3.1.78 \(\int \sqrt {a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [78]

3.1.78.1 Optimal result
3.1.78.2 Mathematica [A] (verified)
3.1.78.3 Rubi [A] (verified)
3.1.78.4 Maple [B] (verified)
3.1.78.5 Fricas [A] (verification not implemented)
3.1.78.6 Sympy [F]
3.1.78.7 Maxima [A] (verification not implemented)
3.1.78.8 Giac [A] (verification not implemented)
3.1.78.9 Mupad [F(-1)]

3.1.78.1 Optimal result

Integrand size = 33, antiderivative size = 96 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a C \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \]

output
2*A*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*a^(1/2)/d+2/3*a*C*s 
in(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*C*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2) 
/d
 
3.1.78.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+C \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{3 d} \]

input
Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x],x]
 
output
(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*A*ArcTanh[Sqrt[2]* 
Sin[(c + d*x)/2]] + C*(3*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2])))/(3*d)
 
3.1.78.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 3525, 27, 3042, 3460, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3525

\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (3 a A+a C \cos (c+d x)) \sec (c+d x)dx}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} (3 a A+a C \cos (c+d x)) \sec (c+d x)dx}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {2 a^2 C \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {6 a^2 A \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {6 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^2 C \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 C \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

input
Int[Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x],x]
 
output
(2*C*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((6*a^(3/2)*A*ArcTanh[ 
(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^2*C*Sin[c + d*x 
])/(d*Sqrt[a + a*Cos[c + d*x]]))/(3*a)
 

3.1.78.3.1 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 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 3460
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]
 

rule 3525
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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1 
)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2))   Int[(a + b*Sin[e + f* 
x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 
)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
 
3.1.78.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(241\) vs. \(2(82)=164\).

Time = 7.43 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.52

method result size
parts \(\frac {A \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )+\ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(242\)
default \(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-4 C \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A \ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +3 A \ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +6 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{3 \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(250\)

input
int((A+C*cos(d*x+c)^2)*sec(d*x+c)*(a+cos(d*x+c)*a)^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
A*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(ln(4/(2*cos(1 
/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*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))*(2^(1 
/2)*a*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)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+2/3*C*cos(1/2*d*x 
+1/2*c)*a*sin(1/2*d*x+1/2*c)*(1+2*cos(1/2*d*x+1/2*c)^2)*2^(1/2)/(a*cos(1/2 
*d*x+1/2*c)^2)^(1/2)/d
 
3.1.78.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.45 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 \, {\left (A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (C \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

input
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)*(a+a*cos(d*x+c))^(1/2),x, algorith 
m="fricas")
 
output
1/6*(3*(A*cos(d*x + c) + A)*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*(C*cos(d*x + c) + 2*C)*sqrt( 
a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c) + d)
 
3.1.78.6 Sympy [F]

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

input
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)*(a+a*cos(d*x+c))**(1/2),x)
 
output
Integral(sqrt(a*(cos(c + d*x) + 1))*(A + C*cos(c + d*x)**2)*sec(c + d*x), 
x)
 
3.1.78.7 Maxima [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {{\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{3 \, d} \]

input
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)*(a+a*cos(d*x+c))^(1/2),x, algorith 
m="maxima")
 
output
1/3*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*C*sqrt 
(a)/d
 
3.1.78.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.19 \[ \int \sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {\sqrt {2} {\left (8 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} A \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 12 \, C \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \]

input
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)*(a+a*cos(d*x+c))^(1/2),x, algorith 
m="giac")
 
output
-1/6*sqrt(2)*(8*C*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 3*sqr 
t(2)*A*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)))*sgn(cos(1/2*d*x + 1/2*c)) - 12*C*sgn(cos(1/2*d*x + 1/2* 
c))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
 
3.1.78.9 Mupad [F(-1)]

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

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