3.55 \(\int x^8 (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\)

Optimal. Leaf size=119 \[ \frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^7}{24 b^3}-\frac{2 a \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^6}{21 b^3}+\frac{a^2 \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{18 b^3} \]

[Out]

(a^2*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(18*b^3) - (2*a*(a + b*x^3)^6*Sqrt[a^2 + 2*a*b*x^3 + b^2*x
^6])/(21*b^3) + ((a + b*x^3)^7*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(24*b^3)

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Rubi [A]  time = 0.0903906, antiderivative size = 119, 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} \[ \frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^7}{24 b^3}-\frac{2 a \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^6}{21 b^3}+\frac{a^2 \sqrt{a^2+2 a b x^3+b^2 x^6} \left (a+b x^3\right )^5}{18 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(a^2*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(18*b^3) - (2*a*(a + b*x^3)^6*Sqrt[a^2 + 2*a*b*x^3 + b^2*x
^6])/(21*b^3) + ((a + b*x^3)^7*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(24*b^3)

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^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int x^8 \left (a b+b^2 x^3\right )^5 \, dx}{b^4 \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^2 \left (a b+b^2 x\right )^5 \, dx,x,x^3\right )}{3 b^4 \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^2 \left (a b+b^2 x\right )^5}{b^2}-\frac{2 a \left (a b+b^2 x\right )^6}{b^3}+\frac{\left (a b+b^2 x\right )^7}{b^4}\right ) \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )}\\ &=\frac{a^2 \left (a+b x^3\right )^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{18 b^3}-\frac{2 a \left (a+b x^3\right )^6 \sqrt{a^2+2 a b x^3+b^2 x^6}}{21 b^3}+\frac{\left (a+b x^3\right )^7 \sqrt{a^2+2 a b x^3+b^2 x^6}}{24 b^3}\\ \end{align*}

Mathematica [A]  time = 0.024054, size = 83, normalized size = 0.7 \[ \frac{x^9 \sqrt{\left (a+b x^3\right )^2} \left (280 a^2 b^3 x^9+336 a^3 b^2 x^6+210 a^4 b x^3+56 a^5+120 a b^4 x^{12}+21 b^5 x^{15}\right )}{504 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(x^9*Sqrt[(a + b*x^3)^2]*(56*a^5 + 210*a^4*b*x^3 + 336*a^3*b^2*x^6 + 280*a^2*b^3*x^9 + 120*a*b^4*x^12 + 21*b^5
*x^15))/(504*(a + b*x^3))

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Maple [A]  time = 0.009, size = 80, normalized size = 0.7 \begin{align*}{\frac{{x}^{9} \left ( 21\,{b}^{5}{x}^{15}+120\,a{b}^{4}{x}^{12}+280\,{a}^{2}{b}^{3}{x}^{9}+336\,{a}^{3}{b}^{2}{x}^{6}+210\,{a}^{4}b{x}^{3}+56\,{a}^{5} \right ) }{504\, \left ( b{x}^{3}+a \right ) ^{5}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)

[Out]

1/504*x^9*(21*b^5*x^15+120*a*b^4*x^12+280*a^2*b^3*x^9+336*a^3*b^2*x^6+210*a^4*b*x^3+56*a^5)*((b*x^3+a)^2)^(5/2
)/(b*x^3+a)^5

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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^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74264, size = 139, normalized size = 1.17 \begin{align*} \frac{1}{24} \, b^{5} x^{24} + \frac{5}{21} \, a b^{4} x^{21} + \frac{5}{9} \, a^{2} b^{3} x^{18} + \frac{2}{3} \, a^{3} b^{2} x^{15} + \frac{5}{12} \, a^{4} b x^{12} + \frac{1}{9} \, a^{5} x^{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/24*b^5*x^24 + 5/21*a*b^4*x^21 + 5/9*a^2*b^3*x^18 + 2/3*a^3*b^2*x^15 + 5/12*a^4*b*x^12 + 1/9*a^5*x^9

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{8} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral(x**8*((a + b*x**3)**2)**(5/2), x)

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Giac [A]  time = 1.12256, size = 142, normalized size = 1.19 \begin{align*} \frac{1}{24} \, b^{5} x^{24} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{5}{21} \, a b^{4} x^{21} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{5}{9} \, a^{2} b^{3} x^{18} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{2}{3} \, a^{3} b^{2} x^{15} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{5}{12} \, a^{4} b x^{12} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{1}{9} \, a^{5} x^{9} \mathrm{sgn}\left (b x^{3} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

1/24*b^5*x^24*sgn(b*x^3 + a) + 5/21*a*b^4*x^21*sgn(b*x^3 + a) + 5/9*a^2*b^3*x^18*sgn(b*x^3 + a) + 2/3*a^3*b^2*
x^15*sgn(b*x^3 + a) + 5/12*a^4*b*x^12*sgn(b*x^3 + a) + 1/9*a^5*x^9*sgn(b*x^3 + a)