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

Optimal. Leaf size=187 \[ \frac {4 (5 \cos (d+e x)-3 \sin (d+e x))}{675 e \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}-\frac {5 \cos (d+e x)-3 \sin (d+e x)}{45 e (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}+\frac {F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{45 \sqrt {2+\sqrt {34}} e}+\frac {4 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{675 e} \]

[Out]

1/45*(-5*cos(e*x+d)+3*sin(e*x+d))/e/(2+3*cos(e*x+d)+5*sin(e*x+d))^(3/2)+4/675*(5*cos(e*x+d)-3*sin(e*x+d))/e/(2
+3*cos(e*x+d)+5*sin(e*x+d))^(1/2)+1/45*(cos(1/2*d+1/2*e*x-1/2*arctan(5/3))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arct
an(5/3))*EllipticF(sin(1/2*d+1/2*e*x-1/2*arctan(5/3)),1/15*(510-30*34^(1/2))^(1/2))/e/(2+34^(1/2))^(1/2)+4/675
*(cos(1/2*d+1/2*e*x-1/2*arctan(5/3))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(5/3))*EllipticE(sin(1/2*d+1/2*e*x-1
/2*arctan(5/3)),1/15*(510-30*34^(1/2))^(1/2))*(2+34^(1/2))^(1/2)/e

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Rubi [A]  time = 0.20, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {4 (5 \cos (d+e x)-3 \sin (d+e x))}{675 e \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}-\frac {5 \cos (d+e x)-3 \sin (d+e x)}{45 e (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}+\frac {F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{45 \sqrt {2+\sqrt {34}} e}+\frac {4 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{675 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2 + Sqrt[34]]*EllipticE[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(675*e) + EllipticF[(d + e
*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15]/(45*Sqrt[2 + Sqrt[34]]*e) - (5*Cos[d + e*x] - 3*Sin[d + e*x])/(45
*e*(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(3/2)) + (4*(5*Cos[d + e*x] - 3*Sin[d + e*x]))/(675*e*Sqrt[2 + 3*Cos[
d + e*x] + 5*Sin[d + e*x]])

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 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]

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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{(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)}{45 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}+\frac {1}{45} \int \frac {-3+\frac {3}{2} \cos (d+e x)+\frac {5}{2} \sin (d+e x)}{(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)}{45 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}+\frac {4 (5 \cos (d+e x)-3 \sin (d+e x))}{675 e \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}+\frac {1}{675} \int \frac {\frac {23}{2}+6 \cos (d+e x)+10 \sin (d+e x)}{\sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}} \, dx\\ &=-\frac {5 \cos (d+e x)-3 \sin (d+e x)}{45 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}+\frac {4 (5 \cos (d+e x)-3 \sin (d+e x))}{675 e \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}+\frac {2}{675} \int \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)} \, dx+\frac {1}{90} \int \frac {1}{\sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}} \, dx\\ &=-\frac {5 \cos (d+e x)-3 \sin (d+e x)}{45 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}+\frac {4 (5 \cos (d+e x)-3 \sin (d+e x))}{675 e \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}+\frac {2}{675} \int \sqrt {2+\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )} \, dx+\frac {1}{90} \int \frac {1}{\sqrt {2+\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )}} \, dx\\ &=\frac {4 \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 )}{675 e}+\frac {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 )}{45 \sqrt {2+\sqrt {34}} e}-\frac {5 \cos (d+e x)-3 \sin (d+e x)}{45 e (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}+\frac {4 (5 \cos (d+e x)-3 \sin (d+e x))}{675 e \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}\\ \end {align*}

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Mathematica [C]  time = 3.18, size = 430, normalized size = 2.30 \[ \frac {23 \sqrt {\frac {10}{3}} \sqrt {\sqrt {34} \sin \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )+2} \sqrt {\cos ^2\left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )} \sec \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{2};\frac {3}{2};\frac {17 \sin \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )+\sqrt {34}}{-17+\sqrt {34}},\frac {17 \sin \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )+\sqrt {34}}{17+\sqrt {34}}\right )-\frac {20 \sqrt {30} \sqrt {\sin ^2\left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )} \csc \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right ) F_1\left (-\frac {1}{2};-\frac {1}{2},-\frac {1}{2};\frac {1}{2};\frac {17 \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+\sqrt {34}}{-17+\sqrt {34}},\frac {17 \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+\sqrt {34}}{17+\sqrt {34}}\right )}{\sqrt {\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+2}}+\frac {100 (17 \sin (d+e x)+5)}{(5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}-\frac {10 (136 \sin (d+e x)+115)}{3 \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}+\frac {272}{3} \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}-24 \sqrt {\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+2}+\frac {20 \sin \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )}{\sqrt {\frac {\cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )}{\sqrt {34}}+\frac {1}{17}}}}{6750 e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-24*Sqrt[2 + Sqrt[34]*Cos[d + e*x - ArcTan[5/3]]] + (272*Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]])/3 + (100*
(5 + 17*Sin[d + e*x]))/(2 + 3*Cos[d + e*x] + 5*Sin[d + e*x])^(3/2) - (10*(115 + 136*Sin[d + e*x]))/(3*Sqrt[2 +
3*Cos[d + e*x] + 5*Sin[d + e*x]]) + 23*Sqrt[10/3]*AppellF1[1/2, 1/2, 1/2, 3/2, (Sqrt[34] + 17*Sin[d + e*x + A
rcTan[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]]] + (20*Sin[d + e*x - A
rcTan[5/3]])/Sqrt[1/17 + Cos[d + e*x - ArcTan[5/3]]/Sqrt[34]] - (20*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] + 17*Cos[d + e*x - ArcTan[5/3]])/(17 + S
qrt[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 - ArcTa
n[5/3]]])/(6750*e)

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fricas [F]  time = 2.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2}}{198 \, \cos \left (e x + d\right )^{3} + 96 \, \cos \left (e x + d\right )^{2} - 5 \, {\left (2 \, \cos \left (e x + d\right )^{2} + 36 \, \cos \left (e x + d\right ) + 37\right )} \sin \left (e x + d\right ) - 261 \, \cos \left (e x + d\right ) - 158}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(3*cos(e*x + d) + 5*sin(e*x + d) + 2)/(198*cos(e*x + d)^3 + 96*cos(e*x + d)^2 - 5*(2*cos(e*x + d
)^2 + 36*cos(e*x + d) + 37)*sin(e*x + d) - 261*cos(e*x + d) - 158), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.61, size = 542, normalized size = 2.90 \[ \frac {\sqrt {-\left (-\sqrt {34}\, \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-2\right ) \left (\cos ^{2}\left (e x +d +\arctan \left (\frac {3}{5}\right )\right )\right )}\, \left (-\frac {\sqrt {34}\, \sqrt {-\left (-\sqrt {34}\, \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-2\right ) \left (\cos ^{2}\left (e x +d +\arctan \left (\frac {3}{5}\right )\right )\right )}}{1530 \left (\sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )+\frac {\sqrt {34}}{17}\right )^{2}}+\frac {68 \sqrt {34}\, \left (\cos ^{2}\left (e x +d +\arctan \left (\frac {3}{5}\right )\right )\right )}{675 \sqrt {-\left (-289 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-17 \sqrt {34}\right ) \sqrt {34}\, \left (\cos ^{2}\left (e x +d +\arctan \left (\frac {3}{5}\right )\right )\right )}}+\frac {23 \left (-1+\frac {\sqrt {34}}{17}\right ) \sqrt {\frac {-17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-\sqrt {34}}{-\sqrt {34}+17}}\, \sqrt {\frac {-17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )+17}{\sqrt {34}+17}}\, \sqrt {\frac {17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )+17}{-\sqrt {34}+17}}\, \EllipticF \left (\sqrt {\frac {-17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-\sqrt {34}}{-\sqrt {34}+17}}, i \sqrt {\frac {-\sqrt {34}+17}{\sqrt {34}+17}}\right )}{675 \sqrt {-\left (-\sqrt {34}\, \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-2\right ) \left (\cos ^{2}\left (e x +d +\arctan \left (\frac {3}{5}\right )\right )\right )}}+\frac {4 \sqrt {34}\, \left (-1+\frac {\sqrt {34}}{17}\right ) \sqrt {\frac {-17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-\sqrt {34}}{-\sqrt {34}+17}}\, \sqrt {\frac {-17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )+17}{\sqrt {34}+17}}\, \sqrt {\frac {17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )+17}{-\sqrt {34}+17}}\, \left (\left (-\frac {\sqrt {34}}{17}-1\right ) \EllipticE \left (\sqrt {\frac {-17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-\sqrt {34}}{-\sqrt {34}+17}}, i \sqrt {\frac {-\sqrt {34}+17}{\sqrt {34}+17}}\right )+\EllipticF \left (\sqrt {\frac {-17 \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-\sqrt {34}}{-\sqrt {34}+17}}, i \sqrt {\frac {-\sqrt {34}+17}{\sqrt {34}+17}}\right )\right )}{675 \sqrt {-\left (-\sqrt {34}\, \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )-2\right ) \left (\cos ^{2}\left (e x +d +\arctan \left (\frac {3}{5}\right )\right )\right )}}\right )}{\cos \left (e x +d +\arctan \left (\frac {3}{5}\right )\right ) \sqrt {\sqrt {34}\, \sin \left (e x +d +\arctan \left (\frac {3}{5}\right )\right )+2}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-34^(1/2)*sin(e*x+d+arctan(3/5))-2)*cos(e*x+d+arctan(3/5))^2)^(1/2)*(-1/1530*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+68/675*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)+23/
675*(-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+arcta
n(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))+4/675*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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