\(\int \frac {1}{x (a^2+2 a b x^3+b^2 x^6)^{3/2}} \, dx\) [103]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {1}{3 a^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{6 a \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a+b x^3\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 147, 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 = {1369, 272, 46} \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {1}{6 a \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{3 a^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\log (x) \left (a+b x^3\right )}{a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

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

[Out]

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(790\) vs. \(2(147)=294\).

Time = 1.10 (sec) , antiderivative size = 790, normalized size of antiderivative = 5.37 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {4 a^4 b x^3+3 a^3 b^2 x^6-a b^4 x^{12}-4 \left (a^2\right )^{3/2} b x^3 \sqrt {\left (a+b x^3\right )^2}+a \sqrt {a^2} b^2 x^6 \sqrt {\left (a+b x^3\right )^2}-\sqrt {a^2} b^3 x^9 \sqrt {\left (a+b x^3\right )^2}+2 \left (\left (a^2\right )^{3/2} b^2 x^6+a^4 \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )+a^3 b x^3 \left (2 \sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right ) \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-2 \left (a^5+2 a^4 b x^3-\left (a^2\right )^{3/2} b x^3 \sqrt {\left (a+b x^3\right )^2}+a^3 \left (b^2 x^6-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )\right ) \log \left (x^3\right )+a^5 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+2 a^4 b x^3 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+a^3 b^2 x^6 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-a^3 \sqrt {a^2} \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-\left (a^2\right )^{3/2} b x^3 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+a^5 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+2 a^4 b x^3 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+a^3 b^2 x^6 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-a^3 \sqrt {a^2} \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-\left (a^2\right )^{3/2} b x^3 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )}{3 a^3 \sqrt {a^2} \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )^2} \]

[In]

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

[Out]

(4*a^4*b*x^3 + 3*a^3*b^2*x^6 - a*b^4*x^12 - 4*(a^2)^(3/2)*b*x^3*Sqrt[(a + b*x^3)^2] + a*Sqrt[a^2]*b^2*x^6*Sqrt
[(a + b*x^3)^2] - Sqrt[a^2]*b^3*x^9*Sqrt[(a + b*x^3)^2] + 2*((a^2)^(3/2)*b^2*x^6 + a^4*(Sqrt[a^2] - Sqrt[(a +
b*x^3)^2]) + a^3*b*x^3*(2*Sqrt[a^2] - Sqrt[(a + b*x^3)^2]))*ArcTanh[(b*x^3)/(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])]
- 2*(a^5 + 2*a^4*b*x^3 - (a^2)^(3/2)*b*x^3*Sqrt[(a + b*x^3)^2] + a^3*(b^2*x^6 - Sqrt[a^2]*Sqrt[(a + b*x^3)^2]
))*Log[x^3] + a^5*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 2*a^4*b*x^3*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a +
b*x^3)^2]] + a^3*b^2*x^6*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] - a^3*Sqrt[a^2]*Sqrt[(a + b*x^3)^2]*Log
[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] - (a^2)^(3/2)*b*x^3*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a^2] - b*x^3 - Sqrt
[(a + b*x^3)^2]] + a^5*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] + 2*a^4*b*x^3*Log[Sqrt[a^2] + b*x^3 - Sqrt
[(a + b*x^3)^2]] + a^3*b^2*x^6*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] - a^3*Sqrt[a^2]*Sqrt[(a + b*x^3)^2
]*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] - (a^2)^(3/2)*b*x^3*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a^2] + b*x^3 -
Sqrt[(a + b*x^3)^2]])/(3*a^3*Sqrt[a^2]*(a^2 + a*b*x^3 - Sqrt[a^2]*Sqrt[(a + b*x^3)^2])^2)

Maple [C] (warning: unable to verify)

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

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.48

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-\ln \left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{2}+\ln \left (b \,x^{3}\right ) \left (b \,x^{3}+a \right )^{2}+a b \,x^{3}+\frac {3 a^{2}}{2}\right )}{3 \left (b \,x^{3}+a \right )^{2} a^{3}}\)	\(70\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {b \,x^{3}}{3 a^{2}}+\frac {1}{2 a}\right )}{\left (b \,x^{3}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right ) a^{3}}\)	\(97\)
default	\(\frac {\left (6 \ln \left (x \right ) b^{2} x^{6}-2 \ln \left (b \,x^{3}+a \right ) b^{2} x^{6}+12 \ln \left (x \right ) a b \,x^{3}-4 \ln \left (b \,x^{3}+a \right ) a b \,x^{3}+2 a b \,x^{3}+6 a^{2} \ln \left (x \right )-2 \ln \left (b \,x^{3}+a \right ) a^{2}+3 a^{2}\right ) \left (b \,x^{3}+a \right )}{6 a^{3} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(107\)

[In]

int(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {2 \, a b x^{3} + 3 \, a^{2} - 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (b x^{3} + a\right ) + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (x\right )}{6 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{3} + a^{5}\right )}} \]

[In]

integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(1/x/(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{3}} + \frac {1}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2}} + \frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a b^{2}} \]

[In]

integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),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^3 + 1/3/(sqrt(b^2*x^6 + 2*a*b*x^3 + a
^2)*a^2) + 1/6/((x^3 + a/b)^2*a*b^2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {\log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{3} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {3 \, b^{2} x^{6} + 8 \, a b x^{3} + 6 \, a^{2}}{6 \, {\left (b x^{3} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{3} + a\right )} \]

[In]

integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

-1/3*log(abs(b*x^3 + a))/(a^3*sgn(b*x^3 + a)) + log(abs(x))/(a^3*sgn(b*x^3 + a)) + 1/6*(3*b^2*x^6 + 8*a*b*x^3
+ 6*a^2)/((b*x^3 + a)^2*a^3*sgn(b*x^3 + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}} \,d x \]

[In]

int(1/(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)),x)

[Out]

int(1/(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)), x)