\(\int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [136]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}-\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}} \] Output:

2*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-5/4*arctanh 
(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/a^(3/2)/d- 
1/2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.14 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.99 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (-4 \sqrt {2} \log \left (i-\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )+4 \sqrt {2} \log \left (i+\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )+10 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )-10 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )-\frac {1}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}+\frac {1}{\left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}\right )}{2 d (a (1+\cos (c+d x)))^{3/2}} \] Input:

Integrate[Sec[c + d*x]/(a + a*Cos[c + d*x])^(3/2),x]
 

Output:

(Cos[(c + d*x)/2]^3*(-4*Sqrt[2]*Log[I - Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^ 
(I*(c + d*x))] + 4*Sqrt[2]*Log[I + Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c 
 + d*x))] + 10*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] - 10*Log[Cos[(c + 
d*x)/4] + Sin[(c + d*x)/4]] - (Cos[(c + d*x)/4] - Sin[(c + d*x)/4])^(-2) + 
 (Cos[(c + d*x)/4] + Sin[(c + d*x)/4])^(-2)))/(2*d*(a*(1 + Cos[c + d*x]))^ 
(3/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3245, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3245

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

\(\displaystyle \frac {4 \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-5 a \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {10 a \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+4 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{4 a^2}-\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {4 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {5 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {-\frac {8 a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {5 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {8 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {5 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

Input:

Int[Sec[c + d*x]/(a + a*Cos[c + d*x])^(3/2),x]
 

Output:

((8*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - 
(5*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[ 
c + d*x]])])/d)/(4*a^2) - Sin[c + d*x]/(2*d*(a + a*Cos[c + d*x])^(3/2))
 

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 3245

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*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( 
a*(2*m + 1)*(b*c - a*d))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + 
f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (Intege 
rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 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 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. \(289\) vs. \(2(93)=186\).

Time = 0.97 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.54

method	result	size
default	\(-\frac {\sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (5 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -4 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -4 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{4 a^{\frac {5}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(290\)

Input:

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

Output:

-1/4/a^(5/2)/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*2^(1/2)* 
ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*co 
s(1/2*d*x+1/2*c)^2*a-4*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))*cos(1 
/2*d*x+1/2*c)^2*a-4*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))*cos(1/2*d 
*x+1/2*c)^2*a+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(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. 254 vs. \(2 (93) = 186\).

Time = 0.10 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.23 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {5 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

1/8*(5*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d 
*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a* 
cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(cos(d*x + 
c)^2 + 2*cos(d*x + c) + 1)*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)*sin( 
d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.20 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {4 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a} + \frac {5 \, \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} - \frac {5 \, \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} - \frac {2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}\right )}}{8 \, \sqrt {a} d \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:

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

Output:

-1/8*sqrt(2)*(4*sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2 
*sqrt(2) + 4*sin(1/2*d*x + 1/2*c)))/a + 5*log(sin(1/2*d*x + 1/2*c) + 1)/a 
- 5*log(-sin(1/2*d*x + 1/2*c) + 1)/a - 2*sin(1/2*d*x + 1/2*c)/((sin(1/2*d* 
x + 1/2*c)^2 - 1)*a))/(sqrt(a)*d*sgn(cos(1/2*d*x + 1/2*c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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