^2+6 c d f m e-c^2 f^2 (1-m) m\right ) b^2-3 a^2 d^2 f^2 (3 d e+c f m) \left (m^2+3 m+2\right ) b+a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )\right ) (a+b x)^m (c+d x)^{-m-1}}{e+f x}dx}{(b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (d e (8 m+15)-c f \left (-2 m^2-2 m+3\right )\right )+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-c d e f (7-8 m)+11 d^2 e^2\right )\right )}{(e+f x) (b e-a f) (d e-c f)}}{2 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)-b c f (2-m)+5 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {\left (a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )-3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (c f m+3 d e)+3 a b^2 d f (m+1) \left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )-\left (b^3 \left (c^3 f^3 m \left (m^2-3 m+2\right )-9 c^2 d e f^2 (1-m) m+18 c d^2 e^2 f m+6 d^3 e^3\right )\right )\right ) \int \frac {(a+b x)^m (c+d x)^{-m-1}}{e+f x}dx}{(b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (d e (8 m+15)-c f \left (-2 m^2-2 m+3\right )\right )+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-c d e f (7-8 m)+11 d^2 e^2\right )\right )}{(e+f x) (b e-a f) (d e-c f)}}{2 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)-b c f (2-m)+5 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 141

\(\displaystyle \frac {\frac {\frac {(a+b x)^m (c+d x)^{-m} \left (a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )-3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (c f m+3 d e)+3 a b^2 d f (m+1) \left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )-\left (b^3 \left (c^3 f^3 m \left (m^2-3 m+2\right )-9 c^2 d e f^2 (1-m) m+18 c d^2 e^2 f m+6 d^3 e^3\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,-m,1-m,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (b e-a f) (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (d e (8 m+15)-c f \left (-2 m^2-2 m+3\right )\right )+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-c d e f (7-8 m)+11 d^2 e^2\right )\right )}{(e+f x) (b e-a f) (d e-c f)}}{2 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)-b c f (2-m)+5 b d e)}{2 (e+f x)^2 (b e-a f) (d e-c f)}}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)}\)

input

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

output

-1/3*(f*(a + b*x)^(1 + m))/((b*e - a*f)*(d*e - c*f)*(c + d*x)^m*(e + f*x)^ 
3) + (-1/2*(f*(5*b*d*e - b*c*f*(2 - m) - a*d*f*(3 + m))*(a + b*x)^(1 + m)) 
/((b*e - a*f)*(d*e - c*f)*(c + d*x)^m*(e + f*x)^2) + (-((f*(a^2*d^2*f^2*(6 
 + 5*m + m^2) - a*b*d*f*(d*e*(15 + 8*m) - c*f*(3 - 2*m - 2*m^2)) + b^2*(11 
*d^2*e^2 - c*d*e*f*(7 - 8*m) + c^2*f^2*(2 - 3*m + m^2)))*(a + b*x)^(1 + m) 
)/((b*e - a*f)*(d*e - c*f)*(c + d*x)^m*(e + f*x))) + ((3*a*b^2*d*f*(1 + m) 
*(6*d^2*e^2 + 6*c*d*e*f*m - c^2*f^2*(1 - m)*m) - 3*a^2*b*d^2*f^2*(3*d*e + 
c*f*m)*(2 + 3*m + m^2) + a^3*d^3*f^3*(6 + 11*m + 6*m^2 + m^3) - b^3*(6*d^3 
*e^3 + 18*c*d^2*e^2*f*m - 9*c^2*d*e*f^2*(1 - m)*m + c^3*f^3*m*(2 - 3*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))])/((b*e - a*f)*(d*e - c*f)^2*m*(c + d*x)^m))/(2*(b 
*e - a*f)*(d*e - c*f)))/(3*(b*e - a*f)*(d*e - c*f))
 

3.31.78.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 114

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

3.31.78.4 Maple [F]

\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-1-m}}{\left (f x +e \right )^{4}}d x\]

input

int((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x)
 

output

int((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x)
 

3.31.78.5 Fricas [F]

\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{4}} \,d x } \]

input

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="fricas")
 

output

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

3.31.78.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\text {Timed out} \]

input

integrate((b*x+a)**m*(d*x+c)**(-1-m)/(f*x+e)**4,x)
 

output

Timed out
 

3.31.78.7 Maxima [F]

\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{4}} \,d x } \]

input

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="maxima")
 

output

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^4, x)
 

3.31.78.8 Giac [F]

\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{4}} \,d x } \]

input

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="giac")
 

output

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^4, x)
 

3.31.78.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^4\,{\left (c+d\,x\right )}^{m+1}} \,d x \]

input

int((a + b*x)^m/((e + f*x)^4*(c + d*x)^(m + 1)),x)
 

output

int((a + b*x)^m/((e + f*x)^4*(c + d*x)^(m + 1)), x)