3.406 \(\int \frac{f+g x}{(d+e x) \sqrt{a+c x^4}} \, dx\)

Optimal. Leaf size=446 \[ \frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right ) \left (\sqrt{a} e g+\sqrt{c} d f\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )}+\frac{(e f-d g) \tan ^{-1}\left (\frac{x \sqrt{-a e^4-c d^4}}{d e \sqrt{a+c x^4}}\right )}{2 \sqrt{-a e^4-c d^4}}-\frac{(e f-d g) \tanh ^{-1}\left (\frac{a e^2+c d^2 x^2}{\sqrt{a+c x^4} \sqrt{a e^4+c d^4}}\right )}{2 \sqrt{a e^4+c d^4}}-\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{c} d^2-\sqrt{a} e^2\right ) (e f-d g) \Pi \left (\frac{\left (\sqrt{c} d^2+\sqrt{a} e^2\right )^2}{4 \sqrt{a} \sqrt{c} d^2 e^2};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )} \]

[Out]

((e*f - d*g)*ArcTan[(Sqrt[-(c*d^4) - a*e^4]*x)/(d*e*Sqrt[a + c*x^4])])/(2*Sqrt[-(c*d^4) - a*e^4]) - ((e*f - d*
g)*ArcTanh[(a*e^2 + c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])])/(2*Sqrt[c*d^4 + a*e^4]) + ((Sqrt[c]*d*f
+ Sqrt[a]*e*g)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4
)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + c*x^4]) - ((Sqrt[c]*d^2 - Sqrt[a]
*e^2)*(e*f - d*g)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + 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)], 1/2])/(4*a^(1/4)*c^(1/4)*d*e*(Sqr
t[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.497405, antiderivative size = 446, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1742, 12, 1248, 725, 206, 1709, 220, 1707} \[ \frac{(e f-d g) \tan ^{-1}\left (\frac{x \sqrt{-a e^4-c d^4}}{d e \sqrt{a+c x^4}}\right )}{2 \sqrt{-a e^4-c d^4}}-\frac{(e f-d g) \tanh ^{-1}\left (\frac{a e^2+c d^2 x^2}{\sqrt{a+c x^4} \sqrt{a e^4+c d^4}}\right )}{2 \sqrt{a e^4+c d^4}}+\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (\sqrt{a} e g+\sqrt{c} d f\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )}-\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{c} d^2-\sqrt{a} e^2\right ) (e f-d g) \Pi \left (\frac{\left (\sqrt{c} d^2+\sqrt{a} e^2\right )^2}{4 \sqrt{a} \sqrt{c} d^2 e^2};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)/((d + e*x)*Sqrt[a + c*x^4]),x]

[Out]

((e*f - d*g)*ArcTan[(Sqrt[-(c*d^4) - a*e^4]*x)/(d*e*Sqrt[a + c*x^4])])/(2*Sqrt[-(c*d^4) - a*e^4]) - ((e*f - d*
g)*ArcTanh[(a*e^2 + c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])])/(2*Sqrt[c*d^4 + a*e^4]) + ((Sqrt[c]*d*f
+ Sqrt[a]*e*g)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4
)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + c*x^4]) - ((Sqrt[c]*d^2 - Sqrt[a]
*e^2)*(e*f - d*g)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + 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)], 1/2])/(4*a^(1/4)*c^(1/4)*d*e*(Sqr
t[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + c*x^4])

Rule 1742

Int[(Px_)/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[P
x, x, 1], C = Coeff[Px, x, 2], D = Coeff[Px, x, 3]}, Int[(x*(B*d - A*e + (d*D - C*e)*x^2))/((d^2 - e^2*x^2)*Sq
rt[a + c*x^4]), x] + Int[(A*d + (C*d - B*e)*x^2 - D*e*x^4)/((d^2 - e^2*x^2)*Sqrt[a + c*x^4]), x]] /; FreeQ[{a,
c, d, e}, x] && PolyQ[Px, x] && LeQ[Expon[Px, x], 3] && NeQ[c*d^4 + a*e^4, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1248

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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 1709

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

Rule 220

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

Rule 1707

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

Rubi steps

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

Mathematica [C]  time = 0.694832, size = 258, normalized size = 0.58 \[ \frac{\frac{(d g-e f) \left (\sqrt [4]{c} d e \sqrt{a+c x^4} \tanh ^{-1}\left (\frac{a e^2+c d^2 x^2}{\sqrt{a+c x^4} \sqrt{a e^4+c d^4}}\right )+2 \sqrt [4]{-1} \sqrt [4]{a} \sqrt{\frac{c x^4}{a}+1} \sqrt{a e^4+c d^4} \Pi \left (\frac{i \sqrt{a} e^2}{\sqrt{c} d^2};\left .\sin ^{-1}\left (\frac{(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )\right )}{\sqrt [4]{c} d \sqrt{a e^4+c d^4}}-\frac{2 i g \sqrt{\frac{c x^4}{a}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{i \sqrt{c}}{\sqrt{a}}}\right ),-1\right )}{\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}}}}{2 e \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)/((d + e*x)*Sqrt[a + c*x^4]),x]

[Out]

(((-2*I)*g*Sqrt[1 + (c*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1])/Sqrt[(I*Sqrt[c])/Sqrt[a]
] + ((-(e*f) + d*g)*(c^(1/4)*d*e*Sqrt[a + c*x^4]*ArcTanh[(a*e^2 + c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x
^4])] + 2*(-1)^(1/4)*a^(1/4)*Sqrt[c*d^4 + a*e^4]*Sqrt[1 + (c*x^4)/a]*EllipticPi[(I*Sqrt[a]*e^2)/(Sqrt[c]*d^2),
ArcSin[((-1)^(3/4)*c^(1/4)*x)/a^(1/4)], -1]))/(c^(1/4)*d*Sqrt[c*d^4 + a*e^4]))/(2*e*Sqrt[a + c*x^4])

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Maple [C]  time = 0.052, size = 251, normalized size = 0.6 \begin{align*}{\frac{g}{e}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}}+{\frac{-dg+ef}{{e}^{2}} \left ( -{\frac{1}{2}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,{\frac{c{d}^{2}{x}^{2}}{{e}^{2}}}+2\,a \right ){\frac{1}{\sqrt{{\frac{c{d}^{4}}{{e}^{4}}}+a}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \right ){\frac{1}{\sqrt{{\frac{c{d}^{4}}{{e}^{4}}}+a}}}}+{\frac{e}{d}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticPi} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},{\frac{-i{e}^{2}}{{d}^{2}}\sqrt{a}{\frac{1}{\sqrt{c}}}},{\sqrt{{-i\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}} \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

g/e/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*
EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+(-d*g+e*f)/e^2*(-1/2/(c*d^4/e^4+a)^(1/2)*arctanh(1/2*(2*c*x^2*d^2/e^2
+2*a)/(c*d^4/e^4+a)^(1/2)/(c*x^4+a)^(1/2))+1/(I/a^(1/2)*c^(1/2))^(1/2)/d*e*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+
I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticPi(x*(I/a^(1/2)*c^(1/2))^(1/2),-I*a^(1/2)/c^(1/2)/d^2*e^2
,(-I/a^(1/2)*c^(1/2))^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{g x + f}{\sqrt{c x^{4} + a}{\left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*x + f)/(sqrt(c*x^4 + a)*(e*x + d)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f + g x}{\sqrt{a + c x^{4}} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((f + g*x)/(sqrt(a + c*x**4)*(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{g x + f}{\sqrt{c x^{4} + a}{\left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*x + f)/(sqrt(c*x^4 + a)*(e*x + d)), x)