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

Optimal. Leaf size=94 \[ -\frac{5 \cos (d+e x)-3 \sin (d+e x)}{15 e \sqrt{5 \sin (d+e x)+3 \cos (d+e x)+2}}-\frac{\sqrt{2+\sqrt{34}} E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )|\frac{2}{15} \left (17-\sqrt{34}\right )\right )}{15 e} \]

[Out]

-(Sqrt[2 + Sqrt[34]]*EllipticE[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(15*e) - (5*Cos[d + e*x] -
3*Sin[d + e*x])/(15*e*Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]])

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Rubi [A]  time = 0.0535315, antiderivative size = 94, 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 = {3128, 3118, 2653} \[ -\frac{5 \cos (d+e x)-3 \sin (d+e x)}{15 e \sqrt{5 \sin (d+e x)+3 \cos (d+e x)+2}}-\frac{\sqrt{2+\sqrt{34}} E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )|\frac{2}{15} \left (17-\sqrt{34}\right )\right )}{15 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[2 + Sqrt[34]]*EllipticE[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(15*e) - (5*Cos[d + e*x] -
3*Sin[d + e*x])/(15*e*Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]])

Rule 3128

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-3/2), x_Symbol] :> Simp[(2*(c*Cos
[d + e*x] - b*Sin[d + e*x]))/(e*(a^2 - b^2 - c^2)*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]), x] + Dist[1/(a^2
- b^2 - c^2), Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 -
b^2 - c^2, 0]

Rule 3118

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{(2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}} \, dx &=-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{15 e \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}}-\frac{1}{30} \int \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)} \, dx\\ &=-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{15 e \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}}-\frac{1}{30} \int \sqrt{2+\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )} \, dx\\ &=-\frac{\sqrt{2+\sqrt{34}} E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )|\frac{2}{15} \left (17-\sqrt{34}\right )\right )}{15 e}-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{15 e \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}}\\ \end{align*}

Mathematica [C]  time = 6.14109, size = 528, normalized size = 5.62 \[ -\frac{17 \left (-\frac{5 \sqrt{\frac{1}{34} \left (17+\sqrt{34}\right )} \sin \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right ) F_1\left (-\frac{1}{2};-\frac{1}{2},-\frac{1}{2};\frac{1}{2};-\frac{\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+2}{\sqrt{34} \left (1-\sqrt{\frac{2}{17}}\right )},-\frac{\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+2}{\sqrt{34} \left (-1-\sqrt{\frac{2}{17}}\right )}\right )}{17 \sqrt{1-\cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )} \sqrt{-\frac{\cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+1}{\sqrt{34}-17}} \sqrt{\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+2}}-\frac{\frac{3}{17} \left (\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+2\right )-\frac{5 \sin \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )}{\sqrt{34}}}{\sqrt{\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+2}}\right )}{75 e}-\frac{\sqrt{\frac{34}{17+\sqrt{34}}} \sqrt{1-\sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )} \sqrt{-\frac{\sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )+1}{\sqrt{34}-17}} \sqrt{\sqrt{34} \sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )+2} \sec \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right ) F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};-\frac{\sqrt{34} \sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )+2}{\sqrt{34} \left (1-\sqrt{\frac{2}{17}}\right )},-\frac{\sqrt{34} \sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )+2}{\sqrt{34} \left (-1-\sqrt{\frac{2}{17}}\right )}\right )}{15 e}+\frac{\sqrt{5 \sin (d+e x)+3 \cos (d+e x)+2} \left (\frac{2 (17 \sin (d+e x)+5)}{45 (5 \sin (d+e x)+3 \cos (d+e x)+2)}-\frac{34}{225}\right )}{e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]]*(-34/225 + (2*(5 + 17*Sin[d + e*x]))/(45*(2 + 3*Cos[d + e*x] + 5*Si
n[d + e*x]))))/e - (Sqrt[34/(17 + Sqrt[34])]*AppellF1[1/2, 1/2, 1/2, 3/2, -((2 + Sqrt[34]*Sin[d + e*x + ArcTan
[3/5]])/(Sqrt[34]*(1 - Sqrt[2/17]))), -((2 + Sqrt[34]*Sin[d + e*x + ArcTan[3/5]])/(Sqrt[34]*(-1 - Sqrt[2/17]))
)]*Sec[d + e*x + ArcTan[3/5]]*Sqrt[1 - Sin[d + e*x + ArcTan[3/5]]]*Sqrt[-((1 + Sin[d + e*x + ArcTan[3/5]])/(-1
7 + Sqrt[34]))]*Sqrt[2 + Sqrt[34]*Sin[d + e*x + ArcTan[3/5]]])/(15*e) - (17*((-5*Sqrt[(17 + Sqrt[34])/34]*Appe
llF1[-1/2, -1/2, -1/2, 1/2, -((2 + Sqrt[34]*Cos[d + e*x - ArcTan[5/3]])/(Sqrt[34]*(1 - Sqrt[2/17]))), -((2 + S
qrt[34]*Cos[d + e*x - ArcTan[5/3]])/(Sqrt[34]*(-1 - Sqrt[2/17])))]*Sin[d + e*x - ArcTan[5/3]])/(17*Sqrt[1 - Co
s[d + e*x - ArcTan[5/3]]]*Sqrt[-((1 + Cos[d + e*x - ArcTan[5/3]])/(-17 + Sqrt[34]))]*Sqrt[2 + Sqrt[34]*Cos[d +
e*x - ArcTan[5/3]]]) - ((3*(2 + Sqrt[34]*Cos[d + e*x - ArcTan[5/3]]))/17 - (5*Sin[d + e*x - ArcTan[5/3]])/Sqr
t[34])/Sqrt[2 + Sqrt[34]*Cos[d + e*x - ArcTan[5/3]]]))/(75*e)

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Maple [C]  time = 3.376, size = 425, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4335*34^(1/2)*(255*((17*sin(e*x+d+arctan(3/5))+34^(1/2))*34^(1/2)*cos(e*x+d+arctan(3/5))^2)^(1/2)*((17*sin(e
*x+d+arctan(3/5))+34^(1/2))/(34^(1/2)+17))^(1/2)*((1+sin(e*x+d+arctan(3/5)))/(-34^(1/2)+17))^(1/2)*(-17*(sin(e
*x+d+arctan(3/5))-1)/(34^(1/2)+17))^(1/2)*EllipticF(((17*sin(e*x+d+arctan(3/5))+34^(1/2))/(34^(1/2)+17))^(1/2)
,I*(1/(-34^(1/2)+17)*(34^(1/2)+17))^(1/2))-255*((17*sin(e*x+d+arctan(3/5))+34^(1/2))*34^(1/2)*cos(e*x+d+arctan
(3/5))^2)^(1/2)*((17*sin(e*x+d+arctan(3/5))+34^(1/2))/(34^(1/2)+17))^(1/2)*((1+sin(e*x+d+arctan(3/5)))/(-34^(1
/2)+17))^(1/2)*(-17*(sin(e*x+d+arctan(3/5))-1)/(34^(1/2)+17))^(1/2)*EllipticE(((17*sin(e*x+d+arctan(3/5))+34^(
1/2))/(34^(1/2)+17))^(1/2),I*(1/(-34^(1/2)+17)*(34^(1/2)+17))^(1/2))+289*((34^(1/2)*sin(e*x+d+arctan(3/5))+2)*
cos(e*x+d+arctan(3/5))^2)^(1/2)*sin(e*x+d+arctan(3/5))^2-289*((34^(1/2)*sin(e*x+d+arctan(3/5))+2)*cos(e*x+d+ar
ctan(3/5))^2)^(1/2))*17^(1/2)/((17*sin(e*x+d+arctan(3/5))+34^(1/2))*34^(1/2)*cos(e*x+d+arctan(3/5))^2)^(1/2)/c
os(e*x+d+arctan(3/5))/(34^(1/2)*sin(e*x+d+arctan(3/5))+2)^(1/2)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2}}{16 \, \cos \left (e x + d\right )^{2} - 10 \,{\left (3 \, \cos \left (e x + d\right ) + 2\right )} \sin \left (e x + d\right ) - 12 \, \cos \left (e x + d\right ) - 29}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(3*cos(e*x + d) + 5*sin(e*x + d) + 2)/(16*cos(e*x + d)^2 - 10*(3*cos(e*x + d) + 2)*sin(e*x + d)
- 12*cos(e*x + d) - 29), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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