\(\int \frac {x^3 (c+d x)^n}{(a+b x^2)^2} \, dx\) [222]

Optimal result

Integrand size = 20, antiderivative size = 297 \[ \int \frac {x^3 (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\frac {a (c-d x) (c+d x)^{1+n}}{2 b \left (b c^2+a d^2\right ) \left (a+b x^2\right )}+\frac {\left (\sqrt {-a} c d n-\frac {2 b c^2+a d^2 (2+n)}{\sqrt {b}}\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{4 b \left (\sqrt {b} c-\sqrt {-a} d\right ) \left (b c^2+a d^2\right ) (1+n)}-\frac {\left (2 b c^2+\sqrt {-a} \sqrt {b} c d n+a d^2 (2+n)\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{4 b^{3/2} \left (\sqrt {b} c+\sqrt {-a} d\right ) \left (b c^2+a d^2\right ) (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=-\frac {(c+d x)^{1+n} \left (\frac {2 a \sqrt {b} (-c+d x)}{a+b x^2}+\frac {\left (2 b c^2-\sqrt {-a} \sqrt {b} c d n+a d^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{\left (\sqrt {b} c-\sqrt {-a} d\right ) (1+n)}+\frac {\left (2 b c^2+\sqrt {-a} \sqrt {b} c d n+a d^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{\left (\sqrt {b} c+\sqrt {-a} d\right ) (1+n)}\right )}{4 b^{3/2} \left (b c^2+a d^2\right )} \] Input:

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

Output:

-1/4*((c + d*x)^(1 + n)*((2*a*Sqrt[b]*(-c + d*x))/(a + b*x^2) + ((2*b*c^2 
- Sqrt[-a]*Sqrt[b]*c*d*n + a*d^2*(2 + n))*Hypergeometric2F1[1, 1 + n, 2 + 
n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c - Sqrt[-a]*d)])/((Sqrt[b]*c - Sqrt[-a]*d 
)*(1 + n)) + ((2*b*c^2 + Sqrt[-a]*Sqrt[b]*c*d*n + a*d^2*(2 + n))*Hypergeom 
etric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/( 
(Sqrt[b]*c + Sqrt[-a]*d)*(1 + n))))/(b^(3/2)*(b*c^2 + a*d^2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {602, 27, 657, 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.

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

\(\Big \downarrow \) 602

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(a*(c - d*x)*(c + d*x)^(1 + n))/(2*b*(b*c^2 + a*d^2)*(a + b*x^2)) - (((2*b 
*c^2 - Sqrt[-a]*Sqrt[b]*c*d*n + a*d^2*(2 + n))*(c + d*x)^(1 + n)*Hypergeom 
etric2F1[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)) + ((2*b*c^2 + Sqrt[-a]*Sqrt[b] 
*c*d*n + a*d^2*(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)))/(2*b*(b*c^2 + a*d^2))
 

Defintions of rubi rules used

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 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 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((c + d*x)**n*a*d*n**2 + 3*(c + d*x)**n*a*d*n - 2*(c + d*x)**n*a*d + 2*(c 
+ d*x)**n*b*c*n*x + 2*(c + d*x)**n*b*d*n*x**2 - 2*(c + d*x)**n*b*d*x**2 - 
int((c + d*x)**n/(a**2*c*n - a**2*c + a**2*d*n*x - a**2*d*x + 2*a*b*c*n*x* 
*2 - 2*a*b*c*x**2 + 2*a*b*d*n*x**3 - 2*a*b*d*x**3 + b**2*c*n*x**4 - b**2*c 
*x**4 + b**2*d*n*x**5 - b**2*d*x**5),x)*a**3*d**2*n**4 - 2*int((c + d*x)** 
n/(a**2*c*n - a**2*c + a**2*d*n*x - a**2*d*x + 2*a*b*c*n*x**2 - 2*a*b*c*x* 
*2 + 2*a*b*d*n*x**3 - 2*a*b*d*x**3 + b**2*c*n*x**4 - b**2*c*x**4 + b**2*d* 
n*x**5 - b**2*d*x**5),x)*a**3*d**2*n**3 + 5*int((c + d*x)**n/(a**2*c*n - a 
**2*c + a**2*d*n*x - a**2*d*x + 2*a*b*c*n*x**2 - 2*a*b*c*x**2 + 2*a*b*d*n* 
x**3 - 2*a*b*d*x**3 + b**2*c*n*x**4 - b**2*c*x**4 + b**2*d*n*x**5 - b**2*d 
*x**5),x)*a**3*d**2*n**2 - 2*int((c + d*x)**n/(a**2*c*n - a**2*c + a**2*d* 
n*x - a**2*d*x + 2*a*b*c*n*x**2 - 2*a*b*c*x**2 + 2*a*b*d*n*x**3 - 2*a*b*d* 
x**3 + b**2*c*n*x**4 - b**2*c*x**4 + b**2*d*n*x**5 - b**2*d*x**5),x)*a**3* 
d**2*n - 2*int((c + d*x)**n/(a**2*c*n - a**2*c + a**2*d*n*x - a**2*d*x + 2 
*a*b*c*n*x**2 - 2*a*b*c*x**2 + 2*a*b*d*n*x**3 - 2*a*b*d*x**3 + b**2*c*n*x* 
*4 - b**2*c*x**4 + b**2*d*n*x**5 - b**2*d*x**5),x)*a**2*b*c**2*n**2 + 2*in 
t((c + d*x)**n/(a**2*c*n - a**2*c + a**2*d*n*x - a**2*d*x + 2*a*b*c*n*x**2 
 - 2*a*b*c*x**2 + 2*a*b*d*n*x**3 - 2*a*b*d*x**3 + b**2*c*n*x**4 - b**2*c*x 
**4 + b**2*d*n*x**5 - b**2*d*x**5),x)*a**2*b*c**2*n - int((c + d*x)**n/(a* 
*2*c*n - a**2*c + a**2*d*n*x - a**2*d*x + 2*a*b*c*n*x**2 - 2*a*b*c*x**2...