\(\int \frac {(a+\frac {b}{c+d x})^p}{x^4} \, dx\) [60]

Optimal result

Integrand size = 17, antiderivative size = 185 \[ \int \frac {\left (a+\frac {b}{c+d x}\right )^p}{x^4} \, dx=\frac {d (6 a c+b (4+p)) (c+d x)^2 \left (a+\frac {b}{c+d x}\right )^{1+p}}{6 c^2 (b+a c)^2 x^2}-\frac {(c+d x)^3 \left (a+\frac {b}{c+d x}\right )^{1+p}}{3 c^2 (b+a c) x^3}-\frac {b d^3 \left (6 a^2 c^2+6 a b c (1+p)+b^2 \left (2+3 p+p^2\right )\right ) \left (a+\frac {b}{c+d x}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {c \left (a+\frac {b}{c+d x}\right )}{b+a c}\right )}{6 c^2 (b+a c)^4 (1+p)} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {896, 941, 948, 100, 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.

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

\(\Big \downarrow \) 896

\(\displaystyle d^3 \int \frac {\left (a+\frac {b}{c+d x}\right )^p}{d^4 x^4}d(c+d x)\)

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

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

Defintions of rubi rules used

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 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(((a*d*x + a*c + b)/(d*x + c))^p/x^4, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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