\(\int x^9 (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\) [23]

Optimal result

Integrand size = 26, antiderivative size = 167 \[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {a^3 x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 \left (a+b x^3\right )}+\frac {3 a^2 b x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {3 a b^2 x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {b^3 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )} \]

[Out]

1/10*a^3*x^10*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+3/13*a^2*b*x^13*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+3/16*a*b^2*x^16*((b*
x^3+a)^2)^(1/2)/(b*x^3+a)+1/19*b^3*x^19*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1369, 276} \[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {3 a b^2 x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {3 a^2 b x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {b^3 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {a^3 x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 \left (a+b x^3\right )} \]

[In]

Int[x^9*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(a^3*x^10*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(10*(a + b*x^3)) + (3*a^2*b*x^13*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(
13*(a + b*x^3)) + (3*a*b^2*x^16*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(16*(a + b*x^3)) + (b^3*x^19*Sqrt[a^2 + 2*a*b
*x^3 + b^2*x^6])/(19*(a + b*x^3))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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^9 \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} \int \left (a^3 b^3 x^9+3 a^2 b^4 x^{12}+3 a b^5 x^{15}+b^6 x^{18}\right ) \, dx}{b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {a^3 x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 \left (a+b x^3\right )}+\frac {3 a^2 b x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {3 a b^2 x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {b^3 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {x^{10} \sqrt {\left (a+b x^3\right )^2} \left (1976 a^3+4560 a^2 b x^3+3705 a b^2 x^6+1040 b^3 x^9\right )}{19760 \left (a+b x^3\right )} \]

[In]

Integrate[x^9*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(x^10*Sqrt[(a + b*x^3)^2]*(1976*a^3 + 4560*a^2*b*x^3 + 3705*a*b^2*x^6 + 1040*b^3*x^9))/(19760*(a + b*x^3))

Maple [A] (verified)

Time = 8.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.35

method	result	size
gosper	\(\frac {x^{10} \left (1040 b^{3} x^{9}+3705 b^{2} x^{6} a +4560 a^{2} b \,x^{3}+1976 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{19760 \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
default	\(\frac {x^{10} \left (1040 b^{3} x^{9}+3705 b^{2} x^{6} a +4560 a^{2} b \,x^{3}+1976 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{19760 \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
risch	\(\frac {a^{3} x^{10} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{10 b \,x^{3}+10 a}+\frac {3 a^{2} b \,x^{13} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{13 \left (b \,x^{3}+a \right )}+\frac {3 a \,b^{2} x^{16} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{16 \left (b \,x^{3}+a \right )}+\frac {b^{3} x^{19} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{19 b \,x^{3}+19 a}\)	\(116\)

[In]

int(x^9*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/19760*x^10*(1040*b^3*x^9+3705*a*b^2*x^6+4560*a^2*b*x^3+1976*a^3)*((b*x^3+a)^2)^(3/2)/(b*x^3+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{19} \, b^{3} x^{19} + \frac {3}{16} \, a b^{2} x^{16} + \frac {3}{13} \, a^{2} b x^{13} + \frac {1}{10} \, a^{3} x^{10} \]

[In]

integrate(x^9*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/19*b^3*x^19 + 3/16*a*b^2*x^16 + 3/13*a^2*b*x^13 + 1/10*a^3*x^10

Sympy [F]

\[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int x^{9} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate(x**9*(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

Integral(x**9*((a + b*x**3)**2)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{19} \, b^{3} x^{19} + \frac {3}{16} \, a b^{2} x^{16} + \frac {3}{13} \, a^{2} b x^{13} + \frac {1}{10} \, a^{3} x^{10} \]

[In]

integrate(x^9*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/19*b^3*x^19 + 3/16*a*b^2*x^16 + 3/13*a^2*b*x^13 + 1/10*a^3*x^10

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.40 \[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{19} \, b^{3} x^{19} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{16} \, a b^{2} x^{16} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{13} \, a^{2} b x^{13} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{10} \, a^{3} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate(x^9*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

1/19*b^3*x^19*sgn(b*x^3 + a) + 3/16*a*b^2*x^16*sgn(b*x^3 + a) + 3/13*a^2*b*x^13*sgn(b*x^3 + a) + 1/10*a^3*x^10
*sgn(b*x^3 + a)

Mupad [F(-1)]

Timed out. \[ \int x^9 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int x^9\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2} \,d x \]

[In]

int(x^9*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)

[Out]

int(x^9*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2), x)