\(\int \frac {\sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [129]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {\sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}} \] Output:

7/4*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d-2^(1/2)*a 
rctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d-1/ 
4*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/2*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos 
(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.18 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.45 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-7 \sqrt {2} \log \left (i-\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )+7 \sqrt {2} \log \left (i+\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )+16 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )+\cos (2 (c+d x)) \left (-7 \sqrt {2} \log \left (i-\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )+7 \sqrt {2} \log \left (i+\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )+16 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )-16 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )\right )-16 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )+20 \sin \left (\frac {1}{2} (c+d x)\right )-4 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{16 d \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sec[c + d*x]^2*(-7*Sqrt[2]*Log[I - Sqrt[2]*E^((I/2)*(c + 
 d*x)) - I*E^(I*(c + d*x))] + 7*Sqrt[2]*Log[I + Sqrt[2]*E^((I/2)*(c + d*x) 
) - I*E^(I*(c + d*x))] + 16*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] + Cos 
[2*(c + d*x)]*(-7*Sqrt[2]*Log[I - Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c 
+ d*x))] + 7*Sqrt[2]*Log[I + Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c + d*x 
))] + 16*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] - 16*Log[Cos[(c + d*x)/4 
] + Sin[(c + d*x)/4]]) - 16*Log[Cos[(c + d*x)/4] + Sin[(c + d*x)/4]] + 20* 
Sin[(c + d*x)/2] - 4*Sin[(3*(c + d*x))/2]))/(16*d*Sqrt[a*(1 + Cos[c + d*x] 
)])
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 3464, 3042, 3128, 219, 3252, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {7 a \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-8 a^2 \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {16 a^2 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+7 a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {7 a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {8 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {14 a^2 \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {14 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

Input:

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

Output:

(Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) - (-1/2*((14*a^ 
(3/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (8*Sqr 
t[2]*a^(3/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d* 
x]])])/d)/a + (a*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a)
 

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 219

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

rule 3042

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

rule 3128

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

rule 3252

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

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 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])
 

rule 3464

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A 
*b - a*B)/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c 
- A*d)/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], 
 x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[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. \(676\) vs. \(2(122)=244\).

Time = 1.07 (sec) , antiderivative size = 677, normalized size of antiderivative = 4.61

method	result	size
default	\(-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-4 a \left (-8 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right )+7 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )+7 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-32 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+28 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +28 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-7 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a -7 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right )}{2 a^{\frac {3}{2}} \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right )^{2} \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right )^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(677\)

Input:

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

Output:

-1/2*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*a*(-8*2^(1/2)*l 
n(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))+7*ln(4/ 
(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*si 
n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))+7*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1 
/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)* 
a^(1/2)-2*a)))*sin(1/2*d*x+1/2*c)^4+(-32*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/2* 
d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a-4*2^(1/2)*(a*sin(1/2*d*x+1/2* 
c)^2)^(1/2)*a^(1/2)+28*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos 
(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+28* 
ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+2^(1/2)* 
(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+8*2^( 
1/2)*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a 
-2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-7*ln(-4/(2*cos(1/2*d*x+1 
/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c) 
^2)^(1/2)*a^(1/2)-2*a))*a-7*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2) 
*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a 
)/a^(3/2)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^ 
2/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (122) = 244\).

Time = 0.09 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.71 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {7 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {8 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:

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

Output:

1/16*(7*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 
7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2) 
*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 4*sqrt(a*cos(d*x 
 + c) + a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*sqrt(2)*(a*cos(d*x + c)^3 + 
 a*cos(d*x + c)^2)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + 
a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x 
+ c) + 1))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
 

Sympy [F]

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

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

Output:

Integral(sec(c + d*x)**3/sqrt(a*(cos(c + d*x) + 1)), x)
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*int((sqrt(cos(c + d*x) + 1)*sec(c + d*x)**3)/(cos(c + d*x) + 1),x 
))/a