\(\int \frac {x^3 (c+d x)^n}{(a-b x^2)^{3/2}} \, dx\) [273]

Optimal result

Integrand size = 23, antiderivative size = 394 \[ \int \frac {x^3 (c+d x)^n}{\left (a-b x^2\right )^{3/2}} \, dx=\frac {a (c-d x) (c+d x)^{1+n}}{b \left (b c^2-a d^2\right ) \sqrt {a-b x^2}}-\frac {c \left (\frac {2 a}{b}-\frac {c^2}{d^2 (1+n)}\right ) (c+d x)^{1+n} \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}} \operatorname {AppellF1}\left (1+n,\frac {1}{2},\frac {1}{2},2+n,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{\left (b c^2-a d^2\right ) \sqrt {a-b x^2}}+\frac {\left (\frac {a}{b}-\frac {c^2}{d^2 (2+n)}\right ) (c+d x)^{2+n} \sqrt {1-\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}}} \sqrt {1-\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}} \operatorname {AppellF1}\left (2+n,\frac {1}{2},\frac {1}{2},3+n,\frac {c+d x}{c-\frac {\sqrt {a} d}{\sqrt {b}}},\frac {c+d x}{c+\frac {\sqrt {a} d}{\sqrt {b}}}\right )}{\left (b c^2-a d^2\right ) \sqrt {a-b x^2}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {602, 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.

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

\(\Big \downarrow \) 602

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 514

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

(a*(c - d*x)*(c + d*x)^(1 + n))/(b*(b*c^2 - a*d^2)*Sqrt[a - b*x^2]) - (-(( 
c*((b*c^2)/d - 2*a*d*(1 + n))*(c + d*x)^(1 + n)*Sqrt[1 - (c + d*x)/(c - (S 
qrt[a]*d)/Sqrt[b])]*Sqrt[1 - (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b])]*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*(1 + n)*Sqrt[a - b*x^2])) + ((b*c^2 - a*d^2*( 
2 + n))*(c + d*x)^(2 + n)*Sqrt[1 - (c + d*x)/(c - (Sqrt[a]*d)/Sqrt[b])]*Sq 
rt[1 - (c + d*x)/(c + (Sqrt[a]*d)/Sqrt[b])]*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[a - b*x^2]))/(b*(b*c^2 - a*d^2))
 

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 602

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomia 
lRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
+ b*x^2, x], x, 1]}, Simp[(-(c + d*x)^(n + 1))*(a + b*x^2)^(p + 1)*((a*(d*e 
 - c*f) + (b*c*e + a*d*f)*x)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2 
*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToS 
um[2*a*(p + 1)*(b*c^2 + a*d^2)*Qx + e*(b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3 
)) - a*c*d*f*n + d*(b*c*e + a*d*f)*(n + 2*p + 4)*x, x], x], x]] /; FreeQ[{a 
, b, c, d, n}, x] && IGtQ[m, 1] && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(x^3*(d*x+c)^n/(-b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(x^3*(d*x+c)^n/(-b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^3*(d*x+c)^n/(-b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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