\(\int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) [97]

Optimal result

Integrand size = 26, antiderivative size = 125 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {-a-b x^3}{3 a x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b \left (a+b x^3\right ) \log (x)}{a^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^2 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

1/3*(-b*x^3-a)/a/x^3/((b*x^3+a)^2)^(1/2)-b*(b*x^3+a)*ln(x)/a^2/((b*x^3+a)^2)^(1/2)+1/3*b*(b*x^3+a)*ln(b*x^3+a)
/a^2/((b*x^3+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 = {1369, 272, 46} \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {a+b x^3}{3 a x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b \log (x) \left (a+b x^3\right )}{a^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^2 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

Int[1/(x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {a^2-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}+2 a b x^3 \log \left (x^3\right )+\left (-a+\sqrt {a^2}\right ) b x^3 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-a b x^3 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-\sqrt {a^2} b x^3 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )}{6 \left (a^2\right )^{3/2} x^3} \]

[In]

Integrate[1/(x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.47 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.35

method	result	size
pseudoelliptic	\(-\frac {\left (\ln \left (b \,x^{3}\right ) b \,x^{3}-b \ln \left (b \,x^{3}+a \right ) x^{3}+a \right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{3 a^{2} x^{3}}\)	\(44\)
default	\(-\frac {\left (b \,x^{3}+a \right ) \left (3 b \ln \left (x \right ) x^{3}-b \ln \left (b \,x^{3}+a \right ) x^{3}+a \right )}{3 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} x^{3}}\)	\(51\)
risch	\(-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}}{3 \left (b \,x^{3}+a \right ) a \,x^{3}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{2}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (-b \,x^{3}-a \right )}{3 \left (b \,x^{3}+a \right ) a^{2}}\)	\(95\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {b x^{3} \log \left (b x^{3} + a\right ) - 3 \, b x^{3} \log \left (x\right ) - a}{3 \, a^{2} x^{3}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {1}{x^{4} \sqrt {\left (a + b x^{3}\right )^{2}}}\, dx \]

[In]

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

[Out]

Integral(1/(x**4*sqrt((a + b*x**3)**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{2}} - \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}}{3 \, a^{2} x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {1}{3} \, {\left (\frac {b \log \left ({\left | b x^{3} + a \right |}\right )}{a^{2}} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {b x^{3} - a}{a^{2} x^{3}}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,a\,x^3}{\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}\right )}{3\,{\left (a^2\right )}^{3/2}}-\frac {\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,a^2\,x^3} \]

[In]

int(1/(x^4*((a + b*x^3)^2)^(1/2)),x)

[Out]

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