\(\int x^8 (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\) [24]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {a^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 b^3}-\frac {2 a \left (a+b x^3\right )^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 b^3}+\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^3} \]

[Out]

1/12*a^2*(b*x^3+a)^3*((b*x^3+a)^2)^(1/2)/b^3-2/15*a*(b*x^3+a)^4*((b*x^3+a)^2)^(1/2)/b^3+1/18*(b*x^3+a)^5*((b*x
^3+a)^2)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.40, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1369, 272, 45} \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {a b^2 x^{15} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {a^2 b x^{12} \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^3 x^{18} \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 \left (a+b x^3\right )}+\frac {a^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )} \]

[In]

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

[Out]

(a^3*x^9*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(9*(a + b*x^3)) + (a^2*b*x^12*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*(a
+ b*x^3)) + (a*b^2*x^15*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*(a + b*x^3)) + (b^3*x^18*Sqrt[a^2 + 2*a*b*x^3 + b
^2*x^6])/(18*(a + b*x^3))

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 1369

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^8 \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} \text {Subst}\left (\int x^2 \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} \text {Subst}\left (\int \left (a^3 b^3 x^2+3 a^2 b^4 x^3+3 a b^5 x^4+b^6 x^5\right ) \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {a^3 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {a^2 b x^{12} \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {a b^2 x^{15} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {b^3 x^{18} \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {x^9 \left (20 a^3+45 a^2 b x^3+36 a b^2 x^6+10 b^3 x^9\right ) \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{180 \left (-a^2-a b x^3+\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]

[In]

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

[Out]

(x^9*(20*a^3 + 45*a^2*b*x^3 + 36*a*b^2*x^6 + 10*b^3*x^9)*(Sqrt[a^2]*b*x^3 + a*(Sqrt[a^2] - Sqrt[(a + b*x^3)^2]
)))/(180*(-a^2 - a*b*x^3 + Sqrt[a^2]*Sqrt[(a + b*x^3)^2]))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{4} \left (10 b^{2} x^{6}-4 a b \,x^{3}+a^{2}\right )}{180 b^{3}}\)	\(42\)
gosper	\(\frac {x^{9} \left (10 b^{3} x^{9}+36 b^{2} x^{6} a +45 a^{2} b \,x^{3}+20 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{180 \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
default	\(\frac {x^{9} \left (10 b^{3} x^{9}+36 b^{2} x^{6} a +45 a^{2} b \,x^{3}+20 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{180 \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} x^{9}}{9 b \,x^{3}+9 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b \,x^{12}}{4 b \,x^{3}+4 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} a \,x^{15}}{5 b \,x^{3}+5 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} x^{18}}{18 b \,x^{3}+18 a}\)	\(116\)

[In]

int(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/180*csgn(b*x^3+a)*(b*x^3+a)^4*(10*b^2*x^6-4*a*b*x^3+a^2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.29 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{18} \, b^{3} x^{18} + \frac {1}{5} \, a b^{2} x^{15} + \frac {1}{4} \, a^{2} b x^{12} + \frac {1}{9} \, a^{3} x^{9} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int x^{8} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{3}}{12 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} x^{3}}{18 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{3}}{12 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a}{90 \, b^{3}} \]

[In]

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

[Out]

1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^2*x^3/b^2 + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*x^3/b^2 + 1/12*(b^
2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^3/b^3 - 7/90*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a/b^3

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.56 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{18} \, b^{3} x^{18} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{5} \, a b^{2} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{4} \, a^{2} b x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{9} \, a^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int x^8\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2} \,d x \]

[In]

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

[Out]

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