3.601 \(\int x^8 (a+b x^3)^p \, dx\)
Optimal. Leaf size=74 \[ \frac{a^2 \left (a+b x^3\right )^{p+1}}{3 b^3 (p+1)}-\frac{2 a \left (a+b x^3\right )^{p+2}}{3 b^3 (p+2)}+\frac{\left (a+b x^3\right )^{p+3}}{3 b^3 (p+3)} \]
[Out]
(a^2*(a + b*x^3)^(1 + p))/(3*b^3*(1 + p)) - (2*a*(a + b*x^3)^(2 + p))/(3*b^3*(2 + p)) + (a + b*x^3)^(3 + p)/(3
*b^3*(3 + p))
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Rubi [A] time = 0.0441732, antiderivative size = 74, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used =
{266, 43} \[ \frac{a^2 \left (a+b x^3\right )^{p+1}}{3 b^3 (p+1)}-\frac{2 a \left (a+b x^3\right )^{p+2}}{3 b^3 (p+2)}+\frac{\left (a+b x^3\right )^{p+3}}{3 b^3 (p+3)} \]
Antiderivative was successfully verified.
[In]
Int[x^8*(a + b*x^3)^p,x]
[Out]
(a^2*(a + b*x^3)^(1 + p))/(3*b^3*(1 + p)) - (2*a*(a + b*x^3)^(2 + p))/(3*b^3*(2 + p)) + (a + b*x^3)^(3 + p)/(3
*b^3*(3 + 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+b x^3\right )^p \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x^2 (a+b x)^p \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{a^2 (a+b x)^p}{b^2}-\frac{2 a (a+b x)^{1+p}}{b^2}+\frac{(a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^3\right )\\ &=\frac{a^2 \left (a+b x^3\right )^{1+p}}{3 b^3 (1+p)}-\frac{2 a \left (a+b x^3\right )^{2+p}}{3 b^3 (2+p)}+\frac{\left (a+b x^3\right )^{3+p}}{3 b^3 (3+p)}\\ \end{align*}
Mathematica [A] time = 0.0299407, size = 64, normalized size = 0.86 \[ \frac{\left (a+b x^3\right )^{p+1} \left (2 a^2-2 a b (p+1) x^3+b^2 \left (p^2+3 p+2\right ) x^6\right )}{3 b^3 (p+1) (p+2) (p+3)} \]
Antiderivative was successfully verified.
[In]
Integrate[x^8*(a + b*x^3)^p,x]
[Out]
((a + b*x^3)^(1 + p)*(2*a^2 - 2*a*b*(1 + p)*x^3 + b^2*(2 + 3*p + p^2)*x^6))/(3*b^3*(1 + p)*(2 + p)*(3 + p))
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Maple [A] time = 0.005, size = 80, normalized size = 1.1 \begin{align*}{\frac{ \left ( b{x}^{3}+a \right ) ^{1+p} \left ({b}^{2}{p}^{2}{x}^{6}+3\,{b}^{2}p{x}^{6}+2\,{b}^{2}{x}^{6}-2\,abp{x}^{3}-2\,{x}^{3}ab+2\,{a}^{2} \right ) }{3\,{b}^{3} \left ({p}^{3}+6\,{p}^{2}+11\,p+6 \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^8*(b*x^3+a)^p,x)
[Out]
1/3*(b*x^3+a)^(1+p)*(b^2*p^2*x^6+3*b^2*p*x^6+2*b^2*x^6-2*a*b*p*x^3-2*a*b*x^3+2*a^2)/b^3/(p^3+6*p^2+11*p+6)
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Maxima [A] time = 0.982983, size = 99, normalized size = 1.34 \begin{align*} \frac{{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{9} +{\left (p^{2} + p\right )} a b^{2} x^{6} - 2 \, a^{2} b p x^{3} + 2 \, a^{3}\right )}{\left (b x^{3} + a\right )}^{p}}{3 \,{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8*(b*x^3+a)^p,x, algorithm="maxima")
[Out]
1/3*((p^2 + 3*p + 2)*b^3*x^9 + (p^2 + p)*a*b^2*x^6 - 2*a^2*b*p*x^3 + 2*a^3)*(b*x^3 + a)^p/((p^3 + 6*p^2 + 11*p
+ 6)*b^3)
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Fricas [A] time = 1.56288, size = 197, normalized size = 2.66 \begin{align*} \frac{{\left ({\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{9} - 2 \, a^{2} b p x^{3} +{\left (a b^{2} p^{2} + a b^{2} p\right )} x^{6} + 2 \, a^{3}\right )}{\left (b x^{3} + a\right )}^{p}}{3 \,{\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8*(b*x^3+a)^p,x, algorithm="fricas")
[Out]
1/3*((b^3*p^2 + 3*b^3*p + 2*b^3)*x^9 - 2*a^2*b*p*x^3 + (a*b^2*p^2 + a*b^2*p)*x^6 + 2*a^3)*(b*x^3 + a)^p/(b^3*p
^3 + 6*b^3*p^2 + 11*b^3*p + 6*b^3)
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Sympy [A] time = 31.4292, size = 1368, normalized size = 18.49 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**8*(b*x**3+a)**p,x)
[Out]
Piecewise((a**p*x**9/9, Eq(b, 0)), (2*a**2*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(6*a**2*b**3 + 12*a*b**
4*x**3 + 6*b**5*x**6) + 2*a**2*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3)
+ 4*x**2)/(6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5*x**6) - 4*a**2*log(2)/(6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5
*x**6) + a**2/(6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5*x**6) + 4*a*b*x**3*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3)
+ x)/(6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5*x**6) + 4*a*b*x**3*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-
1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5*x**6) - 8*a*b*x**3*log(2)/(
6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5*x**6) + 2*b**2*x**6*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(6*a**2*
b**3 + 12*a*b**4*x**3 + 6*b**5*x**6) + 2*b**2*x**6*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**
(1/3)*x*(1/b)**(1/3) + 4*x**2)/(6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5*x**6) - 4*b**2*x**6*log(2)/(6*a**2*b**3
+ 12*a*b**4*x**3 + 6*b**5*x**6) - 2*b**2*x**6/(6*a**2*b**3 + 12*a*b**4*x**3 + 6*b**5*x**6), Eq(p, -3)), (-2*a*
*2*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(3*a*b**3 + 3*b**4*x**3) - 2*a**2*log(4*(-1)**(2/3)*a**(2/3)*(1
/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(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*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(3
*a*b**3 + 3*b**4*x**3) - 2*a*b*x**3*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**
(1/3) + 4*x**2)/(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, -2)), (a**2*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x)/(3*b**3) + a**2*log(4*(-1)**(2/3)*
a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x*(1/b)**(1/3) + 4*x**2)/(3*b**3) - a*x**3/(3*b**2) + x**6/(6*b
), Eq(p, -1)), (2*a**3*(a + b*x**3)**p/(3*b**3*p**3 + 18*b**3*p**2 + 33*b**3*p + 18*b**3) - 2*a**2*b*p*x**3*(a
+ b*x**3)**p/(3*b**3*p**3 + 18*b**3*p**2 + 33*b**3*p + 18*b**3) + a*b**2*p**2*x**6*(a + b*x**3)**p/(3*b**3*p*
*3 + 18*b**3*p**2 + 33*b**3*p + 18*b**3) + a*b**2*p*x**6*(a + b*x**3)**p/(3*b**3*p**3 + 18*b**3*p**2 + 33*b**3
*p + 18*b**3) + b**3*p**2*x**9*(a + b*x**3)**p/(3*b**3*p**3 + 18*b**3*p**2 + 33*b**3*p + 18*b**3) + 3*b**3*p*x
**9*(a + b*x**3)**p/(3*b**3*p**3 + 18*b**3*p**2 + 33*b**3*p + 18*b**3) + 2*b**3*x**9*(a + b*x**3)**p/(3*b**3*p
**3 + 18*b**3*p**2 + 33*b**3*p + 18*b**3), True))
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Giac [B] time = 1.11283, size = 312, normalized size = 4.22 \begin{align*} \frac{{\left (b x^{3} + a\right )}^{3}{\left (b x^{3} + a\right )}^{p} p^{2} - 2 \,{\left (b x^{3} + a\right )}^{2}{\left (b x^{3} + a\right )}^{p} a p^{2} +{\left (b x^{3} + a\right )}{\left (b x^{3} + a\right )}^{p} a^{2} p^{2} + 3 \,{\left (b x^{3} + a\right )}^{3}{\left (b x^{3} + a\right )}^{p} p - 8 \,{\left (b x^{3} + a\right )}^{2}{\left (b x^{3} + a\right )}^{p} a p + 5 \,{\left (b x^{3} + a\right )}{\left (b x^{3} + a\right )}^{p} a^{2} p + 2 \,{\left (b x^{3} + a\right )}^{3}{\left (b x^{3} + a\right )}^{p} - 6 \,{\left (b x^{3} + a\right )}^{2}{\left (b x^{3} + a\right )}^{p} a + 6 \,{\left (b x^{3} + a\right )}{\left (b x^{3} + a\right )}^{p} a^{2}}{3 \,{\left (b^{2} p^{3} + 6 \, b^{2} p^{2} + 11 \, b^{2} p + 6 \, b^{2}\right )} b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8*(b*x^3+a)^p,x, algorithm="giac")
[Out]
1/3*((b*x^3 + a)^3*(b*x^3 + a)^p*p^2 - 2*(b*x^3 + a)^2*(b*x^3 + a)^p*a*p^2 + (b*x^3 + a)*(b*x^3 + a)^p*a^2*p^2
+ 3*(b*x^3 + a)^3*(b*x^3 + a)^p*p - 8*(b*x^3 + a)^2*(b*x^3 + a)^p*a*p + 5*(b*x^3 + a)*(b*x^3 + a)^p*a^2*p + 2
*(b*x^3 + a)^3*(b*x^3 + a)^p - 6*(b*x^3 + a)^2*(b*x^3 + a)^p*a + 6*(b*x^3 + a)*(b*x^3 + a)^p*a^2)/((b^2*p^3 +
6*b^2*p^2 + 11*b^2*p + 6*b^2)*b)