3.600 \(\int x^{11} \left (a+b x^3\right )^p \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^3 \left (a+b x^3\right )^{p+1}}{3 b^4 (p+1)}+\frac{a^2 \left (a+b x^3\right )^{p+2}}{b^4 (p+2)}-\frac{a \left (a+b x^3\right )^{p+3}}{b^4 (p+3)}+\frac{\left (a+b x^3\right )^{p+4}}{3 b^4 (p+4)} \]

[Out]

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

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Rubi [A]  time = 0.127572, antiderivative size = 95, 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 \[ -\frac{a^3 \left (a+b x^3\right )^{p+1}}{3 b^4 (p+1)}+\frac{a^2 \left (a+b x^3\right )^{p+2}}{b^4 (p+2)}-\frac{a \left (a+b x^3\right )^{p+3}}{b^4 (p+3)}+\frac{\left (a+b x^3\right )^{p+4}}{3 b^4 (p+4)} \]

Antiderivative was successfully verified.

[In]  Int[x^11*(a + b*x^3)^p,x]

[Out]

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

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Rubi in Sympy [A]  time = 22.821, size = 78, normalized size = 0.82 \[ - \frac{a^{3} \left (a + b x^{3}\right )^{p + 1}}{3 b^{4} \left (p + 1\right )} + \frac{a^{2} \left (a + b x^{3}\right )^{p + 2}}{b^{4} \left (p + 2\right )} - \frac{a \left (a + b x^{3}\right )^{p + 3}}{b^{4} \left (p + 3\right )} + \frac{\left (a + b x^{3}\right )^{p + 4}}{3 b^{4} \left (p + 4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**11*(b*x**3+a)**p,x)

[Out]

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

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Mathematica [A]  time = 0.0634782, size = 93, normalized size = 0.98 \[ \frac{\left (a+b x^3\right )^{p+1} \left (-6 a^3+6 a^2 b (p+1) x^3-3 a b^2 \left (p^2+3 p+2\right ) x^6+b^3 \left (p^3+6 p^2+11 p+6\right ) x^9\right )}{3 b^4 (p+1) (p+2) (p+3) (p+4)} \]

Antiderivative was successfully verified.

[In]  Integrate[x^11*(a + b*x^3)^p,x]

[Out]

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

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Maple [A]  time = 0.01, size = 132, normalized size = 1.4 \[ -{\frac{ \left ( b{x}^{3}+a \right ) ^{1+p} \left ( -{b}^{3}{p}^{3}{x}^{9}-6\,{b}^{3}{p}^{2}{x}^{9}-11\,{b}^{3}p{x}^{9}-6\,{b}^{3}{x}^{9}+3\,a{b}^{2}{p}^{2}{x}^{6}+9\,a{b}^{2}p{x}^{6}+6\,a{b}^{2}{x}^{6}-6\,{a}^{2}bp{x}^{3}-6\,{a}^{2}b{x}^{3}+6\,{a}^{3} \right ) }{3\,{b}^{4} \left ({p}^{4}+10\,{p}^{3}+35\,{p}^{2}+50\,p+24 \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^11*(b*x^3+a)^p,x)

[Out]

-1/3*(b*x^3+a)^(1+p)*(-b^3*p^3*x^9-6*b^3*p^2*x^9-11*b^3*p*x^9-6*b^3*x^9+3*a*b^2*
p^2*x^6+9*a*b^2*p*x^6+6*a*b^2*x^6-6*a^2*b*p*x^3-6*a^2*b*x^3+6*a^3)/b^4/(p^4+10*p
^3+35*p^2+50*p+24)

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Maxima [A]  time = 1.46397, size = 143, normalized size = 1.51 \[ \frac{{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} x^{12} +{\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} x^{9} - 3 \,{\left (p^{2} + p\right )} a^{2} b^{2} x^{6} + 6 \, a^{3} b p x^{3} - 6 \, a^{4}\right )}{\left (b x^{3} + a\right )}^{p}}{3 \,{\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^3 + a)^p*x^11,x, algorithm="maxima")

[Out]

1/3*((p^3 + 6*p^2 + 11*p + 6)*b^4*x^12 + (p^3 + 3*p^2 + 2*p)*a*b^3*x^9 - 3*(p^2
+ p)*a^2*b^2*x^6 + 6*a^3*b*p*x^3 - 6*a^4)*(b*x^3 + a)^p/((p^4 + 10*p^3 + 35*p^2
+ 50*p + 24)*b^4)

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Fricas [A]  time = 0.245015, size = 200, normalized size = 2.11 \[ \frac{{\left ({\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} x^{12} +{\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} x^{9} + 6 \, a^{3} b p x^{3} - 3 \,{\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{6} - 6 \, a^{4}\right )}{\left (b x^{3} + a\right )}^{p}}{3 \,{\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^3 + a)^p*x^11,x, algorithm="fricas")

[Out]

1/3*((b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*x^12 + (a*b^3*p^3 + 3*a*b^3*p^2 +
2*a*b^3*p)*x^9 + 6*a^3*b*p*x^3 - 3*(a^2*b^2*p^2 + a^2*b^2*p)*x^6 - 6*a^4)*(b*x^3
+ a)^p/(b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**11*(b*x**3+a)**p,x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.218402, size = 597, normalized size = 6.28 \[ \frac{{\left (b x^{3} + a\right )}^{4} p^{3} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} - 3 \,{\left (b x^{3} + a\right )}^{3} a p^{3} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} + 3 \,{\left (b x^{3} + a\right )}^{2} a^{2} p^{3} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} -{\left (b x^{3} + a\right )} a^{3} p^{3} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} + 6 \,{\left (b x^{3} + a\right )}^{4} p^{2} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} - 21 \,{\left (b x^{3} + a\right )}^{3} a p^{2} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} + 24 \,{\left (b x^{3} + a\right )}^{2} a^{2} p^{2} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} - 9 \,{\left (b x^{3} + a\right )} a^{3} p^{2} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} + 11 \,{\left (b x^{3} + a\right )}^{4} p e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} - 42 \,{\left (b x^{3} + a\right )}^{3} a p e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} + 57 \,{\left (b x^{3} + a\right )}^{2} a^{2} p e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} - 26 \,{\left (b x^{3} + a\right )} a^{3} p e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} + 6 \,{\left (b x^{3} + a\right )}^{4} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} - 24 \,{\left (b x^{3} + a\right )}^{3} a e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} + 36 \,{\left (b x^{3} + a\right )}^{2} a^{2} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )} - 24 \,{\left (b x^{3} + a\right )} a^{3} e^{\left (p{\rm ln}\left (b x^{3} + a\right )\right )}}{3 \,{\left (b^{3} p^{4} + 10 \, b^{3} p^{3} + 35 \, b^{3} p^{2} + 50 \, b^{3} p + 24 \, b^{3}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^3 + a)^p*x^11,x, algorithm="giac")

[Out]

1/3*((b*x^3 + a)^4*p^3*e^(p*ln(b*x^3 + a)) - 3*(b*x^3 + a)^3*a*p^3*e^(p*ln(b*x^3
+ a)) + 3*(b*x^3 + a)^2*a^2*p^3*e^(p*ln(b*x^3 + a)) - (b*x^3 + a)*a^3*p^3*e^(p*
ln(b*x^3 + a)) + 6*(b*x^3 + a)^4*p^2*e^(p*ln(b*x^3 + a)) - 21*(b*x^3 + a)^3*a*p^
2*e^(p*ln(b*x^3 + a)) + 24*(b*x^3 + a)^2*a^2*p^2*e^(p*ln(b*x^3 + a)) - 9*(b*x^3
+ a)*a^3*p^2*e^(p*ln(b*x^3 + a)) + 11*(b*x^3 + a)^4*p*e^(p*ln(b*x^3 + a)) - 42*(
b*x^3 + a)^3*a*p*e^(p*ln(b*x^3 + a)) + 57*(b*x^3 + a)^2*a^2*p*e^(p*ln(b*x^3 + a)
) - 26*(b*x^3 + a)*a^3*p*e^(p*ln(b*x^3 + a)) + 6*(b*x^3 + a)^4*e^(p*ln(b*x^3 + a
)) - 24*(b*x^3 + a)^3*a*e^(p*ln(b*x^3 + a)) + 36*(b*x^3 + a)^2*a^2*e^(p*ln(b*x^3
+ a)) - 24*(b*x^3 + a)*a^3*e^(p*ln(b*x^3 + a)))/((b^3*p^4 + 10*b^3*p^3 + 35*b^3
*p^2 + 50*b^3*p + 24*b^3)*b)