3.913 \(\int \frac{\sqrt{2 x^2+\sqrt{3+4 x^4}}}{(c+d x) \sqrt{3+4 x^4}} \, dx\)

Optimal. Leaf size=169 \[ \frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tan ^{-1}\left (\frac{\sqrt{3} d+2 i c x}{\sqrt{\sqrt{3}-2 i x^2} \sqrt{-\sqrt{3} d^2+2 i c^2}}\right )}{\sqrt{-\sqrt{3} d^2+2 i c^2}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{\sqrt{3} d-2 i c x}{\sqrt{\sqrt{3}+2 i x^2} \sqrt{\sqrt{3} d^2+2 i c^2}}\right )}{\sqrt{\sqrt{3} d^2+2 i c^2}} \]

[Out]

((1/2 - I/2)*ArcTan[(Sqrt[3]*d + (2*I)*c*x)/(Sqrt[(2*I)*c^2 - Sqrt[3]*d^2]*Sqrt[Sqrt[3] - (2*I)*x^2])])/Sqrt[(
2*I)*c^2 - Sqrt[3]*d^2] - ((1/2 + I/2)*ArcTanh[(Sqrt[3]*d - (2*I)*c*x)/(Sqrt[(2*I)*c^2 + Sqrt[3]*d^2]*Sqrt[Sqr
t[3] + (2*I)*x^2])])/Sqrt[(2*I)*c^2 + Sqrt[3]*d^2]

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Rubi [A]  time = 0.265768, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2133, 725, 204, 206} \[ \frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tan ^{-1}\left (\frac{\sqrt{3} d+2 i c x}{\sqrt{\sqrt{3}-2 i x^2} \sqrt{-\sqrt{3} d^2+2 i c^2}}\right )}{\sqrt{-\sqrt{3} d^2+2 i c^2}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{\sqrt{3} d-2 i c x}{\sqrt{\sqrt{3}+2 i x^2} \sqrt{\sqrt{3} d^2+2 i c^2}}\right )}{\sqrt{\sqrt{3} d^2+2 i c^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)*Sqrt[3 + 4*x^4]),x]

[Out]

((1/2 - I/2)*ArcTan[(Sqrt[3]*d + (2*I)*c*x)/(Sqrt[(2*I)*c^2 - Sqrt[3]*d^2]*Sqrt[Sqrt[3] - (2*I)*x^2])])/Sqrt[(
2*I)*c^2 - Sqrt[3]*d^2] - ((1/2 + I/2)*ArcTanh[(Sqrt[3]*d - (2*I)*c*x)/(Sqrt[(2*I)*c^2 + Sqrt[3]*d^2]*Sqrt[Sqr
t[3] + (2*I)*x^2])])/Sqrt[(2*I)*c^2 + Sqrt[3]*d^2]

Rule 2133

Int[(((c_.) + (d_.)*(x_))^(m_.)*Sqrt[(b_.)*(x_)^2 + Sqrt[(a_) + (e_.)*(x_)^4]])/Sqrt[(a_) + (e_.)*(x_)^4], x_S
ymbol] :> Dist[(1 - I)/2, Int[(c + d*x)^m/Sqrt[Sqrt[a] - I*b*x^2], x], x] + Dist[(1 + I)/2, Int[(c + d*x)^m/Sq
rt[Sqrt[a] + I*b*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[e, b^2] && GtQ[a, 0]

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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{\sqrt{2 x^2+\sqrt{3+4 x^4}}}{(c+d x) \sqrt{3+4 x^4}} \, dx &=\left (\frac{1}{2}-\frac{i}{2}\right ) \int \frac{1}{(c+d x) \sqrt{\sqrt{3}-2 i x^2}} \, dx+\left (\frac{1}{2}+\frac{i}{2}\right ) \int \frac{1}{(c+d x) \sqrt{\sqrt{3}+2 i x^2}} \, dx\\ &=\left (-\frac{1}{2}-\frac{i}{2}\right ) \operatorname{Subst}\left (\int \frac{1}{2 i c^2+\sqrt{3} d^2-x^2} \, dx,x,\frac{\sqrt{3} d-2 i c x}{\sqrt{\sqrt{3}+2 i x^2}}\right )+\left (-\frac{1}{2}+\frac{i}{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-2 i c^2+\sqrt{3} d^2-x^2} \, dx,x,\frac{\sqrt{3} d+2 i c x}{\sqrt{\sqrt{3}-2 i x^2}}\right )\\ &=\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tan ^{-1}\left (\frac{\sqrt{3} d+2 i c x}{\sqrt{2 i c^2-\sqrt{3} d^2} \sqrt{\sqrt{3}-2 i x^2}}\right )}{\sqrt{2 i c^2-\sqrt{3} d^2}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{\sqrt{3} d-2 i c x}{\sqrt{2 i c^2+\sqrt{3} d^2} \sqrt{\sqrt{3}+2 i x^2}}\right )}{\sqrt{2 i c^2+\sqrt{3} d^2}}\\ \end{align*}

Mathematica [F]  time = 0.109194, size = 0, normalized size = 0. \[ \int \frac{\sqrt{2 x^2+\sqrt{3+4 x^4}}}{(c+d x) \sqrt{3+4 x^4}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)*Sqrt[3 + 4*x^4]),x]

[Out]

Integrate[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)*Sqrt[3 + 4*x^4]), x]

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{dx+c}\sqrt{2\,{x}^{2}+\sqrt{4\,{x}^{4}+3}}{\frac{1}{\sqrt{4\,{x}^{4}+3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x)

[Out]

int((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x^{2} + \sqrt{4 \, x^{4} + 3}}}{\sqrt{4 \, x^{4} + 3}{\left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(2*x^2 + sqrt(4*x^4 + 3))/(sqrt(4*x^4 + 3)*(d*x + c)), 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((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(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{\sqrt{2 x^{2} + \sqrt{4 x^{4} + 3}}}{\left (c + d x\right ) \sqrt{4 x^{4} + 3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+(4*x**4+3)**(1/2))**(1/2)/(d*x+c)/(4*x**4+3)**(1/2),x)

[Out]

Integral(sqrt(2*x**2 + sqrt(4*x**4 + 3))/((c + d*x)*sqrt(4*x**4 + 3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x^{2} + \sqrt{4 \, x^{4} + 3}}}{\sqrt{4 \, x^{4} + 3}{\left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)/(4*x^4+3)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(2*x^2 + sqrt(4*x^4 + 3))/(sqrt(4*x^4 + 3)*(d*x + c)), x)