\(\int x^2 (c+d x)^n \sqrt {a-b x^2} \, dx\) [256]

Optimal result

Integrand size = 23, antiderivative size = 356 \[ \int x^2 (c+d x)^n \sqrt {a-b x^2} \, dx=-\frac {(c+d x)^{1+n} \left (a-b x^2\right )^{3/2}}{b d (4+n)}+\frac {\left (a d+\frac {3 b c^2}{d+d n}\right ) (c+d x)^{1+n} \sqrt {a-b x^2} \operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{b d^2 (4+n) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}}-\frac {3 c (c+d x)^{2+n} \sqrt {a-b x^2} \operatorname {AppellF1}\left (2+n,-\frac {1}{2},-\frac {1}{2},3+n,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{d^3 (2+n) (4+n) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}} \] Output:

-(d*x+c)^(1+n)*(-b*x^2+a)^(3/2)/b/d/(4+n)+(a*d+3*b*c^2/(d*n+d))*(d*x+c)^(1 
+n)*(-b*x^2+a)^(1/2)*AppellF1(1+n,-1/2,-1/2,2+n,(d*x+c)/(c-a^(1/2)*d/b^(1/ 
2)),(d*x+c)/(c+a^(1/2)*d/b^(1/2)))/b/d^2/(4+n)/(1-(d*x+c)/(c-a^(1/2)*d/b^( 
1/2)))^(1/2)/(1-(d*x+c)/(c+a^(1/2)*d/b^(1/2)))^(1/2)-3*c*(d*x+c)^(2+n)*(-b 
*x^2+a)^(1/2)*AppellF1(2+n,-1/2,-1/2,3+n,(d*x+c)/(c-a^(1/2)*d/b^(1/2)),(d* 
x+c)/(c+a^(1/2)*d/b^(1/2)))/d^3/(2+n)/(4+n)/(1-(d*x+c)/(c-a^(1/2)*d/b^(1/2 
)))^(1/2)/(1-(d*x+c)/(c+a^(1/2)*d/b^(1/2)))^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {604, 25, 27, 719, 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.

\(\displaystyle \int x^2 \sqrt {a-b x^2} (c+d x)^n \, dx\)

\(\Big \downarrow \) 604

\(\displaystyle -\frac {\int -d (a d (n+1)-3 b c x) (c+d x)^n \sqrt {a-b x^2}dx}{b d^2 (n+4)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{b d (n+4)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int d (a d (n+1)-3 b c x) (c+d x)^n \sqrt {a-b x^2}dx}{b d^2 (n+4)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{b d (n+4)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a d (n+1)-3 b c x) (c+d x)^n \sqrt {a-b x^2}dx}{b d (n+4)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{b d (n+4)}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {\left (a d^2 (n+1)+3 b c^2\right ) \int (c+d x)^n \sqrt {a-b x^2}dx}{d}-\frac {3 b c \int (c+d x)^{n+1} \sqrt {a-b x^2}dx}{d}}{b d (n+4)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{b d (n+4)}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {\frac {\sqrt {a-b x^2} \left (a d^2 (n+1)+3 b c^2\right ) \int (c+d x)^n \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}d(c+d x)}{d^2 \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}-\frac {3 b c \sqrt {a-b x^2} \int (c+d x)^{n+1} \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}d(c+d x)}{d^2 \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}}{b d (n+4)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{b d (n+4)}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\frac {\sqrt {a-b x^2} (c+d x)^{n+1} \left (a d^2 (n+1)+3 b c^2\right ) \operatorname {AppellF1}\left (n+1,-\frac {1}{2},-\frac {1}{2},n+2,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{d^2 (n+1) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}-\frac {3 b c \sqrt {a-b x^2} (c+d x)^{n+2} \operatorname {AppellF1}\left (n+2,-\frac {1}{2},-\frac {1}{2},n+3,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{d^2 (n+2) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}}{b d (n+4)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{b d (n+4)}\)

Input:

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

Output:

-(((c + d*x)^(1 + n)*(a - b*x^2)^(3/2))/(b*d*(4 + n))) + (((3*b*c^2 + a*d^ 
2*(1 + n))*(c + d*x)^(1 + n)*Sqrt[a - b*x^2]*AppellF1[1 + n, -1/2, -1/2, 2 
 + n, (c + d*x)/(c - (Sqrt[a]*d)/Sqrt[b]), (c + d*x)/(c + (Sqrt[a]*d)/Sqrt 
[b])])/(d^2*(1 + n)*Sqrt[1 - (c + d*x)/(c - (Sqrt[a]*d)/Sqrt[b])]*Sqrt[1 - 
 (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b])]) - (3*b*c*(c + d*x)^(2 + n)*Sqrt[a - 
 b*x^2]*AppellF1[2 + n, -1/2, -1/2, 3 + n, (c + d*x)/(c - (Sqrt[a]*d)/Sqrt 
[b]), (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b])])/(d^2*(2 + n)*Sqrt[1 - (c + d*x 
)/(c - (Sqrt[a]*d)/Sqrt[b])]*Sqrt[1 - (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b])] 
))/(b*d*(4 + n))
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 604

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(x^2*(d*x+c)^n*(-b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**2*(d*x+c)**n*(-b*x**2+a)**(1/2),x)
 

Output:

Integral(x**2*sqrt(a - b*x**2)*(c + d*x)**n, x)
 

Maxima [F]

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

integrate(x^2*(d*x+c)^n*(-b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^2*(d*x+c)^n*(-b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((c + d*x)**n*sqrt(a - b*x**2)*x**2,x)