\(\int \frac {(a+\frac {b}{x})^m}{(c+d x)^4} \, dx\) [603]

Optimal result

Integrand size = 17, antiderivative size = 173 \[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\frac {d^2 \left (a+\frac {b}{x}\right )^{1+m}}{3 c^2 (a c-b d) \left (d+\frac {c}{x}\right )^3}+\frac {\left (a+\frac {b}{x}\right )^{1+m}}{b c^2 (1-m) \left (d+\frac {c}{x}\right )^2}-\frac {b \left (6 a^2 c^2-6 a b c d (1+m)+b^2 d^2 \left (2+3 m+m^2\right )\right ) \left (a+\frac {b}{x}\right )^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{3 c^2 (a c-b d)^4 \left (1-m^2\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\frac {\left (a+\frac {b}{x}\right )^{1+m} \left (\frac {2 d^2 (a c-b d) x^3}{(c+d x)^3}+\frac {d (-6 a c+b d (4+m)) x^2}{(c+d x)^2}-\frac {b \left (6 a^2 c^2-6 a b c d (1+m)+b^2 d^2 \left (2+3 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {b c+a c x}{a c x-b d x}\right )}{(a c-b d)^2 (1+m)}\right )}{6 c^2 (a c-b d)^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {941, 948, 100, 25, 87, 78}

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.

\(\displaystyle \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx\)

\(\Big \downarrow \) 941

\(\displaystyle \int \frac {\left (a+\frac {b}{x}\right )^m}{x^4 \left (\frac {c}{x}+d\right )^4}dx\)

\(\Big \downarrow \) 948

\(\displaystyle -\int \frac {\left (a+\frac {b}{x}\right )^m}{\left (\frac {c}{x}+d\right )^4 x^2}d\frac {1}{x}\)

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {-\frac {\left (6 a^2 c^2-6 a b c d (m+1)+b^2 d^2 \left (m^2+3 m+2\right )\right ) \int \frac {\left (a+\frac {b}{x}\right )^m}{\left (\frac {c}{x}+d\right )^2}d\frac {1}{x}}{2 (a c-b d)}-\frac {d \left (a+\frac {b}{x}\right )^{m+1} (6 a c-b d (m+4))}{2 \left (\frac {c}{x}+d\right )^2 (a c-b d)}}{3 c^2 (a c-b d)}+\frac {d^2 \left (a+\frac {b}{x}\right )^{m+1}}{3 c^2 \left (\frac {c}{x}+d\right )^3 (a c-b d)}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {-\frac {b \left (a+\frac {b}{x}\right )^{m+1} \left (6 a^2 c^2-6 a b c d (m+1)+b^2 d^2 \left (m^2+3 m+2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{2 (m+1) (a c-b d)^3}-\frac {d \left (a+\frac {b}{x}\right )^{m+1} (6 a c-b d (m+4))}{2 \left (\frac {c}{x}+d\right )^2 (a c-b d)}}{3 c^2 (a c-b d)}+\frac {d^2 \left (a+\frac {b}{x}\right )^{m+1}}{3 c^2 \left (\frac {c}{x}+d\right )^3 (a c-b d)}\)

Input:

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

Output:

(d^2*(a + b/x)^(1 + m))/(3*c^2*(a*c - b*d)*(d + c/x)^3) + (-1/2*(d*(6*a*c 
- b*d*(4 + m))*(a + b/x)^(1 + m))/((a*c - b*d)*(d + c/x)^2) - (b*(6*a^2*c^ 
2 - 6*a*b*c*d*(1 + m) + b^2*d^2*(2 + 3*m + m^2))*(a + b/x)^(1 + m)*Hyperge 
ometric2F1[2, 1 + m, 2 + m, (c*(a + b/x))/(a*c - b*d)])/(2*(a*c - b*d)^3*( 
1 + m)))/(3*c^2*(a*c - b*d))
 

Defintions of rubi rules used

rule 25

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

rule 78

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

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(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)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 941

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Sym 
bol] :> Int[(a + b*x^n)^p*((d + c*x^n)^q/x^(n*q)), x] /; FreeQ[{a, b, c, d, 
 n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int \frac {\left (a + \frac {b}{x}\right )^{m}}{\left (c + d x\right )^{4}}\, dx \] Input:

integrate((a+b/x)**m/(d*x+c)**4,x)
 

Output:

Integral((a + b/x)**m/(c + d*x)**4, x)
 

Maxima [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

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

Output:

integrate((a + b/x)^m/(d*x + c)^4, x)
 

Giac [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

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

Output:

integrate((a + b/x)^m/(d*x + c)^4, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^m}{(c+d x)^4} \, dx=\text {too large to display} \] Input:

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

Output:

(24*(a*x + b)**m*a**4*c**3*x + 6*(a*x + b)**m*a**4*c**2*d*x**2 + 2*(a*x + 
b)**m*a**4*c*d**2*x**3 - 30*(a*x + b)**m*a**3*b*c**2*d*m*x - 48*(a*x + b)* 
*m*a**3*b*c**2*d*x - 8*(a*x + b)**m*a**3*b*c*d**2*m*x**2 - 12*(a*x + b)**m 
*a**3*b*c*d**2*x**2 - 2*(a*x + b)**m*a**3*b*d**3*m*x**3 - 4*(a*x + b)**m*a 
**3*b*d**3*x**3 + 10*(a*x + b)**m*a**2*b**2*c*d**2*m**2*x + 40*(a*x + b)** 
m*a**2*b**2*c*d**2*m*x + 24*(a*x + b)**m*a**2*b**2*c*d**2*x + 2*(a*x + b)* 
*m*a**2*b**2*d**3*m**2*x**2 + 4*(a*x + b)**m*a**2*b**2*d**3*m*x**2 - (a*x 
+ b)**m*a*b**3*d**3*m**3*x - 6*(a*x + b)**m*a*b**3*d**3*m**2*x - 11*(a*x + 
 b)**m*a*b**3*d**3*m*x - 6*(a*x + b)**m*a*b**3*d**3*x + 13824*x**m*int((a* 
x + b)**m/(576*x**m*a**9*c**12*x + 2304*x**m*a**9*c**11*d*x**2 + 3456*x**m 
*a**9*c**10*d**2*x**3 + 2304*x**m*a**9*c**9*d**3*x**4 + 576*x**m*a**9*c**8 
*d**4*x**5 + 576*x**m*a**8*b*c**12 - 2880*x**m*a**8*b*c**11*d*m*x - 1728*x 
**m*a**8*b*c**11*d*x - 11520*x**m*a**8*b*c**10*d**2*m*x**2 - 12672*x**m*a* 
*8*b*c**10*d**2*x**2 - 17280*x**m*a**8*b*c**9*d**3*m*x**3 - 21888*x**m*a** 
8*b*c**9*d**3*x**3 - 11520*x**m*a**8*b*c**8*d**4*m*x**4 - 15552*x**m*a**8* 
b*c**8*d**4*x**4 - 2880*x**m*a**8*b*c**7*d**5*m*x**5 - 4032*x**m*a**8*b*c* 
*7*d**5*x**5 - 2880*x**m*a**7*b**2*c**11*d*m - 4032*x**m*a**7*b**2*c**11*d 
 + 5904*x**m*a**7*b**2*c**10*d**2*m**2*x + 5472*x**m*a**7*b**2*c**10*d**2* 
m*x - 4464*x**m*a**7*b**2*c**10*d**2*x + 23616*x**m*a**7*b**2*c**9*d**3*m* 
*2*x**2 + 50688*x**m*a**7*b**2*c**9*d**3*m*x**2 + 22464*x**m*a**7*b**2*...