\(\int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx\) [237]

Optimal result

Integrand size = 23, antiderivative size = 134 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {1+\cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {1+\cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {1+\cos (c+d x)}} \] Output:

-2^(1/2)*arcsin(sin(d*x+c)/(1+cos(d*x+c)))/d+2/5*sin(d*x+c)/d/cos(d*x+c)^( 
5/2)/(1+cos(d*x+c))^(1/2)-2/15*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(1+cos(d*x+c) 
)^(1/2)+26/15*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 7.90 (sec) , antiderivative size = 1538, normalized size of antiderivative = 11.48 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

Integrate[1/(Cos[c + d*x]^(7/2)*Sqrt[1 + Cos[c + d*x]]),x]
 

Output:

(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^6*(4725*Sin[c/2 + (d*x)/2]^2 - 4 
8825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + 
 (d*x)/2]^8 + 655812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 
 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x) 
/2]^10 - 40*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11 
/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2 
]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2, 11/2, 
 Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 
 + 226656*Sin[c/2 + (d*x)/2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/2, Sin 
[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 4 
2048*Sin[c/2 + (d*x)/2]^16 + 440*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + 
 (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 4725*Ar 
cTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c 
/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/ 
2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^2*Sqrt[S 
in[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 291060*ArcTanh[Sqrt[S 
in[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*S 
qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*ArcTanh[S 
qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2 
]^6*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1458000*...
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 3463, 27, 3042, 3260, 223}

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.

\(\displaystyle \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x)+1}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\)

\(\Big \downarrow \) 3258

\(\displaystyle \frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}-\frac {1}{5} \int \frac {1-4 \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}-\frac {1}{5} \int \frac {1-4 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {1}{5} \left (-\frac {2}{3} \int -\frac {13-2 \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}dx-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {13-2 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}dx-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {13-2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (2 \int -\frac {15}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx+\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-15 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-15 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 3260

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {15 \sqrt {2} \int \frac {1}{\sqrt {1-\frac {\sin ^2(c+d x)}{(\cos (c+d x)+1)^2}}}d\left (-\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {15 \sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\)

Input:

Int[1/(Cos[c + d*x]^(7/2)*Sqrt[1 + Cos[c + d*x]]),x]
 

Output:

(2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[1 + Cos[c + d*x]]) + ((-2*Si 
n[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[1 + Cos[c + d*x]]) + ((-15*Sqrt[2 
]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d + (26*Sin[c + d*x])/(d*Sqrt[C 
os[c + d*x]]*Sqrt[1 + Cos[c + d*x]]))/3)/5
 

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 3042

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

rule 3258

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* 
b*(n + 1)*(c^2 - d^2))   Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c 
*(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[e + f*x]]), x], x] 
 /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] 
&& NeQ[c^2 - d^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3260

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-Sqrt[2]/(Sqrt[a]*f)   Subst[Int[1/Sqrt[1 - 
x^2], x], x, b*(Cos[e + f*x]/(a + b*Sin[e + f*x]))], x] /; FreeQ[{a, b, d, 
e, f}, x] && EqQ[a^2 - b^2, 0] && EqQ[d, a/b] && GtQ[a, 0]
 

rule 3463

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 5.86 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.37

method	result	size
default	\(\frac {4 \,\operatorname {csgn}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (52 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-27 \cos \left (d x +c \right )-10\right ) \sqrt {2}+\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-60 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+15 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}\right )\right )}{15 d \sqrt {\cos \left (d x +c \right )}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (d x +c \right )-1\right ) \left (\cos \left (d x +c \right )+1\right )}\)	\(183\)

Input:

int(1/cos(d*x+c)^(7/2)/(cos(d*x+c)+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

4/15/d*csgn(cos(1/2*d*x+1/2*c))/cos(d*x+c)^(1/2)/(4*cos(1/2*d*x+1/2*c)^4-2 
*cos(d*x+c)-1)/(cos(d*x+c)+1)*(cos(1/2*d*x+1/2*c)^2*sin(1/2*d*x+1/2*c)*(52 
*cos(1/2*d*x+1/2*c)^4-27*cos(d*x+c)-10)*2^(1/2)+(cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*(60*cos(1/2*d*x+1/2*c)^7-60*cos(1/2* 
d*x+1/2*c)^5+15*cos(1/2*d*x+1/2*c)^3))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\frac {2 \, {\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{4} + \sqrt {2} \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )}}\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/15*(2*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt(cos(d*x + c) + 1)*sqrt 
(cos(d*x + c))*sin(d*x + c) - 15*(sqrt(2)*cos(d*x + c)^4 + sqrt(2)*cos(d*x 
 + c)^3)*arctan(1/2*sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin( 
d*x + c)/(cos(d*x + c)^2 + cos(d*x + c))))/(d*cos(d*x + c)^4 + d*cos(d*x + 
 c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(1/cos(d*x+c)**(7/2)/(1+cos(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 989, normalized size of antiderivative = 7.38 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/15*(15*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqr 
t(2)*cos(2*d*x + 2*c) + sqrt(2))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 
+ 2*cos(2*d*x + 2*c) + 1)^(1/4)*arctan2(((abs(e^(I*d*x + I*c) + 1)^4 + cos 
(d*x + c)^4 + sin(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos( 
d*x + c) + 1)*abs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + 
 c)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x 
+ c) + 1)^(1/4)*sin(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I 
*d*x + I*c) + 1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x 
 + c)^2 - 2*cos(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sin(d*x + c)) 
/abs(e^(I*d*x + I*c) + 1), ((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + 
 sin(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1) 
*abs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos 
(d*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/ 
4)*cos(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 
 1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*c 
os(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + cos(d*x + c) - 1)/abs(e^(I 
*d*x + I*c) + 1)) - 26*(cos(2*d*x + 2*c)^2*sin(d*x + c) + sin(2*d*x + 2*c) 
^2*sin(d*x + c) + 2*cos(2*d*x + 2*c)*sin(d*x + c) + sin(d*x + c))*cos(1/2* 
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 24*cos(5/2*arctan2(sin( 
2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - 24*(cos(d*x + c) - ...
 

Giac [F]

\[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {\cos \left (d x + c\right ) + 1} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(cos(d*x + c) + 1)*cos(d*x + c)^(7/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \] Input:

int(1/(cos(c + d*x)^(7/2)*(cos(c + d*x) + 1)^(1/2)),x)
 

Output:

int(1/(cos(c + d*x)^(7/2)*(cos(c + d*x) + 1)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}+\cos \left (d x +c \right )^{4}}d x \] Input:

int(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x)
 

Output:

int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**5 + cos(c + 
 d*x)**4),x)