\(\int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx\) [547]

Optimal result

Integrand size = 23, antiderivative size = 111 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {9 \sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{20 d}-\frac {23 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{20 \sqrt {7} d}-\frac {\sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d}+\frac {\cos (c+d x) \sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d} \]

[Out]

-23/140*(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)+9
/20*(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)-1/10*
sin(d*x+c)*(3+4*cos(d*x+c))^(1/2)/d+1/10*cos(d*x+c)*sin(d*x+c)*(3+4*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2872, 3102, 2831, 2740, 2732} \[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=-\frac {23 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{20 \sqrt {7} d}+\frac {9 \sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{20 d}+\frac {\sin (c+d x) \cos (c+d x) \sqrt {4 \cos (c+d x)+3}}{10 d}-\frac {\sin (c+d x) \sqrt {4 \cos (c+d x)+3}}{10 d} \]

[In]

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

[Out]

(9*Sqrt[7]*EllipticE[(c + d*x)/2, 8/7])/(20*d) - (23*EllipticF[(c + d*x)/2, 8/7])/(20*Sqrt[7]*d) - (Sqrt[3 + 4
*Cos[c + d*x]]*Sin[c + d*x])/(10*d) + (Cos[c + d*x]*Sqrt[3 + 4*Cos[c + d*x]]*Sin[c + d*x])/(10*d)

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2831

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 2872

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-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 3102

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*} \text {integral}& = \frac {\cos (c+d x) \sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d}+\frac {1}{10} \int \frac {3+6 \cos (c+d x)-6 \cos ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx \\ & = -\frac {\sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d}+\frac {\cos (c+d x) \sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d}+\frac {1}{60} \int \frac {6+54 \cos (c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx \\ & = -\frac {\sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d}+\frac {\cos (c+d x) \sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d}+\frac {9}{40} \int \sqrt {3+4 \cos (c+d x)} \, dx-\frac {23}{40} \int \frac {1}{\sqrt {3+4 \cos (c+d x)}} \, dx \\ & = \frac {9 \sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{20 d}-\frac {23 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{20 \sqrt {7} d}-\frac {\sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d}+\frac {\cos (c+d x) \sqrt {3+4 \cos (c+d x)} \sin (c+d x)}{10 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {63 \sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )-23 \sqrt {7} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )+7 \sqrt {3+4 \cos (c+d x)} (-2 \sin (c+d x)+\sin (2 (c+d x)))}{140 d} \]

[In]

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

[Out]

(63*Sqrt[7]*EllipticE[(c + d*x)/2, 8/7] - 23*Sqrt[7]*EllipticF[(c + d*x)/2, 8/7] + 7*Sqrt[3 + 4*Cos[c + d*x]]*
(-2*Sin[c + d*x] + Sin[2*(c + d*x)]))/(140*d)

Maple [A] (verified)

Time = 3.84 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.08

method	result	size
default	\(-\frac {\sqrt {\left (8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-64 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+56 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-23 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )-9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )\right )}{20 \sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(231\)

[In]

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

[Out]

-1/20*((8*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-64*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+56*
sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-23*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(8*sin(1/2*d*x+1/2*c)^2-7)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2*2^(1/2))-9*(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(c
os(1/2*d*x+1/2*c),2*2^(1/2)))/(-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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {4 \, \sqrt {4 \, \cos \left (d x + c\right ) + 3} {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 7 i \, \sqrt {2} {\rm weierstrassPInverse}\left (-1, 1, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + \frac {1}{2}\right ) - 7 i \, \sqrt {2} {\rm weierstrassPInverse}\left (-1, 1, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + \frac {1}{2}\right ) + 18 i \, \sqrt {2} {\rm weierstrassZeta}\left (-1, 1, {\rm weierstrassPInverse}\left (-1, 1, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + \frac {1}{2}\right )\right ) - 18 i \, \sqrt {2} {\rm weierstrassZeta}\left (-1, 1, {\rm weierstrassPInverse}\left (-1, 1, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + \frac {1}{2}\right )\right )}{40 \, d} \]

[In]

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

[Out]

1/40*(4*sqrt(4*cos(d*x + c) + 3)*(cos(d*x + c) - 1)*sin(d*x + c) + 7*I*sqrt(2)*weierstrassPInverse(-1, 1, cos(
d*x + c) + I*sin(d*x + c) + 1/2) - 7*I*sqrt(2)*weierstrassPInverse(-1, 1, cos(d*x + c) - I*sin(d*x + c) + 1/2)
+ 18*I*sqrt(2)*weierstrassZeta(-1, 1, weierstrassPInverse(-1, 1, cos(d*x + c) + I*sin(d*x + c) + 1/2)) - 18*I
*sqrt(2)*weierstrassZeta(-1, 1, weierstrassPInverse(-1, 1, cos(d*x + c) - I*sin(d*x + c) + 1/2)))/d

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{3}}{\sqrt {4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{3}}{\sqrt {4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3}{\sqrt {4\,\cos \left (c+d\,x\right )+3}} \,d x \]

[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)