3.239 \(\int \frac {1}{x (a+b x^3+c x^6)^{3/2}} \, dx\)

Optimal. Leaf size=92 \[ \frac {2 \left (-2 a c+b^2+b c x^3\right )}{3 a \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{3 a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1357, 740, 12, 724, 206} \[ \frac {2 \left (-2 a c+b^2+b c x^3\right )}{3 a \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{3 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b^2 - 2*a*c + b*c*x^3))/(3*a*(b^2 - 4*a*c)*Sqrt[a + b*x^3 + c*x^6]) - ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqr
t[a + b*x^3 + c*x^6])]/(3*a^(3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a+b x^3+c x^6\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x \left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^3\right )\\ &=\frac {2 \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 \operatorname {Subst}\left (\int \frac {-\frac {b^2}{2}+2 a c}{x \sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{3 a \left (b^2-4 a c\right )}\\ &=\frac {2 \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{3 a}\\ &=\frac {2 \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^3}{\sqrt {a+b x^3+c x^6}}\right )}{3 a}\\ &=\frac {2 \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{3 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 92, normalized size = 1.00 \[ \frac {1}{3} \left (\frac {2 \left (-2 a c+b^2+b c x^3\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{a^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(b^2 - 2*a*c + b*c*x^3))/(a*(b^2 - 4*a*c)*Sqrt[a + b*x^3 + c*x^6]) - ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt
[a + b*x^3 + c*x^6])]/a^(3/2))/3

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fricas [B]  time = 0.85, size = 389, normalized size = 4.23 \[ \left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{6} + {\left (b^{3} - 4 \, a b c\right )} x^{3} + a b^{2} - 4 \, a^{2} c\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (a b c x^{3} + a b^{2} - 2 \, a^{2} c\right )}}{6 \, {\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{6} + a^{3} b^{2} - 4 \, a^{4} c + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{3}\right )}}, \frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{6} + {\left (b^{3} - 4 \, a b c\right )} x^{3} + a b^{2} - 4 \, a^{2} c\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, \sqrt {c x^{6} + b x^{3} + a} {\left (a b c x^{3} + a b^{2} - 2 \, a^{2} c\right )}}{3 \, {\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{6} + a^{3} b^{2} - 4 \, a^{4} c + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(((b^2*c - 4*a*c^2)*x^6 + (b^3 - 4*a*b*c)*x^3 + a*b^2 - 4*a^2*c)*sqrt(a)*log(-((b^2 + 4*a*c)*x^6 + 8*a*b*
x^3 - 4*sqrt(c*x^6 + b*x^3 + a)*(b*x^3 + 2*a)*sqrt(a) + 8*a^2)/x^6) + 4*sqrt(c*x^6 + b*x^3 + a)*(a*b*c*x^3 + a
*b^2 - 2*a^2*c))/((a^2*b^2*c - 4*a^3*c^2)*x^6 + a^3*b^2 - 4*a^4*c + (a^2*b^3 - 4*a^3*b*c)*x^3), 1/3*(((b^2*c -
4*a*c^2)*x^6 + (b^3 - 4*a*b*c)*x^3 + a*b^2 - 4*a^2*c)*sqrt(-a)*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(b*x^3 + 2*
a)*sqrt(-a)/(a*c*x^6 + a*b*x^3 + a^2)) + 2*sqrt(c*x^6 + b*x^3 + a)*(a*b*c*x^3 + a*b^2 - 2*a^2*c))/((a^2*b^2*c
- 4*a^3*c^2)*x^6 + a^3*b^2 - 4*a^4*c + (a^2*b^3 - 4*a^3*b*c)*x^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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