\(\int x (c+d x) (a+b x^2)^p \, dx\) [5]

Optimal result

Integrand size = 16, antiderivative size = 75 \[ \int x (c+d x) \left (a+b x^2\right )^p \, dx=\frac {c \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+\frac {1}{3} d x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int x (c+d x) \left (a+b x^2\right )^p \, dx=\frac {1}{6} \left (a+b x^2\right )^p \left (\frac {3 c \left (a+b x^2\right )}{b (1+p)}+2 d x^3 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )\right ) \] Input:

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

Output:

((a + b*x^2)^p*((3*c*(a + b*x^2))/(b*(1 + p)) + (2*d*x^3*Hypergeometric2F1 
[3/2, -p, 5/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/6
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {542, 241, 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.

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

\(\Big \downarrow \) 542

\(\displaystyle c \int x \left (b x^2+a\right )^pdx+d \int x^2 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 241

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

(c*(a + b*x^2)^(1 + p))/(2*b*(1 + p)) + (d*x^3*(a + b*x^2)^p*Hypergeometri 
c2F1[3/2, -p, 5/2, -((b*x^2)/a)])/(3*(1 + (b*x^2)/a)^p)
 

Defintions of rubi rules used

rule 241

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

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 3.59 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int x (c+d x) \left (a+b x^2\right )^p \, dx=\frac {a^{p} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + c \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**p*d*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + c*Piece 
wise((a**p*x**2/2, Eq(b, 0)), (Piecewise(((a + b*x**2)**(p + 1)/(p + 1), N 
e(p, -1)), (log(a + b*x**2), True))/(2*b), True))
 

Maxima [F]

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

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

Output:

d*integrate((b*x^2 + a)^p*x^2, x) + 1/2*(b*x^2 + a)^(p + 1)*c/(b*(p + 1))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

\[ \int x (c+d x) \left (a+b x^2\right )^p \, dx=\frac {4 \left (b \,x^{2}+a \right )^{p} a c \,p^{2}+8 \left (b \,x^{2}+a \right )^{p} a c p +3 \left (b \,x^{2}+a \right )^{p} a c +4 \left (b \,x^{2}+a \right )^{p} a d \,p^{2} x +4 \left (b \,x^{2}+a \right )^{p} a d p x +4 \left (b \,x^{2}+a \right )^{p} b c \,p^{2} x^{2}+8 \left (b \,x^{2}+a \right )^{p} b c p \,x^{2}+3 \left (b \,x^{2}+a \right )^{p} b c \,x^{2}+4 \left (b \,x^{2}+a \right )^{p} b d \,p^{2} x^{3}+6 \left (b \,x^{2}+a \right )^{p} b d p \,x^{3}+2 \left (b \,x^{2}+a \right )^{p} b d \,x^{3}-16 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} d \,p^{4}-48 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} d \,p^{3}-44 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} d \,p^{2}-12 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} d p}{2 b \left (4 p^{3}+12 p^{2}+11 p +3\right )} \] Input:

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

Output:

(4*(a + b*x**2)**p*a*c*p**2 + 8*(a + b*x**2)**p*a*c*p + 3*(a + b*x**2)**p* 
a*c + 4*(a + b*x**2)**p*a*d*p**2*x + 4*(a + b*x**2)**p*a*d*p*x + 4*(a + b* 
x**2)**p*b*c*p**2*x**2 + 8*(a + b*x**2)**p*b*c*p*x**2 + 3*(a + b*x**2)**p* 
b*c*x**2 + 4*(a + b*x**2)**p*b*d*p**2*x**3 + 6*(a + b*x**2)**p*b*d*p*x**3 
+ 2*(a + b*x**2)**p*b*d*x**3 - 16*int((a + b*x**2)**p/(4*a*p**2 + 8*a*p + 
3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2),x)*a**2*d*p**4 - 48*int((a + 
b*x**2)**p/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2 
),x)*a**2*d*p**3 - 44*int((a + b*x**2)**p/(4*a*p**2 + 8*a*p + 3*a + 4*b*p* 
*2*x**2 + 8*b*p*x**2 + 3*b*x**2),x)*a**2*d*p**2 - 12*int((a + b*x**2)**p/( 
4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2),x)*a**2*d* 
p)/(2*b*(4*p**3 + 12*p**2 + 11*p + 3))