\(\int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [628]

Optimal result

Integrand size = 26, antiderivative size = 125 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {-a-b x^2}{2 a x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b \left (a+b x^2\right ) \log (x)}{a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98, 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} \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {a+b x^2}{2 a x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b \log (x) \left (a+b x^2\right )}{a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {a^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}+2 a b x^2 \log \left (x^2\right )+\left (-a+\sqrt {a^2}\right ) b x^2 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-a b x^2 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-\sqrt {a^2} b x^2 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )}{4 \left (a^2\right )^{3/2} x^2} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.34

method	result	size
pseudoelliptic	\(-\frac {\left (b \ln \left (x^{2}\right ) x^{2}-\ln \left (b \,x^{2}+a \right ) x^{2} b +a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 x^{2} a^{2}}\)	\(42\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 \ln \left (x \right ) x^{2} b -\ln \left (b \,x^{2}+a \right ) x^{2} b +a \right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} x^{2}}\)	\(51\)
risch	\(-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 \left (b \,x^{2}+a \right ) a \,x^{2}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{2}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (-b \,x^{2}-a \right )}{2 \left (b \,x^{2}+a \right ) a^{2}}\)	\(95\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {b x^{2} \log \left (b x^{2} + a\right ) - 2 \, b x^{2} \log \left (x\right ) - a}{2 \, a^{2} x^{2}} \]

[In]

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

[Out]

1/2*(b*x^2*log(b*x^2 + a) - 2*b*x^2*log(x) - a)/(a^2*x^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {b \log \left (b x^{2} + a\right )}{2 \, a^{2}} - \frac {b \log \left (x^{2}\right )}{2 \, a^{2}} - \frac {1}{2 \, a x^{2}} \]

[In]

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

[Out]

1/2*b*log(b*x^2 + a)/a^2 - 1/2*b*log(x^2)/a^2 - 1/2/(a*x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {1}{2} \, {\left (\frac {b \log \left (x^{2}\right )}{a^{2}} - \frac {b \log \left ({\left | b x^{2} + a \right |}\right )}{a^{2}} - \frac {b x^{2} - a}{a^{2} x^{2}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.49 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,a\,x^2}{\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}\right )}{2\,{\left (a^2\right )}^{3/2}}-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,a^2\,x^2} \]

[In]

int(1/(x^3*((a + b*x^2)^2)^(1/2)),x)

[Out]

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