\(\int x^3 (c+d x)^n \sqrt {a-b x^2} \, dx\) [1796]

Optimal result

Integrand size = 23, antiderivative size = 439 \[ \int x^3 (c+d x)^n \sqrt {a-b x^2} \, dx=\frac {c (8+n) (c+d x)^{1+n} \left (a-b x^2\right )^{3/2}}{b d^2 (4+n) (5+n)}-\frac {(c+d x)^{2+n} \left (a-b x^2\right )^{3/2}}{b d^2 (5+n)}-\frac {c \left (12 b c^2+a d^2 \left (8+9 n+n^2\right )\right ) (c+d x)^{1+n} \sqrt {a-b x^2} \operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{b d^4 (1+n) (4+n) (5+n) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}}+\frac {\left (12 b c^2+a d^2 \left (8+6 n+n^2\right )\right ) (c+d x)^{2+n} \sqrt {a-b x^2} \operatorname {AppellF1}\left (2+n,-\frac {1}{2},-\frac {1}{2},3+n,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{b d^4 (2+n) (4+n) (5+n) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}} \] Output:

c*(8+n)*(d*x+c)^(1+n)*(-b*x^2+a)^(3/2)/b/d^2/(4+n)/(5+n)-(d*x+c)^(2+n)*(-b 
*x^2+a)^(3/2)/b/d^2/(5+n)-c*(12*b*c^2+a*d^2*(n^2+9*n+8))*(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^4/(1+n)/(4+n)/(5+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)+(12*b*c^2+a*d^2* 
(n^2+6*n+8))*(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)))/b/d^4/(2+n)/(4+n 
)/(5+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]

\[ \int x^3 (c+d x)^n \sqrt {a-b x^2} \, dx=\int x^3 (c+d x)^n \sqrt {a-b x^2} \, dx \] Input:

Integrate[x^3*(c + d*x)^n*Sqrt[a - b*x^2],x]
 

Output:

Integrate[x^3*(c + d*x)^n*Sqrt[a - b*x^2], x]
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 425, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {604, 25, 2185, 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.

\(\displaystyle \int x^3 \sqrt {a-b x^2} (c+d x)^n \, dx\)

\(\Big \downarrow \) 604

\(\displaystyle -\frac {\int -(c+d x)^n \sqrt {a-b x^2} \left (-b c (n+8) x^2 d^2+a c (n+2) d^2-\left (3 b c^2-a d^2 (n+2)\right ) x d\right )dx}{b d^3 (n+5)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+2}}{b d^2 (n+5)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int (c+d x)^n \sqrt {a-b x^2} \left (-b c (n+8) x^2 d^2+a c (n+2) d^2-\left (3 b c^2-a d^2 (n+2)\right ) x d\right )dx}{b d^3 (n+5)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+2}}{b d^2 (n+5)}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {c d (n+8) \left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{n+4}-\frac {\int b d^3 (c+d x)^n \left (3 a c d n-\left (12 b c^2+a d^2 \left (n^2+6 n+8\right )\right ) x\right ) \sqrt {a-b x^2}dx}{b d^2 (n+4)}}{b d^3 (n+5)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+2}}{b d^2 (n+5)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {c d (n+8) \left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{n+4}-\frac {d \int (c+d x)^n \left (3 a c d n-\left (12 b c^2+a d^2 \left (n^2+6 n+8\right )\right ) x\right ) \sqrt {a-b x^2}dx}{n+4}}{b d^3 (n+5)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+2}}{b d^2 (n+5)}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {c d (n+8) \left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{n+4}-\frac {d \left (\frac {c \left (a d^2 \left (n^2+9 n+8\right )+12 b c^2\right ) \int (c+d x)^n \sqrt {a-b x^2}dx}{d}-\frac {\left (a d^2 \left (n^2+6 n+8\right )+12 b c^2\right ) \int (c+d x)^{n+1} \sqrt {a-b x^2}dx}{d}\right )}{n+4}}{b d^3 (n+5)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+2}}{b d^2 (n+5)}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {\frac {c d (n+8) \left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{n+4}-\frac {d \left (\frac {c \sqrt {a-b x^2} \left (a d^2 \left (n^2+9 n+8\right )+12 b c^2\right ) \int (c+d x)^n \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}d(c+d x)}{d^2 \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}-\frac {\sqrt {a-b x^2} \left (a d^2 \left (n^2+6 n+8\right )+12 b c^2\right ) \int (c+d x)^{n+1} \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}}d(c+d x)}{d^2 \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}\right )}{n+4}}{b d^3 (n+5)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+2}}{b d^2 (n+5)}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\frac {c d (n+8) \left (a-b x^2\right )^{3/2} (c+d x)^{n+1}}{n+4}-\frac {d \left (\frac {c \sqrt {a-b x^2} (c+d x)^{n+1} \left (a d^2 \left (n^2+9 n+8\right )+12 b c^2\right ) \operatorname {AppellF1}\left (n+1,-\frac {1}{2},-\frac {1}{2},n+2,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{d^2 (n+1) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}-\frac {\sqrt {a-b x^2} (c+d x)^{n+2} \left (a d^2 \left (n^2+6 n+8\right )+12 b c^2\right ) \operatorname {AppellF1}\left (n+2,-\frac {1}{2},-\frac {1}{2},n+3,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{d^2 (n+2) \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{\frac {\sqrt {a} d}{\sqrt {b}}+c}}}\right )}{n+4}}{b d^3 (n+5)}-\frac {\left (a-b x^2\right )^{3/2} (c+d x)^{n+2}}{b d^2 (n+5)}\)

Input:

Int[x^3*(c + d*x)^n*Sqrt[a - b*x^2],x]
 

Output:

-(((c + d*x)^(2 + n)*(a - b*x^2)^(3/2))/(b*d^2*(5 + n))) + ((c*d*(8 + n)*( 
c + d*x)^(1 + n)*(a - b*x^2)^(3/2))/(4 + n) - (d*((c*(12*b*c^2 + a*d^2*(8 
+ 9*n + n^2))*(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)/S 
qrt[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])]) - ((12*b*c^2 + a*d^2*(8 + 6*n + 
n^2))*(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])])))/(4 + n))/(b*d^3*(5 + 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]
 

rule 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [F]

\[\int x^{3} \left (d x +c \right )^{n} \sqrt {-b \,x^{2}+a}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x^3*(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^3, x)
 

Sympy [F]

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

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

Output:

Integral(x**3*sqrt(a - b*x**2)*(c + d*x)**n, x)
 

Maxima [F]

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

integrate(x^3*(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^3, x)
 

Giac [F]

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

integrate(x^3*(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^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 (c+d x)^n \sqrt {a-b x^2} \, dx=\int x^3\,\sqrt {a-b\,x^2}\,{\left (c+d\,x\right )}^n \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int((c + d*x)**n*sqrt(a - b*x**2)*x**3,x)