3.4.92 \(\int x^3 (d+e x)^2 (a+b x^2)^p \, dx\) [392]

3.4.92.1 Optimal result

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

output

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

3.4.92.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.02 \[ \int x^3 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\frac {1}{10} \left (a+b x^2\right )^p \left (\frac {5 d^2 \left (a+b x^2\right ) \left (-a+b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+\frac {5 e^2 \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+4 d e x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \]

input

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

output

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

3.4.92.3 Rubi [A] (verified)

Time = 0.31 (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.

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

\(\Big \downarrow \) 543

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.4.92.3.1 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]
 

3.4.92.4 Maple [F]

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

input

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

output

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

3.4.92.5 Fricas [F]

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

input

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

output

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

3.4.92.6 Sympy [B] (verification not implemented)

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

Time = 9.55 (sec) , antiderivative size = 1294, normalized size of antiderivative = 8.68 \[ \int x^3 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\text {Too large to display} \]

input

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

output

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

3.4.92.7 Maxima [F]

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

input

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

3.4.92.8 Giac [F]

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

input

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

output

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

3.4.92.9 Mupad [F(-1)]

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

input

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

output

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