3.409 \(\int \frac{1}{(2+3 \cos (d+e x)+5 \sin (d+e x))^{7/2}} \, dx\)

Optimal. Leaf size=233 \[ -\frac{8 \text{EllipticF}\left (\frac{1}{2} \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right ),\frac{2}{15} \left (17-\sqrt{34}\right )\right )}{3375 \sqrt{2+\sqrt{34}} e}-\frac{199 (5 \cos (d+e x)-3 \sin (d+e x))}{101250 e \sqrt{5 \sin (d+e x)+3 \cos (d+e x)+2}}+\frac{8 (5 \cos (d+e x)-3 \sin (d+e x))}{3375 e (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (5 \sin (d+e x)+3 \cos (d+e x)+2)^{5/2}}-\frac{199 \sqrt{2+\sqrt{34}} E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )|\frac{2}{15} \left (17-\sqrt{34}\right )\right )}{101250 e} \]

[Out]

(-199*Sqrt[2 + Sqrt[34]]*EllipticE[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(101250*e) - (8*Ellipti
cF[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(3375*Sqrt[2 + Sqrt[34]]*e) - (5*Cos[d + e*x] - 3*Sin[d
+ e*x])/(75*e*(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(5/2)) + (8*(5*Cos[d + e*x] - 3*Sin[d + e*x]))/(3375*e*(2
+ 3*Cos[d + e*x] + 5*Sin[d + e*x])^(3/2)) - (199*(5*Cos[d + e*x] - 3*Sin[d + e*x]))/(101250*e*Sqrt[2 + 3*Cos[
d + e*x] + 5*Sin[d + e*x]])

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Rubi [A]  time = 0.259548, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {3129, 3156, 3149, 3118, 2653, 3126, 2661} \[ -\frac{199 (5 \cos (d+e x)-3 \sin (d+e x))}{101250 e \sqrt{5 \sin (d+e x)+3 \cos (d+e x)+2}}+\frac{8 (5 \cos (d+e x)-3 \sin (d+e x))}{3375 e (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (5 \sin (d+e x)+3 \cos (d+e x)+2)^{5/2}}-\frac{8 F\left (\frac{1}{2} \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )|\frac{2}{15} \left (17-\sqrt{34}\right )\right )}{3375 \sqrt{2+\sqrt{34}} e}-\frac{199 \sqrt{2+\sqrt{34}} E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )|\frac{2}{15} \left (17-\sqrt{34}\right )\right )}{101250 e} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(-7/2),x]

[Out]

(-199*Sqrt[2 + Sqrt[34]]*EllipticE[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(101250*e) - (8*Ellipti
cF[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(3375*Sqrt[2 + Sqrt[34]]*e) - (5*Cos[d + e*x] - 3*Sin[d
+ e*x])/(75*e*(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(5/2)) + (8*(5*Cos[d + e*x] - 3*Sin[d + e*x]))/(3375*e*(2
+ 3*Cos[d + e*x] + 5*Sin[d + e*x])^(3/2)) - (199*(5*Cos[d + e*x] - 3*Sin[d + e*x]))/(101250*e*Sqrt[2 + 3*Cos[
d + e*x] + 5*Sin[d + e*x]])

Rule 3129

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 + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] +
Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c*(n + 2)*Sin[d + e*x])*(a + b*C
os[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n
, -1] && NeQ[n, -3/2]

Rule 3156

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

Rule 3149

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.)
+ (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[B/b, Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]
, x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x], x] /; FreeQ[{a, b, c, d, e
, A, B, C}, x] && EqQ[B*c - b*C, 0] && NeQ[A*b - a*B, 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]

Rule 3126

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] && NeQ[b^2 + c^2, 0] && GtQ[a +
Sqrt[b^2 + c^2], 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), 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))^{7/2}} \, dx &=-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2}}+\frac{1}{75} \int \frac{-5+\frac{9}{2} \cos (d+e x)+\frac{15}{2} \sin (d+e x)}{(2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2}} \, dx\\ &=-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2}}+\frac{8 (5 \cos (d+e x)-3 \sin (d+e x))}{3375 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}+\frac{\int \frac{\frac{183}{2}-12 \cos (d+e x)-20 \sin (d+e x)}{(2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}} \, dx}{3375}\\ &=-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2}}+\frac{8 (5 \cos (d+e x)-3 \sin (d+e x))}{3375 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}-\frac{199 (5 \cos (d+e x)-3 \sin (d+e x))}{101250 e \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}}+\frac{\int \frac{-\frac{319}{2}-\frac{597}{4} \cos (d+e x)-\frac{995}{4} \sin (d+e x)}{\sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}} \, dx}{50625}\\ &=-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2}}+\frac{8 (5 \cos (d+e x)-3 \sin (d+e x))}{3375 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}-\frac{199 (5 \cos (d+e x)-3 \sin (d+e x))}{101250 e \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}}-\frac{199 \int \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)} \, dx}{202500}-\frac{4 \int \frac{1}{\sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}} \, dx}{3375}\\ &=-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2}}+\frac{8 (5 \cos (d+e x)-3 \sin (d+e x))}{3375 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}-\frac{199 (5 \cos (d+e x)-3 \sin (d+e x))}{101250 e \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}}-\frac{199 \int \sqrt{2+\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )} \, dx}{202500}-\frac{4 \int \frac{1}{\sqrt{2+\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )}} \, dx}{3375}\\ &=-\frac{199 \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 )}{101250 e}-\frac{8 F\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 )}{3375 \sqrt{2+\sqrt{34}} e}-\frac{5 \cos (d+e x)-3 \sin (d+e x)}{75 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2}}+\frac{8 (5 \cos (d+e x)-3 \sin (d+e x))}{3375 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}-\frac{199 (5 \cos (d+e x)-3 \sin (d+e x))}{101250 e \sqrt{2+3 \cos (d+e x)+5 \sin (d+e x)}}\\ \end{align*}

Mathematica [C]  time = 3.92931, size = 436, normalized size = 1.87 \[ \frac{-638 \sqrt{30} \sqrt{\sqrt{34} \sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )+2} \sqrt{\cos ^2\left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )} \sec \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right ) F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\frac{17 \sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )+\sqrt{34}}{-17+\sqrt{34}},\frac{17 \sin \left (d+e x+\tan ^{-1}\left (\frac{3}{5}\right )\right )+\sqrt{34}}{17+\sqrt{34}}\right )+\frac{2985 \sqrt{30} \sqrt{\sin ^2\left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )} \csc \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right ) F_1\left (-\frac{1}{2};-\frac{1}{2},-\frac{1}{2};\frac{1}{2};\frac{17 \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+\sqrt{34}}{-17+\sqrt{34}},\frac{17 \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+\sqrt{34}}{17+\sqrt{34}}\right )}{\sqrt{\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+2}}+\frac{27000 (17 \sin (d+e x)+5)}{(5 \sin (d+e x)+3 \cos (d+e x)+2)^{5/2}}-\frac{300 (272 \sin (d+e x)+305)}{(5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}+\frac{20 (3383 \sin (d+e x)+1595)}{\sqrt{5 \sin (d+e x)+3 \cos (d+e x)+2}}-13532 \sqrt{5 \sin (d+e x)+3 \cos (d+e x)+2}+\frac{597 (15 \sin (d+e x)+43 \cos (d+e x)+12)}{\sqrt{\sqrt{34} \cos \left (d+e x-\tan ^{-1}\left (\frac{5}{3}\right )\right )+2}}}{3037500 e} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(-7/2),x]

[Out]

(-13532*Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]] + (597*(12 + 43*Cos[d + e*x] + 15*Sin[d + e*x]))/Sqrt[2 + Sq
rt[34]*Cos[d + e*x - ArcTan[5/3]]] + (27000*(5 + 17*Sin[d + e*x]))/(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(5/2)
- (300*(305 + 272*Sin[d + e*x]))/(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(3/2) + (20*(1595 + 3383*Sin[d + e*x])
)/Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]] - 638*Sqrt[30]*AppellF1[1/2, 1/2, 1/2, 3/2, (Sqrt[34] + 17*Sin[d +
e*x + ArcTan[3/5]])/(-17 + Sqrt[34]), (Sqrt[34] + 17*Sin[d + e*x + ArcTan[3/5]])/(17 + Sqrt[34])]*Sqrt[Cos[d
+ e*x + ArcTan[3/5]]^2]*Sec[d + e*x + ArcTan[3/5]]*Sqrt[2 + Sqrt[34]*Sin[d + e*x + ArcTan[3/5]]] + (2985*Sqrt[
30]*AppellF1[-1/2, -1/2, -1/2, 1/2, (Sqrt[34] + 17*Cos[d + e*x - ArcTan[5/3]])/(-17 + Sqrt[34]), (Sqrt[34] + 1
7*Cos[d + e*x - ArcTan[5/3]])/(17 + Sqrt[34])]*Csc[d + e*x - ArcTan[5/3]]*Sqrt[Sin[d + e*x - ArcTan[5/3]]^2])/
Sqrt[2 + Sqrt[34]*Cos[d + e*x - ArcTan[5/3]]])/(3037500*e)

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Maple [C]  time = 5.834, size = 589, normalized size = 2.5 \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))^(7/2),x)

[Out]

(-(-34^(1/2)*sin(e*x+d+arctan(3/5))-2)*cos(e*x+d+arctan(3/5))^2)^(1/2)*(-1/2550*(-(-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))+1/17*34^(1/2))^3+4/57375*34^(1/2)*(-(-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))+1/17*34^(1/2))^2-3383/1012
50*34^(1/2)*cos(e*x+d+arctan(3/5))^2/(-(-289*sin(e*x+d+arctan(3/5))-17*34^(1/2))*34^(1/2)*cos(e*x+d+arctan(3/5
))^2)^(1/2)-319/50625*(-1+1/17*34^(1/2))*((-17*sin(e*x+d+arctan(3/5))-34^(1/2))/(-34^(1/2)+17))^(1/2)*((-17*si
n(e*x+d+arctan(3/5))+17)/(34^(1/2)+17))^(1/2)*((17*sin(e*x+d+arctan(3/5))+17)/(-34^(1/2)+17))^(1/2)/(-(-34^(1/
2)*sin(e*x+d+arctan(3/5))-2)*cos(e*x+d+arctan(3/5))^2)^(1/2)*EllipticF(((-17*sin(e*x+d+arctan(3/5))-34^(1/2))/
(-34^(1/2)+17))^(1/2),I*((-34^(1/2)+17)/(34^(1/2)+17))^(1/2))-199/101250*34^(1/2)*(-1+1/17*34^(1/2))*((-17*sin
(e*x+d+arctan(3/5))-34^(1/2))/(-34^(1/2)+17))^(1/2)*((-17*sin(e*x+d+arctan(3/5))+17)/(34^(1/2)+17))^(1/2)*((17
*sin(e*x+d+arctan(3/5))+17)/(-34^(1/2)+17))^(1/2)/(-(-34^(1/2)*sin(e*x+d+arctan(3/5))-2)*cos(e*x+d+arctan(3/5)
)^2)^(1/2)*((-1/17*34^(1/2)-1)*EllipticE(((-17*sin(e*x+d+arctan(3/5))-34^(1/2))/(-34^(1/2)+17))^(1/2),I*((-34^
(1/2)+17)/(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*((-34
^(1/2)+17)/(34^(1/2)+17))^(1/2))))/cos(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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((3*cos(e*x + d) + 5*sin(e*x + d) + 2)^(-7/2), 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}}{644 \, \cos \left (e x + d\right )^{4} + 1584 \, \cos \left (e x + d\right )^{3} + 284 \, \cos \left (e x + d\right )^{2} + 20 \,{\left (48 \, \cos \left (e x + d\right )^{3} - 4 \, \cos \left (e x + d\right )^{2} - 111 \, \cos \left (e x + d\right ) - 58\right )} \sin \left (e x + d\right ) - 1896 \, \cos \left (e x + d\right ) - 1241}, 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))^(7/2),x, algorithm="fricas")

[Out]

integral(-sqrt(3*cos(e*x + d) + 5*sin(e*x + d) + 2)/(644*cos(e*x + d)^4 + 1584*cos(e*x + d)^3 + 284*cos(e*x +
d)^2 + 20*(48*cos(e*x + d)^3 - 4*cos(e*x + d)^2 - 111*cos(e*x + d) - 58)*sin(e*x + d) - 1896*cos(e*x + d) - 12
41), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(7/2),x)

[Out]

Timed out

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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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((3*cos(e*x + d) + 5*sin(e*x + d) + 2)^(-7/2), x)