\(\int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx\) [587]

Optimal result

Integrand size = 20, antiderivative size = 194 \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=-\frac {(a+b x)^{1+n} (c+d x)^{1-n}}{3 a c x^3}+\frac {(b c (2-n)+a d (2+n)) (a+b x)^{1+n} (c+d x)^{1-n}}{6 a^2 c^2 x^2}+\frac {(b c-a d) \left (2 a b c d \left (1-n^2\right )+b^2 c^2 \left (2-3 n+n^2\right )+a^2 d^2 \left (2+3 n+n^2\right )\right ) (a+b x)^{1+n} (c+d x)^{-1-n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c (a+b x)}{a (c+d x)}\right )}{6 a^4 c^2 (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx=\frac {(a+b x)^{1+n} (c+d x)^{-1-n} \left (-\frac {2 a^3 c (c+d x)^2}{x^3}+\frac {a^2 (-b c (-2+n)+a d (2+n)) (c+d x)^2}{x^2}+\frac {(b c-a d) \left (-2 a b c d \left (-1+n^2\right )+b^2 c^2 \left (2-3 n+n^2\right )+a^2 d^2 \left (2+3 n+n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c (a+b x)}{a (c+d x)}\right )}{1+n}\right )}{6 a^4 c^2} \] Input:

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

Output:

((a + b*x)^(1 + n)*(c + d*x)^(-1 - n)*((-2*a^3*c*(c + d*x)^2)/x^3 + (a^2*( 
-(b*c*(-2 + n)) + a*d*(2 + n))*(c + d*x)^2)/x^2 + ((b*c - a*d)*(-2*a*b*c*d 
*(-1 + n^2) + b^2*c^2*(2 - 3*n + n^2) + a^2*d^2*(2 + 3*n + n^2))*Hypergeom 
etric2F1[2, 1 + n, 2 + n, (c*(a + b*x))/(a*(c + d*x))])/(1 + n)))/(6*a^4*c 
^2)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {114, 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.

\(\displaystyle \int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx\)

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 141

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

Input:

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

Output:

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

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 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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