\(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\) [256]

Optimal result

Integrand size = 31, antiderivative size = 143 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b^2 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (a^2 A+2 A b^2-2 a b B\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {(A b-a B) \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d} \] Output:

-2*b^2*(A*b-B*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a 
-b)^(1/2)/(a+b)^(1/2)/d+1/2*(A*a^2+2*A*b^2-2*B*a*b)*arctanh(sin(d*x+c))/a^ 
3/d-(A*b-B*a)*tan(d*x+c)/a^2/d+1/2*A*sec(d*x+c)*tan(d*x+c)/a/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(143)=286\).

Time = 3.00 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.10 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {8 b^2 (A b-a B) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-2 \left (a^2 A+2 A b^2-2 a b B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (a^2 A+2 A b^2-2 a b B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (-A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (-A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{4 a^3 d} \] Input:

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

Output:

((8*b^2*(A*b - a*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/ 
Sqrt[-a^2 + b^2] - 2*(a^2*A + 2*A*b^2 - 2*a*b*B)*Log[Cos[(c + d*x)/2] - Si 
n[(c + d*x)/2]] + 2*(a^2*A + 2*A*b^2 - 2*a*b*B)*Log[Cos[(c + d*x)/2] + Sin 
[(c + d*x)/2]] + (a^2*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*a*(- 
(A*b) + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (a^ 
2*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a*(-(A*b) + a*B)*Sin[(c 
+ d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(4*a^3*d)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3479, 25, 3042, 3534, 25, 3042, 3480, 3042, 3138, 218, 4257}

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.

\(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

\(\displaystyle \frac {\int -\frac {\left (-A b \cos ^2(c+d x)-a A \cos (c+d x)+2 (A b-a B)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x)}{2 a d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-A b \cos ^2(c+d x)-a A \cos (c+d x)+2 (A b-a B)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a A \sin \left (c+d x+\frac {\pi }{2}\right )+2 (A b-a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {\left (A a^2-2 b B a+A b \cos (c+d x) a+2 A b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 (A b-a B) \tan (c+d x)}{a d}}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\int \frac {\left (A a^2-2 b B a+A b \cos (c+d x) a+2 A b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\int \frac {A a^2-2 b B a+A b \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}\)

\(\Big \downarrow \) 3480

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 A-2 a b B+2 A b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b^2 (A b-a B) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 A-2 a b B+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^2 (A b-a B) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 A-2 a b B+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 (A b-a B) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 A-2 a b B+2 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {A \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 (A b-a B) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2 A-2 a b B+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b^2 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\)

Input:

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

Output:

(A*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b^2*(A*b - a*B)*ArcTan[(Sq 
rt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + 
((a^2*A + 2*A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/(a*d))/a) + (2*(A*b - 
a*B)*Tan[c + d*x])/(a*d))/(2*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3479

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin 
[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 
1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e 
 + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 
2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) 
*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n 
}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat 
ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(I 
ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0]) 
))
 

rule 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((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/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3534

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

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-A a -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2} A -2 A \,b^{2}+2 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {2 b^{2} \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-A a -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2} A +2 A \,b^{2}-2 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}}{d}\)	\(229\)
default	\(\frac {\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-A a -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2} A -2 A \,b^{2}+2 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}-\frac {2 b^{2} \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-A a -2 A b +2 B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2} A +2 A \,b^{2}-2 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}}{d}\)	\(229\)
risch	\(-\frac {i \left (A a \,{\mathrm e}^{3 i \left (d x +c \right )}+2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-A a \,{\mathrm e}^{i \left (d x +c \right )}+2 A b -2 B a \right )}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{a^{2} d}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{a^{2} d}\)	\(524\)

Input:

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

Output:

1/d*(1/2*A/a/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(-A*a-2*A*b+2*B*a)/a^2/(tan(1/2* 
d*x+1/2*c)-1)+1/2/a^3*(-A*a^2-2*A*b^2+2*B*a*b)*ln(tan(1/2*d*x+1/2*c)-1)-2* 
b^2*(A*b-B*a)/a^3/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a- 
b)*(a+b))^(1/2))-1/2*A/a/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(-A*a-2*A*b+2*B*a)/a 
^2/(tan(1/2*d*x+1/2*c)+1)+1/2*(A*a^2+2*A*b^2-2*B*a*b)/a^3*ln(tan(1/2*d*x+1 
/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (129) = 258\).

Time = 2.27 (sec) , antiderivative size = 589, normalized size of antiderivative = 4.12 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (A a^{4} - 2 \, B a^{3} b + A a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{4} - 2 \, B a^{3} b + A a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{4} - A a^{2} b^{2} + 2 \, {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, \frac {4 \, {\left (B a b^{2} - A b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} + {\left (A a^{4} - 2 \, B a^{3} b + A a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{4} - 2 \, B a^{3} b + A a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{4} - A a^{2} b^{2} + 2 \, {\left (B a^{4} - A a^{3} b - B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="fri 
cas")
 

Output:

[1/4*(2*(B*a*b^2 - A*b^3)*sqrt(-a^2 + b^2)*cos(d*x + c)^2*log((2*a*b*cos(d 
*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c 
) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c 
) + a^2)) + (A*a^4 - 2*B*a^3*b + A*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4)*cos(d*x 
+ c)^2*log(sin(d*x + c) + 1) - (A*a^4 - 2*B*a^3*b + A*a^2*b^2 + 2*B*a*b^3 
- 2*A*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^4 - A*a^2*b^2 + 
2*(B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^ 
5 - a^3*b^2)*d*cos(d*x + c)^2), 1/4*(4*(B*a*b^2 - A*b^3)*sqrt(a^2 - b^2)*a 
rctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 
 + (A*a^4 - 2*B*a^3*b + A*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4)*cos(d*x + c)^2*lo 
g(sin(d*x + c) + 1) - (A*a^4 - 2*B*a^3*b + A*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4 
)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^4 - A*a^2*b^2 + 2*(B*a^4 
- A*a^3*b - B*a^2*b^2 + A*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b 
^2)*d*cos(d*x + c)^2)]
 

Sympy [F]

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

integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(a+b*cos(d*x+c)),x)
 

Output:

Integral((A + B*cos(c + d*x))*sec(c + d*x)**3/(a + b*cos(c + d*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="max 
ima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (129) = 258\).

Time = 0.20 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (A a^{2} - 2 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (A a^{2} - 2 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left (B a b^{2} - A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="gia 
c")
 

Output:

1/2*((A*a^2 - 2*B*a*b + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 
(A*a^2 - 2*B*a*b + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - 4*(B* 
a*b^2 - A*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan( 
-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt 
(a^2 - b^2)*a^3) + 2*(A*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*a*tan(1/2*d*x + 1/2 
*c)^3 + 2*A*b*tan(1/2*d*x + 1/2*c)^3 + A*a*tan(1/2*d*x + 1/2*c) + 2*B*a*ta 
n(1/2*d*x + 1/2*c) - 2*A*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 
- 1)^2*a^2))/d
 

Mupad [B] (verification not implemented)

Time = 45.91 (sec) , antiderivative size = 4051, normalized size of antiderivative = 28.33 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

int((A + B*cos(c + d*x))/(cos(c + d*x)^3*(a + b*cos(c + d*x))),x)
 

Output:

(B*a*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*b 
*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (A*a*sin 
(c + d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*a*atan((sin(c 
/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*d*(a^2 - b^2)*(cos(2*c + 2*d* 
x)/2 + 1/2)) + (B*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/( 
d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*b^2*atan((sin(c/2 + (d*x)/2 
)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2 
)) + (A*b^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(a^3*d*(a 
^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (B*b^3*atan((sin(c/2 + (d*x)/2)*1i 
)/cos(c/2 + (d*x)/2))*1i)/(a^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + 
 (A*b^3*sin(2*c + 2*d*x))/(2*a^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) 
 - (B*b^2*sin(2*c + 2*d*x))/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) 
 - (A*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)* 
1i)/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (B*b*atan((sin(c/2 + (d 
*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(d*(a^2 - b^2)*(cos(2* 
c + 2*d*x)/2 + 1/2)) - (A*b^2*sin(c + d*x))/(2*a*d*(a^2 - b^2)*(cos(2*c + 
2*d*x)/2 + 1/2)) + (A*b^3*atan(((A^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1 
/2) + 8*A^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*A^2*b^9*sin(c/2 + 
 (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/ 
2) + 8*A^2*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^4*b^5...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c)),x)
 

Output:

( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + log(tan((c + d*x)/2) - 1) 
+ log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - log(tan((c + d*x)/2) + 1) - 
sin(c + d*x))/(2*d*(sin(c + d*x)**2 - 1))