3.6.88 \(\int (a+\frac {b}{x})^m (c+d x)^2 \, dx\) [588]

3.6.88.1 Optimal result

Integrand size = 17, antiderivative size = 138 \[ \int \left (a+\frac {b}{x}\right )^m (c+d x)^2 \, dx=\frac {d (6 a c-b d (2-m)) \left (a+\frac {b}{x}\right )^{1+m} x^2}{6 a^2}+\frac {d^2 \left (a+\frac {b}{x}\right )^{1+m} x^3}{3 a}-\frac {b \left (6 a^2 c^2-6 a b c d (1-m)+b^2 d^2 \left (2-3 m+m^2\right )\right ) \left (a+\frac {b}{x}\right )^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,1+\frac {b}{a x}\right )}{6 a^4 (1+m)} \]

output

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

3.6.88.2 Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.81 \[ \int \left (a+\frac {b}{x}\right )^m (c+d x)^2 \, dx=\frac {\left (a+\frac {b}{x}\right )^m (b+a x) \left (a^2 d (1+m) x^2 (b d (-2+m)+2 a (3 c+d x))-b \left (6 a^2 c^2+6 a b c d (-1+m)+b^2 d^2 \left (2-3 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,1+\frac {b}{a x}\right )\right )}{6 a^4 (1+m) x} \]

input

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

output

((a + b/x)^m*(b + a*x)*(a^2*d*(1 + m)*x^2*(b*d*(-2 + m) + 2*a*(3*c + d*x)) 
 - b*(6*a^2*c^2 + 6*a*b*c*d*(-1 + m) + b^2*d^2*(2 - 3*m + m^2))*Hypergeome 
tric2F1[2, 1 + m, 2 + m, 1 + b/(a*x)]))/(6*a^4*(1 + m)*x)
 

3.6.88.3 Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {941, 948, 100, 87, 75}

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.

\(\displaystyle \int (c+d x)^2 \left (a+\frac {b}{x}\right )^m \, dx\)

\(\Big \downarrow \) 941

\(\displaystyle \int x^2 \left (\frac {c}{x}+d\right )^2 \left (a+\frac {b}{x}\right )^mdx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 75

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

input

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

output

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

3.6.88.3.1 Defintions of rubi rules used

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

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]]
 

3.6.88.4 Maple [F]

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

input

int((a+b/x)^m*(d*x+c)^2,x)
 

output

int((a+b/x)^m*(d*x+c)^2,x)
 

3.6.88.5 Fricas [F]

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

input

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

output

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

3.6.88.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.74 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \left (a+\frac {b}{x}\right )^m (c+d x)^2 \, dx=\frac {b^{m} c^{2} x^{1 - m} \Gamma \left (1 - m\right ) {{}_{2}F_{1}\left (\begin {matrix} - m, 1 - m \\ 2 - m \end {matrix}\middle | {\frac {a x e^{i \pi }}{b}} \right )}}{\Gamma \left (2 - m\right )} + \frac {2 b^{m} c d x^{2 - m} \Gamma \left (2 - m\right ) {{}_{2}F_{1}\left (\begin {matrix} - m, 2 - m \\ 3 - m \end {matrix}\middle | {\frac {a x e^{i \pi }}{b}} \right )}}{\Gamma \left (3 - m\right )} + \frac {b^{m} d^{2} x^{3 - m} \Gamma \left (3 - m\right ) {{}_{2}F_{1}\left (\begin {matrix} - m, 3 - m \\ 4 - m \end {matrix}\middle | {\frac {a x e^{i \pi }}{b}} \right )}}{\Gamma \left (4 - m\right )} \]

input

integrate((a+b/x)**m*(d*x+c)**2,x)
 

output

b**m*c**2*x**(1 - m)*gamma(1 - m)*hyper((-m, 1 - m), (2 - m,), a*x*exp_pol 
ar(I*pi)/b)/gamma(2 - m) + 2*b**m*c*d*x**(2 - m)*gamma(2 - m)*hyper((-m, 2 
 - m), (3 - m,), a*x*exp_polar(I*pi)/b)/gamma(3 - m) + b**m*d**2*x**(3 - m 
)*gamma(3 - m)*hyper((-m, 3 - m), (4 - m,), a*x*exp_polar(I*pi)/b)/gamma(4 
 - m)
 

3.6.88.7 Maxima [F]

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

input

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

output

integrate((d*x + c)^2*(a + b/x)^m, x)
 

3.6.88.8 Giac [F]

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

input

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

output

integrate((d*x + c)^2*(a + b/x)^m, x)
 

3.6.88.9 Mupad [F(-1)]

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

input

int((a + b/x)^m*(c + d*x)^2,x)
 

output

int((a + b/x)^m*(c + d*x)^2, x)