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

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

[Out]

ArcTanh[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 = 0.0532055, 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, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sin \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )}{\sqrt{2} \sqrt{\cos \left (d+e x-\tan ^{-1}\left (\frac{3}{4}\right )\right )+1}}\right )}{10 \sqrt{10} e}-\frac{3 \cos (d+e x)-4 \sin (d+e x)}{10 e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[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 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}{(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{\tanh ^{-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 = 0.298685, size = 154, normalized size = 1.6 \[ -\frac{\left (\frac{1}{250}-\frac{i}{125}\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+3 \cos \left (\frac{1}{2} (d+e x)\right )\right ) \left ((5+10 i) \left (\cos \left (\frac{1}{2} (d+e x)\right )-3 \sin \left (\frac{1}{2} (d+e x)\right )\right )-(1-i) \sqrt{20+15 i} \tan ^{-1}\left (\left (\frac{1}{10}+\frac{3 i}{10}\right ) \sqrt{\frac{4}{5}+\frac{3 i}{5}} \left (3 \tan \left (\frac{1}{4} (d+e x)\right )-1\right )\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+3 \cos \left (\frac{1}{2} (d+e x)\right )\right )^2\right )}{e (3 \sin (d+e x)+4 \cos (d+e x)+5)^{3/2}} \]

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

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Maple [A]  time = 1.286, size = 117, normalized size = 1.2 \begin{align*} -{\frac{1}{100\,\cos \left ( ex+d+\arctan \left ( 4/3 \right ) \right ) e} \left ( \sqrt{10}{\it Artanh} \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}{\it Artanh} \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)*arctanh(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)*ar
ctanh(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+
arctan(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 = 1.95377, size = 782, normalized size = 8.15 \begin{align*} \frac{{\left (9 \, \sqrt{10} \cos \left (e x + d\right )^{2} +{\left (13 \, \sqrt{10} \cos \left (e x + d\right ) + 14 \, \sqrt{10}\right )} \sin \left (e x + d\right ) + 27 \, \sqrt{10} \cos \left (e x + d\right ) + 18 \, \sqrt{10}\right )} \log \left (-\frac{9 \, \cos \left (e x + d\right )^{2} +{\left (13 \, \cos \left (e x + d\right ) - 6\right )} \sin \left (e x + d\right ) + 2 \,{\left (\sqrt{10} \cos \left (e x + d\right ) - 3 \, \sqrt{10} \sin \left (e x + d\right ) + \sqrt{10}\right )} \sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5} - 33 \, \cos \left (e x + d\right ) - 42}{9 \, \cos \left (e x + d\right )^{2} +{\left (13 \, \cos \left (e x + d\right ) + 14\right )} \sin \left (e x + d\right ) + 27 \, \cos \left (e x + d\right ) + 18}\right ) - 20 \, \sqrt{4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) + 5}{\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}{200 \,{\left (9 \, e \cos \left (e x + d\right )^{2} + 27 \, e \cos \left (e x + d\right ) +{\left (13 \, e \cos \left (e x + d\right ) + 14 \, e\right )} \sin \left (e x + d\right ) + 18 \, 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]

1/200*((9*sqrt(10)*cos(e*x + d)^2 + (13*sqrt(10)*cos(e*x + d) + 14*sqrt(10))*sin(e*x + d) + 27*sqrt(10)*cos(e*
x + d) + 18*sqrt(10))*log(-(9*cos(e*x + d)^2 + (13*cos(e*x + d) - 6)*sin(e*x + d) + 2*(sqrt(10)*cos(e*x + d) -
3*sqrt(10)*sin(e*x + d) + sqrt(10))*sqrt(4*cos(e*x + d) + 3*sin(e*x + d) + 5) - 33*cos(e*x + d) - 42)/(9*cos(
e*x + d)^2 + (13*cos(e*x + d) + 14)*sin(e*x + d) + 27*cos(e*x + d) + 18)) - 20*sqrt(4*cos(e*x + d) + 3*sin(e*x
+ d) + 5)*(cos(e*x + d) - 3*sin(e*x + d) + 1))/(9*e*cos(e*x + d)^2 + 27*e*cos(e*x + d) + (13*e*cos(e*x + d) +
14*e)*sin(e*x + d) + 18*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 [B]  time = 1.95805, size = 383, normalized size = 3.99 \begin{align*} \frac{1}{100} \,{\left (\frac{\sqrt{10} \log \left (\frac{{\left | -2 \, \sqrt{10} + 2 \, \sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} - 2 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 6 \right |}}{{\left | 2 \, \sqrt{10} + 2 \, \sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} - 2 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 6 \right |}}\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 3\right )} - \frac{20 \,{\left (19 \,{\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} - 51 \,{\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} - 17 \, \sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} + 17 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 3\right )}}{{\left ({\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} - 6 \, \sqrt{\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1} + 6 \, \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - 1\right )}^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + 3\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]

1/100*(sqrt(10)*log(abs(-2*sqrt(10) + 2*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - 2*tan(1/2*x*e + 1/2*d) - 6)/abs(2*s
qrt(10) + 2*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - 2*tan(1/2*x*e + 1/2*d) - 6))/sgn(tan(1/2*x*e + 1/2*d) + 3) - 20
*(19*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^3 - 51*(sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) - tan(
1/2*x*e + 1/2*d))^2 - 17*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) + 17*tan(1/2*x*e + 1/2*d) - 3)/(((sqrt(tan(1/2*x*e +
1/2*d)^2 + 1) - tan(1/2*x*e + 1/2*d))^2 - 6*sqrt(tan(1/2*x*e + 1/2*d)^2 + 1) + 6*tan(1/2*x*e + 1/2*d) - 1)^2*
sgn(tan(1/2*x*e + 1/2*d) + 3)))*e^(-1)