3.1.87 \(\int (a+a \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [87]

3.1.87.1 Optimal result
3.1.87.2 Mathematica [A] (verified)
3.1.87.3 Rubi [A] (verified)
3.1.87.4 Maple [B] (verified)
3.1.87.5 Fricas [A] (verification not implemented)
3.1.87.6 Sympy [F(-1)]
3.1.87.7 Maxima [B] (verification not implemented)
3.1.87.8 Giac [A] (verification not implemented)
3.1.87.9 Mupad [F(-1)]

3.1.87.1 Optimal result

Integrand size = 35, antiderivative size = 136 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {3 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}-\frac {a^2 (3 A-8 C) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}-\frac {a (3 A-2 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d} \]

output
3*a^(3/2)*A*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d-1/3*a^2*( 
3*A-8*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/3*a*(3*A-2*C)*sin(d*x+c)*(a 
+a*cos(d*x+c))^(1/2)/d+A*(a+a*cos(d*x+c))^(3/2)*tan(d*x+c)/d
 
3.1.87.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.78 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (9 \sqrt {2} A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x)+2 (3 A+C+10 C \cos (c+d x)+C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \]

input
Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2 
,x]
 
output
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]*(9*Sqrt[2]*A*A 
rcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x] + 2*(3*A + C + 10*C*Cos[c + 
d*x] + C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(6*d)
 
3.1.87.3 Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3523, 27, 3042, 3455, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

\(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (3 a A-a (3 A-2 C) \cos (c+d x)) \sec (c+d x)dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {2}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (9 a^2 A-a^2 (3 A-8 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \sqrt {\cos (c+d x) a+a} \left (9 a^2 A-a^2 (3 A-8 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (9 a^2 A-a^2 (3 A-8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {\frac {1}{3} \left (9 a^2 A \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-\frac {2 a^3 (3 A-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (9 a^2 A \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^3 (3 A-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {1}{3} \left (-\frac {18 a^3 A \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 a^3 (3 A-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{3} \left (\frac {18 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^3 (3 A-8 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^2 (3 A-2 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}\)

input
Int[(a + a*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]
 
output
((-2*a^2*(3*A - 2*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((18*a 
^(5/2)*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (2* 
a^3*(3*A - 8*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/3)/(2*a) + (A* 
(a + a*Cos[c + d*x])^(3/2)*Tan[c + d*x])/d
 

3.1.87.3.1 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 3455
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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], 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] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 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])
 
3.1.87.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(444\) vs. \(2(120)=240\).

Time = 7.00 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.27

method result size
parts \(\frac {A \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-6 a \left (\ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )+\ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+3 \ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +3 \ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \right )}{\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 C \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(445\)
default \(\frac {\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 (32 C \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 A \ln \left (\frac {2 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+4 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \sqrt {2}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -18 A \ln \left (-\frac {2 \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 ) \sqrt {2}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -112 C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+4 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +9 A \sqrt {2}\, \ln \left (-\frac {2 \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 +12 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+48 C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right ) \sqrt {2}}{6 \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 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(501\)

input
int((a+cos(d*x+c)*a)^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x,method=_RETUR 
NVERBOSE)
 
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)*(-6*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))+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)^2+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a 
^(1/2)+3*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))*a+3*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))*a)/(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+4 
/3*C*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)*(cos(1/2*d*x+1/2*c)^2+2)*2^ 
(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 
3.1.87.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.28 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {9 \, {\left (A a \cos \left (d x + c\right )^{2} + A a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (2 \, C a \cos \left (d x + c\right )^{2} + 10 \, C a \cos \left (d x + c\right ) + 3 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \]

input
integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algori 
thm="fricas")
 
output
1/12*(9*(A*a*cos(d*x + c)^2 + A*a*cos(d*x + c))*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*(2*C*a*c 
os(d*x + c)^2 + 10*C*a*cos(d*x + c) + 3*A*a)*sqrt(a*cos(d*x + c) + a)*sin( 
d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))
 
3.1.87.6 Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \]

input
integrate((a+a*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)
 
output
Timed out
 
3.1.87.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1354 vs. \(2 (120) = 240\).

Time = 0.60 (sec) , antiderivative size = 1354, normalized size of antiderivative = 9.96 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Too large to display} \]

input
integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algori 
thm="maxima")
 
output
1/12*(4*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c) 
)*C*sqrt(a) - 3*(2*sqrt(2)*a*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 6*sqr 
t(2)*a*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + (2*sqrt(2)*a*sin(3/2*d*x + 
3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 3*a*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) + 3*a*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) - 3*a*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) + 3 
*a*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(2*d*x + 2*c)^ 
2 + (2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 
 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*c 
os(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*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) - 3*a*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) + 3*a*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(2*d*x + 2*c)^2 - 4*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 4...
 
3.1.87.8 Giac [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {\sqrt {2} {\left (16 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, \sqrt {2} A a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 48 \, C a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {12 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{12 \, d} \]

input
integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algori 
thm="giac")
 
output
-1/12*sqrt(2)*(16*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 9 
*sqrt(2)*A*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)) - 48*C*a*sgn(cos(1/2*d* 
x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 12*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/ 
2*d*x + 1/2*c)/(2*sin(1/2*d*x + 1/2*c)^2 - 1))*sqrt(a)/d
 
3.1.87.9 Mupad [F(-1)]

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

input
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(3/2))/cos(c + d*x)^2,x)
 
output
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(3/2))/cos(c + d*x)^2, x)