3.5.78 \(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [478]

3.5.78.1 Optimal result

Integrand size = 39, antiderivative size = 154 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(8 A-B-6 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(6 A+8 B+13 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \]

output

-1/7*(A-B+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(8*A-B-6*C)*t 
an(d*x+c)/a/d/(a+a*sec(d*x+c))^3+1/105*(6*A+8*B+13*C)*tan(d*x+c)/d/(a^2+a^ 
2*sec(d*x+c))^2+1/105*(6*A+8*B+13*C)*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))
 

3.5.78.2 Mathematica [A] (verified)

Time = 2.15 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.50 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (70 (9 A+4 B+2 C) \sin \left (\frac {d x}{2}\right )-35 (18 A+5 B+4 C) \sin \left (c+\frac {d x}{2}\right )+441 A \sin \left (c+\frac {3 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )+168 C \sin \left (c+\frac {3 d x}{2}\right )-315 A \sin \left (2 c+\frac {3 d x}{2}\right )-105 B \sin \left (2 c+\frac {3 d x}{2}\right )+147 A \sin \left (2 c+\frac {5 d x}{2}\right )+91 B \sin \left (2 c+\frac {5 d x}{2}\right )+56 C \sin \left (2 c+\frac {5 d x}{2}\right )-105 A \sin \left (3 c+\frac {5 d x}{2}\right )+36 A \sin \left (3 c+\frac {7 d x}{2}\right )+13 B \sin \left (3 c+\frac {7 d x}{2}\right )+8 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{6720 a^4 d} \]

input

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Se 
c[c + d*x])^4,x]
 

output

(Sec[c/2]*Sec[(c + d*x)/2]^7*(70*(9*A + 4*B + 2*C)*Sin[(d*x)/2] - 35*(18*A 
 + 5*B + 4*C)*Sin[c + (d*x)/2] + 441*A*Sin[c + (3*d*x)/2] + 168*B*Sin[c + 
(3*d*x)/2] + 168*C*Sin[c + (3*d*x)/2] - 315*A*Sin[2*c + (3*d*x)/2] - 105*B 
*Sin[2*c + (3*d*x)/2] + 147*A*Sin[2*c + (5*d*x)/2] + 91*B*Sin[2*c + (5*d*x 
)/2] + 56*C*Sin[2*c + (5*d*x)/2] - 105*A*Sin[3*c + (5*d*x)/2] + 36*A*Sin[3 
*c + (7*d*x)/2] + 13*B*Sin[3*c + (7*d*x)/2] + 8*C*Sin[3*c + (7*d*x)/2]))/( 
6720*a^4*d)
 

3.5.78.3 Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {3042, 4566, 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.

\(\displaystyle \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\)

\(\Big \downarrow \) 4566

\(\displaystyle \frac {\int \frac {\sec (c+d x) (a (6 A+B-C)-a (2 A-2 B-5 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a (6 A+B-C)-a (2 A-2 B-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4488

\(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx+\frac {a (8 A-B-6 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx+\frac {a (8 A-B-6 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4283

\(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {a (8 A-B-6 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} (6 A+8 B+13 C) \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {a (8 A-B-6 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {a (8 A-B-6 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {1}{5} (6 A+8 B+13 C) \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{7 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

input

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + 
d*x])^4,x]
 

output

-1/7*((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + 
((a*(8*A - B - 6*C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((6*A + 8 
*B + 13*C)*(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)/(7*a^2)
 

3.5.78.3.1 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)]
 

rule 4566

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

3.5.78.4 Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.56

method	result	size
parallelrisch	\(-\frac {\left (\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {7 \left (-3 A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+7 \left (A +\frac {B}{3}-\frac {C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7 A -7 B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a^{4} d}\)	\(86\)
derivativedivides	\(\frac {\frac {\left (-A +B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (3 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-3 A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(108\)
default	\(\frac {\frac {\left (-A +B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (3 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-3 A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(108\)
norman	\(\frac {-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{56 a d}+\frac {\left (A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (9 A +7 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}+\frac {\left (27 A +11 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {\left (31 A -17 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{280 a d}-\frac {\left (123 A -11 B -31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{420 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{3}}\)	\(181\)
risch	\(\frac {2 i \left (105 A \,{\mathrm e}^{6 i \left (d x +c \right )}+315 A \,{\mathrm e}^{5 i \left (d x +c \right )}+105 B \,{\mathrm e}^{5 i \left (d x +c \right )}+630 A \,{\mathrm e}^{4 i \left (d x +c \right )}+175 B \,{\mathrm e}^{4 i \left (d x +c \right )}+140 C \,{\mathrm e}^{4 i \left (d x +c \right )}+630 A \,{\mathrm e}^{3 i \left (d x +c \right )}+280 B \,{\mathrm e}^{3 i \left (d x +c \right )}+140 C \,{\mathrm e}^{3 i \left (d x +c \right )}+441 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 B \,{\mathrm e}^{2 i \left (d x +c \right )}+168 C \,{\mathrm e}^{2 i \left (d x +c \right )}+147 A \,{\mathrm e}^{i \left (d x +c \right )}+91 B \,{\mathrm e}^{i \left (d x +c \right )}+56 C \,{\mathrm e}^{i \left (d x +c \right )}+36 A +13 B +8 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(213\)

input

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,method 
=_RETURNVERBOSE)
 

output

-1/56*((A-B+C)*tan(1/2*d*x+1/2*c)^6+7/5*(-3*A+B+C)*tan(1/2*d*x+1/2*c)^4+7* 
(A+1/3*B-1/3*C)*tan(1/2*d*x+1/2*c)^2-7*A-7*B-7*C)*tan(1/2*d*x+1/2*c)/a^4/d
 

3.5.78.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.88 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {{\left ({\left (36 \, A + 13 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A + 52 \, B + 32 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (6 \, A + 8 \, B + 13 \, C\right )} \cos \left (d x + c\right ) + 6 \, A + 8 \, B + 13 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, 
 algorithm="fricas")
 

output

1/105*((36*A + 13*B + 8*C)*cos(d*x + c)^3 + (39*A + 52*B + 32*C)*cos(d*x + 
 c)^2 + 4*(6*A + 8*B + 13*C)*cos(d*x + c) + 6*A + 8*B + 13*C)*sin(d*x + c) 
/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 
 4*a^4*d*cos(d*x + c) + a^4*d)
 

3.5.78.6 Sympy [F]

\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4, 
x)
 

output

(Integral(A*sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + 
d*x)**2 + 4*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**2/(sec(c + d* 
x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) + 
Integral(C*sec(c + d*x)**3/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c 
+ d*x)**2 + 4*sec(c + d*x) + 1), x))/a**4
 

3.5.78.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.68 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, 
 algorithm="maxima")
 

output

1/840*(C*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x 
 + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/ 
(cos(d*x + c) + 1)^7)/a^4 + B*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*si 
n(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 
 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 + 3*A*(35*sin(d*x + c)/(cos 
(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^ 
5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d
 

3.5.78.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.11 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 63 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, 
 algorithm="giac")
 

output

-1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 + 15*C*t 
an(1/2*d*x + 1/2*c)^7 - 63*A*tan(1/2*d*x + 1/2*c)^5 + 21*B*tan(1/2*d*x + 1 
/2*c)^5 + 21*C*tan(1/2*d*x + 1/2*c)^5 + 105*A*tan(1/2*d*x + 1/2*c)^3 + 35* 
B*tan(1/2*d*x + 1/2*c)^3 - 35*C*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x 
 + 1/2*c) - 105*B*tan(1/2*d*x + 1/2*c) - 105*C*tan(1/2*d*x + 1/2*c))/(a^4* 
d)
 

3.5.78.9 Mupad [B] (verification not implemented)

Time = 15.93 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.64 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B+C\right )}{8\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-3\,A+C\right )}{40\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A+B-C\right )}{24\,a^4\,d} \]

input

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)*(a + a/cos(c + d 
*x))^4),x)
 

output

(tan(c/2 + (d*x)/2)*(A + B + C))/(8*a^4*d) - (tan(c/2 + (d*x)/2)^7*(A - B 
+ C))/(56*a^4*d) - (tan(c/2 + (d*x)/2)^5*(B - 3*A + C))/(40*a^4*d) - (tan( 
c/2 + (d*x)/2)^3*(3*A + B - C))/(24*a^4*d)