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

Optimal. Leaf size=162 \[ -\frac{b \left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}}{128 a^3 x^6}+\frac{b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{256 a^{7/2}}+\frac{b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac{\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}} \]

[Out]

-(b*(b^2 - 4*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(128*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*Sqr
t[a]*Sqrt[a + b*x^3 + c*x^6])])/(256*a^(7/2))

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Rubi [A]  time = 0.145584, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1357, 730, 720, 724, 206} \[ -\frac{b \left (b^2-4 a c\right ) \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}}{128 a^3 x^6}+\frac{b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{256 a^{7/2}}+\frac{b \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{48 a^2 x^{12}}-\frac{\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)^(3/2)/x^16,x]

[Out]

-(b*(b^2 - 4*a*c)*(2*a + b*x^3)*Sqrt[a + b*x^3 + c*x^6])/(128*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*Sqr
t[a]*Sqrt[a + b*x^3 + c*x^6])])/(256*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 730

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 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^{16}} \, dx &=\frac{1}{3} \operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \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 )}{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]  time = 0.143209, size = 167, normalized size = 1.03 \[ \frac{b \left (16 a^{3/2} \left (2 a+b x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}-3 x^6 \left (b^2-4 a c\right ) \left (2 \sqrt{a} \left (2 a+b x^3\right ) \sqrt{a+b x^3+c x^6}-x^6 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )\right )\right )}{768 a^{7/2} x^{12}}-\frac{\left (a+b x^3+c x^6\right )^{5/2}}{15 a x^{15}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*x^3 + c*x^6)^(5/2)/(15*a*x^15) + (b*(16*a^(3/2)*(2*a + b*x^3)*(a + b*x^3 + c*x^6)^(3/2) - 3*(b^2 - 4*a
*c)*x^6*(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])])))/(768*a^(7/2)*x^12)

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{16}} \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^16,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.73701, size = 890, normalized size = 5.49 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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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^{16}}\, 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**16,x)

[Out]

Integral((a + b*x**3 + c*x**6)**(3/2)/x**16, 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^{16}}\,{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^16,x, algorithm="giac")

[Out]

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