\(\int x^8 (a^2+2 a b x^3+b^2 x^6)^p \, dx\) [126]

Optimal result

Integrand size = 24, antiderivative size = 130 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+2 p)}-\frac {a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+p)}+\frac {\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (3+2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1370, 272, 45} \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+3)}-\frac {a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (p+1)}+\frac {a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+1)} \]

[In]

Int[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

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

Rule 45

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] && NeQ[b*c - a*d, 0] && 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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1370

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

Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \int x^8 \left (1+\frac {b x^3}{a}\right )^{2 p} \, dx \\ & = \frac {1}{3} \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \text {Subst}\left (\int x^2 \left (1+\frac {b x}{a}\right )^{2 p} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \text {Subst}\left (\int \left (\frac {a^2 \left (1+\frac {b x}{a}\right )^{2 p}}{b^2}-\frac {2 a^2 \left (1+\frac {b x}{a}\right )^{1+2 p}}{b^2}+\frac {a^2 \left (1+\frac {b x}{a}\right )^{2+2 p}}{b^2}\right ) \, dx,x,x^3\right ) \\ & = \frac {a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+2 p)}-\frac {a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+p)}+\frac {\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (3+2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.59 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (a^2-a b (1+2 p) x^3+b^2 \left (1+3 p+2 p^2\right ) x^6\right )}{3 b^3 (1+p) (1+2 p) (3+2 p)} \]

[In]

Integrate[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74

method	result	size
gosper	\(\frac {\left (b \,x^{3}+a \right ) \left (2 b^{2} p^{2} x^{6}+3 b^{2} p \,x^{6}+b^{2} x^{6}-2 a b p \,x^{3}-a b \,x^{3}+a^{2}\right ) \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}}{3 b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\)	\(96\)
risch	\(\frac {\left (2 b^{3} p^{2} x^{9}+3 b^{3} p \,x^{9}+b^{3} x^{9}+2 a \,b^{2} p^{2} x^{6}+a \,b^{2} p \,x^{6}-2 a^{2} b p \,x^{3}+a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{p}}{3 \left (1+p \right ) \left (3+2 p \right ) \left (1+2 p \right ) b^{3}}\)	\(98\)
parallelrisch	\(\frac {2 x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3} p^{2}+3 x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3} p +x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3}+2 x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2} b^{2} p^{2}+x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2} b^{2} p -2 x^{3} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{3} b p +\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{4}}{3 \left (3+2 p \right ) \left (1+p \right ) \left (1+2 p \right ) a \,b^{3}}\)	\(238\)

[In]

int(x^8*(b^2*x^6+2*a*b*x^3+a^2)^p,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.83 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (2 \, b^{3} p^{2} + 3 \, b^{3} p + b^{3}\right )} x^{9} - 2 \, a^{2} b p x^{3} + {\left (2 \, a b^{2} p^{2} + a b^{2} p\right )} x^{6} + a^{3}\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{3 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]

[In]

integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\begin {cases} \frac {x^{9} \left (a^{2}\right )^{p}}{9} & \text {for}\: b = 0 \\\int \frac {x^{8}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\- \frac {2 a^{2} \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a^{2} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a^{2}}{3 a b^{3} + 3 b^{4} x^{3}} + \frac {4 a^{2} \log {\left (2 \right )}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a b x^{3} \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a b x^{3} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{3} + 3 b^{4} x^{3}} + \frac {4 a b x^{3} \log {\left (2 \right )}}{3 a b^{3} + 3 b^{4} x^{3}} + \frac {b^{2} x^{6}}{3 a b^{3} + 3 b^{4} x^{3}} & \text {for}\: p = -1 \\\int \frac {x^{8}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} - \frac {2 a^{2} b p x^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {2 a b^{2} p^{2} x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {a b^{2} p x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {2 b^{3} p^{2} x^{9} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {3 b^{3} p x^{9} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {b^{3} x^{9} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**8*(b**2*x**6+2*a*b*x**3+a**2)**p,x)

[Out]

Piecewise((x**9*(a**2)**p/9, Eq(b, 0)), (Integral(x**8/((a + b*x**3)**2)**(3/2), x), Eq(p, -3/2)), (-2*a**2*lo
g(x - (-a/b)**(1/3))/(3*a*b**3 + 3*b**4*x**3) - 2*a**2*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(3*a*
b**3 + 3*b**4*x**3) - 2*a**2/(3*a*b**3 + 3*b**4*x**3) + 4*a**2*log(2)/(3*a*b**3 + 3*b**4*x**3) - 2*a*b*x**3*lo
g(x - (-a/b)**(1/3))/(3*a*b**3 + 3*b**4*x**3) - 2*a*b*x**3*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(
3*a*b**3 + 3*b**4*x**3) + 4*a*b*x**3*log(2)/(3*a*b**3 + 3*b**4*x**3) + b**2*x**6/(3*a*b**3 + 3*b**4*x**3), Eq(
p, -1)), (Integral(x**8/sqrt((a + b*x**3)**2), x), Eq(p, -1/2)), (a**3*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*
b**3*p**3 + 36*b**3*p**2 + 33*b**3*p + 9*b**3) - 2*a**2*b*p*x**3*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**3*p
**3 + 36*b**3*p**2 + 33*b**3*p + 9*b**3) + 2*a*b**2*p**2*x**6*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**3*p**3
+ 36*b**3*p**2 + 33*b**3*p + 9*b**3) + a*b**2*p*x**6*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**3*p**3 + 36*b*
*3*p**2 + 33*b**3*p + 9*b**3) + 2*b**3*p**2*x**9*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**3*p**3 + 36*b**3*p*
*2 + 33*b**3*p + 9*b**3) + 3*b**3*p*x**9*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**3*p**3 + 36*b**3*p**2 + 33*
b**3*p + 9*b**3) + b**3*x**9*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**3*p**3 + 36*b**3*p**2 + 33*b**3*p + 9*b
**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.61 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{9} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{6} - 2 \, a^{2} b p x^{3} + a^{3}\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{3 \, {\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \]

[In]

integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.81 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{3} p^{2} x^{9} + 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{3} p x^{9} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{3} x^{9} + 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{2} p^{2} x^{6} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{2} p x^{6} - 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b p x^{3} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{3}}{3 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]

[In]

integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="giac")

[Out]

1/3*(2*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*b^3*p^2*x^9 + 3*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*b^3*p*x^9 + (b^2*x^6 + 2*a*
b*x^3 + a^2)^p*b^3*x^9 + 2*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^2*p^2*x^6 + (b^2*x^6 + 2*a*b*x^3 + a^2)^p*a*b^2*p
*x^6 - 2*(b^2*x^6 + 2*a*b*x^3 + a^2)^p*a^2*b*p*x^3 + (b^2*x^6 + 2*a*b*x^3 + 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 = 8.36 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx={\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p\,\left (\frac {x^9\,\left (\frac {2\,p^2}{3}+p+\frac {1}{3}\right )}{4\,p^3+12\,p^2+11\,p+3}+\frac {a^3}{3\,b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}-\frac {2\,a^2\,p\,x^3}{3\,b^2\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {a\,p\,x^6\,\left (2\,p+1\right )}{3\,b\,\left (4\,p^3+12\,p^2+11\,p+3\right )}\right ) \]

[In]

int(x^8*(a^2 + b^2*x^6 + 2*a*b*x^3)^p,x)

[Out]

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