\(\int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [128]

Optimal result

Integrand size = 23, antiderivative size = 108 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {\tan (c+d x)}{d \sqrt {a+a \cos (c+d x)}} \] Output:

-arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d+2^(1/2)*arct 
anh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d+tan(d 
*x+c)/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.49 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.78 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} \log \left (i-\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )-\sqrt {2} \log \left (i+\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )-4 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )+4 \sec (c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

(Cos[(c + d*x)/2]*(Sqrt[2]*Log[I - Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c 
 + d*x))] - Sqrt[2]*Log[I + Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c + d*x) 
)] - 4*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] + 4*Log[Cos[(c + d*x)/4] + 
 Sin[(c + d*x)/4]] + 4*Sec[c + d*x]*Sin[(c + d*x)/2]))/(2*d*Sqrt[a*(1 + Co 
s[c + d*x])])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3258, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\)

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\)

Input:

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

Output:

-1/2*((2*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]) 
/d - (2*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a 
*Cos[c + d*x]])])/d)/a + Tan[c + d*x]/(d*Sqrt[a + a*Cos[c + d*x]])
 

Defintions of rubi rules used

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 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. \(469\) vs. \(2(91)=182\).

Time = 1.07 (sec) , antiderivative size = 470, normalized size of antiderivative = 4.35

method	result	size
default	\(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-2 a \left (2 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right )-\ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )-\ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -\ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a -\ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{a^{\frac {3}{2}} \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(470\)

Input:

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

Output:

cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*a*(2*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))-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))-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)^2+2*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-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-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+2*2^(1/2)*(a*sin(1/2* 
d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(3/2)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos 
(1/2*d*x+1/2*c)+2^(1/2))/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. 236 vs. \(2 (91) = 182\).

Time = 0.09 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.19 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sec(c + d*x)**2/sqrt(a*(cos(c + d*x) + 1)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18435 vs. \(2 (91) = 182\).

Time = 0.35 (sec) , antiderivative size = 18435, normalized size of antiderivative = 170.69 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*((2*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si 
n(1/2*d*x + 1/2*c) + 1) - 2*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d 
*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - log(2*cos(1/2*d*x + 1/2*c)^2 
 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*s 
in(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) 
 + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2 
)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1 
/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2 
*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(d*x + c)^4 + (2*sqrt(2)*log 
(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) 
+ 1) - 2*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*s 
in(1/2*d*x + 1/2*c) + 1) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) 
 + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2 
)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1 
/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2 
*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 
 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin 
(1/2*d*x + 1/2*c) + 2))*sin(d*x + c)^4 + 4*sqrt(2)*cos(1/2*d*x + 1/2*c)...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^2/(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. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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