3.6.100 \(\int x^5 (a+b x^3)^p \, dx\) [600]

Optimal. Leaf size=48 \[ -\frac {a \left (a+b x^3\right )^{1+p}}{3 b^2 (1+p)}+\frac {\left (a+b x^3\right )^{2+p}}{3 b^2 (2+p)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 48, 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 {\left (a+b x^3\right )^{p+2}}{3 b^2 (p+2)}-\frac {a \left (a+b x^3\right )^{p+1}}{3 b^2 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^3)^p,x]

[Out]

-1/3*(a*(a + b*x^3)^(1 + p))/(b^2*(1 + p)) + (a + b*x^3)^(2 + p)/(3*b^2*(2 + 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^5 \left (a+b x^3\right )^p \, dx &=\frac {1}{3} \text {Subst}\left (\int x (a+b x)^p \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,x^3\right )\\ &=-\frac {a \left (a+b x^3\right )^{1+p}}{3 b^2 (1+p)}+\frac {\left (a+b x^3\right )^{2+p}}{3 b^2 (2+p)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 40, normalized size = 0.83 \begin {gather*} \frac {\left (a+b x^3\right )^{1+p} \left (-a+b (1+p) x^3\right )}{3 b^2 (1+p) (2+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^3)^p,x]

[Out]

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

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Maple [A]
time = 0.14, size = 42, normalized size = 0.88

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right )^{1+p} \left (-x^{3} p b -b \,x^{3}+a \right )}{3 b^{2} \left (p^{2}+3 p +2\right )}\)	\(42\)
risch	\(-\frac {\left (-b^{2} x^{6} p -b^{2} x^{6}-a p \,x^{3} b +a^{2}\right ) \left (b \,x^{3}+a \right )^{p}}{3 b^{2} \left (2+p \right ) \left (1+p \right )}\)	\(54\)
norman	\(\frac {x^{6} {\mathrm e}^{p \ln \left (b \,x^{3}+a \right )}}{3 p +6}-\frac {a^{2} {\mathrm e}^{p \ln \left (b \,x^{3}+a \right )}}{3 b^{2} \left (p^{2}+3 p +2\right )}+\frac {p a \,x^{3} {\mathrm e}^{p \ln \left (b \,x^{3}+a \right )}}{3 b \left (p^{2}+3 p +2\right )}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^3+a)^p,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 47, normalized size = 0.98 \begin {gather*} \frac {{\left (b^{2} {\left (p + 1\right )} x^{6} + a b p x^{3} - a^{2}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 58, normalized size = 1.21 \begin {gather*} \frac {{\left ({\left (b^{2} p + b^{2}\right )} x^{6} + a b p x^{3} - a^{2}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^p,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (37) = 74\).
time = 1.94, size = 432, normalized size = 9.00 \begin {gather*} \begin {cases} \frac {a^{p} x^{6}}{6} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {a \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} - \frac {2 a \log {\left (2 \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {a}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {b x^{3} \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {b x^{3} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} - \frac {2 b x^{3} \log {\left (2 \right )}}{3 a b^{2} + 3 b^{3} x^{3}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 b^{2}} - \frac {a \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 b^{2}} + \frac {x^{3}}{3 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{3}\right )^{p}}{3 b^{2} p^{2} + 9 b^{2} p + 6 b^{2}} + \frac {a b p x^{3} \left (a + b x^{3}\right )^{p}}{3 b^{2} p^{2} + 9 b^{2} p + 6 b^{2}} + \frac {b^{2} p x^{6} \left (a + b x^{3}\right )^{p}}{3 b^{2} p^{2} + 9 b^{2} p + 6 b^{2}} + \frac {b^{2} x^{6} \left (a + b x^{3}\right )^{p}}{3 b^{2} p^{2} + 9 b^{2} p + 6 b^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**3+a)**p,x)

[Out]

Piecewise((a**p*x**6/6, Eq(b, 0)), (a*log(x - (-a/b)**(1/3))/(3*a*b**2 + 3*b**3*x**3) + a*log(4*x**2 + 4*x*(-a
/b)**(1/3) + 4*(-a/b)**(2/3))/(3*a*b**2 + 3*b**3*x**3) - 2*a*log(2)/(3*a*b**2 + 3*b**3*x**3) + a/(3*a*b**2 + 3
*b**3*x**3) + b*x**3*log(x - (-a/b)**(1/3))/(3*a*b**2 + 3*b**3*x**3) + b*x**3*log(4*x**2 + 4*x*(-a/b)**(1/3) +
4*(-a/b)**(2/3))/(3*a*b**2 + 3*b**3*x**3) - 2*b*x**3*log(2)/(3*a*b**2 + 3*b**3*x**3), Eq(p, -2)), (-a*log(x -
(-a/b)**(1/3))/(3*b**2) - a*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(3*b**2) + x**3/(3*b), Eq(p, -1
)), (-a**2*(a + b*x**3)**p/(3*b**2*p**2 + 9*b**2*p + 6*b**2) + a*b*p*x**3*(a + b*x**3)**p/(3*b**2*p**2 + 9*b**
2*p + 6*b**2) + b**2*p*x**6*(a + b*x**3)**p/(3*b**2*p**2 + 9*b**2*p + 6*b**2) + b**2*x**6*(a + b*x**3)**p/(3*b
**2*p**2 + 9*b**2*p + 6*b**2), True))

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Giac [A]
time = 1.02, size = 51, normalized size = 1.06 \begin {gather*} \frac {{\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p}}{3 \, b^{2} {\left (p + 2\right )}} - \frac {{\left (b x^{3} + a\right )}^{p + 1} a}{3 \, b^{2} {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^p,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.08, size = 68, normalized size = 1.42 \begin {gather*} {\left (b\,x^3+a\right )}^p\,\left (\frac {x^6\,\left (p+1\right )}{3\,\left (p^2+3\,p+2\right )}-\frac {a^2}{3\,b^2\,\left (p^2+3\,p+2\right )}+\frac {a\,p\,x^3}{3\,b\,\left (p^2+3\,p+2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^3)^p,x)

[Out]

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

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