3.27 \(\int x^5 (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\)

Optimal. Leaf size=78 \[ \frac{\left (a+b x^3\right )^4 \sqrt{a^2+2 a b x^3+b^2 x^6}}{15 b^2}-\frac{a \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{12 b^2} \]

[Out]

-(a*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(12*b^2) + ((a + b*x^3)^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/
(15*b^2)

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Rubi [A]  time = 0.050472, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1355, 266, 43} \[ \frac{\left (a+b x^3\right )^4 \sqrt{a^2+2 a b x^3+b^2 x^6}}{15 b^2}-\frac{a \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{12 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(12*b^2) + ((a + b*x^3)^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/
(15*b^2)

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

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^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int x^5 \left (a b+b^2 x^3\right )^3 \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int x \left (a b+b^2 x\right )^3 \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \left (-\frac{a \left (a b+b^2 x\right )^3}{b}+\frac{\left (a b+b^2 x\right )^4}{b^2}\right ) \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=-\frac{a \left (a+b x^3\right )^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{12 b^2}+\frac{\left (a+b x^3\right )^4 \sqrt{a^2+2 a b x^3+b^2 x^6}}{15 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0190837, size = 61, normalized size = 0.78 \[ \frac{x^6 \sqrt{\left (a+b x^3\right )^2} \left (20 a^2 b x^3+10 a^3+15 a b^2 x^6+4 b^3 x^9\right )}{60 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^6*Sqrt[(a + b*x^3)^2]*(10*a^3 + 20*a^2*b*x^3 + 15*a*b^2*x^6 + 4*b^3*x^9))/(60*(a + b*x^3))

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Maple [A]  time = 0.006, size = 58, normalized size = 0.7 \begin{align*}{\frac{{x}^{6} \left ( 4\,{b}^{3}{x}^{9}+15\,a{b}^{2}{x}^{6}+20\,{a}^{2}b{x}^{3}+10\,{a}^{3} \right ) }{60\, \left ( b{x}^{3}+a \right ) ^{3}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

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)^(3/2),x)

[Out]

1/60*x^6*(4*b^3*x^9+15*a*b^2*x^6+20*a^2*b*x^3+10*a^3)*((b*x^3+a)^2)^(3/2)/(b*x^3+a)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65279, size = 84, normalized size = 1.08 \begin{align*} \frac{1}{15} \, b^{3} x^{15} + \frac{1}{4} \, a b^{2} x^{12} + \frac{1}{3} \, a^{2} b x^{9} + \frac{1}{6} \, a^{3} x^{6} \end{align*}

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)^(3/2),x, algorithm="fricas")

[Out]

1/15*b^3*x^15 + 1/4*a*b^2*x^12 + 1/3*a^2*b*x^9 + 1/6*a^3*x^6

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

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)**(3/2),x)

[Out]

Integral(x**5*((a + b*x**3)**2)**(3/2), x)

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Giac [A]  time = 1.09632, size = 61, normalized size = 0.78 \begin{align*} \frac{1}{60} \,{\left (4 \, b^{3} x^{15} + 15 \, a b^{2} x^{12} + 20 \, a^{2} b x^{9} + 10 \, a^{3} x^{6}\right )} \mathrm{sgn}\left (b x^{3} + a\right ) \end{align*}

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)^(3/2),x, algorithm="giac")

[Out]

1/60*(4*b^3*x^15 + 15*a*b^2*x^12 + 20*a^2*b*x^9 + 10*a^3*x^6)*sgn(b*x^3 + a)