\(\int \frac {x (a+b x^2)^p}{c+d x} \, dx\) [1845]

Optimal result

Integrand size = 18, antiderivative size = 131 \[ \int \frac {x \left (a+b x^2\right )^p}{c+d x} \, dx=-\frac {d x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,1,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c^2}+\frac {c \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 \left (b c^2+a d^2\right ) (1+p)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.31 \[ \int \frac {x \left (a+b x^2\right )^p}{c+d x} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {c \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}+2 d x \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{2 d^2} \] Input:

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

Output:

((a + b*x^2)^p*(-((c*AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/ 
(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/(p*((d*(-Sqrt[-(a/b)] + x))/(c 
 + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p)) + (2*d*x*Hypergeometric2 
F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/(2*d^2)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {621, 353, 78, 395, 394}

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.

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

\(\Big \downarrow \) 621

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 78

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle \frac {c \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )}-\frac {d x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,1,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

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 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 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

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

rule 621

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x \left (a+b x^2\right )^p}{c+d x} \, dx=\frac {\left (b \,x^{2}+a \right )^{p} a d +\left (b \,x^{2}+a \right )^{p} b c x -2 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b d p \,x^{3}+2 b c p \,x^{2}+b d \,x^{3}+2 a d p x +b c \,x^{2}+2 a c p +a d x +a c}d x \right ) a b \,c^{2} p -\left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{2 b d p \,x^{3}+2 b c p \,x^{2}+b d \,x^{3}+2 a d p x +b c \,x^{2}+2 a c p +a d x +a c}d x \right ) a b \,c^{2}-4 \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x^{2}}{2 b d p \,x^{3}+2 b c p \,x^{2}+b d \,x^{3}+2 a d p x +b c \,x^{2}+2 a c p +a d x +a c}d x \right ) a b \,d^{2} p^{2}-2 \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x^{2}}{2 b d p \,x^{3}+2 b c p \,x^{2}+b d \,x^{3}+2 a d p x +b c \,x^{2}+2 a c p +a d x +a c}d x \right ) a b \,d^{2} p -4 \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x^{2}}{2 b d p \,x^{3}+2 b c p \,x^{2}+b d \,x^{3}+2 a d p x +b c \,x^{2}+2 a c p +a d x +a c}d x \right ) b^{2} c^{2} p^{2}-4 \left (\int \frac {\left (b \,x^{2}+a \right )^{p} x^{2}}{2 b d p \,x^{3}+2 b c p \,x^{2}+b d \,x^{3}+2 a d p x +b c \,x^{2}+2 a c p +a d x +a c}d x \right ) b^{2} c^{2} p -\left (\int \frac {\left (b \,x^{2}+a \right )^{p} x^{2}}{2 b d p \,x^{3}+2 b c p \,x^{2}+b d \,x^{3}+2 a d p x +b c \,x^{2}+2 a c p +a d x +a c}d x \right ) b^{2} c^{2}}{b c d \left (2 p +1\right )} \] Input:

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

Output:

((a + b*x**2)**p*a*d + (a + b*x**2)**p*b*c*x - 2*int((a + b*x**2)**p/(2*a* 
c*p + a*c + 2*a*d*p*x + a*d*x + 2*b*c*p*x**2 + b*c*x**2 + 2*b*d*p*x**3 + b 
*d*x**3),x)*a*b*c**2*p - int((a + b*x**2)**p/(2*a*c*p + a*c + 2*a*d*p*x + 
a*d*x + 2*b*c*p*x**2 + b*c*x**2 + 2*b*d*p*x**3 + b*d*x**3),x)*a*b*c**2 - 4 
*int(((a + b*x**2)**p*x**2)/(2*a*c*p + a*c + 2*a*d*p*x + a*d*x + 2*b*c*p*x 
**2 + b*c*x**2 + 2*b*d*p*x**3 + b*d*x**3),x)*a*b*d**2*p**2 - 2*int(((a + b 
*x**2)**p*x**2)/(2*a*c*p + a*c + 2*a*d*p*x + a*d*x + 2*b*c*p*x**2 + b*c*x* 
*2 + 2*b*d*p*x**3 + b*d*x**3),x)*a*b*d**2*p - 4*int(((a + b*x**2)**p*x**2) 
/(2*a*c*p + a*c + 2*a*d*p*x + a*d*x + 2*b*c*p*x**2 + b*c*x**2 + 2*b*d*p*x* 
*3 + b*d*x**3),x)*b**2*c**2*p**2 - 4*int(((a + b*x**2)**p*x**2)/(2*a*c*p + 
 a*c + 2*a*d*p*x + a*d*x + 2*b*c*p*x**2 + b*c*x**2 + 2*b*d*p*x**3 + b*d*x* 
*3),x)*b**2*c**2*p - int(((a + b*x**2)**p*x**2)/(2*a*c*p + a*c + 2*a*d*p*x 
 + a*d*x + 2*b*c*p*x**2 + b*c*x**2 + 2*b*d*p*x**3 + b*d*x**3),x)*b**2*c**2 
)/(b*c*d*(2*p + 1))