\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [103]

Optimal result

Integrand size = 29, antiderivative size = 102 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(2 A+3 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(2 A+3 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

1/5*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(2*A+3*B)*tan(d*x+c)/a/d/(a 
+a*sec(d*x+c))^2+1/15*(2*A+3*B)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (5 (8 A+3 B) \sin \left (\frac {d x}{2}\right )-15 (2 A+B) \sin \left (c+\frac {d x}{2}\right )+20 A \sin \left (c+\frac {3 d x}{2}\right )+15 B \sin \left (c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {3 d x}{2}\right )+7 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d (1+\cos (c+d x))^3} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sec[c/2]*(5*(8*A + 3*B)*Sin[(d*x)/2] - 15*(2*A + B)*Sin[ 
c + (d*x)/2] + 20*A*Sin[c + (3*d*x)/2] + 15*B*Sin[c + (3*d*x)/2] - 15*A*Si 
n[2*c + (3*d*x)/2] + 7*A*Sin[2*c + (5*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2])) 
/(30*a^3*d*(1 + Cos[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 4488, 3042, 4283, 3042, 4281}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4488

\(\displaystyle \frac {(2 A+3 B) \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a}+\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(2 A+3 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4283

\(\displaystyle \frac {(2 A+3 B) \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(2 A+3 B) \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(2 A+3 B) \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}\)

Input:

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

Output:

((A - B)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((2*A + 3*B)*(Tan[c 
+ d*x]/(3*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(3*a*d*(a + a*Sec[c + d 
*x]))))/(5*a)
 

Defintions of rubi rules used

rule 3042

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

rule 4281

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} 
, x] && EqQ[a^2 - b^2, 0]
 

rule 4283

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] 
 + Simp[(m + 1)/(a*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 
1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) 
] && IntegerQ[2*m]
 

rule 4488

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)*Cot[e + 
f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 
1))/(a*b*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] 
 /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] 
&& NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {10 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+5 A +5 B \right )}{20 a^{3} d}\)	\(56\)
derivativedivides	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{4 d \,a^{3}}\)	\(64\)
default	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{4 d \,a^{3}}\)	\(64\)
risch	\(\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}+15 B \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{2 i \left (d x +c \right )}+20 \,{\mathrm e}^{i \left (d x +c \right )} A +15 B \,{\mathrm e}^{i \left (d x +c \right )}+7 A +3 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(114\)
norman	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}-\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (5 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (13 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{2}}\)	\(117\)

Input:

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

Output:

1/20*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^4-10/3*A*tan(1/2*d*x+1/2 
*c)^2+5*A+5*B)/a^3/d
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {{\left ({\left (7 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A + 3 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

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

Output:

1/15*((7*A + 3*B)*cos(d*x + c)^2 + 3*(2*A + 3*B)*cos(d*x + c) + 2*A + 3*B) 
*sin(d*x + c)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos 
(d*x + c) + a^3*d)
 

Sympy [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:

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

Output:

(Integral(A*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + 
d*x) + 1), x) + Integral(B*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d* 
x)**2 + 3*sec(c + d*x) + 1), x))/a**3
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, B {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:

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

Output:

1/60*(A*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + 
 c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 + 3*B*(5*sin(d*x + 
 c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \] Input:

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

Output:

1/60*(3*A*tan(1/2*d*x + 1/2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 - 10*A*tan(1 
/2*d*x + 1/2*c)^3 + 15*A*tan(1/2*d*x + 1/2*c) + 15*B*tan(1/2*d*x + 1/2*c)) 
/(a^3*d)
 

Mupad [B] (verification not implemented)

Time = 11.55 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+15\,B-10\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*(15*A + 15*B - 10*A*tan(c/2 + (d*x)/2)^2 + 3*A*tan(c/2 
 + (d*x)/2)^4 - 3*B*tan(c/2 + (d*x)/2)^4))/(60*a^3*d)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b -10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +15 a +15 b \right )}{60 a^{3} d} \] Input:

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

Output:

(tan((c + d*x)/2)*(3*tan((c + d*x)/2)**4*a - 3*tan((c + d*x)/2)**4*b - 10* 
tan((c + d*x)/2)**2*a + 15*a + 15*b))/(60*a**3*d)