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

Optimal. Leaf size=518 \[ \frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}+\frac {e \tan ^{-1}\left (\frac {x \sqrt {-a e^4-b d^2 e^2-c d^4}}{d e \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {-a e^4-b d^2 e^2-c d^4}}-\frac {e \tanh ^{-1}\left (\frac {2 a e^2+x^2 \left (b e^2+2 c d^2\right )+b d^2}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^4+b d^2 e^2+c d^4}}\right )}{2 \sqrt {a e^4+b d^2 e^2+c d^4}} \]

[Out]

1/2*e*arctan(x*(-a*e^4-b*d^2*e^2-c*d^4)^(1/2)/d/e/(c*x^4+b*x^2+a)^(1/2))/(-a*e^4-b*d^2*e^2-c*d^4)^(1/2)-1/2*e*
arctanh(1/2*(b*d^2+2*a*e^2+(b*e^2+2*c*d^2)*x^2)/(a*e^4+b*d^2*e^2+c*d^4)^(1/2)/(c*x^4+b*x^2+a)^(1/2))/(a*e^4+b*
d^2*e^2+c*d^4)^(1/2)+1/2*c^(1/4)*d*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))
*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+b*x
^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+b*x^2+a)^(1/2)-1/4*(cos(2*arctan
(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticPi(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/4
*(e^2*a^(1/2)+d^2*c^(1/2))^2/d^2/e^2/a^(1/2)/c^(1/2),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2
))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(1/4)/d/(e^2*a^(1/2)+d^2*c^
(1/2))/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 518, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1724, 1216, 1103, 1706, 1247, 724, 206} \[ \frac {e \tan ^{-1}\left (\frac {x \sqrt {-a e^4-b d^2 e^2-c d^4}}{d e \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {-a e^4-b d^2 e^2-c d^4}}-\frac {e \tanh ^{-1}\left (\frac {2 a e^2+x^2 \left (b e^2+2 c d^2\right )+b d^2}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^4+b d^2 e^2+c d^4}}\right )}{2 \sqrt {a e^4+b d^2 e^2+c d^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*ArcTan[(Sqrt[-(c*d^4) - b*d^2*e^2 - a*e^4]*x)/(d*e*Sqrt[a + b*x^2 + c*x^4])])/(2*Sqrt[-(c*d^4) - b*d^2*e^2
- a*e^4]) - (e*ArcTanh[(b*d^2 + 2*a*e^2 + (2*c*d^2 + b*e^2)*x^2)/(2*Sqrt[c*d^4 + b*d^2*e^2 + a*e^4]*Sqrt[a + b
*x^2 + c*x^4])])/(2*Sqrt[c*d^4 + b*d^2*e^2 + a*e^4]) + (c^(1/4)*d*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*
x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a^(1
/4)*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + b*x^2 + c*x^4]) - ((Sqrt[c]*d^2 - Sqrt[a]*e^2)*(Sqrt[a] + Sqrt[c]*x^2
)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[(Sqrt[c]*d^2 + Sqrt[a]*e^2)^2/(4*Sqrt[a]*Sqrt
[c]*d^2*e^2), 2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(4*a^(1/4)*c^(1/4)*d*(Sqrt[c]*d^2 +
Sqrt[a]*e^2)*Sqrt[a + b*x^2 + c*x^4])

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]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1216

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Di
st[(c*d + a*e*q)/(c*d^2 - a*e^2), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2)
, Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), 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[c*d^2 - a*e^2, 0] && PosQ[c/a]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^
2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rule 1724

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x
^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; Free
Q[{a, b, c, d, e}, x]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x) \sqrt {a+b x^2+c x^4}} \, dx &=d \int \frac {1}{\left (d^2-e^2 x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx-e \int \frac {x}{\left (d^2-e^2 x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx\\ &=-\left (\frac {1}{2} e \operatorname {Subst}\left (\int \frac {1}{\left (d^2-e^2 x\right ) \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\right )+\frac {\left (\sqrt {c} d\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{\sqrt {c} d^2+\sqrt {a} e^2}+\frac {\left (\sqrt {a} d e^2\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\left (d^2-e^2 x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx}{\sqrt {c} d^2+\sqrt {a} e^2}\\ &=\frac {e \tan ^{-1}\left (\frac {\sqrt {-c d^4-b d^2 e^2-a e^4} x}{d e \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {-c d^4-b d^2 e^2-a e^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+b x^2+c x^4}}-\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+b x^2+c x^4}}+e \operatorname {Subst}\left (\int \frac {1}{4 c d^4+4 b d^2 e^2+4 a e^4-x^2} \, dx,x,\frac {-b d^2-2 a e^2-\left (2 c d^2+b e^2\right ) x^2}{\sqrt {a+b x^2+c x^4}}\right )\\ &=\frac {e \tan ^{-1}\left (\frac {\sqrt {-c d^4-b d^2 e^2-a e^4} x}{d e \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {-c d^4-b d^2 e^2-a e^4}}-\frac {e \tanh ^{-1}\left (\frac {b d^2+2 a e^2+\left (2 c d^2+b e^2\right ) x^2}{2 \sqrt {c d^4+b d^2 e^2+a e^4} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c d^4+b d^2 e^2+a e^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+b x^2+c x^4}}-\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \sqrt {a+b x^2+c x^4}}\\ \end {align*}

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Mathematica [B]  time = 6.90, size = 1725, normalized size = 3.33 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(-(Sqrt[-(b/c) -
Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x)^2*Sqrt[(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(-(Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]
/c]/Sqrt[2]) + x))/((Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*
(-(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x))]*Sqrt[(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(Sqrt[-(b/c) + Sq
rt[b^2 - 4*a*c]/c]/Sqrt[2] + x))/((Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] - Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c
]/c]/Sqrt[2])*(-(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x))]*Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] -
Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*
a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]*((-d + (Sqrt[-
(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2])*EllipticF[ArcSin[Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b +
Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] +
Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]], (Sqrt[(-b - Sqrt[b^2 -
4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2/(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*
a*c])/c])^2] - Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*e*EllipticPi[((Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt
[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(d + (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2]))/((-(Sq
rt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(d - (Sqrt[-(b/c) - Sq
rt[b^2 - 4*a*c]/c]*e)/Sqrt[2])), ArcSin[Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c]
)/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt[(-b + Sqrt[b
^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]], (Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt
[(-b + Sqrt[b^2 - 4*a*c])/c])^2/(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2]))/(Sq
rt[(-b - Sqrt[b^2 - 4*a*c])/c]*(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] - Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c
]/Sqrt[2])*(-d - (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2])*(d - (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/S
qrt[2])*Sqrt[a + b*x^2 + c*x^4])

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 281, normalized size = 0.54 \[ \frac {\frac {\sqrt {2}\, \sqrt {-\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{2 a}+1}\, \sqrt {\frac {\left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{2 a}+1}\, e \EllipticPi \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {2 a \,e^{2}}{\left (-b +\sqrt {-4 a c +b^{2}}\right ) d^{2}}, \frac {\sqrt {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {2}}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}\right )}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, d}-\frac {\arctanh \left (\frac {b \,x^{2}+\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a +\frac {b \,d^{2}}{e^{2}}}{2 \sqrt {a +\frac {b \,d^{2}}{e^{2}}+\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\right )}{2 \sqrt {a +\frac {b \,d^{2}}{e^{2}}+\frac {c \,d^{4}}{e^{4}}}}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(-1/2/(c*d^4/e^4+b*d^2/e^2+a)^(1/2)*arctanh(1/2*(2*c*d^2/e^2*x^2+b*x^2+b*d^2/e^2+2*a)/(c*d^4/e^4+b*d^2/e^2
+a)^(1/2)/(c*x^4+b*x^2+a)^(1/2))+2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)/d*e*(1-1/2*(-b+(-4*a*c+b^2)^(1/2))/
a*x^2)^(1/2)*(1+1/2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticPi(1/2*2^(1/2)*((-b+(-4*
a*c+b^2)^(1/2))/a)^(1/2)*x,2/(-b+(-4*a*c+b^2)^(1/2))*a/d^2*e^2,(-1/2*(b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*2^(1/2)/(
(-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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