\(\int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx\) [521]

Optimal result

Integrand size = 23, antiderivative size = 98 \[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=-\frac {\sqrt {7} E\left (\frac {1}{2} (c+\pi +d x)|\frac {8}{7}\right )}{d}+\frac {3 \operatorname {EllipticF}\left (\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {4 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d} \] Output:

-EllipticE(cos(1/2*d*x+1/2*c),2/7*14^(1/2))*7^(1/2)/d+3/7*InverseJacobiAM( 
1/2*d*x+1/2*Pi+1/2*c,2/7*14^(1/2))*7^(1/2)/d+4/7*EllipticPi(cos(1/2*d*x+1/ 
2*c),2,2/7*14^(1/2))*7^(1/2)/d+(3-4*cos(d*x+c))^(1/2)*tan(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.60 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.82 \[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\frac {-\frac {42 \sqrt {-3+4 \cos (c+d x)} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),8\right )}{\sqrt {3-4 \cos (c+d x)}}-\frac {i \sqrt {7} \left (21 E\left (i \text {arcsinh}\left (\sqrt {3-4 \cos (c+d x)}\right )|-\frac {1}{7}\right )-12 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {3-4 \cos (c+d x)}\right ),-\frac {1}{7}\right )-8 \operatorname {EllipticPi}\left (-\frac {1}{3},i \text {arcsinh}\left (\sqrt {3-4 \cos (c+d x)}\right ),-\frac {1}{7}\right )\right ) \sin (c+d x)}{\sqrt {\sin ^2(c+d x)}}+21 \sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{21 d} \] Input:

Integrate[Sqrt[3 - 4*Cos[c + d*x]]*Sec[c + d*x]^2,x]
 

Output:

((-42*Sqrt[-3 + 4*Cos[c + d*x]]*EllipticPi[2, (c + d*x)/2, 8])/Sqrt[3 - 4* 
Cos[c + d*x]] - (I*Sqrt[7]*(21*EllipticE[I*ArcSinh[Sqrt[3 - 4*Cos[c + d*x] 
]], -1/7] - 12*EllipticF[I*ArcSinh[Sqrt[3 - 4*Cos[c + d*x]]], -1/7] - 8*El 
lipticPi[-1/3, I*ArcSinh[Sqrt[3 - 4*Cos[c + d*x]]], -1/7])*Sin[c + d*x])/S 
qrt[Sin[c + d*x]^2] + 21*Sqrt[3 - 4*Cos[c + d*x]]*Tan[c + d*x])/(21*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3275, 27, 3042, 3539, 3042, 3133, 3481, 3042, 3141, 3285}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\)

\(\Big \downarrow \) 3275

\(\displaystyle \int -\frac {2 \left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}}dx+\frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \int \frac {1-\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3539

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \left (\frac {1}{4} \int \sqrt {3-4 \cos (c+d x)}dx+\frac {1}{4} \int \frac {(4-3 \cos (c+d x)) \sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \left (\frac {1}{4} \int \sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {1}{4} \int \frac {4-3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\)

\(\Big \downarrow \) 3133

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \left (\frac {1}{4} \int \frac {4-3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{2 d}\right )\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \left (\frac {1}{4} \left (4 \int \frac {\sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}}dx-3 \int \frac {1}{\sqrt {3-4 \cos (c+d x)}}dx\right )+\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \left (\frac {1}{4} \left (4 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx-3 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{2 d}\right )\)

\(\Big \downarrow \) 3141

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \left (\frac {1}{4} \left (4 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {6 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}\right )+\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{2 d}\right )\)

\(\Big \downarrow \) 3285

\(\displaystyle \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{d}-2 \left (\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{2 d}+\frac {1}{4} \left (-\frac {6 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {8 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}\right )\right )\)

Input:

Int[Sqrt[3 - 4*Cos[c + d*x]]*Sec[c + d*x]^2,x]
 

Output:

-2*((Sqrt[7]*EllipticE[(c + Pi + d*x)/2, 8/7])/(2*d) + ((-6*EllipticF[(c + 
 Pi + d*x)/2, 8/7])/(Sqrt[7]*d) - (8*EllipticPi[2, (c + Pi + d*x)/2, 8/7]) 
/(Sqrt[7]*d))/4) + (Sqrt[3 - 4*Cos[c + d*x]]*Tan[c + d*x])/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3133

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 3141

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a - b]))*EllipticF[(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 3275

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x] 
)^(m + 1)*((c + d*Sin[e + f*x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m 
 + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^ 
(n - 1)*Simp[a*c*(m + 1) + b*d*n + (a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] 
 - b*d*(m + n + 2)*Sin[e + f*x]^2, 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] && LtQ[m, 
 -1] && LtQ[0, n, 1] && IntegersQ[2*m, 2*n]
 

rule 3285

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]
 

rule 3481

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

rule 3539

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(350\) vs. \(2(95)=190\).

Time = 4.51 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.58

method	result	size
default	\(-\frac {\sqrt {-\left (8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+\frac {3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )}{7 \sqrt {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )}{\sqrt {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \frac {2 \sqrt {14}}{7}\right )}{7 \sqrt {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7}\, d}\)	\(351\)

Input:

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

Output:

-(-(8*cos(1/2*d*x+1/2*c)^2-7)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-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)/(2*cos(1/2*d*x+ 
1/2*c)^2-1)+3/7*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^( 
1/2)/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2 
*d*x+1/2*c),2/7*14^(1/2))-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2 
*c)^2-7)^(1/2)/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti 
cE(cos(1/2*d*x+1/2*c),2/7*14^(1/2))+4/7*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(56*s 
in(1/2*d*x+1/2*c)^2-7)^(1/2)/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2) 
^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2)))/sin(1/2*d*x+1/2*c)/( 
-8*cos(1/2*d*x+1/2*c)^2+7)^(1/2)/d
 

Fricas [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

integral(sqrt(-4*cos(d*x + c) + 3)*sec(d*x + c)^2, x)
 

Sympy [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int \sqrt {3 - 4 \cos {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx \] Input:

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

Output:

Integral(sqrt(3 - 4*cos(c + d*x))*sec(c + d*x)**2, x)
 

Maxima [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

integrate(sqrt(-4*cos(d*x + c) + 3)*sec(d*x + c)^2, x)
 

Giac [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2} \,d x } \] Input:

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

Output:

integrate(sqrt(-4*cos(d*x + c) + 3)*sec(d*x + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int \frac {\sqrt {3-4\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int \sqrt {-4 \cos \left (d x +c \right )+3}\, \sec \left (d x +c \right )^{2}d x \] Input:

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

Output:

int(sqrt( - 4*cos(c + d*x) + 3)*sec(c + d*x)**2,x)