3.516 \(\int \sqrt {3-4 \cos (c+d x)} \cos ^3(c+d x) \, dx\)

Optimal. Leaf size=140 \[ -\frac {59 F\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{60 \sqrt {7} d}-\frac {47 E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{20 \sqrt {7} d}-\frac {\sin (c+d x) \cos (c+d x) (3-4 \cos (c+d x))^{3/2}}{14 d}-\frac {3 \sin (c+d x) (3-4 \cos (c+d x))^{3/2}}{70 d}+\frac {59 \sin (c+d x) \sqrt {3-4 \cos (c+d x)}}{105 d} \]

[Out]

-3/70*(3-4*cos(d*x+c))^(3/2)*sin(d*x+c)/d-1/14*(3-4*cos(d*x+c))^(3/2)*cos(d*x+c)*sin(d*x+c)/d+47/140*(sin(1/2*
d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)*EllipticE(cos(1/2*d*x+1/2*c),2/7*14^(1/2))/d*7^(1/2)+59/420*(sin(1/2*d*
x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)*EllipticF(cos(1/2*d*x+1/2*c),2/7*14^(1/2))/d*7^(1/2)+59/105*sin(d*x+c)*(3
-4*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.19, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2793, 3023, 2753, 2752, 2662, 2654} \[ -\frac {59 F\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{60 \sqrt {7} d}-\frac {47 E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{20 \sqrt {7} d}-\frac {\sin (c+d x) \cos (c+d x) (3-4 \cos (c+d x))^{3/2}}{14 d}-\frac {3 \sin (c+d x) (3-4 \cos (c+d x))^{3/2}}{70 d}+\frac {59 \sin (c+d x) \sqrt {3-4 \cos (c+d x)}}{105 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[3 - 4*Cos[c + d*x]]*Cos[c + d*x]^3,x]

[Out]

(-47*EllipticE[(c + Pi + d*x)/2, 8/7])/(20*Sqrt[7]*d) - (59*EllipticF[(c + Pi + d*x)/2, 8/7])/(60*Sqrt[7]*d) +
(59*Sqrt[3 - 4*Cos[c + d*x]]*Sin[c + d*x])/(105*d) - (3*(3 - 4*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(70*d) - ((3
- 4*Cos[c + d*x])^(3/2)*Cos[c + d*x]*Sin[c + d*x])/(14*d)

Rule 2654

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 2662

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 2752

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

Rule 2753

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

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sqrt {3-4 \cos (c+d x)} \cos ^3(c+d x) \, dx &=-\frac {(3-4 \cos (c+d x))^{3/2} \cos (c+d x) \sin (c+d x)}{14 d}-\frac {1}{14} \int \sqrt {3-4 \cos (c+d x)} \left (3-10 \cos (c+d x)-6 \cos ^2(c+d x)\right ) \, dx\\ &=-\frac {3 (3-4 \cos (c+d x))^{3/2} \sin (c+d x)}{70 d}-\frac {(3-4 \cos (c+d x))^{3/2} \cos (c+d x) \sin (c+d x)}{14 d}+\frac {1}{140} \int \sqrt {3-4 \cos (c+d x)} (6+118 \cos (c+d x)) \, dx\\ &=\frac {59 \sqrt {3-4 \cos (c+d x)} \sin (c+d x)}{105 d}-\frac {3 (3-4 \cos (c+d x))^{3/2} \sin (c+d x)}{70 d}-\frac {(3-4 \cos (c+d x))^{3/2} \cos (c+d x) \sin (c+d x)}{14 d}+\frac {1}{210} \int \frac {-209+141 \cos (c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx\\ &=\frac {59 \sqrt {3-4 \cos (c+d x)} \sin (c+d x)}{105 d}-\frac {3 (3-4 \cos (c+d x))^{3/2} \sin (c+d x)}{70 d}-\frac {(3-4 \cos (c+d x))^{3/2} \cos (c+d x) \sin (c+d x)}{14 d}-\frac {47}{280} \int \sqrt {3-4 \cos (c+d x)} \, dx-\frac {59}{120} \int \frac {1}{\sqrt {3-4 \cos (c+d x)}} \, dx\\ &=-\frac {47 E\left (\frac {1}{2} (c+\pi +d x)|\frac {8}{7}\right )}{20 \sqrt {7} d}-\frac {59 F\left (\frac {1}{2} (c+\pi +d x)|\frac {8}{7}\right )}{60 \sqrt {7} d}+\frac {59 \sqrt {3-4 \cos (c+d x)} \sin (c+d x)}{105 d}-\frac {3 (3-4 \cos (c+d x))^{3/2} \sin (c+d x)}{70 d}-\frac {(3-4 \cos (c+d x))^{3/2} \cos (c+d x) \sin (c+d x)}{14 d}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 114, normalized size = 0.81 \[ \frac {654 \sin (c+d x)-511 \sin (2 (c+d x))+108 \sin (3 (c+d x))-60 \sin (4 (c+d x))-413 \sqrt {4 \cos (c+d x)-3} F\left (\left .\frac {1}{2} (c+d x)\right |8\right )+141 \sqrt {4 \cos (c+d x)-3} E\left (\left .\frac {1}{2} (c+d x)\right |8\right )}{420 d \sqrt {3-4 \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[3 - 4*Cos[c + d*x]]*Cos[c + d*x]^3,x]

[Out]

(141*Sqrt[-3 + 4*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 8] - 413*Sqrt[-3 + 4*Cos[c + d*x]]*EllipticF[(c + d*x)/2
, 8] + 654*Sin[c + d*x] - 511*Sin[2*(c + d*x)] + 108*Sin[3*(c + d*x)] - 60*Sin[4*(c + d*x)])/(420*d*Sqrt[3 - 4
*Cos[c + d*x]])

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fricas [F]  time = 1.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-4 \, \cos \left (d x + c\right ) + 3} \cos \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(3-4*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-4*cos(d*x + c) + 3)*cos(d*x + c)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \cos \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(3-4*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-4*cos(d*x + c) + 3)*cos(d*x + c)^3, x)

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maple [A]  time = 0.75, size = 276, normalized size = 1.97 \[ \frac {\sqrt {-\left (8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (7680 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8064 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5432 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+59 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )+141 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )-568 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 \sqrt {8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(3-4*cos(d*x+c))^(1/2),x)

[Out]

1/420*(-(8*cos(1/2*d*x+1/2*c)^2-7)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(7680*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-8
064*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+5432*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+59*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2/7*14^(1/2))+141*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2/7*14^(1/2))-568*sin(1/2*d*x+1/2*c)
^2*cos(1/2*d*x+1/2*c))/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-8*cos(1/2*d*x+
1/2*c)^2+7)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \cos \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(3-4*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-4*cos(d*x + c) + 3)*cos(d*x + c)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^3\,\sqrt {3-4\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(3 - 4*cos(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^3*(3 - 4*cos(c + d*x))^(1/2), 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(cos(d*x+c)**3*(3-4*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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