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

Optimal. Leaf size=133 \[ \frac{\left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}}{64 a^2 x^6}-\frac{\left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{128 a^{5/2}}-\frac{\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}} \]

[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))

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Rubi [A]  time = 0.1076, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1357, 720, 724, 206} \[ \frac{\left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}}{64 a^2 x^6}-\frac{\left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{128 a^{5/2}}-\frac{\left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 a x^{12}} \]

Antiderivative was successfully verified.

[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 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 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{\left (a+b x^3+c x^6\right )^{3/2}}{x^{13}} \, dx &=\frac{1}{3} \operatorname{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 ) \operatorname{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 \operatorname{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 \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 )}{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]  time = 0.175532, size = 138, normalized size = 1.04 \[ -\frac{\frac{3 \left (b^2-4 a c\right ) \left (x^6 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )-2 \sqrt{a} \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}\right )}{8 a^{3/2} x^6}+\frac{2 \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{x^{12}}}{48 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{13}} \left ( c{x}^{6}+b{x}^{3}+a \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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)^(3/2)/x^13,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.02661, size = 732, normalized size = 5.5 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3} + c x^{6}\right )^{\frac{3}{2}}}{x^{13}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{6} + b x^{3} + a\right )}^{\frac{3}{2}}}{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)^(3/2)/x^13,x, algorithm="giac")

[Out]

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