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

Optimal. Leaf size=109 \[ \frac {1}{3} \sqrt {a+b x^3+c x^6}-\frac {1}{3} \sqrt {a} \tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )+\frac {b \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{6 \sqrt {c}} \]

[Out]

-1/3*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))*a^(1/2)+1/6*b*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x
^6+b*x^3+a)^(1/2))/c^(1/2)+1/3*(c*x^6+b*x^3+a)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1357, 734, 843, 621, 206, 724} \[ \frac {1}{3} \sqrt {a+b x^3+c x^6}-\frac {1}{3} \sqrt {a} \tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )+\frac {b \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{6 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3 + c*x^6]/x,x]

[Out]

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

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 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 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 734

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 + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, 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] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

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 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]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 106, normalized size = 0.97 \[ \frac {1}{6} \left (2 \sqrt {a+b x^3+c x^6}-2 \sqrt {a} \tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )+\frac {b \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{\sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^3 + c*x^6]/x,x]

[Out]

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

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fricas [A]  time = 1.24, size = 566, normalized size = 5.19 \[ \left [\frac {b \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 2 \, \sqrt {a} c \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) + 4 \, \sqrt {c x^{6} + b x^{3} + a} c}{12 \, c}, -\frac {b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - \sqrt {a} c \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) - 2 \, \sqrt {c x^{6} + b x^{3} + a} c}{6 \, c}, \frac {4 \, \sqrt {-a} c \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + b \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, \sqrt {c x^{6} + b x^{3} + a} c}{12 \, c}, \frac {2 \, \sqrt {-a} c \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) - b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) + 2 \, \sqrt {c x^{6} + b x^{3} + a} c}{6 \, c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(b*sqrt(c)*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) +
2*sqrt(a)*c*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*sqrt(c*x^6 + b*x^3 + a)*c)/c, -1/6*(b*sqrt(-c)*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)) - sqrt(a)*c*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) - 2*sqrt(c*x^6 + b*x^3 + a)*c)/c, 1/12*(4*sqrt(-a)*c*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)) + b*sqrt(c)*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sq
rt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(c) - 4*a*c) + 4*sqrt(c*x^6 + b*x^3 + a)*c)/c, 1/6*(2*sqrt(-a)*c*arcta
n(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)) - b*sqrt(-c)*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*sqrt(c*x^6 + b*x^3 + a)*c)/c]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{6} + b x^{3} + a}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)^(1/2)/x,x)

[Out]

int((c*x^6+b*x^3+a)^(1/2)/x,x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 1.36, size = 88, normalized size = 0.81 \[ \frac {\sqrt {c\,x^6+b\,x^3+a}}{3}-\frac {\sqrt {a}\,\ln \left (\frac {b}{2}+\frac {a}{x^3}+\frac {\sqrt {a}\,\sqrt {c\,x^6+b\,x^3+a}}{x^3}\right )}{3}+\frac {b\,\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )}{6\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3 + c*x^6)^(1/2)/x,x)

[Out]

(a + b*x^3 + c*x^6)^(1/2)/3 - (a^(1/2)*log(b/2 + a/x^3 + (a^(1/2)*(a + b*x^3 + c*x^6)^(1/2))/x^3))/3 + (b*log(
(a + b*x^3 + c*x^6)^(1/2) + (b/2 + c*x^3)/c^(1/2)))/(6*c^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b x^{3} + c x^{6}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**6+b*x**3+a)**(1/2)/x,x)

[Out]

Integral(sqrt(a + b*x**3 + c*x**6)/x, x)

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