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

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

[Out]

-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^8-1/480*(-32*a*c+35*b^2)*(c*x^4+b*x^2+a)
^(3/2)/a^3/x^6-1/512*b*(-12*a*c+7*b^2)*(-4*a*c+b^2)*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/a^(
9/2)+1/256*b*(-12*a*c+7*b^2)*(b*x^2+2*a)*(c*x^4+b*x^2+a)^(1/2)/a^4/x^4

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Rubi [A]  time = 0.23, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.12, size = 173, normalized size = 0.87 \[ -\frac {b \left (48 a^2 c^2-40 a b^2 c+7 b^4\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{512 a^{9/2}}-\frac {\sqrt {a+b x^2+c x^4} \left (384 a^4+16 a^3 \left (3 b x^2+8 c x^4\right )-8 a^2 \left (7 b^2 x^4+29 b c x^6+32 c^2 x^8\right )+10 a b^2 x^6 \left (7 b+46 c x^2\right )-105 b^4 x^8\right )}{3840 a^4 x^{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3840*(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)))/(a^4*x^10) - (b*(7*b^4 - 40*a*b^2*c + 48*a^2*c^2)*ArcTa
nh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(512*a^(9/2))

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fricas [A]  time = 1.10, size = 389, normalized size = 1.95 \[ \left [\frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {a} x^{10} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, {\left ({\left (105 \, a b^{4} - 460 \, a^{2} b^{2} c + 256 \, a^{3} c^{2}\right )} x^{8} - 48 \, a^{4} b x^{2} - 2 \, {\left (35 \, a^{2} b^{3} - 116 \, a^{3} b c\right )} x^{6} - 384 \, a^{5} + 8 \, {\left (7 \, a^{3} b^{2} - 16 \, a^{4} c\right )} x^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{15360 \, a^{5} x^{10}}, \frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (105 \, a b^{4} - 460 \, a^{2} b^{2} c + 256 \, a^{3} c^{2}\right )} x^{8} - 48 \, a^{4} b x^{2} - 2 \, {\left (35 \, a^{2} b^{3} - 116 \, a^{3} b c\right )} x^{6} - 384 \, a^{5} + 8 \, {\left (7 \, a^{3} b^{2} - 16 \, a^{4} c\right )} x^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{7680 \, a^{5} x^{10}}\right ] \]

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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giac [B]  time = 0.36, size = 842, normalized size = 4.23 \[ \frac {{\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{256 \, \sqrt {-a} a^{4}} - \frac {105 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{9} b^{5} - 600 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{9} a b^{3} c + 720 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{9} a^{2} b c^{2} - 490 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a b^{5} + 2800 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{2} b^{3} c - 3360 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{3} b c^{2} - 7680 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{6} a^{4} c^{\frac {5}{2}} + 896 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{2} b^{5} - 5120 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{3} b^{3} c - 15360 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{4} b c^{2} - 24320 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{4} b^{2} c^{\frac {3}{2}} - 2560 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{5} c^{\frac {5}{2}} - 790 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{3} b^{5} - 9200 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{4} b^{3} c - 12000 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{5} b c^{2} - 3840 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{4} b^{4} \sqrt {c} - 5120 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{5} b^{2} c^{\frac {3}{2}} - 2560 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{6} c^{\frac {5}{2}} - 105 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{4} b^{5} - 3240 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{5} b^{3} c - 720 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{6} b c^{2} - 1280 \, a^{6} b^{2} c^{\frac {3}{2}} + 512 \, a^{7} c^{\frac {5}{2}}}{3840 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{5} a^{4}} \]

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]

1/256*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*
a^4) - 1/3840*(105*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^9*b^5 - 600*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))
^9*a*b^3*c + 720*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^9*a^2*b*c^2 - 490*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 +
a))^7*a*b^5 + 2800*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a^2*b^3*c - 3360*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x
^2 + a))^7*a^3*b*c^2 - 7680*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^6*a^4*c^(5/2) + 896*(sqrt(c)*x^2 - sqrt(c*
x^4 + b*x^2 + a))^5*a^2*b^5 - 5120*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^3*b^3*c - 15360*(sqrt(c)*x^2 -
sqrt(c*x^4 + b*x^2 + a))^5*a^4*b*c^2 - 24320*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^4*b^2*c^(3/2) - 2560*
(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^5*c^(5/2) - 790*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^3*b^5
- 9200*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^4*b^3*c - 12000*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a
^5*b*c^2 - 3840*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^4*b^4*sqrt(c) - 5120*(sqrt(c)*x^2 - sqrt(c*x^4 + b
*x^2 + a))^2*a^5*b^2*c^(3/2) - 2560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^6*c^(5/2) - 105*(sqrt(c)*x^2 -
sqrt(c*x^4 + b*x^2 + a))*a^4*b^5 - 3240*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^5*b^3*c - 720*(sqrt(c)*x^2
- sqrt(c*x^4 + b*x^2 + a))*a^6*b*c^2 - 1280*a^6*b^2*c^(3/2) + 512*a^7*c^(5/2))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b
*x^2 + a))^2 - a)^5*a^4)

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maple [B]  time = 0.02, size = 442, normalized size = 2.22 \[ -\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} c^{2} x^{2}}{64 a^{4}}+\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{4} c \,x^{2}}{256 a^{5}}-\frac {3 b \,c^{2} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{32 a^{\frac {5}{2}}}+\frac {5 b^{3} c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{64 a^{\frac {7}{2}}}-\frac {7 b^{5} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{512 a^{\frac {9}{2}}}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,c^{2}}{32 a^{3}}-\frac {13 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3} c}{128 a^{4}}+\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{5}}{256 a^{5}}+\frac {3 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{2} c}{64 a^{4} x^{2}}-\frac {7 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{4}}{256 a^{5} x^{2}}-\frac {3 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b c}{32 a^{3} x^{4}}+\frac {7 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}{128 a^{4} x^{4}}+\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} c}{15 a^{2} x^{6}}-\frac {7 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}{96 a^{3} x^{6}}+\frac {7 \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b}{80 a^{2} x^{8}}-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{10 a \,x^{10}} \]

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((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*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((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*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((b*x^2+2*a+2*(c*x^4+b*x^2+a)^
(1/2)*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^11,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 [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,x^4+b\,x^2+a}}{x^{11}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(1/2)/x^11,x)

[Out]

int((a + b*x^2 + c*x^4)^(1/2)/x^11, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{11}}\, dx \]

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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