\(\int x^6 (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\) [57]

Optimal result

Integrand size = 26, antiderivative size = 255 \[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}+\frac {a^4 b x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {5 a b^4 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {b^5 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 \left (a+b x^3\right )} \]

[Out]

1/7*a^5*x^7*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/2*a^4*b*x^10*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+10/13*a^3*b^2*x^13*((b*
x^3+a)^2)^(1/2)/(b*x^3+a)+5/8*a^2*b^3*x^16*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/19*a*b^4*x^19*((b*x^3+a)^2)^(1/2)/(
b*x^3+a)+1/22*b^5*x^22*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 255, 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} \[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {b^5 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 \left (a+b x^3\right )}+\frac {5 a b^4 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {a^5 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}+\frac {a^4 b x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )} \]

[In]

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

[Out]

(a^5*x^7*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*(a + b*x^3)) + (a^4*b*x^10*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(a
+ b*x^3)) + (10*a^3*b^2*x^13*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(13*(a + b*x^3)) + (5*a^2*b^3*x^16*Sqrt[a^2 + 2
*a*b*x^3 + b^2*x^6])/(8*(a + b*x^3)) + (5*a*b^4*x^19*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(19*(a + b*x^3)) + (b^5*
x^22*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(22*(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^6 \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} \int \left (a^5 b^5 x^6+5 a^4 b^6 x^9+10 a^3 b^7 x^{12}+10 a^2 b^8 x^{15}+5 a b^9 x^{18}+b^{10} x^{21}\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {a^5 x^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}+\frac {a^4 b x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {5 a b^4 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {b^5 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^7 \sqrt {\left (a+b x^3\right )^2} \left (21736 a^5+76076 a^4 b x^3+117040 a^3 b^2 x^6+95095 a^2 b^3 x^9+40040 a b^4 x^{12}+6916 b^5 x^{15}\right )}{152152 \left (a+b x^3\right )} \]

[In]

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

[Out]

(x^7*Sqrt[(a + b*x^3)^2]*(21736*a^5 + 76076*a^4*b*x^3 + 117040*a^3*b^2*x^6 + 95095*a^2*b^3*x^9 + 40040*a*b^4*x
^12 + 6916*b^5*x^15))/(152152*(a + b*x^3))

Maple [A] (verified)

Time = 5.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31

method	result	size
gosper	\(\frac {x^{7} \left (6916 b^{5} x^{15}+40040 a \,b^{4} x^{12}+95095 a^{2} b^{3} x^{9}+117040 a^{3} b^{2} x^{6}+76076 a^{4} b \,x^{3}+21736 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{152152 \left (b \,x^{3}+a \right )^{5}}\)	\(80\)
default	\(\frac {x^{7} \left (6916 b^{5} x^{15}+40040 a \,b^{4} x^{12}+95095 a^{2} b^{3} x^{9}+117040 a^{3} b^{2} x^{6}+76076 a^{4} b \,x^{3}+21736 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{152152 \left (b \,x^{3}+a \right )^{5}}\)	\(80\)
risch	\(\frac {a^{5} x^{7} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{7 b \,x^{3}+7 a}+\frac {a^{4} b \,x^{10} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 b \,x^{3}+2 a}+\frac {10 a^{3} b^{2} x^{13} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{13 \left (b \,x^{3}+a \right )}+\frac {5 a^{2} b^{3} x^{16} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{8 \left (b \,x^{3}+a \right )}+\frac {5 a \,b^{4} x^{19} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{19 \left (b \,x^{3}+a \right )}+\frac {b^{5} x^{22} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{22 b \,x^{3}+22 a}\)	\(178\)

[In]

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

[Out]

1/152152*x^7*(6916*b^5*x^15+40040*a*b^4*x^12+95095*a^2*b^3*x^9+117040*a^3*b^2*x^6+76076*a^4*b*x^3+21736*a^5)*(
(b*x^3+a)^2)^(5/2)/(b*x^3+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} + \frac {5}{19} \, a b^{4} x^{19} + \frac {5}{8} \, a^{2} b^{3} x^{16} + \frac {10}{13} \, a^{3} b^{2} x^{13} + \frac {1}{2} \, a^{4} b x^{10} + \frac {1}{7} \, a^{5} x^{7} \]

[In]

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

[Out]

1/22*b^5*x^22 + 5/19*a*b^4*x^19 + 5/8*a^2*b^3*x^16 + 10/13*a^3*b^2*x^13 + 1/2*a^4*b*x^10 + 1/7*a^5*x^7

Sympy [F]

\[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{6} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} + \frac {5}{19} \, a b^{4} x^{19} + \frac {5}{8} \, a^{2} b^{3} x^{16} + \frac {10}{13} \, a^{3} b^{2} x^{13} + \frac {1}{2} \, a^{4} b x^{10} + \frac {1}{7} \, a^{5} x^{7} \]

[In]

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

[Out]

1/22*b^5*x^22 + 5/19*a*b^4*x^19 + 5/8*a^2*b^3*x^16 + 10/13*a^3*b^2*x^13 + 1/2*a^4*b*x^10 + 1/7*a^5*x^7

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{19} \, a b^{4} x^{19} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{8} \, a^{2} b^{3} x^{16} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{13} \, a^{3} b^{2} x^{13} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a^{4} b x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{7} \, a^{5} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

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

[Out]

1/22*b^5*x^22*sgn(b*x^3 + a) + 5/19*a*b^4*x^19*sgn(b*x^3 + a) + 5/8*a^2*b^3*x^16*sgn(b*x^3 + a) + 10/13*a^3*b^
2*x^13*sgn(b*x^3 + a) + 1/2*a^4*b*x^10*sgn(b*x^3 + a) + 1/7*a^5*x^7*sgn(b*x^3 + a)

Mupad [F(-1)]

Timed out. \[ \int x^6 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^6\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]

[In]

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

[Out]

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