\(\int x (a+\frac {b}{(c+d x)^2})^p \, dx\) [98]

Optimal result

Integrand size = 15, antiderivative size = 119 \[ \int x \left (a+\frac {b}{(c+d x)^2}\right )^p \, dx=-\frac {c (c+d x) \left (a+\frac {b}{(c+d x)^2}\right )^p \left (1+\frac {b}{a (c+d x)^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b}{a (c+d x)^2}\right )}{d^2}-\frac {b \left (a+\frac {b}{(c+d x)^2}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,1+\frac {b}{a (c+d x)^2}\right )}{2 a^2 d^2 (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11 \[ \int x \left (a+\frac {b}{(c+d x)^2}\right )^p \, dx=\frac {(c+d x) \left (a+\frac {b}{(c+d x)^2}\right )^p \left (1+\frac {a (c+d x)^2}{b}\right )^{-p} \left (2 c (-1+p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-p,-p,\frac {3}{2}-p,-\frac {a (c+d x)^2}{b}\right )-(-1+2 p) (c+d x) \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {a (c+d x)^2}{b}\right )\right )}{2 d^2 (-1+p) (-1+2 p)} \] Input:

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

Output:

((c + d*x)*(a + b/(c + d*x)^2)^p*(2*c*(-1 + p)*Hypergeometric2F1[1/2 - p, 
-p, 3/2 - p, -((a*(c + d*x)^2)/b)] - (-1 + 2*p)*(c + d*x)*Hypergeometric2F 
1[1 - p, -p, 2 - p, -((a*(c + d*x)^2)/b)]))/(2*d^2*(-1 + p)*(-1 + 2*p)*(1 
+ (a*(c + d*x)^2)/b)^p)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {896, 25, 1774, 1803, 25, 542, 243, 75, 279, 278}

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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1774

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 542

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 75

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 542

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

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 1774

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Sy 
mbol] :> Int[x^(mn*q)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, 
e, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (PosQ[n2] ||  !IntegerQ[p 
])
 

rule 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x 
)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && 
 EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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