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

Optimal. Leaf size=130 \[ \frac{\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+3)}-\frac{a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (p+1)}+\frac{a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+1)} \]

[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))

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Rubi [A]  time = 0.0788709, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1356, 266, 43} \[ \frac{\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+3)}-\frac{a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (p+1)}+\frac{a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+1)} \]

Antiderivative was successfully verified.

[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 1356

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]

Rule 266

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 43

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])

Rubi steps

\begin{align*} \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx &=\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 ) \operatorname{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 ) \operatorname{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]  time = 0.0329366, size = 77, normalized size = 0.59 \[ \frac{\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (a^2-a b (2 p+1) x^3+b^2 \left (2 p^2+3 p+1\right ) x^6\right )}{3 b^3 (p+1) (2 p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.007, size = 96, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2\,{b}^{2}{p}^{2}{x}^{6}+3\,{b}^{2}p{x}^{6}+{b}^{2}{x}^{6}-2\,abp{x}^{3}-ab{x}^{3}+{a}^{2} \right ) \left ( b{x}^{3}+a \right ) \left ({b}^{2}{x}^{6}+2\,ab{x}^{3}+{a}^{2} \right ) ^{p}}{3\,{b}^{3} \left ( 4\,{p}^{3}+12\,{p}^{2}+11\,p+3 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [A]  time = 1.02904, size = 107, normalized size = 0.82 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.62213, size = 223, normalized size = 1.72 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.12567, size = 317, normalized size = 2.44 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)