\(\int \frac {x^5}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [648]

Optimal result

Integrand size = 26, antiderivative size = 74 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x^6}{24 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac {x^6}{8 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[Out]

1/24*x^6/a^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2)+1/8*x^6/a/(b*x^2+a)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1123} \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x^6}{8 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac {x^6}{24 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[In]

Int[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

x^6/(24*a^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)) + x^6/(8*a*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))

Rule 1123

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[2*(d*x)^(m + 1)*((a + b*x^
2 + c*x^4)^(p + 1)/(d*(m + 3)*(2*a + b*x^2))), x] - Simp[(d*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*d*(m
+ 3)*(p + 1))), x] /; FreeQ[{a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[m + 4*p + 5,
0] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^6}{24 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac {x^6}{8 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(206\) vs. \(2(74)=148\).

Time = 0.45 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.78 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x^6 \left (-4 a^6-a^5 b x^2+4 a^4 \sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}-3 a^3 \sqrt {a^2} b x^2 \sqrt {\left (a+b x^2\right )^2}+3 \sqrt {a^2} b^2 x^4 \sqrt {\left (a+b x^2\right )^2} \left (a^2+b^2 x^4\right )+3 a b^3 x^6 \left (b^2 x^4-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 a^6 \left (a+b x^2\right )^3 \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )} \]

[In]

Integrate[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^6*(-4*a^6 - a^5*b*x^2 + 4*a^4*Sqrt[a^2]*Sqrt[(a + b*x^2)^2] - 3*a^3*Sqrt[a^2]*b*x^2*Sqrt[(a + b*x^2)^2] + 3
*Sqrt[a^2]*b^2*x^4*Sqrt[(a + b*x^2)^2]*(a^2 + b^2*x^4) + 3*a*b^3*x^6*(b^2*x^4 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2])
))/(24*a^6*(a + b*x^2)^3*(Sqrt[a^2]*b*x^2 + a*(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 = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.57

method	result	size
pseudoelliptic	\(-\frac {\left (6 b^{2} x^{4}+4 a b \,x^{2}+a^{2}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{24 \left (b \,x^{2}+a \right )^{4} b^{3}}\)	\(42\)
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (6 b^{2} x^{4}+4 a b \,x^{2}+a^{2}\right )}{24 b^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(43\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (6 b^{2} x^{4}+4 a b \,x^{2}+a^{2}\right )}{24 b^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(43\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {x^{4}}{4 b}-\frac {a \,x^{2}}{6 b^{2}}-\frac {a^{2}}{24 b^{3}}\right )}{\left (b \,x^{2}+a \right )^{5}}\)	\(48\)

[In]

int(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {6 \, b^{2} x^{4} + 4 \, a b x^{2} + a^{2}}{24 \, {\left (b^{7} x^{8} + 4 \, a b^{6} x^{6} + 6 \, a^{2} b^{5} x^{4} + 4 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}} \]

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/24*(6*b^2*x^4 + 4*a*b*x^2 + a^2)/(b^7*x^8 + 4*a*b^6*x^6 + 6*a^2*b^5*x^4 + 4*a^3*b^4*x^2 + a^4*b^3)

Sympy [F]

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

[In]

integrate(x**5/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {6 \, b^{2} x^{4} + 4 \, a b x^{2} + a^{2}}{24 \, {\left (b^{7} x^{8} + 4 \, a b^{6} x^{6} + 6 \, a^{2} b^{5} x^{4} + 4 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}} \]

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/24*(6*b^2*x^4 + 4*a*b*x^2 + a^2)/(b^7*x^8 + 4*a*b^6*x^6 + 6*a^2*b^5*x^4 + 4*a^3*b^4*x^2 + a^4*b^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.58 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {6 \, b^{2} x^{4} + 4 \, a b x^{2} + a^{2}}{24 \, {\left (b x^{2} + a\right )}^{4} b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.64 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (a^2+4\,a\,b\,x^2+6\,b^2\,x^4\right )}{24\,b^3\,{\left (b\,x^2+a\right )}^5} \]

[In]

int(x^5/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

-((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*(a^2 + 6*b^2*x^4 + 4*a*b*x^2))/(24*b^3*(a + b*x^2)^5)