3.240 \(\int \frac{1}{x^4 (a+b x^3+c x^6)^{3/2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a+b x^3+c x^6}}{3 a^2 x^3 \left (b^2-4 a c\right )}+\frac{b \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{2 a^{5/2}}+\frac{2 \left (-2 a c+b^2+b c x^3\right )}{3 a x^3 \left (b^2-4 a c\right ) \sqrt{a+b x^3+c x^6}} \]

[Out]

(2*(b^2 - 2*a*c + b*c*x^3))/(3*a*(b^2 - 4*a*c)*x^3*Sqrt[a + b*x^3 + c*x^6]) - ((3*b^2 - 8*a*c)*Sqrt[a + b*x^3
+ c*x^6])/(3*a^2*(b^2 - 4*a*c)*x^3) + (b*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(2*a^(5/2
))

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Rubi [A]  time = 0.124031, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1357, 740, 806, 724, 206} \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a+b x^3+c x^6}}{3 a^2 x^3 \left (b^2-4 a c\right )}+\frac{b \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{2 a^{5/2}}+\frac{2 \left (-2 a c+b^2+b c x^3\right )}{3 a x^3 \left (b^2-4 a c\right ) \sqrt{a+b x^3+c x^6}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(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)*x^3*Sqrt[a + b*x^3 + c*x^6]) - ((3*b^2 - 8*a*c)*Sqrt[a + b*x^3
+ c*x^6])/(3*a^2*(b^2 - 4*a*c)*x^3) + (b*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(2*a^(5/2
))

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]]

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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 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 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])

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a+b x^3+c x^6\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^2 \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 ) x^3 \sqrt{a+b x^3+c x^6}}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (-3 b^2+8 a c\right )-b c x}{x^2 \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 ) x^3 \sqrt{a+b x^3+c x^6}}-\frac{\left (3 b^2-8 a c\right ) \sqrt{a+b x^3+c x^6}}{3 a^2 \left (b^2-4 a c\right ) x^3}-\frac{b \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^3\right )}{2 a^2}\\ &=\frac{2 \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) x^3 \sqrt{a+b x^3+c x^6}}-\frac{\left (3 b^2-8 a c\right ) \sqrt{a+b x^3+c x^6}}{3 a^2 \left (b^2-4 a c\right ) x^3}+\frac{b \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 )}{a^2}\\ &=\frac{2 \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) x^3 \sqrt{a+b x^3+c x^6}}-\frac{\left (3 b^2-8 a c\right ) \sqrt{a+b x^3+c x^6}}{3 a^2 \left (b^2-4 a c\right ) x^3}+\frac{b \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{2 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0845054, size = 137, normalized size = 0.96 \[ \frac{\frac{2 \sqrt{a} \left (-4 a^2 c+a \left (b^2-10 b c x^3-8 c^2 x^6\right )+3 b^2 x^3 \left (b+c x^3\right )\right )}{x^3 \sqrt{a+b x^3+c x^6}}-3 b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{6 a^{5/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a]*(-4*a^2*c + 3*b^2*x^3*(b + c*x^3) + a*(b^2 - 10*b*c*x^3 - 8*c^2*x^6)))/(x^3*Sqrt[a + b*x^3 + c*x^6
]) - 3*b*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(6*a^(5/2)*(-b^2 + 4*a*c))

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Maple [F]  time = 0.038, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( c{x}^{6}+b{x}^{3}+a \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.13855, size = 1026, normalized size = 7.23 \begin{align*} \left [\frac{3 \,{\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{9} +{\left (b^{4} - 4 \, a b^{2} c\right )} x^{6} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\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 \,{\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{6} + a^{2} b^{2} - 4 \, a^{3} c +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{3}\right )} \sqrt{c x^{6} + b x^{3} + a}}{12 \,{\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{9} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{6} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}, -\frac{3 \,{\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{9} +{\left (b^{4} - 4 \, a b^{2} c\right )} x^{6} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\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 \,{\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{6} + a^{2} b^{2} - 4 \, a^{3} c +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{3}\right )} \sqrt{c x^{6} + b x^{3} + a}}{6 \,{\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{9} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{6} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*((b^3*c - 4*a*b*c^2)*x^9 + (b^4 - 4*a*b^2*c)*x^6 + (a*b^3 - 4*a^2*b*c)*x^3)*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*((3*a*b^2*c - 8*a^2*c^2
)*x^6 + a^2*b^2 - 4*a^3*c + (3*a*b^3 - 10*a^2*b*c)*x^3)*sqrt(c*x^6 + b*x^3 + a))/((a^3*b^2*c - 4*a^4*c^2)*x^9
+ (a^3*b^3 - 4*a^4*b*c)*x^6 + (a^4*b^2 - 4*a^5*c)*x^3), -1/6*(3*((b^3*c - 4*a*b*c^2)*x^9 + (b^4 - 4*a*b^2*c)*x
^6 + (a*b^3 - 4*a^2*b*c)*x^3)*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*((3*a*b^2*c - 8*a^2*c^2)*x^6 + a^2*b^2 - 4*a^3*c + (3*a*b^3 - 10*a^2*b*c)*x^3)*sqrt(c*x^6 +
b*x^3 + a))/((a^3*b^2*c - 4*a^4*c^2)*x^9 + (a^3*b^3 - 4*a^4*b*c)*x^6 + (a^4*b^2 - 4*a^5*c)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \left (a + b x^{3} + c x^{6}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac{3}{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(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^4), x)