\(\int \frac {\sec ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx\) [553]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{3 d}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{3 \sqrt {7} d}+\frac {\sqrt {7} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{3 d}-\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {\sqrt {3+4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d} \] Output:

1/3*EllipticE(sin(1/2*d*x+1/2*c),2/7*14^(1/2))*7^(1/2)/d-1/21*InverseJacob 
iAM(1/2*d*x+1/2*c,2/7*14^(1/2))*7^(1/2)/d+1/3*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+1/6*(3 
+4*cos(d*x+c))^(1/2)*sec(d*x+c)*tan(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.59 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {3+4 \cos (c+d x)}} \, dx=\frac {\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7}}+\frac {18 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7}}-\frac {2 i \left (21 E\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right )|-\frac {1}{7}\right )-12 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )-8 \operatorname {EllipticPi}\left (-\frac {1}{3},i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )\right ) \sin (c+d x)}{3 \sqrt {7} \sqrt {\sin ^2(c+d x)}}-(-1+2 \cos (c+d x)) \sqrt {3+4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d} \] Input:

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

Output:

((4*EllipticF[(c + d*x)/2, 8/7])/Sqrt[7] + (18*EllipticPi[2, (c + d*x)/2, 
8/7])/Sqrt[7] - (((2*I)/3)*(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])/( 
Sqrt[7]*Sqrt[Sin[c + d*x]^2]) - (-1 + 2*Cos[c + d*x])*Sqrt[3 + 4*Cos[c + d 
*x]]*Sec[c + d*x]*Tan[c + d*x])/(6*d)
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3281, 25, 3042, 3534, 27, 3042, 3538, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

\(\displaystyle \frac {1}{6} \int -\frac {\left (-2 \cos ^2(c+d x)-3 \cos (c+d x)+6\right ) \sec ^2(c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}-\frac {1}{6} \int \frac {\left (-2 \cos ^2(c+d x)-3 \cos (c+d x)+6\right ) \sec ^2(c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}-\frac {1}{6} \int \frac {-2 \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 \sin \left (c+d x+\frac {\pi }{2}\right )+6}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {1}{6} \left (-\frac {1}{3} \int -\frac {3 \left (4 \cos ^2(c+d x)+2 \cos (c+d x)+7\right ) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx-\frac {2 \sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\int \frac {\left (4 \cos ^2(c+d x)+2 \cos (c+d x)+7\right ) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx-\frac {2 \sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\int \frac {4 \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \sin \left (c+d x+\frac {\pi }{2}\right )+7}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx-\frac {2 \sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {1}{6} \left (\int \sqrt {4 \cos (c+d x)+3}dx-\frac {1}{4} \int -\frac {4 (7-\cos (c+d x)) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx-\frac {2 \sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\int \sqrt {4 \cos (c+d x)+3}dx+\int \frac {(7-\cos (c+d x)) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx-\frac {2 \sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\int \frac {7-\sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx+\int \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}dx-\frac {2 \sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x) \sec (c+d x)}{6 d}\)

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

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 3534
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*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[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3538
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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 - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, 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. \(407\) vs. \(2(122)=244\).

Time = 1.89 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.98

method result size
default \(-\frac {\sqrt {-\left (1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{3 \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{3 \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, 2 \sqrt {2}\right )}{3 \sqrt {-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(408\)

Input:

int(sec(d*x+c)^3/(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)*(-1/3*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)^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)-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)*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)*EllipticE(cos(1/2*d*x+1/2*c),2*2^(1/2) 
)-7/3*(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*si 
n(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]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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