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

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

[Out]

1/2*(-d*g+e*f)*arctan(x*(-a*e^4-c*d^4)^(1/2)/d/e/(c*x^4+a)^(1/2))/(-a*e^4-c*d^4)^(1/2)-1/2*(-d*g+e*f)*arctanh(
(c*d^2*x^2+a*e^2)/(a*e^4+c*d^4)^(1/2)/(c*x^4+a)^(1/2))/(a*e^4+c*d^4)^(1/2)-1/4*(-d*g+e*f)*(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^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((
c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(1/4)/d/e/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)+1/2*(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^(1/2))*(e*g*a^(1/2)+d*f*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^
(1/4)/c^(1/4)/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.50, antiderivative size = 446, normalized size of antiderivative = 1.00, 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 12

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

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 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]

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*}

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Mathematica [C]  time = 0.68, size = 258, normalized size = 0.58 \[ \frac {\frac {(d g-e f) \left (\sqrt [4]{c} d e \sqrt {a+c x^4} \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )+2 \sqrt [4]{-1} \sqrt [4]{a} \sqrt {\frac {c x^4}{a}+1} \sqrt {a e^4+c d^4} \Pi \left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2};\left .\sin ^{-1}\left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )\right )}{\sqrt [4]{c} d \sqrt {a e^4+c d^4}}-\frac {2 i g \sqrt {\frac {c x^4}{a}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}}{2 e \sqrt {a+c x^4}} \]

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

[Out]

Timed out

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

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)

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maple [C]  time = 0.02, size = 251, normalized size = 0.56 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, g \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, e}+\frac {\left (-d g +e f \right ) \left (\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, e \EllipticPi \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, d}-\frac {\arctanh \left (\frac {\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}\right )}{e^{2}} \]

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)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)
*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,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*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(
I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticPi((I/a^(1/2)*c^(1/2))^(1/2)*x,-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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {g x + f}{\sqrt {c x^{4} + a} {\left (e x + d\right )}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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