\(\int \frac {1}{x^4 (a+b x^3+c x^6)^{3/2}} \, dx\) [240]

Optimal result

Integrand size = 20, antiderivative size = 142 \[ \int \frac {1}{x^4 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\frac {2 \left (b^2-2 a c+b c x^3\right )}{3 a \left (b^2-4 a c\right ) x^3 \sqrt {a+b x^3+c x^6}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^3+c x^6}}{3 a^2 \left (b^2-4 a c\right ) x^3}+\frac {b \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{2 a^{5/2}} \]

[Out]

1/2*b*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))/a^(5/2)+2/3*(b*c*x^3-2*a*c+b^2)/a/(-4*a*c+b^2)/x^
3/(c*x^6+b*x^3+a)^(1/2)-1/3*(-8*a*c+3*b^2)*(c*x^6+b*x^3+a)^(1/2)/a^2/(-4*a*c+b^2)/x^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1371, 754, 820, 738, 212} \[ \int \frac {1}{x^4 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\frac {b \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{2 a^{5/2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^3+c x^6}}{3 a^2 x^3 \left (b^2-4 a c\right )}+\frac {2 \left (-2 a c+b^2+b c x^3\right )}{3 a x^3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}} \]

[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 212

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

Rule 738

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 754

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 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(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[S
implify[m + 2*p + 3], 0]

Rule 1371

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*} \text {integral}& = \frac {1}{3} \text {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 \text {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 \text {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 \text {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] (verified)

Time = 0.51 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^4 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\frac {-4 a^2 c+3 b^2 x^3 \left (b+c x^3\right )+a \left (b^2-10 b c x^3-8 c^2 x^6\right )}{3 a^2 \left (-b^2+4 a c\right ) x^3 \sqrt {a+b x^3+c x^6}}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{a^{5/2}} \]

[In]

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

[Out]

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

Maple [F]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 485, normalized size of antiderivative = 3.42 \[ \int \frac {1}{x^4 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{9} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{6} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) - 4 \, {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{6} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{12 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{9} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{6} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}, -\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{9} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{6} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{6} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{6 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{9} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{6} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}\right )}}\right ] \]

[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)]

Sympy [F]

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

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^4 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/x^4/(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 deta

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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