\(\int \frac {(a+b x^3+c x^6)^{3/2}}{x^{16}} \, dx\) [211]

Optimal result

Integrand size = 20, antiderivative size = 162 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {b \left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{128 a^3 x^6}+\frac {b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}}+\frac {b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{256 a^{7/2}} \]

[Out]

1/48*b*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(3/2)/a^2/x^12-1/15*(c*x^6+b*x^3+a)^(5/2)/a/x^15+1/256*b*(-4*a*c+b^2)^2*arc
tanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))/a^(7/2)-1/128*b*(-4*a*c+b^2)*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(
1/2)/a^3/x^6

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1371, 744, 734, 738, 212} \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=\frac {b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{256 a^{7/2}}-\frac {b \left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{128 a^3 x^6}+\frac {b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}} \]

[In]

Int[(a + b*x^3 + c*x^6)^(3/2)/x^16,x]

[Out]

-1/128*(b*(b^2 - 4*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(a^3*x^6) + (b*(2*a + b*x^3)*(a + b*x^3 + c*x^6
)^(3/2))/(48*a^2*x^12) - (a + b*x^3 + c*x^6)^(5/2)/(15*a*x^15) + (b*(b^2 - 4*a*c)^2*ArcTanh[(2*a + b*x^3)/(2*S
qrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(256*a^(7/2))

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 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*
((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[(2*c*d - b*e)/(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, m, p}, 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 + 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 {\left (a+b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^3\right ) \\ & = -\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}}-\frac {b \text {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^3\right )}{6 a} \\ & = \frac {b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}}+\frac {\left (b \left (b^2-4 a c\right )\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^3\right )}{32 a^2} \\ & = -\frac {b \left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{128 a^3 x^6}+\frac {b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}}-\frac {\left (b \left (b^2-4 a c\right )^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{256 a^3} \\ & = -\frac {b \left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{128 a^3 x^6}+\frac {b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}}+\frac {\left (b \left (b^2-4 a c\right )^2\right ) \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 )}{128 a^3} \\ & = -\frac {b \left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{128 a^3 x^6}+\frac {b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}}+\frac {b \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 )}{256 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=\frac {-\frac {\sqrt {a} \sqrt {a+b x^3+c x^6} \left (128 a^4+15 b^4 x^{12}-10 a b^2 x^9 \left (b+10 c x^3\right )+16 a^3 \left (11 b x^3+16 c x^6\right )+8 a^2 x^6 \left (b^2+7 b c x^3+16 c^2 x^6\right )\right )}{x^{15}}-15 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{1920 a^{7/2}} \]

[In]

Integrate[(a + b*x^3 + c*x^6)^(3/2)/x^16,x]

[Out]

(-((Sqrt[a]*Sqrt[a + b*x^3 + c*x^6]*(128*a^4 + 15*b^4*x^12 - 10*a*b^2*x^9*(b + 10*c*x^3) + 16*a^3*(11*b*x^3 +
16*c*x^6) + 8*a^2*x^6*(b^2 + 7*b*c*x^3 + 16*c^2*x^6)))/x^15) - 15*b*(b^2 - 4*a*c)^2*ArcTanh[(Sqrt[c]*x^3 - Sqr
t[a + b*x^3 + c*x^6])/Sqrt[a]])/(1920*a^(7/2))

Maple [F]

\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}{x^{16}}d x\]

[In]

int((c*x^6+b*x^3+a)^(3/2)/x^16,x)

[Out]

int((c*x^6+b*x^3+a)^(3/2)/x^16,x)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.36 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {a} x^{15} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) - 4 \, {\left ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{12} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{9} + 176 \, a^{4} b x^{3} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{6} + 128 \, a^{5}\right )} \sqrt {c x^{6} + b x^{3} + a}}{7680 \, a^{4} x^{15}}, -\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-a} x^{15} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{12} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{9} + 176 \, a^{4} b x^{3} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{6} + 128 \, a^{5}\right )} \sqrt {c x^{6} + b x^{3} + a}}{3840 \, a^{4} x^{15}}\right ] \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^16,x, algorithm="fricas")

[Out]

[1/7680*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(a)*x^15*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*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*x^12 - 2*(5*a^2*
b^3 - 28*a^3*b*c)*x^9 + 176*a^4*b*x^3 + 8*(a^3*b^2 + 32*a^4*c)*x^6 + 128*a^5)*sqrt(c*x^6 + b*x^3 + a))/(a^4*x^
15), -1/3840*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-a)*x^15*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*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*x^12 - 2*(5*a^2*b^3 - 28*
a^3*b*c)*x^9 + 176*a^4*b*x^3 + 8*(a^3*b^2 + 32*a^4*c)*x^6 + 128*a^5)*sqrt(c*x^6 + b*x^3 + a))/(a^4*x^15)]

Sympy [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=\int \frac {\left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}{x^{16}}\, dx \]

[In]

integrate((c*x**6+b*x**3+a)**(3/2)/x**16,x)

[Out]

Integral((a + b*x**3 + c*x**6)**(3/2)/x**16, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^16,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]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}{x^{16}} \,d x } \]

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^16,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^16, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^{16}} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{x^{16}} \,d x \]

[In]

int((a + b*x^3 + c*x^6)^(3/2)/x^16,x)

[Out]

int((a + b*x^3 + c*x^6)^(3/2)/x^16, x)