\(\int \frac {x (a+b x^2)^p}{(c+d x)^2} \, dx\) [41]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 18, antiderivative size = 184 \[ \int \frac {x \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=-\frac {\left (a+b x^2\right )^{1+p}}{2 b p \left (c^2-d^2 x^2\right )}-\frac {2 d x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c^3}+\frac {\left (a d^2+b c^2 (1+2 p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 \left (b c^2+a d^2\right )^2 p (1+p)} \] Output:

-1/2*(b*x^2+a)^(p+1)/b/p/(-d^2*x^2+c^2)-2/3*d*x^3*(b*x^2+a)^p*AppellF1(3/2 
,2,-p,5/2,d^2*x^2/c^2,-b*x^2/a)/c^3/((1+b*x^2/a)^p)+1/2*(a*d^2+b*c^2*(1+2* 
p))*(b*x^2+a)^(p+1)*hypergeom([2, p+1],[2+p],d^2*(b*x^2+a)/(a*d^2+b*c^2))/ 
(a*d^2+b*c^2)^2/p/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 0.15 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.21 \[ \int \frac {x \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\frac {\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (a+b x^2\right )^p \left (-2 c p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )+(-1+2 p) (c+d x) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )\right )}{2 d^2 p (-1+2 p) (c+d x)} \] Input:

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

Output:

((a + b*x^2)^p*(-2*c*p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (c - Sqrt[-(a/b) 
]*d)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)] + (-1 + 2*p)*(c + d*x)*App 
ellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x), (c + Sqrt[-(a 
/b)]*d)/(c + d*x)]))/(2*d^2*p*(-1 + 2*p)*((d*(-Sqrt[-(a/b)] + x))/(c + d*x 
))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p*(c + d*x))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.35, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {594, 25, 719, 238, 237, 504, 334, 333, 353, 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.

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

\(\Big \downarrow \) 594

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-d \int \frac {x \left (b x^2+a\right )^p}{c^2-d^2 x^2}dx\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^2\right )^{p+1}}{(c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {1}{2} d \int \frac {\left (b x^2+a\right )^p}{c^2-d^2 x^2}dx^2\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^2\right )^{p+1}}{(c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \left (\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c}-\frac {d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )}\right )}{d}-\frac {b c (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^2\right )^{p+1}}{(c+d x) \left (a d^2+b c^2\right )}\)

Input:

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

Output:

(c*(a + b*x^2)^(1 + p))/((b*c^2 + a*d^2)*(c + d*x)) + (-((b*c*(1 + 2*p)*x* 
(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(d*(1 + (b*x^ 
2)/a)^p)) + ((a*d^2 + b*c^2*(1 + 2*p))*((x*(a + b*x^2)^p*AppellF1[1/2, -p, 
 1, 3/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c*(1 + (b*x^2)/a)^p) - (d*(a + b*x 
^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b*c^2 + 
a*d^2)])/(2*(b*c^2 + a*d^2)*(1 + p))))/d)/(b*c^2 + a*d^2)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 237
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 333
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 353
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 504
Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 594
Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 719
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(a + b*x**2)**p*a*c*p + (a + b*x**2)**p*a*c - 2*(a + b*x**2)**p*a*d*p*x 
 + (a + b*x**2)**p*a*d*x + 8*int((a + b*x**2)**p/(2*a*c**2*p + a*c**2 + 4* 
a*c*d*p*x + 2*a*c*d*x + 2*a*d**2*p*x**2 + a*d**2*x**2 + 2*b*c**2*p*x**2 + 
b*c**2*x**2 + 4*b*c*d*p*x**3 + 2*b*c*d*x**3 + 2*b*d**2*p*x**4 + b*d**2*x** 
4),x)*a**2*c**2*d*p**2 + 4*int((a + b*x**2)**p/(2*a*c**2*p + a*c**2 + 4*a* 
c*d*p*x + 2*a*c*d*x + 2*a*d**2*p*x**2 + a*d**2*x**2 + 2*b*c**2*p*x**2 + b* 
c**2*x**2 + 4*b*c*d*p*x**3 + 2*b*c*d*x**3 + 2*b*d**2*p*x**4 + b*d**2*x**4) 
,x)*a**2*c**2*d*p + 8*int((a + b*x**2)**p/(2*a*c**2*p + a*c**2 + 4*a*c*d*p 
*x + 2*a*c*d*x + 2*a*d**2*p*x**2 + a*d**2*x**2 + 2*b*c**2*p*x**2 + b*c**2* 
x**2 + 4*b*c*d*p*x**3 + 2*b*c*d*x**3 + 2*b*d**2*p*x**4 + b*d**2*x**4),x)*a 
**2*c*d**2*p**2*x + 4*int((a + b*x**2)**p/(2*a*c**2*p + a*c**2 + 4*a*c*d*p 
*x + 2*a*c*d*x + 2*a*d**2*p*x**2 + a*d**2*x**2 + 2*b*c**2*p*x**2 + b*c**2* 
x**2 + 4*b*c*d*p*x**3 + 2*b*c*d*x**3 + 2*b*d**2*p*x**4 + b*d**2*x**4),x)*a 
**2*c*d**2*p*x + 8*int(((a + b*x**2)**p*x**3)/(2*a*c**2*p + a*c**2 + 4*a*c 
*d*p*x + 2*a*c*d*x + 2*a*d**2*p*x**2 + a*d**2*x**2 + 2*b*c**2*p*x**2 + b*c 
**2*x**2 + 4*b*c*d*p*x**3 + 2*b*c*d*x**3 + 2*b*d**2*p*x**4 + b*d**2*x**4), 
x)*a*b*c*d**2*p**3 - 2*int(((a + b*x**2)**p*x**3)/(2*a*c**2*p + a*c**2 + 4 
*a*c*d*p*x + 2*a*c*d*x + 2*a*d**2*p*x**2 + a*d**2*x**2 + 2*b*c**2*p*x**2 + 
 b*c**2*x**2 + 4*b*c*d*p*x**3 + 2*b*c*d*x**3 + 2*b*d**2*p*x**4 + b*d**2*x* 
*4),x)*a*b*c*d**2*p + 8*int(((a + b*x**2)**p*x**3)/(2*a*c**2*p + a*c**2...