\(\int \frac {x^5}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [635]

Optimal result

Integrand size = 26, antiderivative size = 113 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {a}{b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^2}{4 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 113, 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} \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {a^2}{4 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a}{b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

a/(b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - a^2/(4*b^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a + b*x^
2)*Log[a + b*x^2])/(2*b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.70 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\frac {b x^2 \left (a \sqrt {\left (a+b x^2\right )^2} \left (-2 a^2-a b x^2+b^2 x^4\right )+\sqrt {a^2} \left (2 a^3+3 a^2 b x^2+b^3 x^6\right )\right )}{a^2 \left (a+b x^2\right ) \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )}+2 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-2 \log \left (b^3 \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{4 b^3} \]

[In]

Integrate[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

((b*x^2*(a*Sqrt[(a + b*x^2)^2]*(-2*a^2 - a*b*x^2 + b^2*x^4) + Sqrt[a^2]*(2*a^3 + 3*a^2*b*x^2 + b^3*x^6)))/(a^2
*(a + b*x^2)*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2])) + 2*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]]
- 2*Log[b^3*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/(4*b^3)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.48

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\left (b \,x^{2}+a \right )^{2} \ln \left (b \,x^{2}+a \right )+2 a b \,x^{2}+\frac {3 a^{2}}{2}\right )}{2 \left (b \,x^{2}+a \right )^{2} b^{3}}\)	\(54\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {a \,x^{2}}{b^{2}}+\frac {3 a^{2}}{4 b^{3}}\right )}{\left (b \,x^{2}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{3}}\)	\(73\)
default	\(\frac {\left (2 \ln \left (b \,x^{2}+a \right ) x^{4} b^{2}+4 \ln \left (b \,x^{2}+a \right ) x^{2} a b +4 a b \,x^{2}+2 \ln \left (b \,x^{2}+a \right ) a^{2}+3 a^{2}\right ) \left (b \,x^{2}+a \right )}{4 b^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(81\)

[In]

int(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*csgn(b*x^2+a)*((b*x^2+a)^2*ln(b*x^2+a)+2*a*b*x^2+3/2*a^2)/(b*x^2+a)^2/b^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {4 \, a b x^{2} + 3 \, a^{2} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \]

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**5/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.49 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {4 \, a b x^{2} + 3 \, a^{2}}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{3}} \]

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.55 \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {3 \, b x^{4} + 2 \, a x^{2}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^5}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]

[In]

int(x^5/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int(x^5/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)