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

Optimal. Leaf size=138 \[ \frac {59 F\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{60 \sqrt {7} d}+\frac {47 E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{20 \sqrt {7} d}+\frac {\sin (c+d x) \cos (c+d x) (4 \cos (c+d x)+3)^{3/2}}{14 d}-\frac {3 \sin (c+d x) (4 \cos (c+d x)+3)^{3/2}}{70 d}+\frac {59 \sin (c+d x) \sqrt {4 \cos (c+d x)+3}}{105 d} \]

[Out]

-3/70*(3+4*cos(d*x+c))^(3/2)*sin(d*x+c)/d+1/14*cos(d*x+c)*(3+4*cos(d*x+c))^(3/2)*sin(d*x+c)/d+47/140*(cos(1/2*
d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2/7*14^(1/2))/d*7^(1/2)+59/420*(cos(1/2*d*
x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(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 = 138, 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, 2661, 2653} \[ \frac {59 F\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{60 \sqrt {7} d}+\frac {47 E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{20 \sqrt {7} d}+\frac {\sin (c+d x) \cos (c+d x) (4 \cos (c+d x)+3)^{3/2}}{14 d}-\frac {3 \sin (c+d x) (4 \cos (c+d x)+3)^{3/2}}{70 d}+\frac {59 \sin (c+d x) \sqrt {4 \cos (c+d x)+3}}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 \cos ^3(c+d x) \sqrt {3+4 \cos (c+d x)} \, dx &=\frac {\cos (c+d x) (3+4 \cos (c+d x))^{3/2} \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 {\cos (c+d x) (3+4 \cos (c+d x))^{3/2} \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 {\cos (c+d x) (3+4 \cos (c+d x))^{3/2} \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 {\cos (c+d x) (3+4 \cos (c+d x))^{3/2} \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+d x)|\frac {8}{7}\right )}{20 \sqrt {7} d}+\frac {59 F\left (\frac {1}{2} (c+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 {\cos (c+d x) (3+4 \cos (c+d x))^{3/2} \sin (c+d x)}{14 d}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 92, normalized size = 0.67 \[ \frac {59 \sqrt {7} F\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )+141 \sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )+(212 \sin (c+d x)+9 \sin (2 (c+d x))+30 \sin (3 (c+d x))) \sqrt {4 \cos (c+d x)+3}}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(141*Sqrt[7]*EllipticE[(c + d*x)/2, 8/7] + 59*Sqrt[7]*EllipticF[(c + d*x)/2, 8/7] + Sqrt[3 + 4*Cos[c + d*x]]*(
212*Sin[c + d*x] + 9*Sin[2*(c + d*x)] + 30*Sin[3*(c + d*x)]))/(420*d)

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fricas [F]  time = 0.81, 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 = 0.65, size = 275, normalized size = 1.99 \[ -\frac {\sqrt {\left (8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (7680 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-14976 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12344 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+413 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )-141 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )-4480 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{420 \sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

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)^2-1)*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)^8-1
4976*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+12344*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+413*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(8*sin(1/2*d*x+1/2*c)^2-7)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2*2^(1/2))-141*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(8*sin(1/2*d*x+1/2*c)^2-7)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2*2^(1/2))-4480*sin(1/2*d*x+1/2*c)^2*c
os(1/2*d*x+1/2*c))/(-8*sin(1/2*d*x+1/2*c)^4+7*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)^2-1)^(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 \[ \int {\cos \left (c+d\,x\right )}^3\,\sqrt {4\,\cos \left (c+d\,x\right )+3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(c + d*x)^3*(4*cos(c + d*x) + 3)^(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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