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

Optimal. Leaf size=218 \[ \frac{\sqrt [4]{a} g \sqrt{1-\frac{c x^4}{a}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )}{\sqrt [4]{c} e \sqrt{c x^4-a}}+\frac{(e f-d g) \tanh ^{-1}\left (\frac{a e^2-c d^2 x^2}{\sqrt{c x^4-a} \sqrt{c d^4-a e^4}}\right )}{2 \sqrt{c d^4-a e^4}}+\frac{\sqrt [4]{a} \sqrt{1-\frac{c x^4}{a}} (e f-d g) \Pi \left (\frac{\sqrt{a} e^2}{\sqrt{c} d^2};\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} d e \sqrt{c x^4-a}} \]

[Out]

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

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Rubi [A]  time = 0.294798, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {1742, 12, 1248, 725, 206, 1711, 224, 221, 1219, 1218} \[ \frac{(e f-d g) \tanh ^{-1}\left (\frac{a e^2-c d^2 x^2}{\sqrt{c x^4-a} \sqrt{c d^4-a e^4}}\right )}{2 \sqrt{c d^4-a e^4}}+\frac{\sqrt [4]{a} \sqrt{1-\frac{c x^4}{a}} (e f-d g) \Pi \left (\frac{\sqrt{a} e^2}{\sqrt{c} d^2};\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} d e \sqrt{c x^4-a}}+\frac{\sqrt [4]{a} g \sqrt{1-\frac{c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} e \sqrt{c x^4-a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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]) + (a
^(1/4)*g*Sqrt[1 - (c*x^4)/a]*EllipticF[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(c^(1/4)*e*Sqrt[-a + c*x^4]) + (a^(1/
4)*(e*f - d*g)*Sqrt[1 - (c*x^4)/a]*EllipticPi[(Sqrt[a]*e^2)/(Sqrt[c]*d^2), ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(
c^(1/4)*d*e*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 1711

Int[((A_.) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[B/e, Int[1/Sqr
t[a + c*x^4], x], x] + Dist[(e*A - d*B)/e, Int[1/((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] && NegQ[c/a]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1219

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4]
, Int[1/((d + e*x^2)*Sqrt[1 + (c*x^4)/a]), x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] &&  !GtQ[a, 0]

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 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{g \int \frac{1}{\sqrt{-a+c x^4}} \, dx}{e}+\frac{(d (e f-d g)) \int \frac{1}{\left (d^2-e^2 x^2\right ) \sqrt{-a+c x^4}} \, dx}{e}+(-e f+d g) \int \frac{x}{\left (d^2-e^2 x^2\right ) \sqrt{-a+c x^4}} \, dx\\ &=\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{\left (g \sqrt{1-\frac{c x^4}{a}}\right ) \int \frac{1}{\sqrt{1-\frac{c x^4}{a}}} \, dx}{e \sqrt{-a+c x^4}}+\frac{\left (d (e f-d g) \sqrt{1-\frac{c x^4}{a}}\right ) \int \frac{1}{\left (d^2-e^2 x^2\right ) \sqrt{1-\frac{c x^4}{a}}} \, dx}{e \sqrt{-a+c x^4}}\\ &=\frac{\sqrt [4]{a} g \sqrt{1-\frac{c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} e \sqrt{-a+c x^4}}+\frac{\sqrt [4]{a} (e f-d g) \sqrt{1-\frac{c x^4}{a}} \Pi \left (\frac{\sqrt{a} e^2}{\sqrt{c} d^2};\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} d e \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) \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{\sqrt [4]{a} g \sqrt{1-\frac{c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} e \sqrt{-a+c x^4}}+\frac{\sqrt [4]{a} (e f-d g) \sqrt{1-\frac{c x^4}{a}} \Pi \left (\frac{\sqrt{a} e^2}{\sqrt{c} d^2};\left .\sin ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} d e \sqrt{-a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 1.22663, size = 719, normalized size = 3.3 \[ \frac{\frac{i f \sqrt{-\frac{(1-i) \left (\sqrt [4]{a}-\sqrt [4]{c} x\right )}{\sqrt [4]{c} x+i \sqrt [4]{a}}} \sqrt{\frac{(1+i) \left (\sqrt [4]{a}+i \sqrt [4]{c} x\right ) \left (\sqrt [4]{a}+\sqrt [4]{c} x\right )}{\left (\sqrt [4]{a}-i \sqrt [4]{c} x\right )^2}} \left (\sqrt [4]{a}-i \sqrt [4]{c} x\right )^2 \left (\left (\sqrt [4]{a} e-\sqrt [4]{c} d\right ) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\frac{(1+i) \left (\sqrt [4]{a}+\sqrt [4]{c} x\right )}{2 \sqrt [4]{c} x+2 i \sqrt [4]{a}}}\right ),2\right )-(1-i) \sqrt [4]{a} e \Pi \left (\frac{(1-i) \left (\sqrt [4]{c} d-i \sqrt [4]{a} e\right )}{\sqrt [4]{c} d-\sqrt [4]{a} e};\left .\sin ^{-1}\left (\sqrt{\frac{(1+i) \left (\sqrt [4]{c} x+\sqrt [4]{a}\right )}{2 \sqrt [4]{c} x+2 i \sqrt [4]{a}}}\right )\right |2\right )\right )}{\sqrt [4]{a} \left (\sqrt [4]{a} e-\sqrt [4]{c} d\right ) \left (\sqrt [4]{a} e+i \sqrt [4]{c} d\right )}+\frac{d g \sqrt{-\frac{(1-i) \left (\sqrt [4]{a}-\sqrt [4]{c} x\right )}{\sqrt [4]{c} x+i \sqrt [4]{a}}} \sqrt{\frac{(1+i) \left (\sqrt [4]{a}+i \sqrt [4]{c} x\right ) \left (\sqrt [4]{a}+\sqrt [4]{c} x\right )}{\left (\sqrt [4]{a}-i \sqrt [4]{c} x\right )^2}} \left (\sqrt [4]{a}-i \sqrt [4]{c} x\right )^2 \left (i \left (\sqrt [4]{c} d-\sqrt [4]{a} e\right ) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\frac{(1+i) \left (\sqrt [4]{a}+\sqrt [4]{c} x\right )}{2 \sqrt [4]{c} x+2 i \sqrt [4]{a}}}\right ),2\right )+(1+i) \sqrt [4]{a} e \Pi \left (\frac{(1-i) \left (\sqrt [4]{c} d-i \sqrt [4]{a} e\right )}{\sqrt [4]{c} d-\sqrt [4]{a} e};\left .\sin ^{-1}\left (\sqrt{\frac{(1+i) \left (\sqrt [4]{c} x+\sqrt [4]{a}\right )}{2 \sqrt [4]{c} x+2 i \sqrt [4]{a}}}\right )\right |2\right )\right )}{\sqrt [4]{a} e \left (\sqrt [4]{a} e-\sqrt [4]{c} d\right ) \left (\sqrt [4]{a} e+i \sqrt [4]{c} d\right )}-\frac{i g \sqrt{1-\frac{c x^4}{a}} \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{-\frac{\sqrt{c}}{\sqrt{a}}}\right ),-1\right )}{e \sqrt{-\frac{\sqrt{c}}{\sqrt{a}}}}}{\sqrt{c x^4-a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*g*Sqrt[1 - (c*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x], -1])/(Sqrt[-(Sqrt[c]/Sqrt[a])]*e
) + (I*f*(a^(1/4) - I*c^(1/4)*x)^2*Sqrt[((-1 + I)*(a^(1/4) - c^(1/4)*x))/(I*a^(1/4) + c^(1/4)*x)]*Sqrt[((1 + I
)*(a^(1/4) + I*c^(1/4)*x)*(a^(1/4) + c^(1/4)*x))/(a^(1/4) - I*c^(1/4)*x)^2]*((-(c^(1/4)*d) + a^(1/4)*e)*Ellipt
icF[ArcSin[Sqrt[((1 + I)*(a^(1/4) + c^(1/4)*x))/((2*I)*a^(1/4) + 2*c^(1/4)*x)]], 2] - (1 - I)*a^(1/4)*e*Ellipt
icPi[((1 - I)*(c^(1/4)*d - I*a^(1/4)*e))/(c^(1/4)*d - a^(1/4)*e), ArcSin[Sqrt[((1 + I)*(a^(1/4) + c^(1/4)*x))/
((2*I)*a^(1/4) + 2*c^(1/4)*x)]], 2]))/(a^(1/4)*(-(c^(1/4)*d) + a^(1/4)*e)*(I*c^(1/4)*d + a^(1/4)*e)) + (d*g*(a
^(1/4) - I*c^(1/4)*x)^2*Sqrt[((-1 + I)*(a^(1/4) - c^(1/4)*x))/(I*a^(1/4) + c^(1/4)*x)]*Sqrt[((1 + I)*(a^(1/4)
+ I*c^(1/4)*x)*(a^(1/4) + c^(1/4)*x))/(a^(1/4) - I*c^(1/4)*x)^2]*(I*(c^(1/4)*d - a^(1/4)*e)*EllipticF[ArcSin[S
qrt[((1 + I)*(a^(1/4) + c^(1/4)*x))/((2*I)*a^(1/4) + 2*c^(1/4)*x)]], 2] + (1 + I)*a^(1/4)*e*EllipticPi[((1 - I
)*(c^(1/4)*d - I*a^(1/4)*e))/(c^(1/4)*d - a^(1/4)*e), ArcSin[Sqrt[((1 + I)*(a^(1/4) + c^(1/4)*x))/((2*I)*a^(1/
4) + 2*c^(1/4)*x)]], 2]))/(a^(1/4)*e*(-(c^(1/4)*d) + a^(1/4)*e)*(I*c^(1/4)*d + a^(1/4)*e)))/Sqrt[-a + c*x^4]

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Maple [A]  time = 0.029, size = 247, normalized size = 1.1 \begin{align*}{\frac{g}{e}\sqrt{1+{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1-{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{-{\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{-{\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+{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1-{{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticPi} \left ( x\sqrt{-{\sqrt{c}{\frac{1}{\sqrt{a}}}}},-{\frac{{e}^{2}}{{d}^{2}}\sqrt{a}{\frac{1}{\sqrt{c}}}},{\sqrt{{\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\frac{1}{\sqrt{-{\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}} \right ){\frac{1}{\sqrt{-{\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/(-1/a^(1/2)*c^(1/2))^(1/2)*(1+1/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1-1/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4-a)^(1/2)
*EllipticF(x*(-1/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/(-1/a^(1/2)*c^(1/2))^(1/2)/d*e*(1+1/a^(1/2)*c^(1/2)*x^2)^(1/2)*
(1-1/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4-a)^(1/2)*EllipticPi(x*(-1/a^(1/2)*c^(1/2))^(1/2),-e^2*a^(1/2)/d^2/c^(1/
2),(1/a^(1/2)*c^(1/2))^(1/2)/(-1/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)