\(\int x^5 (a^2+2 a b x^2+b^2 x^4)^p \, dx\) [686]

Optimal result

Integrand size = 24, antiderivative size = 130 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {a^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (1+2 p)}-\frac {a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (1+p)}+\frac {\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (3+2 p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.59 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (a^2-a b (1+2 p) x^2+b^2 \left (1+3 p+2 p^2\right ) x^4\right )}{2 b^3 (1+p) (1+2 p) (3+2 p)} \] Input:

Integrate[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98, 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.

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

\(\Big \downarrow \) 1385

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^5*(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*(-((a^3*(1 + (b*x^2)/a)^(2*(1 + p)))/(b^3*( 
1 + p))) + (a^3*(1 + (b*x^2)/a)^(1 + 2*p))/(b^3*(1 + 2*p)) + (a^3*(1 + (b* 
x^2)/a)^(3 + 2*p))/(b^3*(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 = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right ) \left (2 b^{2} p^{2} x^{4}+3 b^{2} p \,x^{4}+b^{2} x^{4}-2 a b p \,x^{2}-a b \,x^{2}+a^{2}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{2 b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\)	\(96\)
orering	\(\frac {\left (b \,x^{2}+a \right ) \left (2 b^{2} p^{2} x^{4}+3 b^{2} p \,x^{4}+b^{2} x^{4}-2 a b p \,x^{2}-a b \,x^{2}+a^{2}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{2 b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\)	\(96\)
risch	\(\frac {\left (2 b^{3} p^{2} x^{6}+3 b^{3} p \,x^{6}+2 a \,b^{2} p^{2} x^{4}+b^{3} x^{6}+a \,b^{2} p \,x^{4}-2 a^{2} b p \,x^{2}+a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{p}}{2 \left (p +1\right ) \left (3+2 p \right ) \left (1+2 p \right ) b^{3}}\)	\(98\)
norman	\(\frac {x^{6} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 p +6}+\frac {a^{3} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}+\frac {a p \,x^{4} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b \left (2 p^{2}+5 p +3\right )}-\frac {p \,a^{2} x^{2} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{b^{2} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\)	\(178\)
parallelrisch	\(\frac {2 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} b^{3} p^{2}+3 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} b^{3} p +x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} b^{3}+2 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{2} p^{2}+x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{2} p -2 x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b p +\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{3}}{2 b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\)	\(226\)

Input:

int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^p,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.83 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left ({\left (2 \, b^{3} p^{2} + 3 \, b^{3} p + b^{3}\right )} x^{6} - 2 \, a^{2} b p x^{2} + {\left (2 \, a b^{2} p^{2} + a b^{2} p\right )} x^{4} + a^{3}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{2 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \] Input:

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\begin {cases} \frac {x^{6} \left (a^{2}\right )^{p}}{6} & \text {for}\: b = 0 \\\int \frac {x^{5}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\- \frac {2 a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {b^{2} x^{4}}{2 a b^{3} + 2 b^{4} x^{2}} & \text {for}\: p = -1 \\\int \frac {x^{5}}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a^{3} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} - \frac {2 a^{2} b p x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {2 a b^{2} p^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {a b^{2} p x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {2 b^{3} p^{2} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {3 b^{3} p x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {b^{3} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} & \text {otherwise} \end {cases} \] Input:

integrate(x**5*(b**2*x**4+2*a*b*x**2+a**2)**p,x)
 

Output:

Piecewise((x**6*(a**2)**p/6, Eq(b, 0)), (Integral(x**5/((a + b*x**2)**2)** 
(3/2), x), Eq(p, -3/2)), (-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, -1)), (Integral(x**5/sqrt((a + b*x**2)**2), 
 x), Eq(p, -1/2)), (a**3*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 
 24*b**3*p**2 + 22*b**3*p + 6*b**3) - 2*a**2*b*p*x**2*(a**2 + 2*a*b*x**2 + 
 b**2*x**4)**p/(8*b**3*p**3 + 24*b**3*p**2 + 22*b**3*p + 6*b**3) + 2*a*b** 
2*p**2*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24*b**3*p**2 
 + 22*b**3*p + 6*b**3) + a*b**2*p*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/ 
(8*b**3*p**3 + 24*b**3*p**2 + 22*b**3*p + 6*b**3) + 2*b**3*p**2*x**6*(a**2 
 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24*b**3*p**2 + 22*b**3*p + 6* 
b**3) + 3*b**3*p*x**6*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24 
*b**3*p**2 + 22*b**3*p + 6*b**3) + b**3*x**6*(a**2 + 2*a*b*x**2 + b**2*x** 
4)**p/(8*b**3*p**3 + 24*b**3*p**2 + 22*b**3*p + 6*b**3), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.61 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{2 \, {\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \] Input:

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.81 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} p^{2} x^{6} + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} p x^{6} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} p^{2} x^{4} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} x^{6} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} p x^{4} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b p x^{2} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3}}{2 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \] Input:

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 17.90 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx={\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {x^6\,\left (p^2+\frac {3\,p}{2}+\frac {1}{2}\right )}{4\,p^3+12\,p^2+11\,p+3}+\frac {a^3}{2\,b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}-\frac {a^2\,p\,x^2}{b^2\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {a\,p\,x^4\,\left (2\,p+1\right )}{2\,b\,\left (4\,p^3+12\,p^2+11\,p+3\right )}\right ) \] Input:

int(x^5*(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^6*((3*p)/2 + p^2 + 1/2))/(11*p + 12*p^2 
+ 4*p^3 + 3) + a^3/(2*b^3*(11*p + 12*p^2 + 4*p^3 + 3)) - (a^2*p*x^2)/(b^2* 
(11*p + 12*p^2 + 4*p^3 + 3)) + (a*p*x^4*(2*p + 1))/(2*b*(11*p + 12*p^2 + 4 
*p^3 + 3)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.80 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} \left (2 b^{3} p^{2} x^{6}+3 b^{3} p \,x^{6}+2 a \,b^{2} p^{2} x^{4}+b^{3} x^{6}+a \,b^{2} p \,x^{4}-2 a^{2} b p \,x^{2}+a^{3}\right )}{2 b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )} \] Input:

int(x^5*(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*(a**3 - 2*a**2*b*p*x**2 + 2*a*b**2*p** 
2*x**4 + a*b**2*p*x**4 + 2*b**3*p**2*x**6 + 3*b**3*p*x**6 + b**3*x**6))/(2 
*b**3*(4*p**3 + 12*p**2 + 11*p + 3))