3.209 \(\int \frac{(a+b x^3+c x^6)^{3/2}}{x^{10}} \, dx\)

Optimal. Leaf size=163 \[ \frac{b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{48 a^{3/2}}-\frac{\left (x^3 \left (8 a c+b^2\right )+2 a b\right ) \sqrt{a+b x^3+c x^6}}{24 a x^6}+\frac{1}{3} c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{9 x^9} \]

[Out]

-((2*a*b + (b^2 + 8*a*c)*x^3)*Sqrt[a + b*x^3 + c*x^6])/(24*a*x^6) - (a + b*x^3 + c*x^6)^(3/2)/(9*x^9) + (b*(b^
2 - 12*a*c)*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(48*a^(3/2)) + (c^(3/2)*ArcTanh[(b + 2
*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/3

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Rubi [A]  time = 0.17553, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1357, 732, 810, 843, 621, 206, 724} \[ \frac{b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{48 a^{3/2}}-\frac{\left (x^3 \left (8 a c+b^2\right )+2 a b\right ) \sqrt{a+b x^3+c x^6}}{24 a x^6}+\frac{1}{3} c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)^(3/2)/x^10,x]

[Out]

-((2*a*b + (b^2 + 8*a*c)*x^3)*Sqrt[a + b*x^3 + c*x^6])/(24*a*x^6) - (a + b*x^3 + c*x^6)^(3/2)/(9*x^9) + (b*(b^
2 - 12*a*c)*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(48*a^(3/2)) + (c^(3/2)*ArcTanh[(b + 2
*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/3

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 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&&  !IGtQ[m, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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])

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3+c x^6\right )^{3/2}}{x^{10}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^{3/2}}{x^4} \, dx,x,x^3\right )\\ &=-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{9 x^9}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{(b+2 c x) \sqrt{a+b x+c x^2}}{x^3} \, dx,x,x^3\right )\\ &=-\frac{\left (2 a b+\left (b^2+8 a c\right ) x^3\right ) \sqrt{a+b x^3+c x^6}}{24 a x^6}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{9 x^9}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} b \left (b^2-12 a c\right )-8 a c^2 x}{x \sqrt{a+b x+c x^2}} \, dx,x,x^3\right )}{24 a}\\ &=-\frac{\left (2 a b+\left (b^2+8 a c\right ) x^3\right ) \sqrt{a+b x^3+c x^6}}{24 a x^6}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{9 x^9}+\frac{1}{3} c^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^3\right )-\frac{\left (b \left (b^2-12 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^3\right )}{48 a}\\ &=-\frac{\left (2 a b+\left (b^2+8 a c\right ) x^3\right ) \sqrt{a+b x^3+c x^6}}{24 a x^6}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{9 x^9}+\frac{1}{3} \left (2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^3}{\sqrt{a+b x^3+c x^6}}\right )+\frac{\left (b \left (b^2-12 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 )}{24 a}\\ &=-\frac{\left (2 a b+\left (b^2+8 a c\right ) x^3\right ) \sqrt{a+b x^3+c x^6}}{24 a x^6}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{9 x^9}+\frac{b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{48 a^{3/2}}+\frac{1}{3} c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )\\ \end{align*}

Mathematica [A]  time = 0.213963, size = 149, normalized size = 0.91 \[ \frac{1}{144} \left (-\frac{2 \sqrt{a+b x^3+c x^6} \left (8 a^2+14 a b x^3+32 a c x^6+3 b^2 x^6\right )}{a x^9}+\frac{3 b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )}{a^{3/2}}+48 c^{3/2} \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3 + c*x^6)^(3/2)/x^10,x]

[Out]

((-2*Sqrt[a + b*x^3 + c*x^6]*(8*a^2 + 14*a*b*x^3 + 3*b^2*x^6 + 32*a*c*x^6))/(a*x^9) + (3*b*(b^2 - 12*a*c)*ArcT
anh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/a^(3/2) + 48*c^(3/2)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*
Sqrt[a + b*x^3 + c*x^6])])/144

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Maple [F]  time = 0.041, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{10}} \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^10,x)

[Out]

int((c*x^6+b*x^3+a)^(3/2)/x^10,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^10,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.47602, size = 1806, normalized size = 11.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)^(3/2)/x^10,x, algorithm="fricas")

[Out]

[1/288*(48*a^2*c^(3/2)*x^9*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(c)
- 4*a*c) - 3*(b^3 - 12*a*b*c)*sqrt(a)*x^9*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^2 + 32*a^2*c)*x^6 + 14*a^2*b*x^3 + 8*a^3)*sqrt(c*x^6 + b*x^3 + a))
/(a^2*x^9), -1/288*(96*a^2*sqrt(-c)*c*x^9*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(-c)/(c^2*x^6 +
b*c*x^3 + a*c)) + 3*(b^3 - 12*a*b*c)*sqrt(a)*x^9*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^2 + 32*a^2*c)*x^6 + 14*a^2*b*x^3 + 8*a^3)*sqrt(c*x^6 + b*x
^3 + a))/(a^2*x^9), 1/144*(24*a^2*c^(3/2)*x^9*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sqrt(c*x^6 + b*x^3 + a)*(2*
c*x^3 + b)*sqrt(c) - 4*a*c) - 3*(b^3 - 12*a*b*c)*sqrt(-a)*x^9*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^2 + 32*a^2*c)*x^6 + 14*a^2*b*x^3 + 8*a^3)*sqrt(c*x^6 + b*x^3
+ a))/(a^2*x^9), -1/144*(48*a^2*sqrt(-c)*c*x^9*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(-c)/(c^2*
x^6 + b*c*x^3 + a*c)) + 3*(b^3 - 12*a*b*c)*sqrt(-a)*x^9*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^2 + 32*a^2*c)*x^6 + 14*a^2*b*x^3 + 8*a^3)*sqrt(c*x^6 + b*x^3 + a))/
(a^2*x^9)]

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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^{10}}\, 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**10,x)

[Out]

Integral((a + b*x**3 + c*x**6)**(3/2)/x**10, 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^{10}}\,{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^10,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^10, x)