3.427 \(\int \frac{1}{\sqrt{-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx\)

Optimal. Leaf size=49 \[ -\frac{\sqrt{\frac{2}{5}} \tan ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )-1}}\right )}{e} \]

[Out]

-((Sqrt[2/5]*ArcTan[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[-1 + Cos[d + e*x - ArcTan[3/4]]])])/e)

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Rubi [A]  time = 0.0608601, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3115, 2649, 204} \[ -\frac{\sqrt{\frac{2}{5}} \tan ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )-1}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]],x]

[Out]

-((Sqrt[2/5]*ArcTan[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[-1 + Cos[d + e*x - ArcTan[3/4]]])])/e)

Rule 3115

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Int[1/Sqrt[a +
Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx &=\int \frac{1}{\sqrt{-5+5 \cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}} \, dx\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-10-x^2} \, dx,x,-\frac{5 \sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{-5+5 \cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}}\right )}{e}\\ &=-\frac{\sqrt{\frac{2}{5}} \tan ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{-1+\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}}\right )}{e}\\ \end{align*}

Mathematica [C]  time = 0.0891382, size = 99, normalized size = 2.02 \[ \frac{\left (\frac{2}{5}+\frac{6 i}{5}\right ) \sqrt{-\frac{4}{5}-\frac{3 i}{5}} \left (\cos \left (\frac{1}{2} (d+e x)\right )-3 \sin \left (\frac{1}{2} (d+e x)\right )\right ) \tanh ^{-1}\left (\left (\frac{1}{10}+\frac{3 i}{10}\right ) \sqrt{-\frac{4}{5}-\frac{3 i}{5}} \left (\tan \left (\frac{1}{4} (d+e x)\right )+3\right )\right )}{e \sqrt{3 \sin (d+e x)+4 \cos (d+e x)-5}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]],x]

[Out]

((2/5 + (6*I)/5)*Sqrt[-4/5 - (3*I)/5]*ArcTanh[(1/10 + (3*I)/10)*Sqrt[-4/5 - (3*I)/5]*(3 + Tan[(d + e*x)/4])]*(
Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2]))/(e*Sqrt[-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x]])

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Maple [A]  time = 1.171, size = 77, normalized size = 1.6 \begin{align*}{\frac{ \left ( \sin \left ( ex+d+\arctan \left ({\frac{4}{3}} \right ) \right ) -1 \right ) \sqrt{10}}{5\,\cos \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) e}\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) -5}\arctan \left ({\frac{\sqrt{10}}{10}\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) -5}} \right ){\frac{1}{\sqrt{-5+5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2),x)

[Out]

1/5*(sin(e*x+d+arctan(4/3))-1)*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2)*arctan(1/10*(-5*sin(e*x+d+arctan(4
/3))-5)^(1/2)*10^(1/2))/cos(e*x+d+arctan(4/3))/(-5+5*sin(e*x+d+arctan(4/3)))^(1/2)/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5), x)

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Fricas [B]  time = 1.85431, size = 269, normalized size = 5.49 \begin{align*} \frac{\sqrt{5} \sqrt{2} \arctan \left (-\frac{{\left (3 \, \sqrt{5} \sqrt{2} \cos \left (e x + d\right ) + \sqrt{5} \sqrt{2} \sin \left (e x + d\right ) + 3 \, \sqrt{5} \sqrt{2}\right )} \sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{10 \,{\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}\right )}{5 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2),x, algorithm="fricas")

[Out]

1/5*sqrt(5)*sqrt(2)*arctan(-1/10*(3*sqrt(5)*sqrt(2)*cos(e*x + d) + sqrt(5)*sqrt(2)*sin(e*x + d) + 3*sqrt(5)*sq
rt(2))*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5)/(cos(e*x + d) - 3*sin(e*x + d) + 1))/e

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{3 \sin{\left (d + e x \right )} + 4 \cos{\left (d + e x \right )} - 5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+4*cos(e*x+d)+3*sin(e*x+d))**(1/2),x)

[Out]

Integral(1/sqrt(3*sin(d + e*x) + 4*cos(d + e*x) - 5), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5), x)