\(\int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) [94]

Optimal result

Integrand size = 26, antiderivative size = 80 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (a+b x^3\right ) \log (x)}{a \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1369, 272, 36, 29, 31} \[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\log (x) \left (a+b x^3\right )}{a \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

Int[1/(x*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a b+b^2 x^3\right ) \int \frac {1}{x \left (a b+b^2 x^3\right )} \, dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (a b+b^2 x^3\right ) \text {Subst}\left (\int \frac {1}{x \left (a b+b^2 x\right )} \, dx,x,x^3\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (a b+b^2 x^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^3\right )}{3 a b \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (b \left (a b+b^2 x^3\right )\right ) \text {Subst}\left (\int \frac {1}{a b+b^2 x} \, dx,x,x^3\right )}{3 a \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (a+b x^3\right ) \log (x)}{a \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.68 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {-2 a \log \left (x^3\right )+\left (a-\sqrt {a^2}\right ) \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+a \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+\sqrt {a^2} \log \left (a \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{6 a \sqrt {a^2}} \]

[In]

Integrate[1/(x*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.39

method	result	size
pseudoelliptic	\(-\frac {\left (\ln \left (b \,x^{3}+a \right )-\ln \left (b \,x^{3}\right )\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{3 a}\)	\(31\)
default	\(\frac {\left (b \,x^{3}+a \right ) \left (3 \ln \left (x \right )-\ln \left (b \,x^{3}+a \right )\right )}{3 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a}\)	\(39\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right ) a}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {\log \left (b x^{3} + a\right ) - 3 \, \log \left (x\right )}{3 \, a} \]

[In]

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

[Out]

-1/3*(log(b*x^3 + a) - 3*log(x))/a

Sympy [F]

\[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {1}{x \sqrt {\left (a + b x^{3}\right )^{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {1}{3} \, {\left (\frac {\log \left ({\left | b x^{3} + a \right |}\right )}{a} - \frac {3 \, \log \left ({\left | x \right |}\right )}{a}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.57 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {\ln \left (a\,b+\frac {a^2}{x^3}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{x^3}\right )}{3\,\sqrt {a^2}} \]

[In]

int(1/(x*((a + b*x^3)^2)^(1/2)),x)

[Out]

-log(a*b + a^2/x^3 + ((a^2)^(1/2)*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/x^3)/(3*(a^2)^(1/2))