3.925 \(\int \frac{\sqrt{a+b x^2+c x^4}}{x^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac{\sqrt{a+b x^2+c x^4}}{2 x^2}-\frac{b \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.103766, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1114, 732, 843, 621, 206, 724} \[ -\frac{\sqrt{a+b x^2+c x^4}}{2 x^2}-\frac{b \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^3,x]

[Out]

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

Rule 1114

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

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

Mathematica [A]  time = 0.0472109, size = 112, normalized size = 1. \[ -\frac{\sqrt{a+b x^2+c x^4}}{2 x^2}-\frac{b \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2 + c*x^4]/x^3,x]

[Out]

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

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Maple [A]  time = 0.162, size = 140, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,a{x}^{2}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{b}{2\,a}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{b}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+b{x}^{2}+2\,\sqrt{a}\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{c{x}^{2}}{2\,a}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{1}{2}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(1/2)/x^3,x)

[Out]

-1/2/a/x^2*(c*x^4+b*x^2+a)^(3/2)+1/2*b/a*(c*x^4+b*x^2+a)^(1/2)-1/4*b/a^(1/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*
x^2+a)^(1/2))/x^2)+1/2*c/a*(c*x^4+b*x^2+a)^(1/2)*x^2+1/2*c^(1/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2
))

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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^4+b*x^2+a)^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86422, size = 1408, normalized size = 12.57 \begin{align*} \left [\frac{2 \, a \sqrt{c} x^{2} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 4 \, a c\right ) + \sqrt{a} b x^{2} \log \left (-\frac{{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, \sqrt{c x^{4} + b x^{2} + a} a}{8 \, a x^{2}}, -\frac{4 \, a \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - \sqrt{a} b x^{2} \log \left (-\frac{{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, \sqrt{c x^{4} + b x^{2} + a} a}{8 \, a x^{2}}, \frac{\sqrt{-a} b x^{2} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + a \sqrt{c} x^{2} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 4 \, a c\right ) - 2 \, \sqrt{c x^{4} + b x^{2} + a} a}{4 \, a x^{2}}, \frac{\sqrt{-a} b x^{2} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (b x^{2} + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - 2 \, a \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, \sqrt{c x^{4} + b x^{2} + a} a}{4 \, a x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/8*(2*a*sqrt(c)*x^2*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a
*c) + sqrt(a)*b*x^2*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*
a^2)/x^4) - 4*sqrt(c*x^4 + b*x^2 + a)*a)/(a*x^2), -1/8*(4*a*sqrt(-c)*x^2*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2
*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) - sqrt(a)*b*x^2*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c
*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) + 4*sqrt(c*x^4 + b*x^2 + a)*a)/(a*x^2), 1/4*(sqrt(-a)*b*
x^2*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + a*sqrt(c)*x^2*log(-
8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) - 2*sqrt(c*x^4 + b*x^2
+ a)*a)/(a*x^2), 1/4*(sqrt(-a)*b*x^2*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*
x^2 + a^2)) - 2*a*sqrt(-c)*x^2*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 +
a*c)) - 2*sqrt(c*x^4 + b*x^2 + a)*a)/(a*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b x^{2} + c x^{4}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(1/2)/x**3,x)

[Out]

Integral(sqrt(a + b*x**2 + c*x**4)/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{4} + b x^{2} + a}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^3,x, algorithm="giac")

[Out]

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