3.2.27 \(\int x^5 (a^2+2 a b x^3+b^2 x^6)^p \, dx\) [127]

Optimal. Leaf size=84 \[ -\frac {a \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^2 (1+2 p)}+\frac {\left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^2 (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 = {1370, 272, 45} \begin {gather*} \frac {\left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^2 (p+1)}-\frac {a \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^2 (2 p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

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

Rule 1370

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]

Rubi steps

\begin {align*} \int x^5 \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^5 \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 ) \text {Subst}\left (\int x \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 ) \text {Subst}\left (\int \left (-\frac {a \left (1+\frac {b x}{a}\right )^{2 p}}{b}+\frac {a \left (1+\frac {b x}{a}\right )^{1+2 p}}{b}\right ) \, dx,x,x^3\right )\\ &=-\frac {a \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^2 (1+2 p)}+\frac {\left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^2 (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 51, normalized size = 0.61 \begin {gather*} \frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (-a+b (1+2 p) x^3\right )}{6 b^2 (1+p) (1+2 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a^2 + 2*a*b*x^3 + b^2*x^6)^p,x]

[Out]

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

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Maple [A]
time = 0.03, size = 58, normalized size = 0.69

method	result	size
risch	\(-\frac {\left (-2 b^{2} p \,x^{6}-b^{2} x^{6}-2 a b p \,x^{3}+a^{2}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{p}}{6 b^{2} \left (1+p \right ) \left (1+2 p \right )}\)	\(58\)
gosper	\(-\frac {\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} \left (-2 x^{3} p b -b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )}{6 b^{2} \left (2 p^{2}+3 p +1\right )}\)	\(60\)
norman	\(\frac {x^{6} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{6 p +6}-\frac {a^{2} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{6 b^{2} \left (2 p^{2}+3 p +1\right )}+\frac {p a \,x^{3} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{3 b \left (2 p^{2}+3 p +1\right )}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b^2*x^6+2*a*b*x^3+a^2)^p,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 54, normalized size = 0.64 \begin {gather*} \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{6} + 2 \, a b p x^{3} - a^{2}\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{6 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 70, normalized size = 0.83 \begin {gather*} \frac {{\left ({\left (2 \, b^{2} p + b^{2}\right )} x^{6} + 2 \, a b p x^{3} - a^{2}\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{6 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {x^{6} \left (a^{2}\right )^{p}}{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 = -1 \\\int \frac {x^{5}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\- \frac {a^{2} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {2 a b p x^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {2 b^{2} p x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {b^{2} x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 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**2*x**6+2*a*b*x**3+a**2)**p,x)

[Out]

Piecewise((x**6*(a**2)**p/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, -1)), (Integ
ral(x**5/sqrt((a + b*x**3)**2), x), Eq(p, -1/2)), (-a**2*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**2*p**2 + 18
*b**2*p + 6*b**2) + 2*a*b*p*x**3*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**2*p**2 + 18*b**2*p + 6*b**2) + 2*b*
*2*p*x**6*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(12*b**2*p**2 + 18*b**2*p + 6*b**2) + b**2*x**6*(a**2 + 2*a*b*x**
3 + b**2*x**6)**p/(12*b**2*p**2 + 18*b**2*p + 6*b**2), True))

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Giac [A]
time = 3.03, size = 132, normalized size = 1.57 \begin {gather*} \frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{2} p x^{6} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{2} x^{6} + 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b p x^{3} - {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2}}{6 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^p,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.19, size = 85, normalized size = 1.01 \begin {gather*} {\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p\,\left (\frac {x^6\,\left (2\,p+1\right )}{6\,\left (2\,p^2+3\,p+1\right )}-\frac {a^2}{6\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {a\,p\,x^3}{3\,b\,\left (2\,p^2+3\,p+1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^p,x)

[Out]

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

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