3.5.35 \(\int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [435]

3.5.35.1 Optimal result
3.5.35.2 Mathematica [C] (warning: unable to verify)
3.5.35.3 Rubi [A] (verified)
3.5.35.4 Maple [B] (verified)
3.5.35.5 Fricas [C] (verification not implemented)
3.5.35.6 Sympy [F(-1)]
3.5.35.7 Maxima [F]
3.5.35.8 Giac [F]
3.5.35.9 Mupad [B] (verification not implemented)

3.5.35.1 Optimal result

Integrand size = 41, antiderivative size = 101 \[ \int \frac {(a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 a (A-B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (3 A+3 B+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 a C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \]

output
-2*a*(A-B-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin 
(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*(3*A+3*B+C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/ 
cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2*a*A*sin(d*x+c 
)/d/cos(d*x+c)^(1/2)+2/3*a*C*sin(d*x+c)*cos(d*x+c)^(1/2)/d
 
3.5.35.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.41 (sec) , antiderivative size = 1173, normalized size of antiderivative = 11.61 \[ \int \frac {(a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \]

input
Integrate[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/C 
os[c + d*x]^(3/2),x]
 
output
a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(-1/2*((-2*A 
 + B + C + B*Cos[2*c] + C*Cos[2*c])*Csc[c]*Sec[c])/d + (C*Cos[d*x]*Sin[c]) 
/(3*d) + (C*Cos[c]*Sin[d*x])/(3*d) + (A*Sec[c]*Sec[c + d*x]*Sin[d*x])/d) - 
 (A*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x 
 - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[ 
1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - 
ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*Sqrt[1 + Cot[c]^ 
2]) - (B*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Si 
n[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]* 
Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d 
*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*Sqrt[1 + Co 
t[c]^2]) - (C*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4 
}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[ 
c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]* 
Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt 
[1 + Cot[c]^2]) + (A*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((Hype 
rgeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + 
ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[ 
d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[ 
c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1...
 
3.5.35.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 3510, 27, 3042, 3502, 27, 3042, 3227, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{3} \left (a (3 (A+B)+C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a (A-B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 a C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{3} \left (\frac {2 a (3 (A+B)+C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a (A-B-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 a C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

input
Int[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + 
 d*x]^(3/2),x]
 
output
((-6*a*(A - B - C)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(3*(A + B) + C)*Ell 
ipticF[(c + d*x)/2, 2])/d)/3 + (2*a*A*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) 
 + (2*a*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)
 

3.5.35.3.1 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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3502
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3510
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 
3.5.35.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(380\) vs. \(2(145)=290\).

Time = 12.50 (sec) , antiderivative size = 381, normalized size of antiderivative = 3.77

method result size
default \(\frac {2 a \left (-4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(381\)
parts \(\frac {2 \left (a A +B a \right ) \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d}+\frac {2 \left (B a +a C \right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 a A \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 a C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(528\)

input
int((a+cos(d*x+c)*a)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2),x,me 
thod=_RETURNVERBOSE)
 
output
2/3*a*(-4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*A*cos(1/2*d*x+1/2*c) 
*sin(1/2*d*x+1/2*c)^2-3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* 
c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2 
^(1/2))-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)* 
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* 
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*co 
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*(sin 
(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/ 
2*d*x+1/2*c),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2) 
/d
 
3.5.35.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.00 \[ \int \frac {(a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-i \, \sqrt {2} {\left (3 \, A + 3 \, B + C\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (3 \, A + 3 \, B + C\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (A - B - C\right )} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (A - B - C\right )} a \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (C a \cos \left (d x + c\right ) + 3 \, A a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \]

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2 
),x, algorithm="fricas")
 
output
1/3*(-I*sqrt(2)*(3*A + 3*B + C)*a*cos(d*x + c)*weierstrassPInverse(-4, 0, 
cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*(3*A + 3*B + C)*a*cos(d*x + c)* 
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(A 
 - B - C)*a*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
 cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(A - B - C)*a*cos(d*x + c)* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
 + c))) + 2*(C*a*cos(d*x + c) + 3*A*a)*sqrt(cos(d*x + c))*sin(d*x + c))/(d 
*cos(d*x + c))
 
3.5.35.6 Sympy [F(-1)]

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

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3 
/2),x)
 
output
Timed out
 
3.5.35.7 Maxima [F]

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

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2 
),x, algorithm="maxima")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/cos 
(d*x + c)^(3/2), x)
 
3.5.35.8 Giac [F]

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

input
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2 
),x, algorithm="giac")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/cos 
(d*x + c)^(3/2), x)
 
3.5.35.9 Mupad [B] (verification not implemented)

Time = 2.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.45 \[ \int \frac {(a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2\,C\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int(((a + a*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + 
 d*x)^(3/2),x)
 
output
(2*C*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3 
*d) + (2*A*a*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*a*ellipticE(c/2 + (d*x) 
/2, 2))/d + (2*B*a*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a*ellipticE(c/2 + 
 (d*x)/2, 2))/d + (2*A*a*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + 
d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))