\(\int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx\) [552]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{3 d}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {4 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{3 \sqrt {7} d}+\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d} \] Output:

-1/3*EllipticE(sin(1/2*d*x+1/2*c),2/7*14^(1/2))*7^(1/2)/d+1/7*InverseJacob 
iAM(1/2*d*x+1/2*c,2/7*14^(1/2))*7^(1/2)/d-4/21*EllipticPi(sin(1/2*d*x+1/2* 
c),2,2/7*14^(1/2))*7^(1/2)/d+1/3*(3+4*cos(d*x+c))^(1/2)*tan(d*x+c)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.56 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {-\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7}}+\frac {i \left (21 E\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right )|-\frac {1}{7}\right )-12 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )-8 \operatorname {EllipticPi}\left (-\frac {1}{3},i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )\right ) \sin (c+d x)}{3 \sqrt {7} \sqrt {\sin ^2(c+d x)}}+\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d} \] Input:

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

Output:

((-6*EllipticPi[2, (c + d*x)/2, 8/7])/Sqrt[7] + ((I/3)*(21*EllipticE[I*Arc 
Sinh[Sqrt[3 + 4*Cos[c + d*x]]], -1/7] - 12*EllipticF[I*ArcSinh[Sqrt[3 + 4* 
Cos[c + d*x]]], -1/7] - 8*EllipticPi[-1/3, I*ArcSinh[Sqrt[3 + 4*Cos[c + d* 
x]]], -1/7])*Sin[c + d*x])/(Sqrt[7]*Sqrt[Sin[c + d*x]^2]) + Sqrt[3 + 4*Cos 
[c + d*x]]*Tan[c + d*x])/(3*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3281, 27, 3042, 3539, 25, 3042, 3132, 3481, 3042, 3140, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3539

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3132

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 3140

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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[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] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

rule 3284

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. \(349\) vs. \(2(93)=186\).

Time = 1.41 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.47

method	result	size
default	\(-\frac {\sqrt {-\left (1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{\sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{3 \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, 2 \sqrt {2}\right )}{3 \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(350\)

Input:

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

Output:

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

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {4\,\cos \left (c+d\,x\right )+3}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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