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

Optimal. Leaf size=161 \[ -\frac{\left (5 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}}{192 a^3 x^6}+\frac{\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{384 a^{7/2}}+\frac{5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}} \]

[Out]

-((5*b^2 - 4*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(192*a^3*x^6) - (a + b*x^3 + c*x^6)^(3/2)/(12*a*x^12)
+ (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*Sqrt
[a]*Sqrt[a + b*x^3 + c*x^6])])/(384*a^(7/2))

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Rubi [A]  time = 0.147009, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1357, 744, 806, 720, 724, 206} \[ -\frac{\left (5 b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}}{192 a^3 x^6}+\frac{\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{384 a^{7/2}}+\frac{5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 a^2 x^9}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{12 a x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3 + c*x^6]/x^13,x]

[Out]

-((5*b^2 - 4*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(192*a^3*x^6) - (a + b*x^3 + c*x^6)^(3/2)/(12*a*x^12)
+ (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*Sqrt
[a]*Sqrt[a + b*x^3 + c*x^6])])/(384*a^(7/2))

Rule 1357

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]]

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[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 806

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

Rule 720

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] /; FreeQ[
{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 724

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 206

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x^3+c x^6}}{x^{13}} \, dx &=\frac{1}{3} \operatorname{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{\operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.102086, size = 139, normalized size = 0.86 \[ \frac{\left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{384 a^{7/2}}-\frac{\sqrt{a+b x^3+c x^6} \left (8 a^2 x^3 \left (b+3 c x^3\right )+48 a^3-2 a b x^6 \left (5 b+26 c x^3\right )+15 b^3 x^9\right )}{576 a^3 x^{12}} \]

Antiderivative was successfully verified.

[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 + 15*b^3*x^9 + 8*a^2*x^3*(b + 3*c*x^3) - 2*a*b*x^6*(5*b + 26*c*x^3)))/(576*a
^3*x^12) + ((5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(384
*a^(7/2))

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{13}}\sqrt{c{x}^{6}+b{x}^{3}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.06822, size = 749, normalized size = 4.65 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)
]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b x^{3} + c x^{6}}}{x^{13}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{6} + b x^{3} + a}}{x^{13}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)