\(\int x^4 (c+d x)^2 (a+b x^2)^p \, dx\) [1825]

Optimal result

Integrand size = 20, antiderivative size = 177 \[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {a^2 c d \left (a+b x^2\right )^{1+p}}{b^3 (1+p)}+\frac {d^2 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}-\frac {2 a c d \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac {c d \left (a+b x^2\right )^{3+p}}{b^3 (3+p)}-\frac {\left (5 a d^2-b c^2 (7+2 p)\right ) x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )}{5 b (7+2 p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.88 \[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {1}{35} \left (a+b x^2\right )^p \left (\frac {35 c d \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+7 c^2 x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )+5 d^2 x^7 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-p,\frac {9}{2},-\frac {b x^2}{a}\right )\right ) \] Input:

Integrate[x^4*(c + d*x)^2*(a + b*x^2)^p,x]
 

Output:

((a + b*x^2)^p*((35*c*d*(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)) + (7*c^2*x^5*Hypergeometric 
2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (5*d^2*x^7*Hypergeome 
tric2F1[7/2, -p, 9/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/35
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 167, 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, 27, 243, 53, 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.

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

\(\Big \downarrow \) 543

\(\displaystyle \int x^4 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+\int 2 c d x^5 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 27

\(\displaystyle \int x^4 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+2 c d \int x^5 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 363

\(\displaystyle c d \int \left (\frac {a^2 \left (b x^2+a\right )^p}{b^2}-\frac {2 a \left (b x^2+a\right )^{p+1}}{b^2}+\frac {\left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+\left (c^2-\frac {5 a d^2}{2 b p+7 b}\right ) \int x^4 \left (b x^2+a\right )^pdx+\frac {d^2 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

\(\Big \downarrow \) 279

\(\displaystyle c d \int \left (\frac {a^2 \left (b x^2+a\right )^p}{b^2}-\frac {2 a \left (b x^2+a\right )^{p+1}}{b^2}+\frac {\left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c^2-\frac {5 a d^2}{2 b p+7 b}\right ) \int x^4 \left (\frac {b x^2}{a}+1\right )^pdx+\frac {d^2 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

\(\Big \downarrow \) 278

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^4*(c + d*x)^2*(a + b*x^2)^p,x]
 

Output:

(d^2*x^5*(a + b*x^2)^(1 + p))/(b*(7 + 2*p)) + c*d*((a^2*(a + b*x^2)^(1 + p 
))/(b^3*(1 + p)) - (2*a*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (a + b*x^2)^( 
3 + p)/(b^3*(3 + p))) + ((c^2 - (5*a*d^2)/(7*b + 2*b*p))*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 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 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]

\[\int x^{4} \left (d x +c \right )^{2} \left (b \,x^{2}+a \right )^{p}d x\]

Input:

int(x^4*(d*x+c)^2*(b*x^2+a)^p,x)
 

Output:

int(x^4*(d*x+c)^2*(b*x^2+a)^p,x)
 

Fricas [F]

\[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:

integrate(x^4*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="fricas")
 

Output:

integral((d^2*x^6 + 2*c*d*x^5 + c^2*x^4)*(b*x^2 + a)^p, x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 925 vs. \(2 (151) = 302\).

Time = 20.36 (sec) , antiderivative size = 986, normalized size of antiderivative = 5.57 \[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx =\text {Too large to display} \] Input:

integrate(x**4*(d*x+c)**2*(b*x**2+a)**p,x)
 

Output:

a**p*c**2*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + a**p 
*d**2*x**7*hyper((7/2, -p), (9/2,), b*x**2*exp_polar(I*pi)/a)/7 + 2*c*d*Pi 
ecewise((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*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(-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**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2 
*a*b*x**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a*b* 
*3 + 2*b**4*x**2), Eq(p, -2)), (a**2*log(x - sqrt(-a/b))/(2*b**3) + a**2*l 
og(x + sqrt(-a/b))/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)), (2 
*a**3*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) - 
 2*a**2*b*p*x**2*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 
 12*b**3) + a*b**2*p**2*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 
 22*b**3*p + 12*b**3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12...
 

Maxima [F]

\[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:

integrate(x^4*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^2*(b*x^2 + a)^p*x^4, x)
 

Giac [F]

\[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \] Input:

integrate(x^4*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="giac")
 

Output:

integrate((d*x + c)^2*(b*x^2 + a)^p*x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int x^4\,{\left (b\,x^2+a\right )}^p\,{\left (c+d\,x\right )}^2 \,d x \] Input:

int(x^4*(a + b*x^2)^p*(c + d*x)^2,x)
 

Output:

int(x^4*(a + b*x^2)^p*(c + d*x)^2, x)
 

Reduce [F]

\[ \int x^4 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\text {too large to display} \] Input:

int(x^4*(d*x+c)^2*(b*x^2+a)^p,x)
 

Output:

(32*(a + b*x**2)**p*a**3*c*d*p**4 + 256*(a + b*x**2)**p*a**3*c*d*p**3 + 68 
8*(a + b*x**2)**p*a**3*c*d*p**2 + 704*(a + b*x**2)**p*a**3*c*d*p + 210*(a 
+ b*x**2)**p*a**3*c*d + 30*(a + b*x**2)**p*a**3*d**2*p**4*x + 180*(a + b*x 
**2)**p*a**3*d**2*p**3*x + 330*(a + b*x**2)**p*a**3*d**2*p**2*x + 180*(a + 
 b*x**2)**p*a**3*d**2*p*x - 12*(a + b*x**2)**p*a**2*b*c**2*p**5*x - 114*(a 
 + b*x**2)**p*a**2*b*c**2*p**4*x - 384*(a + b*x**2)**p*a**2*b*c**2*p**3*x 
- 534*(a + b*x**2)**p*a**2*b*c**2*p**2*x - 252*(a + b*x**2)**p*a**2*b*c**2 
*p*x - 32*(a + b*x**2)**p*a**2*b*c*d*p**5*x**2 - 256*(a + b*x**2)**p*a**2* 
b*c*d*p**4*x**2 - 688*(a + b*x**2)**p*a**2*b*c*d*p**3*x**2 - 704*(a + b*x* 
*2)**p*a**2*b*c*d*p**2*x**2 - 210*(a + b*x**2)**p*a**2*b*c*d*p*x**2 - 20*( 
a + b*x**2)**p*a**2*b*d**2*p**5*x**3 - 130*(a + b*x**2)**p*a**2*b*d**2*p** 
4*x**3 - 280*(a + b*x**2)**p*a**2*b*d**2*p**3*x**3 - 230*(a + b*x**2)**p*a 
**2*b*d**2*p**2*x**3 - 60*(a + b*x**2)**p*a**2*b*d**2*p*x**3 + 8*(a + b*x* 
*2)**p*a*b**2*c**2*p**6*x**3 + 80*(a + b*x**2)**p*a*b**2*c**2*p**5*x**3 + 
294*(a + b*x**2)**p*a*b**2*c**2*p**4*x**3 + 484*(a + b*x**2)**p*a*b**2*c** 
2*p**3*x**3 + 346*(a + b*x**2)**p*a*b**2*c**2*p**2*x**3 + 84*(a + b*x**2)* 
*p*a*b**2*c**2*p*x**3 + 16*(a + b*x**2)**p*a*b**2*c*d*p**6*x**4 + 144*(a + 
 b*x**2)**p*a*b**2*c*d*p**5*x**4 + 472*(a + b*x**2)**p*a*b**2*c*d*p**4*x** 
4 + 696*(a + b*x**2)**p*a*b**2*c*d*p**3*x**4 + 457*(a + b*x**2)**p*a*b**2* 
c*d*p**2*x**4 + 105*(a + b*x**2)**p*a*b**2*c*d*p*x**4 + 8*(a + b*x**2)*...