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

Optimal. Leaf size=405 \[ \frac{\sqrt [4]{c} d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )}+\frac{e \tan ^{-1}\left (\frac{x \sqrt{-a e^4-c d^4}}{d e \sqrt{a+c x^4}}\right )}{2 \sqrt{-a e^4-c d^4}}-\frac{e \tanh ^{-1}\left (\frac{a e^2+c d^2 x^2}{\sqrt{a+c x^4} \sqrt{a e^4+c d^4}}\right )}{2 \sqrt{a e^4+c d^4}}-\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{c} d^2-\sqrt{a} e^2\right ) \Pi \left (\frac{\left (\sqrt{c} d^2+\sqrt{a} e^2\right )^2}{4 \sqrt{a} \sqrt{c} d^2 e^2};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.389379, antiderivative size = 405, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {1725, 1217, 220, 1707, 1248, 725, 206} \[ \frac{e \tan ^{-1}\left (\frac{x \sqrt{-a e^4-c d^4}}{d e \sqrt{a+c x^4}}\right )}{2 \sqrt{-a e^4-c d^4}}-\frac{e \tanh ^{-1}\left (\frac{a e^2+c d^2 x^2}{\sqrt{a+c x^4} \sqrt{a e^4+c d^4}}\right )}{2 \sqrt{a e^4+c d^4}}+\frac{\sqrt [4]{c} d \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )}-\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{c} d^2-\sqrt{a} e^2\right ) \Pi \left (\frac{\left (\sqrt{c} d^2+\sqrt{a} e^2\right )^2}{4 \sqrt{a} \sqrt{c} d^2 e^2};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d \sqrt{a+c x^4} \left (\sqrt{a} e^2+\sqrt{c} d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*Sqrt[a + c*x^4]),x]

[Out]

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

Rule 1725

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

Rule 1217

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

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]

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])

Rubi steps

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

Mathematica [C]  time = 0.277587, size = 200, normalized size = 0.49 \[ \frac{\sqrt{\frac{c x^4}{a}+1} \left (\sqrt [4]{c} d \log \left (\frac{e^2 x^2-d^2}{a e^2 \left (\sqrt{\frac{c x^4}{a}+1} \sqrt{\frac{c d^4}{a e^4}+1}+1\right )+c d^2 x^2}\right )-2 \sqrt [4]{-1} \sqrt [4]{a} e \sqrt{\frac{c d^4}{a e^4}+1} \Pi \left (\frac{i \sqrt{a} e^2}{\sqrt{c} d^2};\left .\sin ^{-1}\left (\frac{(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )\right )}{2 \sqrt [4]{c} d e \sqrt{a+c x^4} \sqrt{\frac{c d^4}{a e^4}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + (c*x^4)/a]*(-2*(-1)^(1/4)*a^(1/4)*Sqrt[1 + (c*d^4)/(a*e^4)]*e*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*Log[(-d^2 + e^2*x^2)/(c*d^2*x^2 + a*e^2*(1 + Sqrt[
1 + (c*d^4)/(a*e^4)]*Sqrt[1 + (c*x^4)/a]))]))/(2*c^(1/4)*d*Sqrt[1 + (c*d^4)/(a*e^4)]*e*Sqrt[a + c*x^4])

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Maple [C]  time = 0.013, size = 169, normalized size = 0.4 \begin{align*}{\frac{1}{e} \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(1/(e*x+d)/(c*x^4+a)^(1/2),x)

[Out]

1/e*(-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)*Ellipti
cPi(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{1}{\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(1/(e*x+d)/(c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(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(1/(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{1}{\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(1/(e*x+d)/(c*x**4+a)**(1/2),x)

[Out]

Integral(1/(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{1}{\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(1/(e*x+d)/(c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

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