\(\int x^2 (a+\frac {b}{c+d x})^p \, dx\) [54]

Optimal result

Integrand size = 17, antiderivative size = 158 \[ \int x^2 \left (a+\frac {b}{c+d x}\right )^p \, dx=-\frac {(6 a c+b (2-p)) (c+d x)^2 \left (a+\frac {b}{c+d x}\right )^{1+p}}{6 a^2 d^3}+\frac {(c+d x)^3 \left (a+\frac {b}{c+d x}\right )^{1+p}}{3 a d^3}-\frac {b \left (6 a^2 c^2+b (6 a c+b (2-p)) (1-p)\right ) \left (a+\frac {b}{c+d x}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b}{a (c+d x)}\right )}{6 a^4 d^3 (1+p)} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {896, 941, 948, 100, 25, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**2*((a*c + a*d*x + b)/(c + d*x))**p, x)
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int x^2 \left (a+\frac {b}{c+d x}\right )^p \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{1,[0 
,2,1,0]%%%} / %%%{1,[0,0,0,2]%%%} Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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