\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [110]

Optimal result

Integrand size = 35, antiderivative size = 243 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {(107 A+112 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 \sqrt {a} d}-\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {(21 A+16 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {(43 A+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}-\frac {A \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^3(c+d x) \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}} \] Output:

1/64*(107*A+112*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1 
/2)/d-2^(1/2)*(A+C)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c) 
)^(1/2))/a^(1/2)/d-1/64*(21*A+16*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/ 
96*(43*A+48*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/24*A*sec(d 
*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/4*A*sec(d*x+c)^3*tan(d*x+c)/ 
d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 2.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.72 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (768 (A+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)-6 \sqrt {2} (107 A+112 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)+(-364 A-192 C+(221 A+144 C) \cos (c+d x)-4 (43 A+48 C) \cos (2 (c+d x))+63 A \cos (3 (c+d x))+48 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{384 d \sqrt {a (1+\cos (c+d x))}} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5)/Sqrt[a + a*Cos[c + d*x]] 
,x]
 

Output:

-1/384*(Cos[(c + d*x)/2]*Sec[c + d*x]^4*(768*(A + C)*ArcTanh[Sin[(c + d*x) 
/2]]*Cos[c + d*x]^4 - 6*Sqrt[2]*(107*A + 112*C)*ArcTanh[Sqrt[2]*Sin[(c + d 
*x)/2]]*Cos[c + d*x]^4 + (-364*A - 192*C + (221*A + 144*C)*Cos[c + d*x] - 
4*(43*A + 48*C)*Cos[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] + 48*C*Cos[3*(c + 
 d*x)])*Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 1.76 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.14, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {3042, 3523, 27, 3042, 3463, 27, 3042, 3463, 27, 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.

\(\displaystyle \int \frac {\sec ^5(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a \cos (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5 \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 3523

\(\displaystyle \frac {\int -\frac {(a A-a (7 A+8 C) \cos (c+d x)) \sec ^4(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {(a A-a (7 A+8 C) \cos (c+d x)) \sec ^4(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{8 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {\left (a^2 (43 A+48 C)-5 a^2 A \cos (c+d x)\right ) \sec ^3(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^2 (43 A+48 C)-5 a^2 A \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^2 (43 A+48 C)-5 a^2 A \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{6 a}}{8 a}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {3 \left (a^3 (21 A+16 C)-a^3 (43 A+48 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}}{6 a}}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {\left (a^3 (21 A+16 C)-a^3 (43 A+48 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {a^3 (21 A+16 C)-a^3 (43 A+48 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {\int -\frac {\left (a^4 (107 A+112 C)-a^4 (21 A+16 C) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^4 (107 A+112 C)-a^4 (21 A+16 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^4 (107 A+112 C)-a^4 (21 A+16 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (107 A+112 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-128 a^4 (A+C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (107 A+112 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-128 a^4 (A+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (107 A+112 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {256 a^4 (A+C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (107 A+112 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {128 \sqrt {2} a^{7/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a^4 (107 A+112 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {128 \sqrt {2} a^{7/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (43 A+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A+16 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^{7/2} (107 A+112 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {128 \sqrt {2} a^{7/2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}}{8 a}\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5)/Sqrt[a + a*Cos[c + d*x]],x]
 

Output:

(A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d*Sqrt[a + a*Cos[c + d*x]]) - ((a*A*Sec 
[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) - ((a^2*(43*A + 4 
8*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) - (3*(-1/2* 
((2*a^(7/2)*(107*A + 112*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[ 
c + d*x]]])/d - (128*Sqrt[2]*a^(7/2)*(A + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x] 
)/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a + (a^3*(21*A + 16*C)*Tan[c + d 
*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/(4*a))/(6*a))/(8*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 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]
 

rule 3523

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1775\) vs. \(2(210)=420\).

Time = 0.54 (sec) , antiderivative size = 1776, normalized size of antiderivative = 7.31

method	result	size
parts	\(\text {Expression too large to display}\)	\(1776\)
default	\(\text {Expression too large to display}\)	\(2069\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+a*cos(d*x+c))^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/24*A*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-48*a*(-128*2^( 
1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))+1 
07*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)-a^(1 
/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))+107*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)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+ 
1/2*c)^2*a)^(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^8+48*(-21*a^(1/2)*2^(1/2)*(sin 
(1/2*d*x+1/2*c)^2*a)^(1/2)-256*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(s 
in(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a+214*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)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^( 
1/2)-2*a))*a+214*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)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a)*sin(1/2*d 
*x+1/2*c)^6+(9216*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/ 
2*c)^2*a)^(1/2)+a))*a+824*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-7 
704*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)-a^( 
1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a-7704*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)+a^(1/2)*2^(1/2)*(sin(1/2* 
d*x+1/2*c)^2*a)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^4+4*(-25*a^(1/2)*2^(1/2) 
*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-768*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/ 
2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a+642*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)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c...
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.28 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {3 \, {\left ({\left (107 \, A + 112 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (107 \, A + 112 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left (3 \, {\left (21 \, A + 16 \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (43 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, A \cos \left (d x + c\right ) - 48 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {384 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{5} + {\left (A + C\right )} a \cos \left (d x + c\right )^{4}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{768 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+a*cos(d*x+c))^(1/2),x, algori 
thm="fricas")
 

Output:

1/768*(3*((107*A + 112*C)*cos(d*x + c)^5 + (107*A + 112*C)*cos(d*x + c)^4) 
*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 + co 
s(d*x + c)^2)) - 4*(3*(21*A + 16*C)*cos(d*x + c)^3 - 2*(43*A + 48*C)*cos(d 
*x + c)^2 + 8*A*cos(d*x + c) - 48*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) 
 + 384*sqrt(2)*((A + C)*a*cos(d*x + c)^5 + (A + C)*a*cos(d*x + c)^4)*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)^5 + a*d*cos(d*x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**5/(a+a*cos(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+a*cos(d*x+c))^(1/2),x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+a*cos(d*x+c))^(1/2),x, algori 
thm="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. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^5\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^5*(a + a*cos(c + d*x))^(1/2)),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^5*(a + a*cos(c + d*x))^(1/2)), x)
 

Reduce [F]

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

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+a*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**5)/(co 
s(c + d*x) + 1),x)*c + int((sqrt(cos(c + d*x) + 1)*sec(c + d*x)**5)/(cos(c 
 + d*x) + 1),x)*a))/a