\(\int x^7 (a^2+2 a b x^2+b^2 x^4)^p \, dx\) [685]

Optimal result

Integrand size = 24, antiderivative size = 174 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=-\frac {a^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (1+2 p)}+\frac {3 a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (1+p)}-\frac {3 a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (3+2 p)}+\frac {\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (2+p)} \] Output:

-1/2*a^3*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p/b^4/(1+2*p)+3/4*a^2*(b*x^2+a) 
^2*(b^2*x^4+2*a*b*x^2+a^2)^p/b^4/(p+1)-3/2*a*(b*x^2+a)^3*(b^2*x^4+2*a*b*x^ 
2+a^2)^p/b^4/(3+2*p)+1/4*(b*x^2+a)^4*(b^2*x^4+2*a*b*x^2+a^2)^p/b^4/(2+p)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.63 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (-3 a^3+3 a^2 b (1+2 p) x^2-3 a b^2 \left (1+3 p+2 p^2\right ) x^4+b^3 \left (3+11 p+12 p^2+4 p^3\right ) x^6\right )}{4 b^4 (1+p) (2+p) (1+2 p) (3+2 p)} \] Input:

Integrate[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]
 

Output:

((a + b*x^2)*((a + b*x^2)^2)^p*(-3*a^3 + 3*a^2*b*(1 + 2*p)*x^2 - 3*a*b^2*( 
1 + 3*p + 2*p^2)*x^4 + b^3*(3 + 11*p + 12*p^2 + 4*p^3)*x^6))/(4*b^4*(1 + p 
)*(2 + p)*(1 + 2*p)*(3 + 2*p))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1385, 243, 53, 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.

\(\displaystyle \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx\)

\(\Big \downarrow \) 1385

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

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int x^6 \left (\frac {b x^2}{a}+1\right )^{2 p}dx^2\)

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]
 

Output:

((a^2 + 2*a*b*x^2 + b^2*x^4)^p*((3*a^4*(1 + (b*x^2)/a)^(2*(1 + p)))/(2*b^4 
*(1 + p)) + (a^4*(1 + (b*x^2)/a)^(2*(2 + p)))/(2*b^4*(2 + p)) - (a^4*(1 + 
(b*x^2)/a)^(1 + 2*p))/(b^4*(1 + 2*p)) - (3*a^4*(1 + (b*x^2)/a)^(3 + 2*p))/ 
(b^4*(3 + 2*p))))/(2*(1 + (b*x^2)/a)^(2*p))
 

Defintions of rubi rules used

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 1385

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* 
FracPart[p]))   Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, 
 p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p] && NeQ[u, 
 x^(n - 1)] && NeQ[u, x^(2*n - 1)]
 

rule 2009

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.86

method	result	size
gosper	\(-\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} \left (-4 b^{3} p^{3} x^{6}-12 b^{3} p^{2} x^{6}-11 b^{3} p \,x^{6}+6 a \,b^{2} p^{2} x^{4}-3 b^{3} x^{6}+9 a \,b^{2} p \,x^{4}+3 b^{2} x^{4} a -6 a^{2} b p \,x^{2}-3 a^{2} b \,x^{2}+3 a^{3}\right ) \left (b \,x^{2}+a \right )}{4 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}\)	\(150\)
orering	\(-\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} \left (-4 b^{3} p^{3} x^{6}-12 b^{3} p^{2} x^{6}-11 b^{3} p \,x^{6}+6 a \,b^{2} p^{2} x^{4}-3 b^{3} x^{6}+9 a \,b^{2} p \,x^{4}+3 b^{2} x^{4} a -6 a^{2} b p \,x^{2}-3 a^{2} b \,x^{2}+3 a^{3}\right ) \left (b \,x^{2}+a \right )}{4 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}\)	\(150\)
risch	\(-\frac {\left (-4 b^{4} p^{3} x^{8}-12 b^{4} p^{2} x^{8}-4 a \,b^{3} p^{3} x^{6}-11 b^{4} p \,x^{8}-6 a \,b^{3} p^{2} x^{6}-3 b^{4} x^{8}-2 a p \,x^{6} b^{3}+6 a^{2} b^{2} p^{2} x^{4}+3 a^{2} p \,x^{4} b^{2}-6 a^{3} p \,x^{2} b +3 a^{4}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{p}}{4 \left (3+2 p \right ) \left (2+p \right ) \left (p +1\right ) \left (1+2 p \right ) b^{4}}\)	\(156\)
norman	\(\frac {x^{8} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 p +8}-\frac {3 a^{4} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}+\frac {a p \,x^{6} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b \left (2 p^{2}+7 p +6\right )}-\frac {3 a^{2} p \,x^{4} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 b^{2} \left (2 p^{3}+9 p^{2}+13 p +6\right )}+\frac {3 p \,a^{3} x^{2} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b^{3} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}\)	\(237\)
parallelrisch	\(\frac {4 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4} p^{3}+12 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4} p^{2}+11 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4} p +4 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b^{3} p^{3}+3 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4}+6 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b^{3} p^{2}+2 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b^{3} p -6 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{3} b^{2} p^{2}-3 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{3} b^{2} p +6 x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{4} b p -3 a^{5} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 \left (2+p \right ) \left (3+2 p \right ) \left (p +1\right ) \left (1+2 p \right ) b^{4} a}\)	\(378\)

Input:

int(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x,method=_RETURNVERBOSE)
 

Output:

-1/4*(b^2*x^4+2*a*b*x^2+a^2)^p*(-4*b^3*p^3*x^6-12*b^3*p^2*x^6-11*b^3*p*x^6 
+6*a*b^2*p^2*x^4-3*b^3*x^6+9*a*b^2*p*x^4+3*a*b^2*x^4-6*a^2*b*p*x^2-3*a^2*b 
*x^2+3*a^3)*(b*x^2+a)/b^4/(4*p^4+20*p^3+35*p^2+25*p+6)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left ({\left (4 \, b^{4} p^{3} + 12 \, b^{4} p^{2} + 11 \, b^{4} p + 3 \, b^{4}\right )} x^{8} + 6 \, a^{3} b p x^{2} + 2 \, {\left (2 \, a b^{3} p^{3} + 3 \, a b^{3} p^{2} + a b^{3} p\right )} x^{6} - 3 \, {\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{4} - 3 \, a^{4}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \] Input:

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")
 

Output:

1/4*((4*b^4*p^3 + 12*b^4*p^2 + 11*b^4*p + 3*b^4)*x^8 + 6*a^3*b*p*x^2 + 2*( 
2*a*b^3*p^3 + 3*a*b^3*p^2 + a*b^3*p)*x^6 - 3*(2*a^2*b^2*p^2 + a^2*b^2*p)*x 
^4 - 3*a^4)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p/(4*b^4*p^4 + 20*b^4*p^3 + 35*b^4 
*p^2 + 25*b^4*p + 6*b^4)
 

Sympy [F]

\[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\text {Too large to display} \] Input:

integrate(x**7*(b**2*x**4+2*a*b*x**2+a**2)**p,x)
 

Output:

Piecewise((x**8*(a**2)**p/8, Eq(b, 0)), (6*a**3*log(x - sqrt(-a/b))/(12*a* 
*3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 6*a**3*log( 
x + sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b* 
*7*x**6) + 11*a**3/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12 
*b**7*x**6) + 18*a**2*b*x**2*log(x - sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b 
**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a**2*b*x**2*log(x + sqrt(-a 
/b))/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 
27*a**2*b*x**2/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b** 
7*x**6) + 18*a*b**2*x**4*log(x - sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5* 
x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a*b**2*x**4*log(x + sqrt(-a/b)) 
/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a 
*b**2*x**4/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x* 
*6) + 6*b**3*x**6*log(x - sqrt(-a/b))/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 
36*a*b**6*x**4 + 12*b**7*x**6) + 6*b**3*x**6*log(x + sqrt(-a/b))/(12*a**3* 
b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6), Eq(p, -2)), (In 
tegral(x**7/((a + b*x**2)**2)**(3/2), x), Eq(p, -3/2)), (6*a**3*log(x - sq 
rt(-a/b))/(4*a*b**4 + 4*b**5*x**2) + 6*a**3*log(x + sqrt(-a/b))/(4*a*b**4 
+ 4*b**5*x**2) + 6*a**3/(4*a*b**4 + 4*b**5*x**2) + 6*a**2*b*x**2*log(x - s 
qrt(-a/b))/(4*a*b**4 + 4*b**5*x**2) + 6*a**2*b*x**2*log(x + sqrt(-a/b))/(4 
*a*b**4 + 4*b**5*x**2) - 3*a*b**2*x**4/(4*a*b**4 + 4*b**5*x**2) + b**3*...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.66 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{8} + 2 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{6} - 3 \, {\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{4} + 6 \, a^{3} b p x^{2} - 3 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{4 \, {\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{4}} \] Input:

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")
 

Output:

1/4*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^8 + 2*(2*p^3 + 3*p^2 + p)*a*b^3*x^6 
 - 3*(2*p^2 + p)*a^2*b^2*x^4 + 6*a^3*b*p*x^2 - 3*a^4)*(b*x^2 + a)^(2*p)/(( 
4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (166) = 332\).

Time = 0.16 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.16 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p^{3} x^{8} + 12 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p^{2} x^{8} + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p^{3} x^{6} + 11 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p x^{8} + 6 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p^{2} x^{6} + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} x^{8} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p x^{6} - 6 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p^{2} x^{4} - 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p x^{4} + 6 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3} b p x^{2} - 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{4}}{4 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \] Input:

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")
 

Output:

1/4*(4*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*p^3*x^8 + 12*(b^2*x^4 + 2*a*b*x^2 
 + a^2)^p*b^4*p^2*x^8 + 4*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b^3*p^3*x^6 + 11 
*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*p*x^8 + 6*(b^2*x^4 + 2*a*b*x^2 + a^2)^p 
*a*b^3*p^2*x^6 + 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*x^8 + 2*(b^2*x^4 + 2* 
a*b*x^2 + a^2)^p*a*b^3*p*x^6 - 6*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^2*b^2*p^2 
*x^4 - 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^2*b^2*p*x^4 + 6*(b^2*x^4 + 2*a*b* 
x^2 + a^2)^p*a^3*b*p*x^2 - 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^4)/(4*b^4*p^4 
 + 20*b^4*p^3 + 35*b^4*p^2 + 25*b^4*p + 6*b^4)
 

Mupad [B] (verification not implemented)

Time = 17.77 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx={\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {x^8\,\left (p^3+3\,p^2+\frac {11\,p}{4}+\frac {3}{4}\right )}{4\,p^4+20\,p^3+35\,p^2+25\,p+6}-\frac {3\,a^4}{4\,b^4\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {3\,a^3\,p\,x^2}{2\,b^3\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {a\,p\,x^6\,\left (2\,p^2+3\,p+1\right )}{2\,b\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}-\frac {3\,a^2\,p\,x^4\,\left (2\,p+1\right )}{4\,b^2\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}\right ) \] Input:

int(x^7*(a^2 + b^2*x^4 + 2*a*b*x^2)^p,x)
 

Output:

(a^2 + b^2*x^4 + 2*a*b*x^2)^p*((x^8*((11*p)/4 + 3*p^2 + p^3 + 3/4))/(25*p 
+ 35*p^2 + 20*p^3 + 4*p^4 + 6) - (3*a^4)/(4*b^4*(25*p + 35*p^2 + 20*p^3 + 
4*p^4 + 6)) + (3*a^3*p*x^2)/(2*b^3*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) + 
 (a*p*x^6*(3*p + 2*p^2 + 1))/(2*b*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)) - 
(3*a^2*p*x^4*(2*p + 1))/(4*b^2*(25*p + 35*p^2 + 20*p^3 + 4*p^4 + 6)))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.93 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} \left (4 b^{4} p^{3} x^{8}+12 b^{4} p^{2} x^{8}+4 a \,b^{3} p^{3} x^{6}+11 b^{4} p \,x^{8}+6 a \,b^{3} p^{2} x^{6}+3 b^{4} x^{8}+2 a \,b^{3} p \,x^{6}-6 a^{2} b^{2} p^{2} x^{4}-3 a^{2} b^{2} p \,x^{4}+6 a^{3} b p \,x^{2}-3 a^{4}\right )}{4 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )} \] Input:

int(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x)
 

Output:

((a**2 + 2*a*b*x**2 + b**2*x**4)**p*( - 3*a**4 + 6*a**3*b*p*x**2 - 6*a**2* 
b**2*p**2*x**4 - 3*a**2*b**2*p*x**4 + 4*a*b**3*p**3*x**6 + 6*a*b**3*p**2*x 
**6 + 2*a*b**3*p*x**6 + 4*b**4*p**3*x**8 + 12*b**4*p**2*x**8 + 11*b**4*p*x 
**8 + 3*b**4*x**8))/(4*b**4*(4*p**4 + 20*p**3 + 35*p**2 + 25*p + 6))