\(\int x (c+d x)^n (a+b x^2)^p \, dx\) [1938]

Optimal result

Integrand size = 18, antiderivative size = 192 \[ \int x (c+d x)^n \left (a+b x^2\right )^p \, dx=\frac {(c+d x)^n \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {(c+d x)^n \left (a+b x^2\right )^{1+p} \left (1-\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \left (1-\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )^{-1-p} \operatorname {AppellF1}\left (n,-1-p,-1-p,1+n,\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b (1+p)} \] Output:

1/2*(d*x+c)^n*(b*x^2+a)^(p+1)/b/(p+1)-1/2*(d*x+c)^n*(b*x^2+a)^(p+1)*(1-(d* 
x+c)/(c-(-a)^(1/2)*d/b^(1/2)))^(-1-p)*(1-(d*x+c)/(c+(-a)^(1/2)*d/b^(1/2))) 
^(-1-p)*AppellF1(n,-1-p,-1-p,1+n,(d*x+c)/(c-(-a)^(1/2)*d/b^(1/2)),(d*x+c)/ 
(c+(-a)^(1/2)*d/b^(1/2)))/b/(p+1)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.63 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.60, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {624, 514, 150}

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.

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

\(\Big \downarrow \) 624

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

\(\Big \downarrow \) 514

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 514

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

rule 624

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[1/d   Int[x^(m - 1)*(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] - Si 
mp[c/d   Int[x^(m - 1)*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, 
 d, n, p}, x] && IGtQ[m, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(c + d*x)**n*(a + b*x**2)**p*a*d*n + 2*(c + d*x)**n*(a + b*x**2)**p*a*d 
*p + (c + d*x)**n*(a + b*x**2)**p*a*d + (c + d*x)**n*(a + b*x**2)**p*b*c*n 
*x + (c + d*x)**n*(a + b*x**2)**p*b*d*n*x**2 + 2*(c + d*x)**n*(a + b*x**2) 
**p*b*d*p*x**2 + (c + d*x)**n*(a + b*x**2)**p*b*d*x**2 - 2*int(((c + d*x)* 
*n*(a + b*x**2)**p*x**2)/(a*c*n**2 + 4*a*c*n*p + 3*a*c*n + 4*a*c*p**2 + 6* 
a*c*p + 2*a*c + a*d*n**2*x + 4*a*d*n*p*x + 3*a*d*n*x + 4*a*d*p**2*x + 6*a* 
d*p*x + 2*a*d*x + b*c*n**2*x**2 + 4*b*c*n*p*x**2 + 3*b*c*n*x**2 + 4*b*c*p* 
*2*x**2 + 6*b*c*p*x**2 + 2*b*c*x**2 + b*d*n**2*x**3 + 4*b*d*n*p*x**3 + 3*b 
*d*n*x**3 + 4*b*d*p**2*x**3 + 6*b*d*p*x**3 + 2*b*d*x**3),x)*a*b*d**2*n**4 
- 12*int(((c + d*x)**n*(a + b*x**2)**p*x**2)/(a*c*n**2 + 4*a*c*n*p + 3*a*c 
*n + 4*a*c*p**2 + 6*a*c*p + 2*a*c + a*d*n**2*x + 4*a*d*n*p*x + 3*a*d*n*x + 
 4*a*d*p**2*x + 6*a*d*p*x + 2*a*d*x + b*c*n**2*x**2 + 4*b*c*n*p*x**2 + 3*b 
*c*n*x**2 + 4*b*c*p**2*x**2 + 6*b*c*p*x**2 + 2*b*c*x**2 + b*d*n**2*x**3 + 
4*b*d*n*p*x**3 + 3*b*d*n*x**3 + 4*b*d*p**2*x**3 + 6*b*d*p*x**3 + 2*b*d*x** 
3),x)*a*b*d**2*n**3*p - 7*int(((c + d*x)**n*(a + b*x**2)**p*x**2)/(a*c*n** 
2 + 4*a*c*n*p + 3*a*c*n + 4*a*c*p**2 + 6*a*c*p + 2*a*c + a*d*n**2*x + 4*a* 
d*n*p*x + 3*a*d*n*x + 4*a*d*p**2*x + 6*a*d*p*x + 2*a*d*x + b*c*n**2*x**2 + 
 4*b*c*n*p*x**2 + 3*b*c*n*x**2 + 4*b*c*p**2*x**2 + 6*b*c*p*x**2 + 2*b*c*x* 
*2 + b*d*n**2*x**3 + 4*b*d*n*p*x**3 + 3*b*d*n*x**3 + 4*b*d*p**2*x**3 + 6*b 
*d*p*x**3 + 2*b*d*x**3),x)*a*b*d**2*n**3 - 24*int(((c + d*x)**n*(a + b*...