\(\int \frac {1}{x^3 (a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [652]

Optimal result

Integrand size = 26, antiderivative size = 267 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {2 b}{a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{8 a^2 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{3 a^3 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 b}{4 a^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{2 a^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 b \left (a+b x^2\right ) \log (x)}{a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 267, 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 = {1126, 272, 46} \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {b}{8 a^2 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 b \log (x) \left (a+b x^2\right )}{a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {2 b}{a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{2 a^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 b}{4 a^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{3 a^3 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

Int[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

(-2*b)/(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - b/(8*a^2*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - b/(3*
a^3*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (3*b)/(4*a^4*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
- (a + b*x^2)/(2*a^5*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (5*b*(a + b*x^2)*Log[x])/(a^6*Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4]) + (5*b*(a + b*x^2)*Log[a + b*x^2])/(2*a^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && 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^2\right )\right ) \int \frac {1}{x^3 \left (a b+b^2 x^2\right )^5} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (a b+b^2 x\right )^5} \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \left (\frac {1}{a^5 b^5 x^2}-\frac {5}{a^6 b^4 x}+\frac {1}{a^2 b^3 (a+b x)^5}+\frac {2}{a^3 b^3 (a+b x)^4}+\frac {3}{a^4 b^3 (a+b x)^3}+\frac {4}{a^5 b^3 (a+b x)^2}+\frac {5}{a^6 b^3 (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {2 b}{a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{8 a^2 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{3 a^3 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 b}{4 a^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{2 a^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 b \left (a+b x^2\right ) \log (x)}{a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-a \left (12 a^4+125 a^3 b x^2+260 a^2 b^2 x^4+210 a b^3 x^6+60 b^4 x^8\right )-120 b x^2 \left (a+b x^2\right )^4 \log (x)+60 b x^2 \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 a^6 x^2 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \]

[In]

Integrate[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

(-(a*(12*a^4 + 125*a^3*b*x^2 + 260*a^2*b^2*x^4 + 210*a*b^3*x^6 + 60*b^4*x^8)) - 120*b*x^2*(a + b*x^2)^4*Log[x]
+ 60*b*x^2*(a + b*x^2)^4*Log[a + b*x^2])/(24*a^6*x^2*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

Maple [C] (warning: unable to verify)

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

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.42

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-5 b \,x^{2} \left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )+5 b \,x^{2} \left (b \,x^{2}+a \right )^{4} \ln \left (x^{2}\right )+a \left (5 b^{4} x^{8}+\frac {35}{2} a \,b^{3} x^{6}+\frac {65}{3} a^{2} b^{2} x^{4}+\frac {125}{12} a^{3} b \,x^{2}+a^{4}\right )\right )}{2 \left (b \,x^{2}+a \right )^{4} x^{2} a^{6}}\)	\(112\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {5 b^{4} x^{8}}{2 a^{5}}-\frac {35 b^{3} x^{6}}{4 a^{4}}-\frac {65 b^{2} x^{4}}{6 a^{3}}-\frac {125 b \,x^{2}}{24 a^{2}}-\frac {1}{2 a}\right )}{\left (b \,x^{2}+a \right )^{5} x^{2}}-\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{6}}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (-b \,x^{2}-a \right )}{2 \left (b \,x^{2}+a \right ) a^{6}}\)	\(139\)
default	\(-\frac {\left (120 \ln \left (x \right ) x^{10} b^{5}-60 \ln \left (b \,x^{2}+a \right ) x^{10} b^{5}+480 \ln \left (x \right ) x^{8} a \,b^{4}-240 \ln \left (b \,x^{2}+a \right ) x^{8} a \,b^{4}+60 a \,x^{8} b^{4}+720 \ln \left (x \right ) x^{6} a^{2} b^{3}-360 \ln \left (b \,x^{2}+a \right ) x^{6} a^{2} b^{3}+210 a^{2} x^{6} b^{3}+480 \ln \left (x \right ) x^{4} a^{3} b^{2}-240 \ln \left (b \,x^{2}+a \right ) x^{4} a^{3} b^{2}+260 a^{3} x^{4} b^{2}+120 \ln \left (x \right ) x^{2} a^{4} b -60 \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b +125 x^{2} a^{4} b +12 a^{5}\right ) \left (b \,x^{2}+a \right )}{24 x^{2} a^{6} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(219\)

[In]

int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*csgn(b*x^2+a)*(-5*b*x^2*(b*x^2+a)^4*ln(b*x^2+a)+5*b*x^2*(b*x^2+a)^4*ln(x^2)+a*(5*b^4*x^8+35/2*a*b^3*x^6+6
5/3*a^2*b^2*x^4+125/12*a^3*b*x^2+a^4))/(b*x^2+a)^4/x^2/a^6

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {60 \, a b^{4} x^{8} + 210 \, a^{2} b^{3} x^{6} + 260 \, a^{3} b^{2} x^{4} + 125 \, a^{4} b x^{2} + 12 \, a^{5} - 60 \, {\left (b^{5} x^{10} + 4 \, a b^{4} x^{8} + 6 \, a^{2} b^{3} x^{6} + 4 \, a^{3} b^{2} x^{4} + a^{4} b x^{2}\right )} \log \left (b x^{2} + a\right ) + 120 \, {\left (b^{5} x^{10} + 4 \, a b^{4} x^{8} + 6 \, a^{2} b^{3} x^{6} + 4 \, a^{3} b^{2} x^{4} + a^{4} b x^{2}\right )} \log \left (x\right )}{24 \, {\left (a^{6} b^{4} x^{10} + 4 \, a^{7} b^{3} x^{8} + 6 \, a^{8} b^{2} x^{6} + 4 \, a^{9} b x^{4} + a^{10} x^{2}\right )}} \]

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/24*(60*a*b^4*x^8 + 210*a^2*b^3*x^6 + 260*a^3*b^2*x^4 + 125*a^4*b*x^2 + 12*a^5 - 60*(b^5*x^10 + 4*a*b^4*x^8
+ 6*a^2*b^3*x^6 + 4*a^3*b^2*x^4 + a^4*b*x^2)*log(b*x^2 + a) + 120*(b^5*x^10 + 4*a*b^4*x^8 + 6*a^2*b^3*x^6 + 4*
a^3*b^2*x^4 + a^4*b*x^2)*log(x))/(a^6*b^4*x^10 + 4*a^7*b^3*x^8 + 6*a^8*b^2*x^6 + 4*a^9*b*x^4 + a^10*x^2)

Sympy [F]

\[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/x**3/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(1/(x**3*((a + b*x**2)**2)**(5/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {60 \, b^{4} x^{8} + 210 \, a b^{3} x^{6} + 260 \, a^{2} b^{2} x^{4} + 125 \, a^{3} b x^{2} + 12 \, a^{4}}{24 \, {\left (a^{5} b^{4} x^{10} + 4 \, a^{6} b^{3} x^{8} + 6 \, a^{7} b^{2} x^{6} + 4 \, a^{8} b x^{4} + a^{9} x^{2}\right )}} + \frac {5 \, b \log \left (b x^{2} + a\right )}{2 \, a^{6}} - \frac {5 \, b \log \left (x\right )}{a^{6}} \]

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/24*(60*b^4*x^8 + 210*a*b^3*x^6 + 260*a^2*b^2*x^4 + 125*a^3*b*x^2 + 12*a^4)/(a^5*b^4*x^10 + 4*a^6*b^3*x^8 +
6*a^7*b^2*x^6 + 4*a^8*b*x^4 + a^9*x^2) + 5/2*b*log(b*x^2 + a)/a^6 - 5*b*log(x)/a^6

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {5 \, b \log \left (x^{2}\right )}{2 \, a^{6} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {5 \, b \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{6} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {5 \, b x^{2} - a}{2 \, a^{6} x^{2} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {125 \, b^{5} x^{8} + 548 \, a b^{4} x^{6} + 912 \, a^{2} b^{3} x^{4} + 688 \, a^{3} b^{2} x^{2} + 202 \, a^{4} b}{24 \, {\left (b x^{2} + a\right )}^{4} a^{6} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

-5/2*b*log(x^2)/(a^6*sgn(b*x^2 + a)) + 5/2*b*log(abs(b*x^2 + a))/(a^6*sgn(b*x^2 + a)) + 1/2*(5*b*x^2 - a)/(a^6
*x^2*sgn(b*x^2 + a)) - 1/24*(125*b^5*x^8 + 548*a*b^4*x^6 + 912*a^2*b^3*x^4 + 688*a^3*b^2*x^2 + 202*a^4*b)/((b*
x^2 + a)^4*a^6*sgn(b*x^2 + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x^3\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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