3.393 \(\int \frac {1}{x^6 \sqrt {d+e x^2} (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=443 \[ -\frac {\left (b^2-a c\right ) \sqrt {d+e x^2}}{a^3 d x}-\frac {c \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{a^3 \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}-\frac {c \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{a^3 \sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}-\frac {2 b e \sqrt {d+e x^2}}{3 a^2 d^2 x}+\frac {b \sqrt {d+e x^2}}{3 a^2 d x^3}-\frac {8 e^2 \sqrt {d+e x^2}}{15 a d^3 x}+\frac {4 e \sqrt {d+e x^2}}{15 a d^2 x^3}-\frac {\sqrt {d+e x^2}}{5 a d x^5} \]

[Out]

-1/5*(e*x^2+d)^(1/2)/a/d/x^5+1/3*b*(e*x^2+d)^(1/2)/a^2/d/x^3+4/15*e*(e*x^2+d)^(1/2)/a/d^2/x^3-(-a*c+b^2)*(e*x^
2+d)^(1/2)/a^3/d/x-2/3*b*e*(e*x^2+d)^(1/2)/a^2/d^2/x-8/15*e^2*(e*x^2+d)^(1/2)/a/d^3/x-c*arctan(x*(2*c*d-e*(b-(
-4*a*c+b^2)^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*(b^2-a*c+b*(-3*a*c+b^2)/(-4*a*c+b^2)^(
1/2))/a^3/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-c*arctan(x*(2*c*d-e*(b+(-4*a*c+b
^2)^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*(b^2-a*c-b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/a^
3/(b+(-4*a*c+b^2)^(1/2))^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)

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Rubi [A]  time = 1.43, antiderivative size = 443, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1303, 271, 264, 1692, 377, 205} \[ -\frac {\left (b^2-a c\right ) \sqrt {d+e x^2}}{a^3 d x}-\frac {c \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{a^3 \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}-\frac {c \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{a^3 \sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}-\frac {2 b e \sqrt {d+e x^2}}{3 a^2 d^2 x}+\frac {b \sqrt {d+e x^2}}{3 a^2 d x^3}-\frac {8 e^2 \sqrt {d+e x^2}}{15 a d^3 x}+\frac {4 e \sqrt {d+e x^2}}{15 a d^2 x^3}-\frac {\sqrt {d+e x^2}}{5 a d x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)),x]

[Out]

-Sqrt[d + e*x^2]/(5*a*d*x^5) + (b*Sqrt[d + e*x^2])/(3*a^2*d*x^3) + (4*e*Sqrt[d + e*x^2])/(15*a*d^2*x^3) - ((b^
2 - a*c)*Sqrt[d + e*x^2])/(a^3*d*x) - (2*b*e*Sqrt[d + e*x^2])/(3*a^2*d^2*x) - (8*e^2*Sqrt[d + e*x^2])/(15*a*d^
3*x) - (c*(b^2 - a*c + (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)
/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(a^3*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 -
4*a*c])*e]) - (c*(b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c
])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(a^3*Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sq
rt[b^2 - 4*a*c])*e])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1303

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x^2)^q, (f*x)^m/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[b^2
- 4*a*c, 0] &&  !IntegerQ[q] && IntegerQ[m]

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx &=\int \left (\frac {1}{a x^6 \sqrt {d+e x^2}}-\frac {b}{a^2 x^4 \sqrt {d+e x^2}}+\frac {b^2-a c}{a^3 x^2 \sqrt {d+e x^2}}+\frac {-b \left (b^2-2 a c\right )-c \left (b^2-a c\right ) x^2}{a^3 \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=\frac {\int \frac {-b \left (b^2-2 a c\right )-c \left (b^2-a c\right ) x^2}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{a^3}+\frac {\int \frac {1}{x^6 \sqrt {d+e x^2}} \, dx}{a}-\frac {b \int \frac {1}{x^4 \sqrt {d+e x^2}} \, dx}{a^2}+\frac {\left (b^2-a c\right ) \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx}{a^3}\\ &=-\frac {\sqrt {d+e x^2}}{5 a d x^5}+\frac {b \sqrt {d+e x^2}}{3 a^2 d x^3}-\frac {\left (b^2-a c\right ) \sqrt {d+e x^2}}{a^3 d x}+\frac {\int \left (\frac {-\frac {b c \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-c \left (b^2-a c\right )}{\left (b-\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}}+\frac {\frac {b c \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-c \left (b^2-a c\right )}{\left (b+\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}}\right ) \, dx}{a^3}-\frac {(4 e) \int \frac {1}{x^4 \sqrt {d+e x^2}} \, dx}{5 a d}+\frac {(2 b e) \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx}{3 a^2 d}\\ &=-\frac {\sqrt {d+e x^2}}{5 a d x^5}+\frac {b \sqrt {d+e x^2}}{3 a^2 d x^3}+\frac {4 e \sqrt {d+e x^2}}{15 a d^2 x^3}-\frac {\left (b^2-a c\right ) \sqrt {d+e x^2}}{a^3 d x}-\frac {2 b e \sqrt {d+e x^2}}{3 a^2 d^2 x}-\frac {\left (c \left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\left (b+\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}} \, dx}{a^3}-\frac {\left (c \left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\left (b-\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}} \, dx}{a^3}+\frac {\left (8 e^2\right ) \int \frac {1}{x^2 \sqrt {d+e x^2}} \, dx}{15 a d^2}\\ &=-\frac {\sqrt {d+e x^2}}{5 a d x^5}+\frac {b \sqrt {d+e x^2}}{3 a^2 d x^3}+\frac {4 e \sqrt {d+e x^2}}{15 a d^2 x^3}-\frac {\left (b^2-a c\right ) \sqrt {d+e x^2}}{a^3 d x}-\frac {2 b e \sqrt {d+e x^2}}{3 a^2 d^2 x}-\frac {8 e^2 \sqrt {d+e x^2}}{15 a d^3 x}-\frac {\left (c \left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b+\sqrt {b^2-4 a c}-\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{a^3}-\frac {\left (c \left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b-\sqrt {b^2-4 a c}-\left (-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{a^3}\\ &=-\frac {\sqrt {d+e x^2}}{5 a d x^5}+\frac {b \sqrt {d+e x^2}}{3 a^2 d x^3}+\frac {4 e \sqrt {d+e x^2}}{15 a d^2 x^3}-\frac {\left (b^2-a c\right ) \sqrt {d+e x^2}}{a^3 d x}-\frac {2 b e \sqrt {d+e x^2}}{3 a^2 d^2 x}-\frac {8 e^2 \sqrt {d+e x^2}}{15 a d^3 x}-\frac {c \left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \tan ^{-1}\left (\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{a^3 \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}-\frac {c \left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \tan ^{-1}\left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{a^3 \sqrt {b+\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}\\ \end {align*}

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Mathematica [A]  time = 1.64, size = 383, normalized size = 0.86 \[ -\frac {\frac {a^2 \sqrt {d+e x^2} \left (3 d^2-4 d e x^2+8 e^2 x^4\right )}{d^3 x^5}+\frac {15 \left (b^2-a c\right ) \sqrt {d+e x^2}}{d x}+\frac {15 c \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {e \sqrt {b^2-4 a c}-b e+2 c d}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {e \left (\sqrt {b^2-4 a c}-b\right )+2 c d}}+\frac {15 c \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}-\frac {5 a b \left (d-2 e x^2\right ) \sqrt {d+e x^2}}{d^2 x^3}}{15 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)),x]

[Out]

-1/15*((15*(b^2 - a*c)*Sqrt[d + e*x^2])/(d*x) - (5*a*b*(d - 2*e*x^2)*Sqrt[d + e*x^2])/(d^2*x^3) + (a^2*Sqrt[d
+ e*x^2]*(3*d^2 - 4*d*e*x^2 + 8*e^2*x^4))/(d^3*x^5) + (15*c*(b^2 - a*c + (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*
ArcTan[(Sqrt[2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b - S
qrt[b^2 - 4*a*c]]*Sqrt[2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e]) + (15*c*(b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 -
4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sq
rt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]))/a^3

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.04, size = 350, normalized size = 0.79 \[ -\frac {\sqrt {e}\, \left (a \,c^{2} d^{2}-b^{2} c \,d^{2}+\left (a c -b^{2}\right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} c +2 \left (4 a b c e -a \,c^{2} d -2 b^{3} e +b^{2} c d \right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )+\left (-\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )^{2}\right )}{2 a^{3} \left (\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{3} c +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} b e -3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} c d +8 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) a \,e^{2}-4 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) b d e +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) c \,d^{2}+b \,d^{2} e -c \,d^{3}\right )}-\frac {8 \sqrt {e \,x^{2}+d}\, e^{2}}{15 a \,d^{3} x}-\frac {2 \sqrt {e \,x^{2}+d}\, b e}{3 a^{2} d^{2} x}+\frac {4 \sqrt {e \,x^{2}+d}\, e}{15 a \,d^{2} x^{3}}+\frac {\sqrt {e \,x^{2}+d}\, b}{3 a^{2} d \,x^{3}}-\frac {\left (-a c +b^{2}\right ) \sqrt {e \,x^{2}+d}}{a^{3} d x}-\frac {\sqrt {e \,x^{2}+d}}{5 a d \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x)

[Out]

-(-a*c+b^2)*(e*x^2+d)^(1/2)/a^3/d/x-1/2/a^3*e^(1/2)*sum((c*(a*c-b^2)*_R^2+2*(4*a*b*c*e-a*c^2*d-2*b^3*e+b^2*c*d
)*_R+a*c^2*d^2-b^2*c*d^2)/(_R^3*c+3*_R^2*b*e-3*_R^2*c*d+8*_R*a*e^2-4*_R*b*d*e+3*_R*c*d^2+b*d^2*e-c*d^3)*ln(-_R
+(-e^(1/2)*x+(e*x^2+d)^(1/2))^2),_R=RootOf(_Z^4*c+c*d^4+(4*b*e-4*c*d)*_Z^3+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^2+(4*
b*d^2*e-4*c*d^3)*_Z))+1/3*b*(e*x^2+d)^(1/2)/a^2/d/x^3-2/3*b*e*(e*x^2+d)^(1/2)/a^2/d^2/x-1/5*(e*x^2+d)^(1/2)/a/
d/x^5+4/15*e*(e*x^2+d)^(1/2)/a/d^2/x^3-8/15*e^2*(e*x^2+d)^(1/2)/a/d^3/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} \sqrt {e x^{2} + d} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(c*x^4+b*x^2+a)/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2 + a)*sqrt(e*x^2 + d)*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^6\,\sqrt {e\,x^2+d}\,\left (c\,x^4+b\,x^2+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(d + e*x^2)^(1/2)*(a + b*x^2 + c*x^4)),x)

[Out]

int(1/(x^6*(d + e*x^2)^(1/2)*(a + b*x^2 + c*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{6} \sqrt {d + e x^{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(c*x**4+b*x**2+a)/(e*x**2+d)**(1/2),x)

[Out]

Integral(1/(x**6*sqrt(d + e*x**2)*(a + b*x**2 + c*x**4)), x)

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