\(\int x (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\) [62]

Optimal result

Integrand size = 24, antiderivative size = 252 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a b^4 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {b^5 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )} \]

[Out]

1/2*a^5*x^2*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+a^4*b*x^5*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/4*a^3*b^2*x^8*((b*x^3+a)^2
)^(1/2)/(b*x^3+a)+10/11*a^2*b^3*x^11*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/14*a*b^4*x^14*((b*x^3+a)^2)^(1/2)/(b*x^3+
a)+1/17*b^5*x^17*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1369, 276} \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {b^5 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {5 a b^4 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {a^5 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )} \]

[In]

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

[Out]

(a^5*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(a + b*x^3)) + (a^4*b*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a + b
*x^3) + (5*a^3*b^2*x^8*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*(a + b*x^3)) + (10*a^2*b^3*x^11*Sqrt[a^2 + 2*a*b*x^
3 + b^2*x^6])/(11*(a + b*x^3)) + (5*a*b^4*x^14*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(14*(a + b*x^3)) + (b^5*x^17*S
qrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(17*(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 \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+5 a^4 b^6 x^4+10 a^3 b^7 x^7+10 a^2 b^8 x^{10}+5 a b^9 x^{13}+b^{10} x^{16}\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {a^5 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {5 a^3 b^2 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a b^4 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {b^5 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^2 \sqrt {\left (a+b x^3\right )^2} \left (2618 a^5+5236 a^4 b x^3+6545 a^3 b^2 x^6+4760 a^2 b^3 x^9+1870 a b^4 x^{12}+308 b^5 x^{15}\right )}{5236 \left (a+b x^3\right )} \]

[In]

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

[Out]

(x^2*Sqrt[(a + b*x^3)^2]*(2618*a^5 + 5236*a^4*b*x^3 + 6545*a^3*b^2*x^6 + 4760*a^2*b^3*x^9 + 1870*a*b^4*x^12 +
308*b^5*x^15))/(5236*(a + b*x^3))

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32

method	result	size
gosper	\(\frac {x^{2} \left (308 b^{5} x^{15}+1870 a \,b^{4} x^{12}+4760 a^{2} b^{3} x^{9}+6545 a^{3} b^{2} x^{6}+5236 a^{4} b \,x^{3}+2618 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{5236 \left (b \,x^{3}+a \right )^{5}}\)	\(80\)
default	\(\frac {x^{2} \left (308 b^{5} x^{15}+1870 a \,b^{4} x^{12}+4760 a^{2} b^{3} x^{9}+6545 a^{3} b^{2} x^{6}+5236 a^{4} b \,x^{3}+2618 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{5236 \left (b \,x^{3}+a \right )^{5}}\)	\(80\)
risch	\(\frac {a^{5} x^{2} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 b \,x^{3}+2 a}+\frac {a^{4} b \,x^{5} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}+\frac {5 a^{3} b^{2} x^{8} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{4 \left (b \,x^{3}+a \right )}+\frac {10 a^{2} b^{3} x^{11} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{11 \left (b \,x^{3}+a \right )}+\frac {5 a \,b^{4} x^{14} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{14 \left (b \,x^{3}+a \right )}+\frac {b^{5} x^{17} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{17 b \,x^{3}+17 a}\)	\(177\)

[In]

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

[Out]

1/5236*x^2*(308*b^5*x^15+1870*a*b^4*x^12+4760*a^2*b^3*x^9+6545*a^3*b^2*x^6+5236*a^4*b*x^3+2618*a^5)*((b*x^3+a)
^2)^(5/2)/(b*x^3+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{17} \, b^{5} x^{17} + \frac {5}{14} \, a b^{4} x^{14} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {5}{4} \, a^{3} b^{2} x^{8} + a^{4} b x^{5} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

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

[Out]

1/17*b^5*x^17 + 5/14*a*b^4*x^14 + 10/11*a^2*b^3*x^11 + 5/4*a^3*b^2*x^8 + a^4*b*x^5 + 1/2*a^5*x^2

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{17} \, b^{5} x^{17} + \frac {5}{14} \, a b^{4} x^{14} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {5}{4} \, a^{3} b^{2} x^{8} + a^{4} b x^{5} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

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

[Out]

1/17*b^5*x^17 + 5/14*a*b^4*x^14 + 10/11*a^2*b^3*x^11 + 5/4*a^3*b^2*x^8 + a^4*b*x^5 + 1/2*a^5*x^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.41 \[ \int x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{17} \, b^{5} x^{17} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{14} \, a b^{4} x^{14} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{11} \, a^{2} b^{3} x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{4} \, a^{3} b^{2} x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{4} b x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a^{5} x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

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

[Out]

1/17*b^5*x^17*sgn(b*x^3 + a) + 5/14*a*b^4*x^14*sgn(b*x^3 + a) + 10/11*a^2*b^3*x^11*sgn(b*x^3 + a) + 5/4*a^3*b^
2*x^8*sgn(b*x^3 + a) + a^4*b*x^5*sgn(b*x^3 + a) + 1/2*a^5*x^2*sgn(b*x^3 + a)

Mupad [F(-1)]

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

[In]

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

[Out]

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