\(\int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx\) [194]

Optimal result

Integrand size = 20, antiderivative size = 161 \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=-\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{192 a^3 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}}+\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}+\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{384 a^{7/2}} \]

[Out]

-1/12*(c*x^6+b*x^3+a)^(3/2)/a/x^12+5/72*b*(c*x^6+b*x^3+a)^(3/2)/a^2/x^9+1/384*(-4*a*c+b^2)*(-4*a*c+5*b^2)*arct
anh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))/a^(7/2)-1/192*(-4*a*c+5*b^2)*(b*x^3+2*a)*(c*x^6+b*x^3+a)^(1
/2)/a^3/x^6

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1371, 758, 820, 734, 738, 212} \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{384 a^{7/2}}-\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{192 a^3 x^6}+\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}} \]

[In]

Int[Sqrt[a + b*x^3 + c*x^6]/x^13,x]

[Out]

-1/192*((5*b^2 - 4*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(a^3*x^6) - (a + b*x^3 + c*x^6)^(3/2)/(12*a*x^1
2) + (5*b*(a + b*x^3 + c*x^6)^(3/2))/(72*a^2*x^9) + ((b^2 - 4*a*c)*(5*b^2 - 4*a*c)*ArcTanh[(2*a + b*x^3)/(2*Sq
rt[a]*Sqrt[a + b*x^3 + c*x^6])])/(384*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 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 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^5} \, dx,x,x^3\right ) \\ & = -\frac {\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}}-\frac {\text {Subst}\left (\int \frac {\left (\frac {5 b}{2}+c x\right ) \sqrt {a+b x+c x^2}}{x^4} \, dx,x,x^3\right )}{12 a} \\ & = -\frac {\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}}+\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}+\frac {\left (5 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 )}{48 a^2} \\ & = -\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{192 a^3 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}}+\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}-\frac {\left (\left (b^2-4 a c\right ) \left (5 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 )}{384 a^3} \\ & = -\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{192 a^3 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}}+\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}+\frac {\left (\left (b^2-4 a c\right ) \left (5 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 )}{192 a^3} \\ & = -\frac {\left (5 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt {a+b x^3+c x^6}}{192 a^3 x^6}-\frac {\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}}+\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}+\frac {\left (b^2-4 a c\right ) \left (5 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 )}{384 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=\frac {\sqrt {a+b x^3+c x^6} \left (-48 a^3-8 a^2 b x^3+10 a b^2 x^6-24 a^2 c x^6-15 b^3 x^9+52 a b c x^9\right )}{576 a^3 x^{12}}+\frac {\left (-5 b^4+24 a b^2 c-16 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{192 a^{7/2}} \]

[In]

Integrate[Sqrt[a + b*x^3 + c*x^6]/x^13,x]

[Out]

(Sqrt[a + b*x^3 + c*x^6]*(-48*a^3 - 8*a^2*b*x^3 + 10*a*b^2*x^6 - 24*a^2*c*x^6 - 15*b^3*x^9 + 52*a*b*c*x^9))/(5
76*a^3*x^12) + ((-5*b^4 + 24*a*b^2*c - 16*a^2*c^2)*ArcTanh[(Sqrt[c]*x^3 - Sqrt[a + b*x^3 + c*x^6])/Sqrt[a]])/(
192*a^(7/2))

Maple [F]

\[\int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}}{x^{13}}d x\]

[In]

int((c*x^6+b*x^3+a)^(1/2)/x^13,x)

[Out]

int((c*x^6+b*x^3+a)^(1/2)/x^13,x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.02 \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=\left [\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{12} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) - 4 \, {\left ({\left (15 \, a b^{3} - 52 \, a^{2} b c\right )} x^{9} + 8 \, a^{3} b x^{3} - 2 \, {\left (5 \, a^{2} b^{2} - 12 \, a^{3} c\right )} x^{6} + 48 \, a^{4}\right )} \sqrt {c x^{6} + b x^{3} + a}}{2304 \, a^{4} x^{12}}, -\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{12} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (15 \, a b^{3} - 52 \, a^{2} b c\right )} x^{9} + 8 \, a^{3} b x^{3} - 2 \, {\left (5 \, a^{2} b^{2} - 12 \, a^{3} c\right )} x^{6} + 48 \, a^{4}\right )} \sqrt {c x^{6} + b x^{3} + a}}{1152 \, a^{4} x^{12}}\right ] \]

[In]

integrate((c*x^6+b*x^3+a)^(1/2)/x^13,x, algorithm="fricas")

[Out]

[1/2304*(3*(5*b^4 - 24*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*((15*a*b^3 - 52*a^2*b*c)*x^9 + 8*a^3*b*x^3 - 2*(5*a^2*b^2
- 12*a^3*c)*x^6 + 48*a^4)*sqrt(c*x^6 + b*x^3 + a))/(a^4*x^12), -1/1152*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*sq
rt(-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*((15*a*b^
3 - 52*a^2*b*c)*x^9 + 8*a^3*b*x^3 - 2*(5*a^2*b^2 - 12*a^3*c)*x^6 + 48*a^4)*sqrt(c*x^6 + b*x^3 + a))/(a^4*x^12)
]

Sympy [F]

\[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=\int \frac {\sqrt {a + b x^{3} + c x^{6}}}{x^{13}}\, dx \]

[In]

integrate((c*x**6+b*x**3+a)**(1/2)/x**13,x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

\[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=\int { \frac {\sqrt {c x^{6} + b x^{3} + a}}{x^{13}} \,d x } \]

[In]

integrate((c*x^6+b*x^3+a)^(1/2)/x^13,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^3+c x^6}}{x^{13}} \, dx=\int \frac {\sqrt {c\,x^6+b\,x^3+a}}{x^{13}} \,d x \]

[In]

int((a + b*x^3 + c*x^6)^(1/2)/x^13,x)

[Out]

int((a + b*x^3 + c*x^6)^(1/2)/x^13, x)