\(\int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [295]

Optimal result

Integrand size = 34, antiderivative size = 123 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b^3 B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (a^2+2 b^2\right ) B \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {b B \tan (c+d x)}{a^2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.94 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {B \left (\frac {8 b^3 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-4 a b \tan (c+d x)\right )}{4 a^3 d} \] Input:

Integrate[((a*B + b*B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + b*Cos[c + d*x])^2 
,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.382, Rules used = {2011, 3042, 3281, 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.

\(\displaystyle \int \frac {\sec ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

\(\displaystyle B \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

\(\displaystyle B \left (\frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \sec (c+d x)dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{2 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle B \left (\frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\right )\)

\(\Big \downarrow \) 3138

\(\displaystyle B \left (\frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^3 \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{2 a}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle B \left (\frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^3 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\right )\)

\(\Big \downarrow \) 4257

\(\displaystyle B \left (\frac {\tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 b \tan (c+d x)}{a d}-\frac {\frac {\left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b^3 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\right )\)

Input:

Int[((a*B + b*B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + b*Cos[c + d*x])^2,x]
 

Output:

B*((Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-(((-4*b^3*ArcTan[(Sqrt[a - b]*T 
an[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 + 2*b 
^2)*ArcTanh[Sin[c + d*x]])/(a*d))/a) + (2*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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[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 = 3.14 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.58

method	result	size
derivativedivides	\(\frac {2 B \left (-\frac {b^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a -2 b}{4 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 a^{3}}+\frac {1}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{4 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 a^{3}}\right )}{d}\)	\(194\)
default	\(\frac {2 B \left (-\frac {b^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-a -2 b}{4 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 a^{3}}+\frac {1}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{4 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 a^{3}}\right )}{d}\)	\(194\)
risch	\(-\frac {i B \left ({\mathrm e}^{3 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{a^{3} d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{a^{3} d}-\frac {b^{3} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b^{3} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}\)	\(306\)

Input:

int((B*a+b*B*cos(d*x+c))*sec(d*x+c)^3/(a+cos(d*x+c)*b)^2,x,method=_RETURNV 
ERBOSE)
 

Output:

2/d*B*(-b^3/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))-1/4/a/(tan(1/2*d*x+1/2*c)+1)^2-1/4*(-a-2*b)/a^2/(tan(1/2*d* 
x+1/2*c)+1)+1/4*(a^2+2*b^2)/a^3*ln(tan(1/2*d*x+1/2*c)+1)+1/4/a/(tan(1/2*d* 
x+1/2*c)-1)^2-1/4*(-a-2*b)/a^2/(tan(1/2*d*x+1/2*c)-1)+1/4/a^3*(-a^2-2*b^2) 
*ln(tan(1/2*d*x+1/2*c)-1))
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 487, normalized size of antiderivative = 3.96 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {2 \, \sqrt {-a^{2} + b^{2}} B b^{3} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{4} - B a^{2} b^{2} - 2 \, {\left (B a^{3} b - B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, \sqrt {a^{2} - b^{2}} B b^{3} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{4} - B a^{2} b^{2} - 2 \, {\left (B a^{3} b - B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \] Input:

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c))^2,x, algorith 
m="fricas")
 

Output:

[-1/4*(2*sqrt(-a^2 + b^2)*B*b^3*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)) - 
 (B*a^4 + B*a^2*b^2 - 2*B*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (B*a 
^4 + B*a^2*b^2 - 2*B*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(B*a^4 
 - B*a^2*b^2 - 2*(B*a^3*b - B*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*sqrt(a^2 - b^2)*B*b^3*arctan(-(a*cos(d* 
x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 - (B*a^4 + B*a^ 
2*b^2 - 2*B*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (B*a^4 + B*a^2*b^2 
 - 2*B*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(B*a^4 - B*a^2*b^2 - 
 2*(B*a^3*b - B*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]

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

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)**3/(a+b*cos(d*x+c))**2,x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c))^2,x, algorith 
m="maxima")
 

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. 221 vs. \(2 (110) = 220\).

Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.80 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} B b^{3}}{\sqrt {a^{2} - b^{2}} a^{3}} - \frac {{\left (B a^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {{\left (B a^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \] Input:

integrate((a*B+b*B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c))^2,x, algorith 
m="giac")
 

Output:

-1/2*(4*(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)))*B*b^3/(sqrt(a^ 
2 - b^2)*a^3) - (B*a^2 + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + 
 (B*a^2 + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - 2*(B*a*tan(1/2 
*d*x + 1/2*c)^3 + 2*B*b*tan(1/2*d*x + 1/2*c)^3 + B*a*tan(1/2*d*x + 1/2*c) 
- 2*B*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 = 25.22 (sec) , antiderivative size = 1099, normalized size of antiderivative = 8.93 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int((B*a + B*b*cos(c + d*x))/(cos(c + d*x)^3*(a + b*cos(c + d*x))^2),x)
 

Output:

((B*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (B*b^2*sin(c + d 
*x))/2 + (B*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d 
*x))/2)/(a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (a*((B*sin(c + d*x) 
)/2 + (B*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (B*atanh(sin(c/ 
2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2))/(d*(a^2 - b^2)*(cos 
(2*c + 2*d*x)/2 + 1/2)) - (B*b*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos(2*c 
 + 2*d*x)/2 + 1/2)) - (B*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))) 
/(a^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (B*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^3*atan(((a^9*sin( 
c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3 
/2) - 8*b^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)^(1/2) + 3*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 
 3*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 2*a^6*b^3*sin(c/2 + (d*x 
)/2)*(b^2 - a^2)^(1/2) + 2*a^7*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 
a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 
 - a^3)*(a^7 - 3*a^3*b^4 + 2*a^5*b^2)))*1i)/(a^3*d*(b^2 - a^2)^(1/2)*(cos( 
2*c + 2*d*x)/2 + 1/2)) - (B*b^3*atan(((a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^ 
(1/2) + 8*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*b^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)^(1/2) + 3 
*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 3*a^5*b^4*sin(c/2 + (d*...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 480, normalized size of antiderivative = 3.90 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {b \left (-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} b^{3}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) b^{3}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}-\sin \left (d x +c \right ) a^{4}+\sin \left (d x +c \right ) a^{2} b^{2}\right )}{2 a^{3} d \left (\sin \left (d x +c \right )^{2} a^{2}-\sin \left (d x +c \right )^{2} b^{2}-a^{2}+b^{2}\right )} \] Input:

int((a*B+b*B*cos(d*x+c))*sec(d*x+c)^3/(a+b*cos(d*x+c))^2,x)
 

Output:

(b*( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ 
sqrt(a**2 - b**2))*sin(c + d*x)**2*b**3 + 4*sqrt(a**2 - b**2)*atan((tan((c 
 + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*b**3 + 2*cos(c + d*x 
)*sin(c + d*x)*a**3*b - 2*cos(c + d*x)*sin(c + d*x)*a*b**3 - log(tan((c + 
d*x)/2) - 1)*sin(c + d*x)**2*a**4 - log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**2*a**2*b**2 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**4 + log(tan 
((c + d*x)/2) - 1)*a**4 + log(tan((c + d*x)/2) - 1)*a**2*b**2 - 2*log(tan( 
(c + d*x)/2) - 1)*b**4 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 + 
log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 - 2*log(tan((c + d*x)/ 
2) + 1)*sin(c + d*x)**2*b**4 - log(tan((c + d*x)/2) + 1)*a**4 - log(tan((c 
 + d*x)/2) + 1)*a**2*b**2 + 2*log(tan((c + d*x)/2) + 1)*b**4 - sin(c + d*x 
)*a**4 + sin(c + d*x)*a**2*b**2))/(2*a**3*d*(sin(c + d*x)**2*a**2 - sin(c 
+ d*x)**2*b**2 - a**2 + b**2))