\(\int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [541]

Optimal result

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

[Out]

1/6*a*x^6*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/8*b*x^8*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {b x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {a x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )} \]

[In]

Int[x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(a*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*(a + b*x^2)) + (b*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*(a + b*x^
2))

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 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int x^2 \left (a b+b^2 x\right ) \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (a b x^2+b^2 x^3\right ) \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )} \\ & = \frac {a x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {b x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {x^6 \left (4 a+3 b x^2\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(x^6*(4*a + 3*b*x^2)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])))/(24*(-a^2 - a*b*x^2 + Sqrt[a^2]*
Sqrt[(a + b*x^2)^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 = 24, normalized size of antiderivative = 0.30

method	result	size
pseudoelliptic	\(\frac {x^{6} \left (3 b \,x^{2}+4 a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{24}\)	\(24\)
gosper	\(\frac {x^{6} \left (3 b \,x^{2}+4 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{24 b \,x^{2}+24 a}\)	\(36\)
default	\(\frac {x^{6} \left (3 b \,x^{2}+4 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{24 b \,x^{2}+24 a}\)	\(36\)
risch	\(\frac {a \,x^{6} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{6 b \,x^{2}+6 a}+\frac {b \,x^{8} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{8 b \,x^{2}+8 a}\)	\(54\)

[In]

int(x^5*((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*x^6*(3*b*x^2+4*a)*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.16 \[ \int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{8} \, b x^{8} + \frac {1}{6} \, a x^{6} \]

[In]

integrate(x^5*((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/8*b*x^8 + 1/6*a*x^6

Sympy [F]

\[ \int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\int x^{5} \sqrt {\left (a + b x^{2}\right )^{2}}\, dx \]

[In]

integrate(x**5*((b*x**2+a)**2)**(1/2),x)

[Out]

Integral(x**5*sqrt((a + b*x**2)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.16 \[ \int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{8} \, b x^{8} + \frac {1}{6} \, a x^{6} \]

[In]

integrate(x^5*((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/8*b*x^8 + 1/6*a*x^6

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.37 \[ \int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{8} \, b x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{6} \, a x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^5*((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/8*b*x^8*sgn(b*x^2 + a) + 1/6*a*x^6*sgn(b*x^2 + a)

Mupad [B] (verification not implemented)

Time = 13.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int x^5 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (a^3-4\,a^2\,b\,x^2-5\,a\,b^2\,x^4+3\,b\,x^2\,\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )\right )}{24\,b^3} \]

[In]

int(x^5*((a + b*x^2)^2)^(1/2),x)

[Out]

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