\(\int \frac {(a+b x^3+c x^6)^{3/2}}{x^{13}} \, dx\) [210]

Optimal result

Integrand size = 20, antiderivative size = 133 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=\frac {\left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{64 a^2 x^6}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}}-\frac {\left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{128 a^{5/2}} \]

[Out]

-1/24*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(3/2)/a/x^12-1/128*(-4*a*c+b^2)^2*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x
^3+a)^(1/2))/a^(5/2)+1/64*(-4*a*c+b^2)*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(1/2)/a^2/x^6

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1371, 734, 738, 212} \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{128 a^{5/2}}+\frac {\left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{64 a^2 x^6}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}} \]

[In]

Int[(a + b*x^3 + c*x^6)^(3/2)/x^13,x]

[Out]

((b^2 - 4*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(64*a^2*x^6) - ((2*a + b*x^3)*(a + b*x^3 + c*x^6)^(3/2))
/(24*a*x^12) - ((b^2 - 4*a*c)^2*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(128*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 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))
*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[p*((b^2
- 4*a*c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; Free
Q[{a, b, c, d, e}, 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] && EqQ[m
+ 2*p + 2, 0] && GtQ[p, 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 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 {\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^3\right ) \\ & = -\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}}-\frac {\left (b^2-4 a c\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^3\right )}{16 a} \\ & = \frac {\left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{64 a^2 x^6}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}}+\frac {\left (b^2-4 a c\right )^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{128 a^2} \\ & = \frac {\left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{64 a^2 x^6}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}}-\frac {\left (b^2-4 a c\right )^2 \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 )}{64 a^2} \\ & = \frac {\left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{64 a^2 x^6}-\frac {\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}}-\frac {\left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{128 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {\left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6} \left (8 a^2+8 a b x^3-3 b^2 x^6+20 a c x^6\right )}{192 a^2 x^{12}}+\frac {\left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{64 a^{5/2}} \]

[In]

Integrate[(a + b*x^3 + c*x^6)^(3/2)/x^13,x]

[Out]

-1/192*((2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6]*(8*a^2 + 8*a*b*x^3 - 3*b^2*x^6 + 20*a*c*x^6))/(a^2*x^12) + ((b^2
- 4*a*c)^2*ArcTanh[(Sqrt[c]*x^3 - Sqrt[a + b*x^3 + c*x^6])/Sqrt[a]])/(64*a^(5/2))

Maple [F]

\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}{x^{13}}d x\]

[In]

int((c*x^6+b*x^3+a)^(3/2)/x^13,x)

[Out]

int((c*x^6+b*x^3+a)^(3/2)/x^13,x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.40 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{12} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) + 4 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{9} - 24 \, a^{3} b x^{3} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{6} - 16 \, a^{4}\right )} \sqrt {c x^{6} + b x^{3} + a}}{768 \, a^{3} x^{12}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{12} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{9} - 24 \, a^{3} b x^{3} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{6} - 16 \, a^{4}\right )} \sqrt {c x^{6} + b x^{3} + a}}{384 \, a^{3} x^{12}}\right ] \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^13,x, algorithm="fricas")

[Out]

[1/768*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(a)*x^12*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^3 - 20*a^2*b*c)*x^9 - 24*a^3*b*x^3 - 2*(a^2*b^2 + 20*a
^3*c)*x^6 - 16*a^4)*sqrt(c*x^6 + b*x^3 + a))/(a^3*x^12), 1/384*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-a)*x^12
*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^3 - 20*a^2*b
*c)*x^9 - 24*a^3*b*x^3 - 2*(a^2*b^2 + 20*a^3*c)*x^6 - 16*a^4)*sqrt(c*x^6 + b*x^3 + a))/(a^3*x^12)]

Sympy [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=\int \frac {\left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}{x^{13}}\, dx \]

[In]

integrate((c*x**6+b*x**3+a)**(3/2)/x**13,x)

[Out]

Integral((a + b*x**3 + c*x**6)**(3/2)/x**13, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^13,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]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}{x^{13}} \,d x } \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^13,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^13, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{x^{13}} \,d x \]

[In]

int((a + b*x^3 + c*x^6)^(3/2)/x^13,x)

[Out]

int((a + b*x^3 + c*x^6)^(3/2)/x^13, x)