\(\int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx\) [7]

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {a x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {b x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )} \]

[Out]

1/5*a*x^5*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/8*b*x^8*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 79, 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, 14} \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {b x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {a x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )} \]

[In]

Int[x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.49 \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (8 a x^5+5 b x^8\right )}{40 \left (a+b x^3\right )} \]

[In]

Integrate[x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

(Sqrt[(a + b*x^3)^2]*(8*a*x^5 + 5*b*x^8))/(40*(a + b*x^3))

Maple [A] (verified)

Time = 4.60 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.46

method	result	size
gosper	\(\frac {x^{5} \left (5 b \,x^{3}+8 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{40 b \,x^{3}+40 a}\)	\(36\)
default	\(\frac {x^{5} \left (5 b \,x^{3}+8 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{40 b \,x^{3}+40 a}\)	\(36\)
risch	\(\frac {a \,x^{5} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{5 b \,x^{3}+5 a}+\frac {b \,x^{8} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{8 b \,x^{3}+8 a}\)	\(54\)

[In]

int(x^4*((b*x^3+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/40*x^5*(5*b*x^3+8*a)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.16 \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {1}{8} \, b x^{8} + \frac {1}{5} \, a x^{5} \]

[In]

integrate(x^4*((b*x^3+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/8*b*x^8 + 1/5*a*x^5

Sympy [F(-1)]

Timed out. \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\text {Timed out} \]

[In]

integrate(x**4*((b*x**3+a)**2)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.16 \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {1}{8} \, b x^{8} + \frac {1}{5} \, a x^{5} \]

[In]

integrate(x^4*((b*x^3+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/8*b*x^8 + 1/5*a*x^5

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.37 \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {1}{8} \, b x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{5} \, a x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate(x^4*((b*x^3+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/8*b*x^8*sgn(b*x^3 + a) + 1/5*a*x^5*sgn(b*x^3 + a)

Mupad [F(-1)]

Timed out. \[ \int x^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\int x^4\,\sqrt {{\left (b\,x^3+a\right )}^2} \,d x \]

[In]

int(x^4*((a + b*x^3)^2)^(1/2),x)

[Out]

int(x^4*((a + b*x^3)^2)^(1/2), x)