3.125 \(\int x^{11} (a^2+2 a b x^3+b^2 x^6)^p \, dx\)

Optimal. Leaf size=172 \[ \frac{\left (a+b x^3\right )^4 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^4 (p+2)}-\frac{a \left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^4 (2 p+3)}+\frac{a^2 \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{2 b^4 (p+1)}-\frac{a^3 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^4 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.112672, antiderivative size = 172, 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} \[ \frac{\left (a+b x^3\right )^4 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^4 (p+2)}-\frac{a \left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^4 (2 p+3)}+\frac{a^2 \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{2 b^4 (p+1)}-\frac{a^3 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^4 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

-(a^3*(a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^p)/(3*b^4*(1 + 2*p)) + (a^2*(a + b*x^3)^2*(a^2 + 2*a*b*x^3 + b^2
*x^6)^p)/(2*b^4*(1 + p)) - (a*(a + b*x^3)^3*(a^2 + 2*a*b*x^3 + b^2*x^6)^p)/(b^4*(3 + 2*p)) + ((a + b*x^3)^4*(a
^2 + 2*a*b*x^3 + b^2*x^6)^p)/(6*b^4*(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^{11} \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^{11} \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^3 \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^3 \left (1+\frac{b x}{a}\right )^{2 p}}{b^3}+\frac{3 a^3 \left (1+\frac{b x}{a}\right )^{1+2 p}}{b^3}-\frac{3 a^3 \left (1+\frac{b x}{a}\right )^{2+2 p}}{b^3}+\frac{a^3 \left (1+\frac{b x}{a}\right )^{3+2 p}}{b^3}\right ) \, dx,x,x^3\right )\\ &=-\frac{a^3 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^4 (1+2 p)}+\frac{a^2 \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{2 b^4 (1+p)}-\frac{a \left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{b^4 (3+2 p)}+\frac{\left (a+b x^3\right )^4 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^4 (2+p)}\\ \end{align*}

Mathematica [A]  time = 0.0558852, size = 110, normalized size = 0.64 \[ \frac{\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (3 a^2 b (2 p+1) x^3-3 a^3-3 a b^2 \left (2 p^2+3 p+1\right ) x^6+b^3 \left (4 p^3+12 p^2+11 p+3\right ) x^9\right )}{6 b^4 (p+1) (p+2) (2 p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

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

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Maple [A]  time = 0.008, size = 150, normalized size = 0.9 \begin{align*} -{\frac{ \left ({b}^{2}{x}^{6}+2\,ab{x}^{3}+{a}^{2} \right ) ^{p} \left ( -4\,{b}^{3}{p}^{3}{x}^{9}-12\,{b}^{3}{p}^{2}{x}^{9}-11\,{b}^{3}p{x}^{9}-3\,{b}^{3}{x}^{9}+6\,a{b}^{2}{p}^{2}{x}^{6}+9\,a{b}^{2}p{x}^{6}+3\,a{b}^{2}{x}^{6}-6\,{a}^{2}bp{x}^{3}-3\,{a}^{2}b{x}^{3}+3\,{a}^{3} \right ) \left ( b{x}^{3}+a \right ) }{6\,{b}^{4} \left ( 4\,{p}^{4}+20\,{p}^{3}+35\,{p}^{2}+25\,p+6 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x)

[Out]

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

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Maxima [A]  time = 1.15209, size = 155, normalized size = 0.9 \begin{align*} \frac{{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{12} + 2 \,{\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{9} - 3 \,{\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{6} + 6 \, a^{3} b p x^{3} - 3 \, a^{4}\right )}{\left (b x^{3} + a\right )}^{2 \, p}}{6 \,{\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58433, size = 336, normalized size = 1.95 \begin{align*} \frac{{\left ({\left (4 \, b^{4} p^{3} + 12 \, b^{4} p^{2} + 11 \, b^{4} p + 3 \, b^{4}\right )} x^{12} + 2 \,{\left (2 \, a b^{3} p^{3} + 3 \, a b^{3} p^{2} + a b^{3} p\right )} x^{9} + 6 \, a^{3} b p x^{3} - 3 \,{\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{6} - 3 \, a^{4}\right )}{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{6 \,{\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="fricas")

[Out]

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

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

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.13232, size = 506, normalized size = 2.94 \begin{align*} \frac{4 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p^{3} x^{12} + 12 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p^{2} x^{12} + 11 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} p x^{12} + 4 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p^{3} x^{9} + 3 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{4} x^{12} + 6 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p^{2} x^{9} + 2 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{3} p x^{9} - 6 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b^{2} p^{2} x^{6} - 3 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b^{2} p x^{6} + 6 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{3} b p x^{3} - 3 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{4}}{6 \,{\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="giac")

[Out]

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