\(\int \frac {x^3 (c+d x)^n}{a+b x^2} \, dx\) [215]

Optimal result

Integrand size = 20, antiderivative size = 191 \[ \int \frac {x^3 (c+d x)^n}{a+b x^2} \, dx=-\frac {c (c+d x)^{1+n}}{b d^2 (1+n)}+\frac {(c+d x)^{2+n}}{b d^2 (2+n)}+\frac {a (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b^2 \left (c-\frac {\sqrt {-a} d}{\sqrt {b}}\right ) (1+n)}+\frac {a (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 b^2 \left (c+\frac {\sqrt {-a} d}{\sqrt {b}}\right ) (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 (c+d x)^n}{a+b x^2} \, dx=\frac {(c+d x)^{1+n} \left (-\frac {2 \sqrt {b} (c-d (1+n) x)}{d^2 (2+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{\sqrt {b} c-\sqrt {-a} d}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{\sqrt {b} c+\sqrt {-a} d}\right )}{2 b^{3/2} (1+n)} \] Input:

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

Output:

((c + d*x)^(1 + n)*((-2*Sqrt[b]*(c - d*(1 + n)*x))/(d^2*(2 + n)) + (a*Hype 
rgeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c - Sqrt[-a]*d 
)])/(Sqrt[b]*c - Sqrt[-a]*d) + (a*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt 
[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/(Sqrt[b]*c + Sqrt[-a]*d)))/(2*b^ 
(3/2)*(1 + n))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {604, 25, 2160, 2009}

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.

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

\(\Big \downarrow \) 604

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2160

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c + d*x)^(2 + n)/(b*d^2*(2 + n)) - ((c*d*(2 + n)*(c + d*x)^(1 + n))/(1 + 
n) - (a*d^3*(2 + n)*(c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ( 
Sqrt[b]*(c + d*x))/(Sqrt[b]*c - Sqrt[-a]*d)])/(2*Sqrt[b]*(Sqrt[b]*c - Sqrt 
[-a]*d)*(1 + n)) - (a*d^3*(2 + n)*(c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 
 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/(2*Sqrt[b]*(Sq 
rt[b]*c + Sqrt[-a]*d)*(1 + n)))/(b*d^3*(2 + n))
 

Defintions of rubi rules used

rule 25

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

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2160

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, 
 d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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