3.2.6 \(\int \frac {1}{x^4 (a^2+2 a b x^3+b^2 x^6)^{3/2}} \, dx\) [106]

Optimal. Leaf size=188 \[ -\frac {2 b}{3 a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{6 a^2 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{3 a^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {3 b \left (a+b x^3\right ) \log (x)}{a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1369, 272, 46} \begin {gather*} -\frac {b}{6 a^2 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {3 b \log (x) \left (a+b x^3\right )}{a^4 \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 )}{a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {2 b}{3 a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{3 a^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.02, size = 97, normalized size = 0.52 \begin {gather*} \frac {-a \left (2 a^2+9 a b x^3+6 b^2 x^6\right )-18 b x^3 \left (a+b x^3\right )^2 \log (x)+6 b x^3 \left (a+b x^3\right )^2 \log \left (a+b x^3\right )}{6 a^4 x^3 \left (a+b x^3\right ) \sqrt {\left (a+b x^3\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*(2*a^2 + 9*a*b*x^3 + 6*b^2*x^6)) - 18*b*x^3*(a + b*x^3)^2*Log[x] + 6*b*x^3*(a + b*x^3)^2*Log[a + b*x^3])/
(6*a^4*x^3*(a + b*x^3)*Sqrt[(a + b*x^3)^2])

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Maple [A]
time = 0.04, size = 133, normalized size = 0.71

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {b^{2} x^{6}}{a^{3}}-\frac {3 b \,x^{3}}{2 a^{2}}-\frac {1}{3 a}\right )}{\left (b \,x^{3}+a \right )^{3} x^{3}}-\frac {3 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{4}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (-b \,x^{3}-a \right )}{\left (b \,x^{3}+a \right ) a^{4}}\)	\(116\)
default	\(\frac {\left (6 \ln \left (b \,x^{3}+a \right ) b^{3} x^{9}-18 b^{3} \ln \left (x \right ) x^{9}+12 \ln \left (b \,x^{3}+a \right ) a \,b^{2} x^{6}-36 \ln \left (x \right ) a \,b^{2} x^{6}-6 a \,b^{2} x^{6}+6 \ln \left (b \,x^{3}+a \right ) a^{2} b \,x^{3}-18 a^{2} b \ln \left (x \right ) x^{3}-9 a^{2} b \,x^{3}-2 a^{3}\right ) \left (b \,x^{3}+a \right )}{6 x^{3} a^{4} \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(6*ln(b*x^3+a)*b^3*x^9-18*b^3*ln(x)*x^9+12*ln(b*x^3+a)*a*b^2*x^6-36*ln(x)*a*b^2*x^6-6*a*b^2*x^6+6*ln(b*x^3
+a)*a^2*b*x^3-18*a^2*b*ln(x)*x^3-9*a^2*b*x^3-2*a^3)*(b*x^3+a)/x^3/a^4/((b*x^3+a)^2)^(3/2)

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Maxima [A]
time = 0.28, size = 117, normalized size = 0.62 \begin {gather*} \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 )}{a^{4}} - \frac {b}{\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{3}} - \frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a^{2} b} - \frac {1}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 119, normalized size = 0.63 \begin {gather*} -\frac {6 \, a b^{2} x^{6} + 9 \, a^{2} b x^{3} + 2 \, a^{3} - 6 \, {\left (b^{3} x^{9} + 2 \, a b^{2} x^{6} + a^{2} b x^{3}\right )} \log \left (b x^{3} + a\right ) + 18 \, {\left (b^{3} x^{9} + 2 \, a b^{2} x^{6} + a^{2} b x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{4} b^{2} x^{9} + 2 \, a^{5} b x^{6} + a^{6} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/6*(6*a*b^2*x^6 + 9*a^2*b*x^3 + 2*a^3 - 6*(b^3*x^9 + 2*a*b^2*x^6 + a^2*b*x^3)*log(b*x^3 + a) + 18*(b^3*x^9 +
2*a*b^2*x^6 + a^2*b*x^3)*log(x))/(a^4*b^2*x^9 + 2*a^5*b*x^6 + a^6*x^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

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

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Giac [A]
time = 5.02, size = 120, normalized size = 0.64 \begin {gather*} \frac {b \log \left ({\left | b x^{3} + a \right |}\right )}{a^{4} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{4} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {9 \, b^{3} x^{6} + 22 \, a b^{2} x^{3} + 14 \, a^{2} b}{6 \, {\left (b x^{3} + a\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {3 \, b x^{3} - a}{3 \, a^{4} x^{3} \mathrm {sgn}\left (b x^{3} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

b*log(abs(b*x^3 + a))/(a^4*sgn(b*x^3 + a)) - 3*b*log(abs(x))/(a^4*sgn(b*x^3 + a)) - 1/6*(9*b^3*x^6 + 22*a*b^2*
x^3 + 14*a^2*b)/((b*x^3 + a)^2*a^4*sgn(b*x^3 + a)) + 1/3*(3*b*x^3 - a)/(a^4*x^3*sgn(b*x^3 + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^4\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)),x)

[Out]

int(1/(x^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)), x)

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