3.7.2 \(\int x^{11} (a+b x^3)^p \, dx\) [602]
Optimal. Leaf size=95 \[ -\frac {a^3 \left (a+b x^3\right )^{1+p}}{3 b^4 (1+p)}+\frac {a^2 \left (a+b x^3\right )^{2+p}}{b^4 (2+p)}-\frac {a \left (a+b x^3\right )^{3+p}}{b^4 (3+p)}+\frac {\left (a+b x^3\right )^{4+p}}{3 b^4 (4+p)} \]
[Out]
-1/3*a^3*(b*x^3+a)^(1+p)/b^4/(1+p)+a^2*(b*x^3+a)^(2+p)/b^4/(2+p)-a*(b*x^3+a)^(3+p)/b^4/(3+p)+1/3*(b*x^3+a)^(4+
p)/b^4/(4+p)
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Rubi [A]
time = 0.04, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 = {272, 45}
\begin {gather*} -\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)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^11*(a + b*x^3)^p,x]
[Out]
-1/3*(a^3*(a + b*x^3)^(1 + p))/(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))
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]]
Rubi steps
\begin {align*} \int x^{11} \left (a+b x^3\right )^p \, dx &=\frac {1}{3} \text {Subst}\left (\int x^3 (a+b x)^p \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {a^3 (a+b x)^p}{b^3}+\frac {3 a^2 (a+b x)^{1+p}}{b^3}-\frac {3 a (a+b x)^{2+p}}{b^3}+\frac {(a+b x)^{3+p}}{b^3}\right ) \, dx,x,x^3\right )\\ &=-\frac {a^3 \left (a+b x^3\right )^{1+p}}{3 b^4 (1+p)}+\frac {a^2 \left (a+b x^3\right )^{2+p}}{b^4 (2+p)}-\frac {a \left (a+b x^3\right )^{3+p}}{b^4 (3+p)}+\frac {\left (a+b x^3\right )^{4+p}}{3 b^4 (4+p)}\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 95, normalized size = 1.00 \begin {gather*} -\frac {a^3 \left (a+b x^3\right )^{1+p}}{3 b^4 (1+p)}+\frac {a^2 \left (a+b x^3\right )^{2+p}}{b^4 (2+p)}-\frac {a \left (a+b x^3\right )^{3+p}}{b^4 (3+p)}+\frac {\left (a+b x^3\right )^{4+p}}{3 b^4 (4+p)} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^11*(a + b*x^3)^p,x]
[Out]
-1/3*(a^3*(a + b*x^3)^(1 + p))/(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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Maple [A]
time = 0.14, size = 132, normalized size = 1.39
| | |
| method |
result |
size |
| | |
| gosper |
\(-\frac {\left (b \,x^{3}+a \right )^{1+p} \left (-b^{3} p^{3} x^{9}-6 b^{3} p^{2} x^{9}-11 b^{3} p \,x^{9}-6 b^{3} x^{9}+3 a \,b^{2} p^{2} x^{6}+9 a p \,x^{6} b^{2}+6 a \,b^{2} x^{6}-6 a^{2} p \,x^{3} b -6 a^{2} b \,x^{3}+6 a^{3}\right )}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) |
\(132\) |
| risch |
\(-\frac {\left (-b^{4} p^{3} x^{12}-6 b^{4} p^{2} x^{12}-11 b^{4} p \,x^{12}-a \,b^{3} p^{3} x^{9}-6 b^{4} x^{12}-3 a \,b^{3} p^{2} x^{9}-2 a p \,x^{9} b^{3}+3 a^{2} b^{2} p^{2} x^{6}+3 a^{2} p \,x^{6} b^{2}-6 a^{3} p \,x^{3} b +6 a^{4}\right ) \left (b \,x^{3}+a \right )^{p}}{3 \left (3+p \right ) \left (4+p \right ) \left (2+p \right ) \left (1+p \right ) b^{4}}\) |
\(150\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^11*(b*x^3+a)^p,x,method=_RETURNVERBOSE)
[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 = 0.31, size = 106, normalized size = 1.12 \begin {gather*} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11*(b*x^3+a)^p,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 = 0.38, size = 148, normalized size = 1.56 \begin {gather*} \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 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11*(b*x^3+a)^p,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 [B] Leaf count of result is larger than twice the leaf count of optimal. 2467 vs.
\(2 (78) = 156\).
time = 22.44, size = 2467, normalized size = 25.97 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**11*(b*x**3+a)**p,x)
[Out]
Piecewise((a**p*x**12/12, Eq(b, 0)), (6*a**3*log(x - (-a/b)**(1/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b
**6*x**6 + 18*b**7*x**9) + 6*a**3*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(18*a**3*b**4 + 54*a**2*b*
*5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 12*a**3*log(2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 +
18*b**7*x**9) + 11*a**3/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a**2*b*x**3*l
og(x - (-a/b)**(1/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a**2*b*x**3*log(
4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**
9) - 36*a**2*b*x**3*log(2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 27*a**2*b*x**3
/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*b**2*x**6*log(x - (-a/b)**(1/3))/(1
8*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*b**2*x**6*log(4*x**2 + 4*x*(-a/b)**(1/
3) + 4*(-a/b)**(2/3))/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 36*a*b**2*x**6*log(
2)/(18*a**3*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 18*a*b**2*x**6/(18*a**3*b**4 + 54*a**2
*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 6*b**3*x**9*log(x - (-a/b)**(1/3))/(18*a**3*b**4 + 54*a**2*b**5*
x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) + 6*b**3*x**9*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(18*a**3
*b**4 + 54*a**2*b**5*x**3 + 54*a*b**6*x**6 + 18*b**7*x**9) - 12*b**3*x**9*log(2)/(18*a**3*b**4 + 54*a**2*b**5*
x**3 + 54*a*b**6*x**6 + 18*b**7*x**9), Eq(p, -4)), (-6*a**3*log(x - (-a/b)**(1/3))/(6*a**2*b**4 + 12*a*b**5*x*
*3 + 6*b**6*x**6) - 6*a**3*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(6*a**2*b**4 + 12*a*b**5*x**3 + 6
*b**6*x**6) - 9*a**3/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b**6*x**6) + 12*a**3*log(2)/(6*a**2*b**4 + 12*a*b**5*x*
*3 + 6*b**6*x**6) - 12*a**2*b*x**3*log(x - (-a/b)**(1/3))/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b**6*x**6) - 12*a*
*2*b*x**3*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b**6*x**6) - 12*
a**2*b*x**3/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b**6*x**6) + 24*a**2*b*x**3*log(2)/(6*a**2*b**4 + 12*a*b**5*x**3
+ 6*b**6*x**6) - 6*a*b**2*x**6*log(x - (-a/b)**(1/3))/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b**6*x**6) - 6*a*b**2
*x**6*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b**6*x**6) + 12*a*b*
*2*x**6*log(2)/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b**6*x**6) + 2*b**3*x**9/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b*
*6*x**6), Eq(p, -3)), (6*a**3*log(x - (-a/b)**(1/3))/(6*a*b**4 + 6*b**5*x**3) + 6*a**3*log(4*x**2 + 4*x*(-a/b)
**(1/3) + 4*(-a/b)**(2/3))/(6*a*b**4 + 6*b**5*x**3) - 12*a**3*log(2)/(6*a*b**4 + 6*b**5*x**3) + 6*a**3/(6*a*b*
*4 + 6*b**5*x**3) + 6*a**2*b*x**3*log(x - (-a/b)**(1/3))/(6*a*b**4 + 6*b**5*x**3) + 6*a**2*b*x**3*log(4*x**2 +
4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(6*a*b**4 + 6*b**5*x**3) - 12*a**2*b*x**3*log(2)/(6*a*b**4 + 6*b**5*x**3
) - 3*a*b**2*x**6/(6*a*b**4 + 6*b**5*x**3) + b**3*x**9/(6*a*b**4 + 6*b**5*x**3), Eq(p, -2)), (-a**3*log(x - (-
a/b)**(1/3))/(3*b**4) - a**3*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(3*b**4) + a**2*x**3/(3*b**3) -
a*x**6/(6*b**2) + x**9/(9*b), Eq(p, -1)), (-6*a**4*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**
2 + 150*b**4*p + 72*b**4) + 6*a**3*b*p*x**3*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*
b**4*p + 72*b**4) - 3*a**2*b**2*p**2*x**6*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b*
*4*p + 72*b**4) - 3*a**2*b**2*p*x**6*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p
+ 72*b**4) + a*b**3*p**3*x**9*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p + 72*b*
*4) + 3*a*b**3*p**2*x**9*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p + 72*b**4) +
2*a*b**3*p*x**9*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p + 72*b**4) + b**4*p*
*3*x**12*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p + 72*b**4) + 6*b**4*p**2*x**
12*(a + b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p + 72*b**4) + 11*b**4*p*x**12*(a +
b*x**3)**p/(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p + 72*b**4) + 6*b**4*x**12*(a + b*x**3)**p/
(3*b**4*p**4 + 30*b**4*p**3 + 105*b**4*p**2 + 150*b**4*p + 72*b**4), True))
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 260 vs.
\(2 (91) = 182\).
time = 1.19, size = 260, normalized size = 2.74 \begin {gather*} \frac {{\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} p^{2} - 3 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a p^{2} + 3 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2} p^{2} + 5 \, {\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} p - 18 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a p + 21 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2} p + 6 \, {\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} - 24 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a + 36 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2}}{3 \, {\left (b^{4} p^{3} + 9 \, b^{4} p^{2} + 26 \, b^{4} p + 24 \, b^{4}\right )}} - \frac {{\left (b x^{3} + a\right )}^{p + 1} a^{3}}{3 \, b^{4} {\left (p + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11*(b*x^3+a)^p,x, algorithm="giac")
[Out]
1/3*((b*x^3 + a)^4*(b*x^3 + a)^p*p^2 - 3*(b*x^3 + a)^3*(b*x^3 + a)^p*a*p^2 + 3*(b*x^3 + a)^2*(b*x^3 + a)^p*a^2
*p^2 + 5*(b*x^3 + a)^4*(b*x^3 + a)^p*p - 18*(b*x^3 + a)^3*(b*x^3 + a)^p*a*p + 21*(b*x^3 + a)^2*(b*x^3 + a)^p*a
^2*p + 6*(b*x^3 + a)^4*(b*x^3 + a)^p - 24*(b*x^3 + a)^3*(b*x^3 + a)^p*a + 36*(b*x^3 + a)^2*(b*x^3 + a)^p*a^2)/
(b^4*p^3 + 9*b^4*p^2 + 26*b^4*p + 24*b^4) - 1/3*(b*x^3 + a)^(p + 1)*a^3/(b^4*(p + 1))
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Mupad [B]
time = 1.12, size = 183, normalized size = 1.93 \begin {gather*} {\left (b\,x^3+a\right )}^p\,\left (\frac {x^{12}\,\left (p^3+6\,p^2+11\,p+6\right )}{3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {2\,a^4}{b^4\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {2\,a^3\,p\,x^3}{b^3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {a\,p\,x^9\,\left (p^2+3\,p+2\right )}{3\,b\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {a^2\,p\,x^6\,\left (p+1\right )}{b^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^11*(a + b*x^3)^p,x)
[Out]
(a + b*x^3)^p*((x^12*(11*p + 6*p^2 + p^3 + 6))/(3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (2*a^4)/(b^4*(50*p +
35*p^2 + 10*p^3 + p^4 + 24)) + (2*a^3*p*x^3)/(b^3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (a*p*x^9*(3*p + p^2 +
2))/(3*b*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (a^2*p*x^6*(p + 1))/(b^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24))
)
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