3.927 \(\int \frac{\sqrt{a+b x^2+c x^4}}{x^7} \, dx\)

Optimal. Leaf size=116 \[ -\frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{32 a^{5/2}}+\frac{b \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{16 a^2 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6} \]

[Out]

(b*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*a^2*x^4) - (a + b*x^2 + c*x^4)^(3/2)/(6*a*x^6) - (b*(b^2 - 4*a*c
)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(5/2))

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Rubi [A]  time = 0.0964375, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1114, 730, 720, 724, 206} \[ -\frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{32 a^{5/2}}+\frac{b \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{16 a^2 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^7,x]

[Out]

(b*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*a^2*x^4) - (a + b*x^2 + c*x^4)^(3/2)/(6*a*x^6) - (b*(b^2 - 4*a*c
)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(5/2))

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 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{\sqrt{a+b x^2+c x^4}}{x^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{a+b x+c x^2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}-\frac{b \operatorname{Subst}\left (\int \frac{\sqrt{a+b x+c x^2}}{x^3} \, dx,x,x^2\right )}{4 a}\\ &=\frac{b \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{16 a^2 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}+\frac{\left (b \left (b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{32 a^2}\\ &=\frac{b \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{16 a^2 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}-\frac{\left (b \left (b^2-4 a c\right )\right ) \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 )}{16 a^2}\\ &=\frac{b \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{16 a^2 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 a x^6}-\frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{32 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0756663, size = 108, normalized size = 0.93 \[ -\frac{\sqrt{a+b x^2+c x^4} \left (8 a^2+2 a x^2 \left (b+4 c x^2\right )-3 b^2 x^4\right )}{48 a^2 x^6}-\frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{32 a^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2 + c*x^4]/x^7,x]

[Out]

-(Sqrt[a + b*x^2 + c*x^4]*(8*a^2 - 3*b^2*x^4 + 2*a*x^2*(b + 4*c*x^2)))/(48*a^2*x^6) - (b*(b^2 - 4*a*c)*ArcTanh
[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(5/2))

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Maple [B]  time = 0.164, size = 222, normalized size = 1.9 \begin{align*} -{\frac{1}{6\,a{x}^{6}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{b}{8\,{a}^{2}{x}^{4}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}}{16\,{x}^{2}{a}^{3}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{3}}{16\,{a}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{{b}^{3}}{32}\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 ){a}^{-{\frac{5}{2}}}}+{\frac{{b}^{2}c{x}^{2}}{16\,{a}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{bc}{8\,{a}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{bc}{8}\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 ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(1/2)/x^7,x)

[Out]

-1/6*(c*x^4+b*x^2+a)^(3/2)/a/x^6+1/8*b/a^2/x^4*(c*x^4+b*x^2+a)^(3/2)-1/16*b^2/a^3/x^2*(c*x^4+b*x^2+a)^(3/2)+1/
16*b^3/a^3*(c*x^4+b*x^2+a)^(1/2)-1/32*b^3/a^(5/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)+1/16*b^2
/a^3*c*(c*x^4+b*x^2+a)^(1/2)*x^2-1/8*b/a^2*c*(c*x^4+b*x^2+a)^(1/2)+1/8*b/a^(3/2)*c*ln((2*a+b*x^2+2*a^(1/2)*(c*
x^4+b*x^2+a)^(1/2))/x^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^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78987, size = 594, normalized size = 5.12 \begin{align*} \left [-\frac{3 \,{\left (b^{3} - 4 \, a b c\right )} \sqrt{a} x^{6} \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 \,{\left (2 \, a^{2} b x^{2} -{\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt{c x^{4} + b x^{2} + a}}{192 \, a^{3} x^{6}}, \frac{3 \,{\left (b^{3} - 4 \, a b c\right )} \sqrt{-a} x^{6} \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 \,{\left (2 \, a^{2} b x^{2} -{\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt{c x^{4} + b x^{2} + a}}{96 \, a^{3} x^{6}}\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^7,x, algorithm="fricas")

[Out]

[-1/192*(3*(b^3 - 4*a*b*c)*sqrt(a)*x^6*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*(2*a^2*b*x^2 - (3*a*b^2 - 8*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/(a^3
*x^6), 1/96*(3*(b^3 - 4*a*b*c)*sqrt(-a)*x^6*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*(2*a^2*b*x^2 - (3*a*b^2 - 8*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/(a^3*x^6)]

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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^{7}}\, 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**7,x)

[Out]

Integral(sqrt(a + b*x**2 + c*x**4)/x**7, 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^{7}}\,{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^7,x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2 + a)/x^7, x)