\(\int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [104]

Optimal result

Integrand size = 23, antiderivative size = 108 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {A x}{a^3}-\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {2 (11 A-B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

A*x/a^3-1/5*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(7*A-2*B)*tan(d*x+c 
)/a/d/(a+a*sec(d*x+c))^2-2/15*(11*A-B)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(108)=216\).

Time = 0.92 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.23 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (150 A d x \cos \left (\frac {d x}{2}\right )+150 A d x \cos \left (c+\frac {d x}{2}\right )+75 A d x \cos \left (c+\frac {3 d x}{2}\right )+75 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-370 A \sin \left (\frac {d x}{2}\right )+80 B \sin \left (\frac {d x}{2}\right )+270 A \sin \left (c+\frac {d x}{2}\right )-60 B \sin \left (c+\frac {d x}{2}\right )-230 A \sin \left (c+\frac {3 d x}{2}\right )+40 B \sin \left (c+\frac {3 d x}{2}\right )+90 A \sin \left (2 c+\frac {3 d x}{2}\right )-30 B \sin \left (2 c+\frac {3 d x}{2}\right )-64 A \sin \left (2 c+\frac {5 d x}{2}\right )+14 B \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{480 a^3 d} \] Input:

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

Output:

(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*A*d*x*Cos[(d*x)/2] + 150*A*d*x*Cos[c + ( 
d*x)/2] + 75*A*d*x*Cos[c + (3*d*x)/2] + 75*A*d*x*Cos[2*c + (3*d*x)/2] + 15 
*A*d*x*Cos[2*c + (5*d*x)/2] + 15*A*d*x*Cos[3*c + (5*d*x)/2] - 370*A*Sin[(d 
*x)/2] + 80*B*Sin[(d*x)/2] + 270*A*Sin[c + (d*x)/2] - 60*B*Sin[c + (d*x)/2 
] - 230*A*Sin[c + (3*d*x)/2] + 40*B*Sin[c + (3*d*x)/2] + 90*A*Sin[2*c + (3 
*d*x)/2] - 30*B*Sin[2*c + (3*d*x)/2] - 64*A*Sin[2*c + (5*d*x)/2] + 14*B*Si 
n[2*c + (5*d*x)/2]))/(480*a^3*d)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4410, 25, 3042, 4410, 25, 3042, 4407, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4410

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4410

\(\displaystyle \frac {-\frac {\int -\frac {15 a^2 A-a^2 (7 A-2 B) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-2 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-2 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (7 A-2 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {\frac {15 a A x-2 a^2 (11 A-B) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-2 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {15 a A x-2 a^2 (11 A-B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (7 A-2 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {15 a A x-\frac {2 a^2 (11 A-B) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (7 A-2 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

Input:

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

Output:

-1/5*((A - B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (-1/3*(a*(7*A - 2 
*B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (15*a*A*x - (2*a^2*(11*A - 
B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2)
 

Defintions of rubi rules used

rule 25

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

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 4407

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0]
 

rule 4410

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(\frac {3 \left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+10 \left (2 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+15 \left (-7 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 A x d}{60 a^{3} d}\)	\(69\)
norman	\(\frac {\frac {A x}{a}-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 a d}+\frac {\left (2 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {\left (7 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}}{a^{2}}\)	\(86\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(102\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(102\)
risch	\(\frac {A x}{a^{3}}-\frac {2 i \left (45 A \,{\mathrm e}^{4 i \left (d x +c \right )}-15 B \,{\mathrm e}^{4 i \left (d x +c \right )}+135 A \,{\mathrm e}^{3 i \left (d x +c \right )}-30 B \,{\mathrm e}^{3 i \left (d x +c \right )}+185 A \,{\mathrm e}^{2 i \left (d x +c \right )}-40 B \,{\mathrm e}^{2 i \left (d x +c \right )}+115 \,{\mathrm e}^{i \left (d x +c \right )} A -20 B \,{\mathrm e}^{i \left (d x +c \right )}+32 A -7 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(133\)

Input:

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

Output:

1/60*(3*(-A+B)*tan(1/2*d*x+1/2*c)^5+10*(2*A-B)*tan(1/2*d*x+1/2*c)^3+15*(-7 
*A+B)*tan(1/2*d*x+1/2*c)+60*A*x*d)/a^3/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.28 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, A d x \cos \left (d x + c\right )^{3} + 45 \, A d x \cos \left (d x + c\right )^{2} + 45 \, A d x \cos \left (d x + c\right ) + 15 \, A d x - {\left ({\left (32 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 22 \, A - 2 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

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

Output:

1/15*(15*A*d*x*cos(d*x + c)^3 + 45*A*d*x*cos(d*x + c)^2 + 45*A*d*x*cos(d*x 
 + c) + 15*A*d*x - ((32*A - 7*B)*cos(d*x + c)^2 + 3*(17*A - 2*B)*cos(d*x + 
 c) + 22*A - 2*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]

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

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

Output:

(Integral(A/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) 
 + Integral(B*sec(c + d*x)/(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 = 0.11 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.48 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:

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

Output:

-1/60*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*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 - 120*arctan(si 
n(d*x + c)/(cos(d*x + c) + 1))/a^3) - B*(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)/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} A}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:

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

Output:

1/60*(60*(d*x + c)*A/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan 
(1/2*d*x + 1/2*c)^5 - 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 10*B*a^12*tan(1/2 
*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 15*B*a^12*tan(1/2*d*x 
+ 1/2*c))/a^15)/d
 

Mupad [B] (verification not implemented)

Time = 11.75 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.23 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {A\,x}{a^3}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )-\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \] Input:

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

Output:

(A*x)/a^3 + (cos(c/2 + (d*x)/2)^2*((A*sin(c/2 + (d*x)/2)^3)/3 - (B*sin(c/2 
 + (d*x)/2)^3)/6) - cos(c/2 + (d*x)/2)^4*((7*A*sin(c/2 + (d*x)/2))/4 - (B* 
sin(c/2 + (d*x)/2))/4) - (A*sin(c/2 + (d*x)/2)^5)/20 + (B*sin(c/2 + (d*x)/ 
2)^5)/20)/(a^3*d*cos(c/2 + (d*x)/2)^5)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.87 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +60 a d x}{60 a^{3} d} \] Input:

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

Output:

( - 3*tan((c + d*x)/2)**5*a + 3*tan((c + d*x)/2)**5*b + 20*tan((c + d*x)/2 
)**3*a - 10*tan((c + d*x)/2)**3*b - 105*tan((c + d*x)/2)*a + 15*tan((c + d 
*x)/2)*b + 60*a*d*x)/(60*a**3*d)