3.929 \(\int \frac{\sqrt{a+b x^2+c x^4}}{x^{11}} \, dx\)

Optimal. Leaf size=199 \[ -\frac{\left (35 b^2-32 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 a^3 x^6}+\frac{b \left (7 b^2-12 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^4 x^4}-\frac{b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{512 a^{9/2}}+\frac{7 b \left (a+b x^2+c x^4\right )^{3/2}}{80 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}} \]

[Out]

(b*(7*b^2 - 12*a*c)*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*a^4*x^4) - (a + b*x^2 + c*x^4)^(3/2)/(10*a*x^1
0) + (7*b*(a + b*x^2 + c*x^4)^(3/2))/(80*a^2*x^8) - ((35*b^2 - 32*a*c)*(a + b*x^2 + c*x^4)^(3/2))/(480*a^3*x^6
) - (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(512*a^(9/2)
)

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Rubi [A]  time = 0.229035, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1114, 744, 834, 806, 720, 724, 206} \[ -\frac{\left (35 b^2-32 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 a^3 x^6}+\frac{b \left (7 b^2-12 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^4 x^4}-\frac{b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{512 a^{9/2}}+\frac{7 b \left (a+b x^2+c x^4\right )^{3/2}}{80 a^2 x^8}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^11,x]

[Out]

(b*(7*b^2 - 12*a*c)*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*a^4*x^4) - (a + b*x^2 + c*x^4)^(3/2)/(10*a*x^1
0) + (7*b*(a + b*x^2 + c*x^4)^(3/2))/(80*a^2*x^8) - ((35*b^2 - 32*a*c)*(a + b*x^2 + c*x^4)^(3/2))/(480*a^3*x^6
) - (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(512*a^(9/2)
)

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 744

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[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 834

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

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[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^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{a+b x+c x^2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}}-\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{7 b}{2}+2 c x\right ) \sqrt{a+b x+c x^2}}{x^5} \, dx,x,x^2\right )}{10 a}\\ &=-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}}+\frac{7 b \left (a+b x^2+c x^4\right )^{3/2}}{80 a^2 x^8}+\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{1}{4} \left (35 b^2-32 a c\right )+\frac{7 b c x}{2}\right ) \sqrt{a+b x+c x^2}}{x^4} \, dx,x,x^2\right )}{40 a^2}\\ &=-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}}+\frac{7 b \left (a+b x^2+c x^4\right )^{3/2}}{80 a^2 x^8}-\frac{\left (35 b^2-32 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 a^3 x^6}-\frac{\left (b \left (7 b^2-12 a c\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x+c x^2}}{x^3} \, dx,x,x^2\right )}{64 a^3}\\ &=\frac{b \left (7 b^2-12 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^4 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}}+\frac{7 b \left (a+b x^2+c x^4\right )^{3/2}}{80 a^2 x^8}-\frac{\left (35 b^2-32 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 a^3 x^6}+\frac{\left (b \left (7 b^2-12 a c\right ) \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 )}{512 a^4}\\ &=\frac{b \left (7 b^2-12 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^4 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}}+\frac{7 b \left (a+b x^2+c x^4\right )^{3/2}}{80 a^2 x^8}-\frac{\left (35 b^2-32 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 a^3 x^6}-\frac{\left (b \left (7 b^2-12 a c\right ) \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 )}{256 a^4}\\ &=\frac{b \left (7 b^2-12 a c\right ) \left (2 a+b x^2\right ) \sqrt{a+b x^2+c x^4}}{256 a^4 x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{10 a x^{10}}+\frac{7 b \left (a+b x^2+c x^4\right )^{3/2}}{80 a^2 x^8}-\frac{\left (35 b^2-32 a c\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 a^3 x^6}-\frac{b \left (7 b^2-12 a c\right ) \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 )}{512 a^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.121544, size = 173, normalized size = 0.87 \[ -\frac{\sqrt{a+b x^2+c x^4} \left (-8 a^2 \left (7 b^2 x^4+29 b c x^6+32 c^2 x^8\right )+16 a^3 \left (3 b x^2+8 c x^4\right )+384 a^4+10 a b^2 x^6 \left (7 b+46 c x^2\right )-105 b^4 x^8\right )}{3840 a^4 x^{10}}-\frac{b \left (48 a^2 c^2-40 a b^2 c+7 b^4\right ) \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{512 a^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2 + c*x^4]/x^11,x]

[Out]

-(Sqrt[a + b*x^2 + c*x^4]*(384*a^4 - 105*b^4*x^8 + 10*a*b^2*x^6*(7*b + 46*c*x^2) + 16*a^3*(3*b*x^2 + 8*c*x^4)
- 8*a^2*(7*b^2*x^4 + 29*b*c*x^6 + 32*c^2*x^8)))/(3840*a^4*x^10) - (b*(7*b^4 - 40*a*b^2*c + 48*a^2*c^2)*ArcTanh
[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(512*a^(9/2))

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Maple [B]  time = 0.165, size = 442, normalized size = 2.2 \begin{align*} -{\frac{1}{10\,a{x}^{10}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{7\,b}{80\,{a}^{2}{x}^{8}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{2}}{96\,{a}^{3}{x}^{6}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{b}^{3}}{128\,{a}^{4}{x}^{4}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{4}}{256\,{a}^{5}{x}^{2}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{b}^{5}}{256\,{a}^{5}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{7\,{b}^{5}}{512}\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{9}{2}}}}+{\frac{7\,c{b}^{4}{x}^{2}}{256\,{a}^{5}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{13\,{b}^{3}c}{128\,{a}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,{b}^{3}c}{64}\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{7}{2}}}}-{\frac{3\,bc}{32\,{a}^{3}{x}^{4}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{b}^{2}c}{64\,{a}^{4}{x}^{2}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{2}{c}^{2}{x}^{2}}{64\,{a}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,b{c}^{2}}{32\,{a}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,b{c}^{2}}{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{c}{15\,{a}^{2}{x}^{6}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\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^11,x)

[Out]

-1/10*(c*x^4+b*x^2+a)^(3/2)/a/x^10+7/80*b*(c*x^4+b*x^2+a)^(3/2)/a^2/x^8-7/96*b^2/a^3/x^6*(c*x^4+b*x^2+a)^(3/2)
+7/128*b^3/a^4/x^4*(c*x^4+b*x^2+a)^(3/2)-7/256*b^4/a^5/x^2*(c*x^4+b*x^2+a)^(3/2)+7/256*b^5/a^5*(c*x^4+b*x^2+a)
^(1/2)-7/512*b^5/a^(9/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/2))/x^2)+7/256*b^4/a^5*c*(c*x^4+b*x^2+a)^(
1/2)*x^2-13/128*b^3/a^4*c*(c*x^4+b*x^2+a)^(1/2)+5/64*b^3/a^(7/2)*c*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1/
2))/x^2)-3/32*b/a^3*c/x^4*(c*x^4+b*x^2+a)^(3/2)+3/64*b^2/a^4*c/x^2*(c*x^4+b*x^2+a)^(3/2)-3/64*b^2/a^4*c^2*(c*x
^4+b*x^2+a)^(1/2)*x^2+3/32*b/a^3*c^2*(c*x^4+b*x^2+a)^(1/2)-3/32*b/a^(5/2)*c^2*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b
*x^2+a)^(1/2))/x^2)+1/15*c/a^2/x^6*(c*x^4+b*x^2+a)^(3/2)

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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^11,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.66864, size = 906, normalized size = 4.55 \begin{align*} \left [\frac{15 \,{\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt{a} x^{10} \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 ({\left (105 \, a b^{4} - 460 \, a^{2} b^{2} c + 256 \, a^{3} c^{2}\right )} x^{8} - 48 \, a^{4} b x^{2} - 2 \,{\left (35 \, a^{2} b^{3} - 116 \, a^{3} b c\right )} x^{6} - 384 \, a^{5} + 8 \,{\left (7 \, a^{3} b^{2} - 16 \, a^{4} c\right )} x^{4}\right )} \sqrt{c x^{4} + b x^{2} + a}}{15360 \, a^{5} x^{10}}, \frac{15 \,{\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt{-a} x^{10} \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 ({\left (105 \, a b^{4} - 460 \, a^{2} b^{2} c + 256 \, a^{3} c^{2}\right )} x^{8} - 48 \, a^{4} b x^{2} - 2 \,{\left (35 \, a^{2} b^{3} - 116 \, a^{3} b c\right )} x^{6} - 384 \, a^{5} + 8 \,{\left (7 \, a^{3} b^{2} - 16 \, a^{4} c\right )} x^{4}\right )} \sqrt{c x^{4} + b x^{2} + a}}{7680 \, a^{5} x^{10}}\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^11,x, algorithm="fricas")

[Out]

[1/15360*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sqrt(a)*x^10*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*((105*a*b^4 - 460*a^2*b^2*c + 256*a^3*c^2)*x^8 - 48*a^
4*b*x^2 - 2*(35*a^2*b^3 - 116*a^3*b*c)*x^6 - 384*a^5 + 8*(7*a^3*b^2 - 16*a^4*c)*x^4)*sqrt(c*x^4 + b*x^2 + a))/
(a^5*x^10), 1/7680*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sqrt(-a)*x^10*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*((105*a*b^4 - 460*a^2*b^2*c + 256*a^3*c^2)*x^8 - 48*a^4*b*
x^2 - 2*(35*a^2*b^3 - 116*a^3*b*c)*x^6 - 384*a^5 + 8*(7*a^3*b^2 - 16*a^4*c)*x^4)*sqrt(c*x^4 + b*x^2 + a))/(a^5
*x^10)]

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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^{11}}\, 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**11,x)

[Out]

Integral(sqrt(a + b*x**2 + c*x**4)/x**11, 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^{11}}\,{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^11,x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2 + a)/x^11, x)