\(\int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx\) [195]

Optimal result

Integrand size = 20, antiderivative size = 199 \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=\frac {b \left (7 b^2-12 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{384 a^4 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}}+\frac {7 b \left (a+b x^3+c x^6\right )^{3/2}}{120 a^2 x^{12}}-\frac {\left (35 b^2-32 a c\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 a^3 x^9}-\frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{768 a^{9/2}} \]

[Out]

-1/15*(c*x^6+b*x^3+a)^(3/2)/a/x^15+7/120*b*(c*x^6+b*x^3+a)^(3/2)/a^2/x^12-1/720*(-32*a*c+35*b^2)*(c*x^6+b*x^3+
a)^(3/2)/a^3/x^9-1/768*b*(-12*a*c+7*b^2)*(-4*a*c+b^2)*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))/a
^(9/2)+1/384*b*(-12*a*c+7*b^2)*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(1/2)/a^4/x^6

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1371, 758, 848, 820, 734, 738, 212} \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=-\frac {b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{768 a^{9/2}}+\frac {b \left (7 b^2-12 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{384 a^4 x^6}-\frac {\left (35 b^2-32 a c\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 a^3 x^9}+\frac {7 b \left (a+b x^3+c x^6\right )^{3/2}}{120 a^2 x^{12}}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}} \]

[In]

Int[Sqrt[a + b*x^3 + c*x^6]/x^16,x]

[Out]

(b*(7*b^2 - 12*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(384*a^4*x^6) - (a + b*x^3 + c*x^6)^(3/2)/(15*a*x^1
5) + (7*b*(a + b*x^3 + c*x^6)^(3/2))/(120*a^2*x^12) - ((35*b^2 - 32*a*c)*(a + b*x^3 + c*x^6)^(3/2))/(720*a^3*x
^9) - (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(768*a^(9/
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 758

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[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

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 848

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)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

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 {\sqrt {a+b x+c x^2}}{x^6} \, dx,x,x^3\right ) \\ & = -\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}}-\frac {\text {Subst}\left (\int \frac {\left (\frac {7 b}{2}+2 c x\right ) \sqrt {a+b x+c x^2}}{x^5} \, dx,x,x^3\right )}{15 a} \\ & = -\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}}+\frac {7 b \left (a+b x^3+c x^6\right )^{3/2}}{120 a^2 x^{12}}+\frac {\text {Subst}\left (\int \frac {\left (\frac {1}{4} \left (35 b^2-32 a c\right )+\frac {7 b c x}{2}\right ) \sqrt {a+b x+c x^2}}{x^4} \, dx,x,x^3\right )}{60 a^2} \\ & = -\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}}+\frac {7 b \left (a+b x^3+c x^6\right )^{3/2}}{120 a^2 x^{12}}-\frac {\left (35 b^2-32 a c\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 a^3 x^9}-\frac {\left (b \left (7 b^2-12 a c\right )\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^3\right )}{96 a^3} \\ & = \frac {b \left (7 b^2-12 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{384 a^4 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}}+\frac {7 b \left (a+b x^3+c x^6\right )^{3/2}}{120 a^2 x^{12}}-\frac {\left (35 b^2-32 a c\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 a^3 x^9}+\frac {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{768 a^4} \\ & = \frac {b \left (7 b^2-12 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{384 a^4 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}}+\frac {7 b \left (a+b x^3+c x^6\right )^{3/2}}{120 a^2 x^{12}}-\frac {\left (35 b^2-32 a c\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 a^3 x^9}-\frac {\left (b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right )\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 )}{384 a^4} \\ & = \frac {b \left (7 b^2-12 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{384 a^4 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{15 a x^{15}}+\frac {7 b \left (a+b x^3+c x^6\right )^{3/2}}{120 a^2 x^{12}}-\frac {\left (35 b^2-32 a c\right ) \left (a+b x^3+c x^6\right )^{3/2}}{720 a^3 x^9}-\frac {b \left (7 b^2-12 a c\right ) \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 )}{768 a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=\frac {\sqrt {a+b x^3+c x^6} \left (-384 a^4-48 a^3 b x^3+56 a^2 b^2 x^6-128 a^3 c x^6-70 a b^3 x^9+232 a^2 b c x^9+105 b^4 x^{12}-460 a b^2 c x^{12}+256 a^2 c^2 x^{12}\right )}{5760 a^4 x^{15}}+\frac {\left (7 b^5-40 a b^3 c+48 a^2 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{384 a^{9/2}} \]

[In]

Integrate[Sqrt[a + b*x^3 + c*x^6]/x^16,x]

[Out]

(Sqrt[a + b*x^3 + c*x^6]*(-384*a^4 - 48*a^3*b*x^3 + 56*a^2*b^2*x^6 - 128*a^3*c*x^6 - 70*a*b^3*x^9 + 232*a^2*b*
c*x^9 + 105*b^4*x^12 - 460*a*b^2*c*x^12 + 256*a^2*c^2*x^12))/(5760*a^4*x^15) + ((7*b^5 - 40*a*b^3*c + 48*a^2*b
*c^2)*ArcTanh[(Sqrt[c]*x^3 - Sqrt[a + b*x^3 + c*x^6])/Sqrt[a]])/(384*a^(9/2))

Maple [F]

\[\int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}}{x^{16}}d x\]

[In]

int((c*x^6+b*x^3+a)^(1/2)/x^16,x)

[Out]

int((c*x^6+b*x^3+a)^(1/2)/x^16,x)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.95 \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=\left [\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {a} x^{15} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) + 4 \, {\left ({\left (105 \, a b^{4} - 460 \, a^{2} b^{2} c + 256 \, a^{3} c^{2}\right )} x^{12} - 2 \, {\left (35 \, a^{2} b^{3} - 116 \, a^{3} b c\right )} x^{9} - 48 \, a^{4} b x^{3} + 8 \, {\left (7 \, a^{3} b^{2} - 16 \, a^{4} c\right )} x^{6} - 384 \, a^{5}\right )} \sqrt {c x^{6} + b x^{3} + a}}{23040 \, a^{5} x^{15}}, \frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {-a} x^{15} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (105 \, a b^{4} - 460 \, a^{2} b^{2} c + 256 \, a^{3} c^{2}\right )} x^{12} - 2 \, {\left (35 \, a^{2} b^{3} - 116 \, a^{3} b c\right )} x^{9} - 48 \, a^{4} b x^{3} + 8 \, {\left (7 \, a^{3} b^{2} - 16 \, a^{4} c\right )} x^{6} - 384 \, a^{5}\right )} \sqrt {c x^{6} + b x^{3} + a}}{11520 \, a^{5} x^{15}}\right ] \]

[In]

integrate((c*x^6+b*x^3+a)^(1/2)/x^16,x, algorithm="fricas")

[Out]

[1/23040*(15*(7*b^5 - 40*a*b^3*c + 48*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*((105*a*b^4 - 460*a^2*b^2*c + 256*a^3*c^2)*x^12 - 2*(3
5*a^2*b^3 - 116*a^3*b*c)*x^9 - 48*a^4*b*x^3 + 8*(7*a^3*b^2 - 16*a^4*c)*x^6 - 384*a^5)*sqrt(c*x^6 + b*x^3 + a))
/(a^5*x^15), 1/11520*(15*(7*b^5 - 40*a*b^3*c + 48*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*((105*a*b^4 - 460*a^2*b^2*c + 256*a^3*c^2)*x^12 - 2*(35*
a^2*b^3 - 116*a^3*b*c)*x^9 - 48*a^4*b*x^3 + 8*(7*a^3*b^2 - 16*a^4*c)*x^6 - 384*a^5)*sqrt(c*x^6 + b*x^3 + a))/(
a^5*x^15)]

Sympy [F]

\[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=\int \frac {\sqrt {a + b x^{3} + c x^{6}}}{x^{16}}\, dx \]

[In]

integrate((c*x**6+b*x**3+a)**(1/2)/x**16,x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

\[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=\int { \frac {\sqrt {c x^{6} + b x^{3} + a}}{x^{16}} \,d x } \]

[In]

integrate((c*x^6+b*x^3+a)^(1/2)/x^16,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{16}} \, dx=\int \frac {\sqrt {c\,x^6+b\,x^3+a}}{x^{16}} \,d x \]

[In]

int((a + b*x^3 + c*x^6)^(1/2)/x^16,x)

[Out]

int((a + b*x^3 + c*x^6)^(1/2)/x^16, x)