\(\int \frac {1}{x (a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [651]

Optimal result

Integrand size = 26, antiderivative size = 223 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {1}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{8 a \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{6 a^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log (x)}{a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 223, 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} \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {1}{6 a^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{8 a \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\log (x) \left (a+b x^2\right )}{a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

1/(2*a^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(8*a*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(6*a^2*(
a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a +
b*x^2)*Log[x])/(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((a + b*x^2)*Log[a + b*x^2])/(2*a^5*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 \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 \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}-\frac {1}{a b^4 (a+b x)^5}-\frac {1}{a^2 b^4 (a+b x)^4}-\frac {1}{a^3 b^4 (a+b x)^3}-\frac {1}{a^4 b^4 (a+b x)^2}-\frac {1}{a^5 b^4 (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {1}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{8 a \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{6 a^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a^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 (x)}{a^5 \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 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {a \left (25 a^3+52 a^2 b x^2+42 a b^2 x^4+12 b^3 x^6\right )+24 \left (a+b x^2\right )^4 \log (x)-12 \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 a^5 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \]

[In]

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

[Out]

(a*(25*a^3 + 52*a^2*b*x^2 + 42*a*b^2*x^4 + 12*b^3*x^6) + 24*(a + b*x^2)^4*Log[x] - 12*(a + b*x^2)^4*Log[a + b*
x^2])/(24*a^5*(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 = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.40

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-\left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )+\left (b \,x^{2}+a \right )^{4} \ln \left (x^{2}\right )+a \,b^{3} x^{6}+\frac {7 a^{2} b^{2} x^{4}}{2}+\frac {13 a^{3} b \,x^{2}}{3}+\frac {25 a^{4}}{12}\right )}{2 \left (b \,x^{2}+a \right )^{4} a^{5}}\)	\(90\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b^{3} x^{6}}{2 a^{4}}+\frac {7 b^{2} x^{4}}{4 a^{3}}+\frac {13 b \,x^{2}}{6 a^{2}}+\frac {25}{24 a}\right )}{\left (b \,x^{2}+a \right )^{5}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{5}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{5}}\)	\(119\)
default	\(\frac {\left (24 b^{4} \ln \left (x \right ) x^{8}-12 \ln \left (b \,x^{2}+a \right ) x^{8} b^{4}+96 a \,b^{3} \ln \left (x \right ) x^{6}-48 \ln \left (b \,x^{2}+a \right ) x^{6} a \,b^{3}+12 a \,b^{3} x^{6}+144 a^{2} b^{2} \ln \left (x \right ) x^{4}-72 \ln \left (b \,x^{2}+a \right ) x^{4} a^{2} b^{2}+42 a^{2} b^{2} x^{4}+96 a^{3} b \ln \left (x \right ) x^{2}-48 \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b +52 a^{3} b \,x^{2}+24 a^{4} \ln \left (x \right )-12 \ln \left (b \,x^{2}+a \right ) a^{4}+25 a^{4}\right ) \left (b \,x^{2}+a \right )}{24 a^{5} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(193\)

[In]

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

[Out]

1/2*csgn(b*x^2+a)*(-(b*x^2+a)^4*ln(b*x^2+a)+(b*x^2+a)^4*ln(x^2)+a*b^3*x^6+7/2*a^2*b^2*x^4+13/3*a^3*b*x^2+25/12
*a^4)/(b*x^2+a)^4/a^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {12 \, a b^{3} x^{6} + 42 \, a^{2} b^{2} x^{4} + 52 \, a^{3} b x^{2} + 25 \, a^{4} - 12 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right ) + 24 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (x\right )}{24 \, {\left (a^{5} b^{4} x^{8} + 4 \, a^{6} b^{3} x^{6} + 6 \, a^{7} b^{2} x^{4} + 4 \, a^{8} b x^{2} + a^{9}\right )}} \]

[In]

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

[Out]

1/24*(12*a*b^3*x^6 + 42*a^2*b^2*x^4 + 52*a^3*b*x^2 + 25*a^4 - 12*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^
3*b*x^2 + a^4)*log(b*x^2 + a) + 24*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*log(x))/(a^5*b^
4*x^8 + 4*a^6*b^3*x^6 + 6*a^7*b^2*x^4 + 4*a^8*b*x^2 + a^9)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {12 \, b^{3} x^{6} + 42 \, a b^{2} x^{4} + 52 \, a^{2} b x^{2} + 25 \, a^{3}}{24 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{5}} + \frac {\log \left (x\right )}{a^{5}} \]

[In]

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

[Out]

1/24*(12*b^3*x^6 + 42*a*b^2*x^4 + 52*a^2*b*x^2 + 25*a^3)/(a^4*b^4*x^8 + 4*a^5*b^3*x^6 + 6*a^6*b^2*x^4 + 4*a^7*
b*x^2 + a^8) - 1/2*log(b*x^2 + a)/a^5 + log(x)/a^5

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\log \left (x^{2}\right )}{2 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {25 \, b^{4} x^{8} + 112 \, a b^{3} x^{6} + 192 \, a^{2} b^{2} x^{4} + 152 \, a^{3} b x^{2} + 50 \, a^{4}}{24 \, {\left (b x^{2} + a\right )}^{4} a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

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

[Out]

1/2*log(x^2)/(a^5*sgn(b*x^2 + a)) - 1/2*log(abs(b*x^2 + a))/(a^5*sgn(b*x^2 + a)) + 1/24*(25*b^4*x^8 + 112*a*b^
3*x^6 + 192*a^2*b^2*x^4 + 152*a^3*b*x^2 + 50*a^4)/((b*x^2 + a)^4*a^5*sgn(b*x^2 + a))

Mupad [F(-1)]

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

[In]

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

[Out]

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