3.428 \(\int \frac{1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )-1}}\right )}{10 \sqrt{10} e} \]

[Out]

ArcTan[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[-1 + Cos[d + e*x - ArcTan[3/4]]])]/(10*Sqrt[10]*e) + (3*Cos[d
+ e*x] - 4*Sin[d + e*x])/(10*e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

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Rubi [A]  time = 0.0523508, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3116, 3115, 2649, 204} \[ \frac{3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )-1}}\right )}{10 \sqrt{10} e} \]

Antiderivative was successfully verified.

[In]

Int[(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(-3/2),x]

[Out]

ArcTan[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[-1 + Cos[d + e*x - ArcTan[3/4]]])]/(10*Sqrt[10]*e) + (3*Cos[d
+ e*x] - 4*Sin[d + e*x])/(10*e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

Rule 3116

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

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}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}} \, dx &=\frac{3 \cos (d+e x)-4 \sin (d+e x)}{10 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}-\frac{1}{20} \int \frac{1}{\sqrt{-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx\\ &=\frac{3 \cos (d+e x)-4 \sin (d+e x)}{10 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}-\frac{1}{20} \int \frac{1}{\sqrt{-5+5 \cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}} \, dx\\ &=\frac{3 \cos (d+e x)-4 \sin (d+e x)}{10 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}+\frac{\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 )}{10 e}\\ &=\frac{\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 )}{10 \sqrt{10} e}+\frac{3 \cos (d+e x)-4 \sin (d+e x)}{10 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.29477, size = 152, normalized size = 1.58 \[ \frac{\left (\frac{1}{250}-\frac{i}{125}\right ) \left (\cos \left (\frac{1}{2} (d+e x)\right )-3 \sin \left (\frac{1}{2} (d+e x)\right )\right ) \left ((5+10 i) \left (\sin \left (\frac{1}{2} (d+e x)\right )+3 \cos \left (\frac{1}{2} (d+e x)\right )\right )-(1-i) \sqrt{-20-15 i} \left (\cos \left (\frac{1}{2} (d+e x)\right )-3 \sin \left (\frac{1}{2} (d+e x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{10}+\frac{3 i}{10}\right ) \sqrt{-\frac{4}{5}-\frac{3 i}{5}} \left (\tan \left (\frac{1}{4} (d+e x)\right )+3\right )\right )\right )}{e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(-3/2),x]

[Out]

((1/250 - I/125)*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])*((-1 + I)*Sqrt[-20 - 15*I]*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])^2 + (5 + 10*I)*(3*Cos[(d
+ e*x)/2] + Sin[(d + e*x)/2])))/(e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

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Maple [A]  time = 1.116, size = 118, normalized size = 1.2 \begin{align*}{\frac{1}{100\,\cos \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) e} \left ( -\sqrt{10}\arctan \left ({\frac{\sqrt{10}}{10}\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) -5}} \right ) \sin \left ( ex+d+\arctan \left ({\frac{4}{3}} \right ) \right ) +\sqrt{10}\arctan \left ({\frac{\sqrt{10}}{10}\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) -5}} \right ) +2\,\sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) -5} \right ) \sqrt{-5\,\sin \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) -5}{\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))^(3/2),x)

[Out]

1/100*(-10^(1/2)*arctan(1/10*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2))*sin(e*x+d+arctan(4/3))+10^(1/2)*arc
tan(1/10*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2))+2*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2))*(-5*sin(e*x+d+ar
ctan(4/3))-5)^(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}{{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.78628, size = 614, normalized size = 6.4 \begin{align*} -\frac{{\left (13 \, \sqrt{10} \cos \left (e x + d\right )^{2} - 9 \,{\left (\sqrt{10} \cos \left (e x + d\right ) - 2 \, \sqrt{10}\right )} \sin \left (e x + d\right ) - \sqrt{10} \cos \left (e x + d\right ) - 14 \, \sqrt{10}\right )} \arctan \left (-\frac{{\left (3 \, \sqrt{10} \cos \left (e x + d\right ) + \sqrt{10} \sin \left (e x + d\right ) + 3 \, \sqrt{10}\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 ) + 10 \, \sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}{\left (3 \, \cos \left (e x + d\right ) + \sin \left (e x + d\right ) + 3\right )}}{100 \,{\left (13 \, e \cos \left (e x + d\right )^{2} - e \cos \left (e x + d\right ) - 9 \,{\left (e \cos \left (e x + d\right ) - 2 \, e\right )} \sin \left (e x + d\right ) - 14 \, e\right )}} \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))^(3/2),x, algorithm="fricas")

[Out]

-1/100*((13*sqrt(10)*cos(e*x + d)^2 - 9*(sqrt(10)*cos(e*x + d) - 2*sqrt(10))*sin(e*x + d) - sqrt(10)*cos(e*x +
d) - 14*sqrt(10))*arctan(-1/10*(3*sqrt(10)*cos(e*x + d) + sqrt(10)*sin(e*x + d) + 3*sqrt(10))*sqrt(4*cos(e*x
+ d) + 3*sin(e*x + d) - 5)/(cos(e*x + d) - 3*sin(e*x + d) + 1)) + 10*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) - 5)
*(3*cos(e*x + d) + sin(e*x + d) + 3))/(13*e*cos(e*x + d)^2 - e*cos(e*x + d) - 9*(e*cos(e*x + d) - 2*e)*sin(e*x
+ d) - 14*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (3 \sin{\left (d + e x \right )} + 4 \cos{\left (d + e x \right )} - 5\right )^{\frac{3}{2}}}\, 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))**(3/2),x)

[Out]

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

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Giac [C]  time = 1.52085, size = 336, normalized size = 3.5 \begin{align*} -\frac{1}{450} \,{\left (\frac{9 \, \sqrt{10} \arctan \left (\frac{1}{10} \, \sqrt{10}{\left (-3 i \, \sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} + 3 i \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - i\right )}\right )}{\mathrm{sgn}\left (-3 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 1\right )} + \frac{10 \,{\left (33 i \,{\left (\sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} - \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )\right )}^{3} - 7 i \,{\left (\sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} - \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )\right )}^{2} + 21 i \, \sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} - 21 i \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 9 i\right )}}{{\left (-3 i \,{\left (\sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} - \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )\right )}^{2} - 2 i \, \sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} + 2 i \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 3 i\right )}^{2} \mathrm{sgn}\left (-3 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 1\right )}\right )} e^{\left (-1\right )} \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))^(3/2),x, algorithm="giac")

[Out]

-1/450*(9*sqrt(10)*arctan(1/10*sqrt(10)*(-3*I*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) + 3*I*tan(1/2*x*e + 1/2*d) - I)
)/sgn(-3*tan(1/2*x*e + 1/2*d) + 1) + 10*(33*I*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^3 - 7*
I*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^2 + 21*I*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - 21*I*t
an(1/2*x*e + 1/2*d) + 9*I)/((-3*I*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^2 - 2*I*sqrt(tan(1
/2*x*e + 1/2*d)^2 + 1) + 2*I*tan(1/2*x*e + 1/2*d) + 3*I)^2*sgn(-3*tan(1/2*x*e + 1/2*d) + 1)))*e^(-1)