3.207 \(\int \frac{(a+b x^3+c x^6)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=150 \[ \frac{\left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{8 \sqrt{c}}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{3 x^3}+\frac{1}{4} \left (3 b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}-\frac{1}{2} \sqrt{a} b \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.169818, antiderivative size = 150, 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, 814, 843, 621, 206, 724} \[ \frac{\left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{8 \sqrt{c}}-\frac{\left (a+b x^3+c x^6\right )^{3/2}}{3 x^3}+\frac{1}{4} \left (3 b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}-\frac{1}{2} \sqrt{a} b \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)^(3/2)/x^4,x]

[Out]

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

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 814

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

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

Mathematica [A]  time = 0.110244, size = 134, normalized size = 0.89 \[ \frac{1}{24} \left (\frac{3 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{\sqrt{c}}+\frac{2 \sqrt{a+b x^3+c x^6} \left (-4 a+5 b x^3+2 c x^6\right )}{x^3}-12 \sqrt{a} b \tanh ^{-1}\left (\frac{2 a+b x^3}{2 \sqrt{a} \sqrt{a+b x^3+c x^6}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3 + c*x^6)^(3/2)/x^4,x]

[Out]

((2*Sqrt[a + b*x^3 + c*x^6]*(-4*a + 5*b*x^3 + 2*c*x^6))/x^3 - 12*Sqrt[a]*b*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sq
rt[a + b*x^3 + c*x^6])] + (3*(b^2 + 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/Sqrt[c]
)/24

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \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^4,x)

[Out]

int((c*x^6+b*x^3+a)^(3/2)/x^4,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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.25671, size = 1667, normalized size = 11.11 \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^4,x, algorithm="fricas")

[Out]

[1/48*(12*sqrt(a)*b*c*x^3*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) + 3*(b^2 + 4*a*c)*sqrt(c)*x^3*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) + 4*(2*c^2*x^6 + 5*b*c*x^3 - 4*a*c)*sqrt(c*x^6 + b*x^3 + a))/(c*x^3), 1/24*(6*sqrt
(a)*b*c*x^3*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
) - 3*(b^2 + 4*a*c)*sqrt(-c)*x^3*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)) + 2*(2*c^2*x^6 + 5*b*c*x^3 - 4*a*c)*sqrt(c*x^6 + b*x^3 + a))/(c*x^3), 1/48*(24*sqrt(-a)*b*c*x^3*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)) + 3*(b^2 + 4*a*c)*sqrt(c)*x^3*l
og(-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) + 4*(2*c^2*x^6 + 5*
b*c*x^3 - 4*a*c)*sqrt(c*x^6 + b*x^3 + a))/(c*x^3), 1/24*(12*sqrt(-a)*b*c*x^3*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)) - 3*(b^2 + 4*a*c)*sqrt(-c)*x^3*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)) + 2*(2*c^2*x^6 + 5*b*c*x^3 - 4*a*c)*sqrt(c*x^6 + b*x
^3 + a))/(c*x^3)]

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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^{4}}\, 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**4,x)

[Out]

Integral((a + b*x**3 + c*x**6)**(3/2)/x**4, 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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^4, x)