\(\int x^3 (c+d x)^2 (a+b x^2)^p \, dx\) [1826]

Optimal result

Integrand size = 20, antiderivative size = 149 \[ \int x^3 (c+d x)^2 \left (a+b x^2\right )^p \, dx=-\frac {a \left (b c^2-a d^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {\left (b c^2-2 a d^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {d^2 \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac {2}{5} c d x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \] Output:

-1/2*a*(-a*d^2+b*c^2)*(b*x^2+a)^(p+1)/b^3/(p+1)+1/2*(-2*a*d^2+b*c^2)*(b*x^ 
2+a)^(2+p)/b^3/(2+p)+1/2*d^2*(b*x^2+a)^(3+p)/b^3/(3+p)+2/5*c*d*x^5*(b*x^2+ 
a)^p*hypergeom([5/2, -p],[7/2],-b*x^2/a)/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.02 \[ \int x^3 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {1}{10} \left (a+b x^2\right )^p \left (\frac {5 c^2 \left (a+b x^2\right ) \left (-a+b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+\frac {5 d^2 \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+4 c d x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \] Input:

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

Output:

((a + b*x^2)^p*((5*c^2*(a + b*x^2)*(-a + b*(1 + p)*x^2))/(b^2*(1 + p)*(2 + 
 p)) + (5*d^2*(a + b*x^2)*(2*a^2 - 2*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2) 
*x^4))/(b^3*(1 + p)*(2 + p)*(3 + p)) + (4*c*d*x^5*Hypergeometric2F1[5/2, - 
p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/10
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {543, 27, 279, 278, 354, 86, 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.

\(\displaystyle \int x^3 (c+d x)^2 \left (a+b x^2\right )^p \, dx\)

\(\Big \downarrow \) 543

\(\displaystyle \int x^3 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+\int 2 c d x^4 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 27

\(\displaystyle \int x^3 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+2 c d \int x^4 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

\(\displaystyle \int x^3 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+\frac {2}{5} c d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int x^2 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx^2+\frac {2}{5} c d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {1}{2} \int \left (\frac {a \left (a d^2-b c^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (b c^2-2 a d^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {d^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+\frac {2}{5} c d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {a \left (b c^2-a d^2\right ) \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}+\frac {\left (b c^2-2 a d^2\right ) \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {d^2 \left (a+b x^2\right )^{p+3}}{b^3 (p+3)}\right )+\frac {2}{5} c d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\)

Input:

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

Output:

(-((a*(b*c^2 - a*d^2)*(a + b*x^2)^(1 + p))/(b^3*(1 + p))) + ((b*c^2 - 2*a* 
d^2)*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (d^2*(a + b*x^2)^(3 + p))/(b^3*( 
3 + p)))/2 + (2*c*d*x^5*(a + b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b 
*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)
 

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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 543

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), 
 {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ 
(n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; 
 FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] &&  !IntegerQ[2*p] 
&&  !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^2*x^5 + 2*c*d*x^4 + c^2*x^3)*(b*x^2 + a)^p, x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (124) = 248\).

Time = 12.07 (sec) , antiderivative size = 1294, normalized size of antiderivative = 8.68 \[ \int x^3 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\text {Too large to display} \] Input:

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

Output:

2*a**p*c*d*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + c** 
2*Piecewise((a**p*x**4/4, Eq(b, 0)), (a*log(x - sqrt(-a/b))/(2*a*b**2 + 2* 
b**3*x**2) + a*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 
+ 2*b**3*x**2) + b*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x 
**2*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(x - 
sqrt(-a/b))/(2*b**2) - a*log(x + sqrt(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, 
-1)), (-a**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x** 
2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x 
**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b 
**2*p**2 + 6*b**2*p + 4*b**2), True)) + d**2*Piecewise((a**p*x**6/6, Eq(b, 
 0)), (2*a**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x* 
*4) + 2*a**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x** 
4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(x 
 - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*lo 
g(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2 
/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x - sqrt(-a 
/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x + sqr 
t(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(p, -3)), (-2*a**2 
*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(x + sqrt(-a/b)) 
/(2*a*b**3 + 2*b**4*x**2) - 2*a**2/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**...
 

Maxima [F]

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

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

Output:

1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2)*(b*x^2 + a)^p*c^2/((p^2 + 3*p + 2) 
*b^2) + integrate((d^2*x^5 + 2*c*d*x^4)*(b*x^2 + a)^p, x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(16*(a + b*x**2)**p*a**3*d**2*p**3 + 72*(a + b*x**2)**p*a**3*d**2*p**2 + 9 
2*(a + b*x**2)**p*a**3*d**2*p + 30*(a + b*x**2)**p*a**3*d**2 - 8*(a + b*x* 
*2)**p*a**2*b*c**2*p**4 - 60*(a + b*x**2)**p*a**2*b*c**2*p**3 - 154*(a + b 
*x**2)**p*a**2*b*c**2*p**2 - 153*(a + b*x**2)**p*a**2*b*c**2*p - 45*(a + b 
*x**2)**p*a**2*b*c**2 - 24*(a + b*x**2)**p*a**2*b*c*d*p**4*x - 144*(a + b* 
x**2)**p*a**2*b*c*d*p**3*x - 264*(a + b*x**2)**p*a**2*b*c*d*p**2*x - 144*( 
a + b*x**2)**p*a**2*b*c*d*p*x - 16*(a + b*x**2)**p*a**2*b*d**2*p**4*x**2 - 
 72*(a + b*x**2)**p*a**2*b*d**2*p**3*x**2 - 92*(a + b*x**2)**p*a**2*b*d**2 
*p**2*x**2 - 30*(a + b*x**2)**p*a**2*b*d**2*p*x**2 + 8*(a + b*x**2)**p*a*b 
**2*c**2*p**5*x**2 + 60*(a + b*x**2)**p*a*b**2*c**2*p**4*x**2 + 154*(a + b 
*x**2)**p*a*b**2*c**2*p**3*x**2 + 153*(a + b*x**2)**p*a*b**2*c**2*p**2*x** 
2 + 45*(a + b*x**2)**p*a*b**2*c**2*p*x**2 + 16*(a + b*x**2)**p*a*b**2*c*d* 
p**5*x**3 + 104*(a + b*x**2)**p*a*b**2*c*d*p**4*x**3 + 224*(a + b*x**2)**p 
*a*b**2*c*d*p**3*x**3 + 184*(a + b*x**2)**p*a*b**2*c*d*p**2*x**3 + 48*(a + 
 b*x**2)**p*a*b**2*c*d*p*x**3 + 8*(a + b*x**2)**p*a*b**2*d**2*p**5*x**4 + 
44*(a + b*x**2)**p*a*b**2*d**2*p**4*x**4 + 82*(a + b*x**2)**p*a*b**2*d**2* 
p**3*x**4 + 61*(a + b*x**2)**p*a*b**2*d**2*p**2*x**4 + 15*(a + b*x**2)**p* 
a*b**2*d**2*p*x**4 + 8*(a + b*x**2)**p*b**3*c**2*p**5*x**4 + 68*(a + b*x** 
2)**p*b**3*c**2*p**4*x**4 + 214*(a + b*x**2)**p*b**3*c**2*p**3*x**4 + 307* 
(a + b*x**2)**p*b**3*c**2*p**2*x**4 + 198*(a + b*x**2)**p*b**3*c**2*p*x...