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

Optimal. Leaf size=142 \[ -\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3 \tan ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )-1}}\right )}{400 \sqrt {10} e} \]

[Out]

1/20*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-5+4*cos(e*x+d)+3*sin(e*x+d))^(5/2)-3/400*(3*cos(e*x+d)-4*sin(e*x+d))/e/(-
5+4*cos(e*x+d)+3*sin(e*x+d))^(3/2)-3/4000*arctan(1/2*sin(d+e*x-arctan(3/4))*2^(1/2)/(-1+cos(d+e*x-arctan(3/4))
)^(1/2))*10^(1/2)/e

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Rubi [A]  time = 0.08, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3116, 3115, 2649, 204} \[ -\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{3/2}}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}}-\frac {3 \tan ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )-1}}\right )}{400 \sqrt {10} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTan[Sin[d + e*x - ArcTan[3/4]]/(Sqrt[2]*Sqrt[-1 + Cos[d + e*x - ArcTan[3/4]]])])/(400*Sqrt[10]*e) + (3*
Cos[d + e*x] - 4*Sin[d + e*x])/(20*e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(5/2)) - (3*(3*Cos[d + e*x] - 4*Si
n[d + e*x]))/(400*e*(-5 + 4*Cos[d + e*x] + 3*Sin[d + e*x])^(3/2))

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{(-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}} \, dx &=\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3}{40} \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)}{20 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}+\frac {3}{800} \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)}{20 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}+\frac {3}{800} \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)}{20 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}-\frac {3 \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 )}{400 e}\\ &=-\frac {3 \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 )}{400 \sqrt {10} e}+\frac {3 \cos (d+e x)-4 \sin (d+e x)}{20 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{5/2}}-\frac {3 (3 \cos (d+e x)-4 \sin (d+e x))}{400 e (-5+4 \cos (d+e x)+3 \sin (d+e x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.38, size = 178, normalized size = 1.25 \[ \frac {\left (\frac {1}{10000}+\frac {i}{20000}\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right ) \left ((10-5 i) \left (55 \sin \left (\frac {1}{2} (d+e x)\right )-39 \sin \left (\frac {3}{2} (d+e x)\right )+165 \cos \left (\frac {1}{2} (d+e x)\right )-27 \cos \left (\frac {3}{2} (d+e x)\right )\right )+(6+6 i) \sqrt {-20-15 i} \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \left (\tan \left (\frac {1}{4} (d+e x)\right )+3\right )\right )\right )}{e (3 \sin (d+e x)+4 \cos (d+e x)-5)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/10000 + I/20000)*(Cos[(d + e*x)/2] - 3*Sin[(d + e*x)/2])*((6 + 6*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])^4 + (10 - 5*I)*(165*
Cos[(d + e*x)/2] - 27*Cos[(3*(d + e*x))/2] + 55*Sin[(d + e*x)/2] - 39*Sin[(3*(d + e*x))/2])))/(e*(-5 + 4*Cos[d
+ e*x] + 3*Sin[d + e*x])^(5/2))

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fricas [B]  time = 0.89, size = 280, normalized size = 1.97 \[ \frac {3 \, {\left (79 \, \sqrt {10} \cos \left (e x + d\right )^{3} - 123 \, \sqrt {10} \cos \left (e x + d\right )^{2} + 3 \, {\left (\sqrt {10} \cos \left (e x + d\right )^{2} + 38 \, \sqrt {10} \cos \left (e x + d\right ) - 44 \, \sqrt {10}\right )} \sin \left (e x + d\right ) - 78 \, \sqrt {10} \cos \left (e x + d\right ) + 124 \, \sqrt {10}\right )} \arctan \left (-\frac {{\left (3 \, \sqrt {10} \cos \left (e x + d\right ) + \sqrt {10} \sin \left (e x + d\right ) + 3 \, \sqrt {10}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{10 \, {\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}\right ) + 10 \, {\left (27 \, \cos \left (e x + d\right )^{2} + {\left (39 \, \cos \left (e x + d\right ) - 8\right )} \sin \left (e x + d\right ) - 69 \, \cos \left (e x + d\right ) - 96\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{4000 \, {\left (79 \, e \cos \left (e x + d\right )^{3} - 123 \, e \cos \left (e x + d\right )^{2} - 78 \, e \cos \left (e x + d\right ) + 3 \, {\left (e \cos \left (e x + d\right )^{2} + 38 \, e \cos \left (e x + d\right ) - 44 \, e\right )} \sin \left (e x + d\right ) + 124 \, e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4000*(3*(79*sqrt(10)*cos(e*x + d)^3 - 123*sqrt(10)*cos(e*x + d)^2 + 3*(sqrt(10)*cos(e*x + d)^2 + 38*sqrt(10)
*cos(e*x + d) - 44*sqrt(10))*sin(e*x + d) - 78*sqrt(10)*cos(e*x + d) + 124*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*(27*cos(e*x + d)^2 + (39*cos(e*x + d) - 8)*sin(e*x + d) - 69*cos(e*x + d) - 96)*sqrt(
4*cos(e*x + d) + 3*sin(e*x + d) - 5))/(79*e*cos(e*x + d)^3 - 123*e*cos(e*x + d)^2 - 78*e*cos(e*x + d) + 3*(e*c
os(e*x + d)^2 + 38*e*cos(e*x + d) - 44*e)*sin(e*x + d) + 124*e)

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giac [C]  time = 0.80, size = 381, normalized size = 2.68 \[ -\frac {1}{162000} \, {\left (\frac {243 \, \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 (15039 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 )}^{7} + 6291 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 )}^{6} - 579 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 )}^{5} + 1645 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 )}^{4} + 25365 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} - 11367 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} + 4887 i \, \sqrt {\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1} - 4887 i \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 3807 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 )}^{4} \mathrm {sgn}\left (-3 \, \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 1\right )}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/162000*(243*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*(15039*I*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))
^7 + 6291*I*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^6 - 579*I*(sqrt(tan(1/2*x*e + 1/2*d)^2 +
1) - tan(1/2*x*e + 1/2*d))^5 + 1645*I*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^4 + 25365*I*(
sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^3 - 11367*I*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1
/2*x*e + 1/2*d))^2 + 4887*I*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - 4887*I*tan(1/2*x*e + 1/2*d) + 3807*I)/((3*I*(sq
rt(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)^4*sgn(-3*tan(1/2*x*e + 1/2*d) + 1)))*e^(-1)

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maple [A]  time = 0.34, size = 190, normalized size = 1.34 \[ \frac {\left (3 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right ) \left (\sin ^{2}\left (e x +d +\arctan \left (\frac {4}{3}\right )\right )\right )-6 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right ) \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+3 \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right )-6 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )+14 \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}}{4000 \left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4000*(3*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))^2-6*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))+3*10^(1/2)*arctan(1/10*(-5
*sin(e*x+d+arctan(4/3))-5)^(1/2)*10^(1/2))-6*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)*sin(e*x+d+arctan(4/3))+14*(-5
*sin(e*x+d+arctan(4/3))-5)^(1/2))*(-5*sin(e*x+d+arctan(4/3))-5)^(1/2)/(sin(e*x+d+arctan(4/3))-1)/cos(e*x+d+arc
tan(4/3))/(-5+5*sin(e*x+d+arctan(4/3)))^(1/2)/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} - 5\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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