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

Optimal. Leaf size=185 \[ -\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}{5 e}-\frac {32 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{15 e}+\frac {64 F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{\sqrt {2+\sqrt {34}} e}+\frac {796 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{15 e} \]

[Out]

-2/5*(5*cos(e*x+d)-3*sin(e*x+d))*(2+3*cos(e*x+d)+5*sin(e*x+d))^(3/2)/e-32/15*(5*cos(e*x+d)-3*sin(e*x+d))*(2+3*
cos(e*x+d)+5*sin(e*x+d))^(1/2)/e+64*(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))*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)+796/15*(
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 = 0.27, antiderivative size = 185, 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 = {3120, 3146, 3149, 3118, 2653, 3126, 2661} \[ -\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) (5 \sin (d+e x)+3 \cos (d+e x)+2)^{3/2}}{5 e}-\frac {32 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2}}{15 e}+\frac {64 F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{\sqrt {2+\sqrt {34}} e}+\frac {796 \sqrt {2+\sqrt {34}} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )|\frac {2}{15} \left (17-\sqrt {34}\right )\right )}{15 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(796*Sqrt[2 + Sqrt[34]]*EllipticE[(d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(15*e) + (64*EllipticF[(
d + e*x - ArcTan[5/3])/2, (2*(17 - Sqrt[34]))/15])/(Sqrt[2 + Sqrt[34]]*e) - (32*(5*Cos[d + e*x] - 3*Sin[d + e*
x])*Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]])/(15*e) - (2*(5*Cos[d + e*x] - 3*Sin[d + e*x])*(2 + 3*Cos[d + e*
x] + 5*Sin[d + e*x])^(3/2))/(5*e)

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 3120

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

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 3146

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

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]

Rubi steps

\begin {align*} \int (2+3 \cos (d+e x)+5 \sin (d+e x))^{5/2} \, dx &=-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}{5 e}+\frac {2}{5} \int \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)} (61+24 \cos (d+e x)+40 \sin (d+e x)) \, dx\\ &=-\frac {32 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}{15 e}-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}{5 e}+\frac {2}{15} \int \frac {638+597 \cos (d+e x)+995 \sin (d+e x)}{\sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}} \, dx\\ &=-\frac {32 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}{15 e}-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}{5 e}+\frac {398}{15} \int \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)} \, dx+32 \int \frac {1}{\sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}} \, dx\\ &=-\frac {32 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}{15 e}-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}{5 e}+\frac {398}{15} \int \sqrt {2+\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )} \, dx+32 \int \frac {1}{\sqrt {2+\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )}} \, dx\\ &=\frac {796 \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 {64 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 )}{\sqrt {2+\sqrt {34}} e}-\frac {32 (5 \cos (d+e x)-3 \sin (d+e x)) \sqrt {2+3 \cos (d+e x)+5 \sin (d+e x)}}{15 e}-\frac {2 (5 \cos (d+e x)-3 \sin (d+e x)) (2+3 \cos (d+e x)+5 \sin (d+e x))^{3/2}}{5 e}\\ \end {align*}

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Mathematica [C]  time = 5.91, size = 399, normalized size = 2.16 \[ \frac {1276 \sqrt {\frac {10}{3}} \sqrt {\sqrt {34} \sin \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )+2} \sqrt {\cos ^2\left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )} \sec \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{2};\frac {3}{2};\frac {17 \sin \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )+\sqrt {34}}{-17+\sqrt {34}},\frac {17 \sin \left (d+e x+\tan ^{-1}\left (\frac {3}{5}\right )\right )+\sqrt {34}}{17+\sqrt {34}}\right )-\frac {1990 \sqrt {30} \sqrt {\sin ^2\left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )} \csc \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right ) F_1\left (-\frac {1}{2};-\frac {1}{2},-\frac {1}{2};\frac {1}{2};\frac {17 \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+\sqrt {34}}{-17+\sqrt {34}},\frac {17 \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+\sqrt {34}}{17+\sqrt {34}}\right )}{\sqrt {\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+2}}-2 \sqrt {5 \sin (d+e x)+3 \cos (d+e x)+2} (550 \cos (d+e x)+3 (-110 \sin (d+e x)+40 \sin (2 (d+e x))+75 \cos (2 (d+e x))-398))-2388 \sqrt {\sqrt {34} \cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )+2}+\frac {1990 \sin \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )}{\sqrt {\frac {\cos \left (d+e x-\tan ^{-1}\left (\frac {5}{3}\right )\right )}{\sqrt {34}}+\frac {1}{17}}}}{75 e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2388*Sqrt[2 + Sqrt[34]*Cos[d + e*x - ArcTan[5/3]]] - 2*Sqrt[2 + 3*Cos[d + e*x] + 5*Sin[d + e*x]]*(550*Cos[d
+ e*x] + 3*(-398 + 75*Cos[2*(d + e*x)] - 110*Sin[d + e*x] + 40*Sin[2*(d + e*x)])) + 1276*Sqrt[10/3]*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 + A
rcTan[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]]] + (1990*Sin[d + e*x - ArcTan[5/3]])/Sqrt[1/17 + Cos[d + e*x - ArcTan[5/3]]/Sqrt[34
]] - (1990*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 + Sqrt[34])]*Csc[d + e*x - ArcTan[5/3]]*Sqrt[Sin[d + e*x - A
rcTan[5/3]]^2])/Sqrt[2 + Sqrt[34]*Cos[d + e*x - ArcTan[5/3]]])/(75*e)

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fricas [F]  time = 1.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (16 \, \cos \left (e x + d\right )^{2} - 10 \, {\left (3 \, \cos \left (e x + d\right ) + 2\right )} \sin \left (e x + d\right ) - 12 \, \cos \left (e x + d\right ) - 29\right )} \sqrt {3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.66, size = 701, normalized size = 3.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16/17*((17*sin(e*x+d+arctan(3/5))+34^(1/2))/(34^(1/2)+17))^(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))*34^(1/2)+16*((17*sin(e*x+d+arctan(3/5))+3
4^(1/2))/(34^(1/2)+17))^(1/2)*17^(1/2)*((1+sin(e*x+d+arctan(3/5)))/(-34^(1/2)+17))^(1/2)*(-17*(sin(e*x+d+arcta
n(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))-44*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)*(-(17*sin(e*x+d+arctan(3/5))+34^(1/2))/(-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))*34^(1/2)+7
96/17*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)*(-(17*sin(e*x+d+arctan(3/5))+34^(1/2))/(-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*((-34^(1/2)+17)/(34^(1/2)+17))^(1/2))*34^(1/2)-48*(-(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))-1)/(34^(1/2)+17))^(1/2)*17^(1/2)*((1+sin(e*x+d+a
rctan(3/5)))/(-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))+68/5*34^(1/2)*sin(e*x+d+arctan(3/5))^4-116/15*34^(1/2)*sin(e*x+d+arctan(3/
5))^2+1904/15*sin(e*x+d+arctan(3/5))^3-1904/15*sin(e*x+d+arctan(3/5))-88/15*34^(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 \[ \int {\left (3 \, \cos \left (e x + d\right ) + 5 \, \sin \left (e x + d\right ) + 2\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 \[ \int {\left (3\,\cos \left (d+e\,x\right )+5\,\sin \left (d+e\,x\right )+2\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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