\(\int x^2 (c+d x)^3 (a+b x^2)^p \, dx\) [22]

Optimal result

Integrand size = 20, antiderivative size = 210 \[ \int x^2 (c+d x)^3 \left (a+b x^2\right )^p \, dx=-\frac {a d \left (3 b c^2-a d^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {3 c d^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {d \left (3 b c^2-2 a d^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {d^3 \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}-\frac {c \left (9 a d^2-b c^2 (5+2 p)\right ) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )}{3 b (5+2 p)} \] Output:

-1/2*a*d*(-a*d^2+3*b*c^2)*(b*x^2+a)^(p+1)/b^3/(p+1)+3*c*d^2*x^3*(b*x^2+a)^ 
(p+1)/b/(5+2*p)+1/2*d*(-2*a*d^2+3*b*c^2)*(b*x^2+a)^(2+p)/b^3/(2+p)+1/2*d^3 
*(b*x^2+a)^(3+p)/b^3/(3+p)-1/3*c*(9*a*d^2-b*c^2*(5+2*p))*x^3*(b*x^2+a)^p*h 
ypergeom([3/2, -p],[5/2],-b*x^2/a)/b/(5+2*p)/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.93 \[ \int x^2 (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {1}{30} \left (a+b x^2\right )^p \left (\frac {45 c^2 d \left (a+b x^2\right ) \left (-a+b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+\frac {15 d^3 \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+10 c^3 x^3 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+18 c d^2 x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \] Input:

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

Output:

((a + b*x^2)^p*((45*c^2*d*(a + b*x^2)*(-a + b*(1 + p)*x^2))/(b^2*(1 + p)*( 
2 + p)) + (15*d^3*(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)) + (10*c^3*x^3*Hypergeometric2F1[3 
/2, -p, 5/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (18*c*d^2*x^5*Hypergeometr 
ic2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/30
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {543, 354, 27, 86, 363, 279, 278, 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.

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

\(\Big \downarrow \) 543

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 363

\(\displaystyle \frac {1}{2} d \int \left (\frac {a \left (a d^2-3 b c^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (3 b c^2-2 a d^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {d^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+c \left (c^2-\frac {9 a d^2}{2 b p+5 b}\right ) \int x^2 \left (b x^2+a\right )^pdx+\frac {3 c d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {1}{2} d \int \left (\frac {a \left (a d^2-3 b c^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (3 b c^2-2 a d^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {d^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+c \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c^2-\frac {9 a d^2}{2 b p+5 b}\right ) \int x^2 \left (\frac {b x^2}{a}+1\right )^pdx+\frac {3 c d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {1}{2} d \int \left (\frac {a \left (a d^2-3 b c^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (3 b c^2-2 a d^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {d^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+\frac {1}{3} c x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c^2-\frac {9 a d^2}{2 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+\frac {3 c d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} d \left (-\frac {a \left (3 b c^2-a d^2\right ) \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}+\frac {\left (3 b c^2-2 a d^2\right ) \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {d^2 \left (a+b x^2\right )^{p+3}}{b^3 (p+3)}\right )+\frac {1}{3} c x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c^2-\frac {9 a d^2}{2 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+\frac {3 c d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

Input:

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

Output:

(3*c*d^2*x^3*(a + b*x^2)^(1 + p))/(b*(5 + 2*p)) + (d*(-((a*(3*b*c^2 - a*d^ 
2)*(a + b*x^2)^(1 + p))/(b^3*(1 + p))) + ((3*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 + (c* 
(c^2 - (9*a*d^2)/(5*b + 2*b*p))*x^3*(a + b*x^2)^p*Hypergeometric2F1[3/2, - 
p, 5/2, -((b*x^2)/a)])/(3*(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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 923 vs. \(2 (180) = 360\).

Time = 11.76 (sec) , antiderivative size = 1329, normalized size of antiderivative = 6.33 \[ \int x^2 (c+d x)^3 \left (a+b x^2\right )^p \, dx=\text {Too large to display} \] Input:

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

Output:

a**p*c**3*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + 3*a* 
*p*c*d**2*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + 3*c* 
*2*d*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**3*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*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 
*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/(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 + 
sqrt(-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(...
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(16*(a + b*x**2)**p*a**3*d**3*p**3 + 72*(a + b*x**2)**p*a**3*d**3*p**2 + 9 
2*(a + b*x**2)**p*a**3*d**3*p + 30*(a + b*x**2)**p*a**3*d**3 - 24*(a + b*x 
**2)**p*a**2*b*c**2*d*p**4 - 180*(a + b*x**2)**p*a**2*b*c**2*d*p**3 - 462* 
(a + b*x**2)**p*a**2*b*c**2*d*p**2 - 459*(a + b*x**2)**p*a**2*b*c**2*d*p - 
 135*(a + b*x**2)**p*a**2*b*c**2*d - 36*(a + b*x**2)**p*a**2*b*c*d**2*p**4 
*x - 216*(a + b*x**2)**p*a**2*b*c*d**2*p**3*x - 396*(a + b*x**2)**p*a**2*b 
*c*d**2*p**2*x - 216*(a + b*x**2)**p*a**2*b*c*d**2*p*x - 16*(a + b*x**2)** 
p*a**2*b*d**3*p**4*x**2 - 72*(a + b*x**2)**p*a**2*b*d**3*p**3*x**2 - 92*(a 
 + b*x**2)**p*a**2*b*d**3*p**2*x**2 - 30*(a + b*x**2)**p*a**2*b*d**3*p*x** 
2 + 8*(a + b*x**2)**p*a*b**2*c**3*p**5*x + 68*(a + b*x**2)**p*a*b**2*c**3* 
p**4*x + 208*(a + b*x**2)**p*a*b**2*c**3*p**3*x + 268*(a + b*x**2)**p*a*b* 
*2*c**3*p**2*x + 120*(a + b*x**2)**p*a*b**2*c**3*p*x + 24*(a + b*x**2)**p* 
a*b**2*c**2*d*p**5*x**2 + 180*(a + b*x**2)**p*a*b**2*c**2*d*p**4*x**2 + 46 
2*(a + b*x**2)**p*a*b**2*c**2*d*p**3*x**2 + 459*(a + b*x**2)**p*a*b**2*c** 
2*d*p**2*x**2 + 135*(a + b*x**2)**p*a*b**2*c**2*d*p*x**2 + 24*(a + b*x**2) 
**p*a*b**2*c*d**2*p**5*x**3 + 156*(a + b*x**2)**p*a*b**2*c*d**2*p**4*x**3 
+ 336*(a + b*x**2)**p*a*b**2*c*d**2*p**3*x**3 + 276*(a + b*x**2)**p*a*b**2 
*c*d**2*p**2*x**3 + 72*(a + b*x**2)**p*a*b**2*c*d**2*p*x**3 + 8*(a + b*x** 
2)**p*a*b**2*d**3*p**5*x**4 + 44*(a + b*x**2)**p*a*b**2*d**3*p**4*x**4 + 8 
2*(a + b*x**2)**p*a*b**2*d**3*p**3*x**4 + 61*(a + b*x**2)**p*a*b**2*d**...