\(\int \sqrt {3+4 \cos (c+d x)} \sec (c+d x) \, dx\) [513]

Optimal result

Integrand size = 21, antiderivative size = 48 \[ \int \sqrt {3+4 \cos (c+d x)} \sec (c+d x) \, dx=\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d} \]

[Out]

8/7*(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)+6/7*(
cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2/7*14^(1/2))/d*7^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2882, 2740, 2884} \[ \int \sqrt {3+4 \cos (c+d x)} \sec (c+d x) \, dx=\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d} \]

[In]

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

[Out]

(8*EllipticF[(c + d*x)/2, 8/7])/(Sqrt[7]*d) + (6*EllipticPi[2, (c + d*x)/2, 8/7])/(Sqrt[7]*d)

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 2882

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

Rule 2884

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

Rubi steps \begin{align*} \text {integral}& = 3 \int \frac {\sec (c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx+4 \int \frac {1}{\sqrt {3+4 \cos (c+d x)}} \, dx \\ & = \frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \sqrt {3+4 \cos (c+d x)} \sec (c+d x) \, dx=\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )+6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d} \]

[In]

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

[Out]

(8*EllipticF[(c + d*x)/2, 8/7] + 6*EllipticPi[2, (c + d*x)/2, 8/7])/(Sqrt[7]*d)

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 3.29

method	result	size
default	\(-\frac {2 \sqrt {\left (8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )-3 \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, 2 \sqrt {2}\right )\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(158\)

[In]

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

[Out]

-2*((8*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-8*cos(1/2*d*x+1/2*c
)^2)^(1/2)*(4*EllipticF(cos(1/2*d*x+1/2*c),2*2^(1/2))-3*EllipticPi(cos(1/2*d*x+1/2*c),2,2*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 [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(4*cos(c + d*x) + 3)*sec(c + d*x), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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