\(\int \frac {x^5}{(a^2+2 a b x^3+b^2 x^6)^{5/2}} \, dx\) [108]

Optimal result

Integrand size = 26, antiderivative size = 78 \[ \int \frac {x^5}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {a}{12 b^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {1}{9 b^2 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

1/12*a/b^2/(b*x^3+a)^3/((b*x^3+a)^2)^(1/2)-1/9/b^2/(b*x^3+a)^2/((b*x^3+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, 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 = {1369, 272, 45} \[ \int \frac {x^5}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {a}{12 b^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {1}{9 b^2 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

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

[Out]

a/(12*b^2*(a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) - 1/(9*b^2*(a + b*x^3)^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x
^6])

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 {\left (b^4 \left (a b+b^2 x^3\right )\right ) \int \frac {x^5}{\left (a b+b^2 x^3\right )^5} \, dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^3\right )\right ) \text {Subst}\left (\int \frac {x}{\left (a b+b^2 x\right )^5} \, dx,x,x^3\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^3\right )\right ) \text {Subst}\left (\int \left (-\frac {a}{b^6 (a+b x)^5}+\frac {1}{b^6 (a+b x)^4}\right ) \, dx,x,x^3\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {a}{12 b^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {1}{9 b^2 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(232\) vs. \(2(78)=156\).

Time = 0.50 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.97 \[ \int \frac {x^5}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {x^6 \left (3 \sqrt {a^2} b^6 x^{18}+3 a^3 b^3 x^9 \sqrt {\left (a+b x^3\right )^2}-3 a^2 b^4 x^{12} \sqrt {\left (a+b x^3\right )^2}+3 a b^5 x^{15} \sqrt {\left (a+b x^3\right )^2}+a^4 b^2 x^6 \left (\sqrt {a^2}-3 \sqrt {\left (a+b x^3\right )^2}\right )+6 a^6 \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )+2 a^5 b x^3 \left (2 \sqrt {a^2}+\sqrt {\left (a+b x^3\right )^2}\right )\right )}{36 a^7 \left (a+b x^3\right )^3 \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]

[In]

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

[Out]

-1/36*(x^6*(3*Sqrt[a^2]*b^6*x^18 + 3*a^3*b^3*x^9*Sqrt[(a + b*x^3)^2] - 3*a^2*b^4*x^12*Sqrt[(a + b*x^3)^2] + 3*
a*b^5*x^15*Sqrt[(a + b*x^3)^2] + a^4*b^2*x^6*(Sqrt[a^2] - 3*Sqrt[(a + b*x^3)^2]) + 6*a^6*(Sqrt[a^2] - Sqrt[(a
+ b*x^3)^2]) + 2*a^5*b*x^3*(2*Sqrt[a^2] + Sqrt[(a + b*x^3)^2])))/(a^7*(a + b*x^3)^3*(a^2 + a*b*x^3 - Sqrt[a^2]
*Sqrt[(a + b*x^3)^2]))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.40

method	result	size
pseudoelliptic	\(-\frac {\left (4 b \,x^{3}+a \right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{36 b^{2} \left (b \,x^{3}+a \right )^{4}}\)	\(31\)
gosper	\(-\frac {\left (b \,x^{3}+a \right ) \left (4 b \,x^{3}+a \right )}{36 b^{2} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(32\)
default	\(-\frac {\left (b \,x^{3}+a \right ) \left (4 b \,x^{3}+a \right )}{36 b^{2} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(32\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {x^{3}}{9 b}-\frac {a}{36 b^{2}}\right )}{\left (b \,x^{3}+a \right )^{5}}\)	\(37\)

[In]

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

[Out]

-1/36*(4*b*x^3+a)*csgn(b*x^3+a)/b^2/(b*x^3+a)^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {x^5}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {4 \, b x^{3} + a}{36 \, {\left (b^{6} x^{12} + 4 \, a b^{5} x^{9} + 6 \, a^{2} b^{4} x^{6} + 4 \, a^{3} b^{3} x^{3} + a^{4} b^{2}\right )}} \]

[In]

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

[Out]

-1/36*(4*b*x^3 + a)/(b^6*x^12 + 4*a*b^5*x^9 + 6*a^2*b^4*x^6 + 4*a^3*b^3*x^3 + a^4*b^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.55 \[ \int \frac {x^5}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {1}{9 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {a}{12 \, {\left (x^{3} + \frac {a}{b}\right )}^{4} b^{6}} \]

[In]

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

[Out]

-1/9/((b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^2) + 1/12*a/((x^3 + a/b)^4*b^6)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.41 \[ \int \frac {x^5}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {4 \, b x^{3} + a}{36 \, {\left (b x^{3} + a\right )}^{4} b^{2} \mathrm {sgn}\left (b x^{3} + a\right )} \]

[In]

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

[Out]

-1/36*(4*b*x^3 + a)/((b*x^3 + a)^4*b^2*sgn(b*x^3 + a))

Mupad [B] (verification not implemented)

Time = 8.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.54 \[ \int \frac {x^5}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {\left (4\,b\,x^3+a\right )\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{36\,b^2\,{\left (b\,x^3+a\right )}^5} \]

[In]

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

[Out]

-((a + 4*b*x^3)*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(36*b^2*(a + b*x^3)^5)