\(\int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) [1547]

Optimal result

Integrand size = 31, antiderivative size = 420 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\left (5 a^3 A+a^2 b (20 A+B)+a b^2 (29 A+4 B)+b^3 (16 A+5 B)\right ) \log (1-\sin (c+d x))}{32 (a+b)^4 d}+\frac {\left (5 a^3 A-b^3 (16 A-5 B)+a b^2 (29 A-4 B)-a^2 b (20 A-B)\right ) \log (1+\sin (c+d x))}{32 (a-b)^4 d}+\frac {b^6 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}+\frac {A+B}{48 (a+b) d (1-\sin (c+d x))^3}+\frac {2 a A+3 A b+a B+2 b B}{32 (a+b)^2 d (1-\sin (c+d x))^2}+\frac {a^2 (5 A+B)+2 a b (7 A+2 B)+b^2 (11 A+5 B)}{32 (a+b)^3 d (1-\sin (c+d x))}-\frac {A-B}{48 (a-b) d (1+\sin (c+d x))^3}-\frac {2 a A-3 A b-a B+2 b B}{32 (a-b)^2 d (1+\sin (c+d x))^2}-\frac {b^2 (11 A-5 B)+a^2 (5 A-B)-a (14 A b-4 b B)}{32 (a-b)^3 d (1+\sin (c+d x))} \] Output:

-1/32*(5*a^3*A+a^2*b*(20*A+B)+a*b^2*(29*A+4*B)+b^3*(16*A+5*B))*ln(1-sin(d* 
x+c))/(a+b)^4/d+1/32*(5*a^3*A-b^3*(16*A-5*B)+a*b^2*(29*A-4*B)-a^2*b*(20*A- 
B))*ln(1+sin(d*x+c))/(a-b)^4/d+b^6*(A*b-B*a)*ln(a+b*sin(d*x+c))/(a^2-b^2)^ 
4/d+1/48*(A+B)/(a+b)/d/(1-sin(d*x+c))^3+1/32*(2*A*a+3*A*b+B*a+2*B*b)/(a+b) 
^2/d/(1-sin(d*x+c))^2+1/32*(a^2*(5*A+B)+2*a*b*(7*A+2*B)+b^2*(11*A+5*B))/(a 
+b)^3/d/(1-sin(d*x+c))-1/48*(A-B)/(a-b)/d/(1+sin(d*x+c))^3-1/32*(2*A*a-3*A 
*b-B*a+2*B*b)/(a-b)^2/d/(1+sin(d*x+c))^2-1/32*(b^2*(11*A-5*B)+a^2*(5*A-B)- 
a*(14*A*b-4*B*b))/(a-b)^3/d/(1+sin(d*x+c))
 

Mathematica [A] (verified)

Time = 6.41 (sec) , antiderivative size = 583, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {b^7 \left (-\frac {\sec ^6(c+d x) \left (-A b^2+a b B-b (-a A+b B) \sin (c+d x)\right )}{6 b^8 \left (-a^2+b^2\right )}+\frac {-\frac {\sec ^4(c+d x) \left (-5 a b^2 (a A-b B)-b^2 \left (-5 a^2 A+6 A b^2-a b B\right )-b \left (-5 b^2 (a A-b B)-a \left (-5 a^2 A+6 A b^2-a b B\right )\right ) \sin (c+d x)\right )}{4 b^6 \left (-a^2+b^2\right )}+\frac {\frac {3 \left (\frac {(a-b)^3 \left (5 a^3 A+a^2 b (20 A+B)+a b^2 (29 A+4 B)+b^3 (16 A+5 B)\right ) \log (1-\sin (c+d x))}{2 b (a+b)}-\frac {(a+b)^3 \left (5 a^3 A-b^3 (16 A-5 B)+a b^2 (29 A-4 B)-a^2 b (20 A-B)\right ) \log (1+\sin (c+d x))}{2 (a-b) b}-\frac {16 b^5 (A b-a B) \log (a+b \sin (c+d x))}{(a-b) (a+b)}\right )}{2 b^2 \left (-a^2+b^2\right )}-\frac {\sec ^2(c+d x) \left (3 a b^2 \left (5 a^3 A-11 a A b^2+a^2 b B+5 b^3 B\right )-3 b^2 \left (5 a^4 A-11 a^2 A b^2+8 A b^4+a^3 b B-3 a b^3 B\right )-b \left (3 b^2 \left (5 a^3 A-11 a A b^2+a^2 b B+5 b^3 B\right )-3 a \left (5 a^4 A-11 a^2 A b^2+8 A b^4+a^3 b B-3 a b^3 B\right )\right ) \sin (c+d x)\right )}{2 b^4 \left (-a^2+b^2\right )}}{4 b^2 \left (-a^2+b^2\right )}}{6 b^2 \left (-a^2+b^2\right )}\right )}{d} \] Input:

Integrate[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
 

Output:

(b^7*(-1/6*(Sec[c + d*x]^6*(-(A*b^2) + a*b*B - b*(-(a*A) + b*B)*Sin[c + d* 
x]))/(b^8*(-a^2 + b^2)) + (-1/4*(Sec[c + d*x]^4*(-5*a*b^2*(a*A - b*B) - b^ 
2*(-5*a^2*A + 6*A*b^2 - a*b*B) - b*(-5*b^2*(a*A - b*B) - a*(-5*a^2*A + 6*A 
*b^2 - a*b*B))*Sin[c + d*x]))/(b^6*(-a^2 + b^2)) + ((3*(((a - b)^3*(5*a^3* 
A + a^2*b*(20*A + B) + a*b^2*(29*A + 4*B) + b^3*(16*A + 5*B))*Log[1 - Sin[ 
c + d*x]])/(2*b*(a + b)) - ((a + b)^3*(5*a^3*A - b^3*(16*A - 5*B) + a*b^2* 
(29*A - 4*B) - a^2*b*(20*A - B))*Log[1 + Sin[c + d*x]])/(2*(a - b)*b) - (1 
6*b^5*(A*b - a*B)*Log[a + b*Sin[c + d*x]])/((a - b)*(a + b))))/(2*b^2*(-a^ 
2 + b^2)) - (Sec[c + d*x]^2*(3*a*b^2*(5*a^3*A - 11*a*A*b^2 + a^2*b*B + 5*b 
^3*B) - 3*b^2*(5*a^4*A - 11*a^2*A*b^2 + 8*A*b^4 + a^3*b*B - 3*a*b^3*B) - b 
*(3*b^2*(5*a^3*A - 11*a*A*b^2 + a^2*b*B + 5*b^3*B) - 3*a*(5*a^4*A - 11*a^2 
*A*b^2 + 8*A*b^4 + a^3*b*B - 3*a*b^3*B))*Sin[c + d*x]))/(2*b^4*(-a^2 + b^2 
)))/(4*b^2*(-a^2 + b^2)))/(6*b^2*(-a^2 + b^2))))/d
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 663, 2009}

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.

\(\displaystyle \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x)^7 (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {b^7 \int \frac {A b+B \sin (c+d x) b}{b (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^4}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^6 \int \frac {A b+B \sin (c+d x) b}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^4}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 663

\(\displaystyle \frac {b^6 \int \left (\frac {5 A a^3-b (20 A-B) a^2+b^2 (29 A-4 B) a-b^3 (16 A-5 B)}{32 (a-b)^4 b^6 (\sin (c+d x) b+b)}+\frac {5 A a^3+b (20 A+B) a^2+b^2 (29 A+4 B) a+b^3 (16 A+5 B)}{32 b^6 (a+b)^4 (b-b \sin (c+d x))}+\frac {A b-a B}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac {(5 A+B) a^2+2 b (7 A+2 B) a+b^2 (11 A+5 B)}{32 b^5 (a+b)^3 (b-b \sin (c+d x))^2}+\frac {(5 A-B) a^2-(14 A b-4 b B) a+b^2 (11 A-5 B)}{32 (a-b)^3 b^5 (\sin (c+d x) b+b)^2}+\frac {2 a A+3 b A+a B+2 b B}{16 b^4 (a+b)^2 (b-b \sin (c+d x))^3}+\frac {2 a A-3 b A-a B+2 b B}{16 (a-b)^2 b^4 (\sin (c+d x) b+b)^3}+\frac {A+B}{16 b^3 (a+b) (b-b \sin (c+d x))^4}+\frac {A-B}{16 (a-b) b^3 (\sin (c+d x) b+b)^4}\right )d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^6 \left (\frac {(A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}+\frac {a^2 (5 A+B)+2 a b (7 A+2 B)+b^2 (11 A+5 B)}{32 b^5 (a+b)^3 (b-b \sin (c+d x))}-\frac {a^2 (5 A-B)-a (14 A b-4 b B)+b^2 (11 A-5 B)}{32 b^5 (a-b)^3 (b \sin (c+d x)+b)}-\frac {\left (5 a^3 A+a^2 b (20 A+B)+a b^2 (29 A+4 B)+b^3 (16 A+5 B)\right ) \log (b-b \sin (c+d x))}{32 b^6 (a+b)^4}+\frac {\left (5 a^3 A-a^2 b (20 A-B)+a b^2 (29 A-4 B)-b^3 (16 A-5 B)\right ) \log (b \sin (c+d x)+b)}{32 b^6 (a-b)^4}+\frac {2 a A+a B+3 A b+2 b B}{32 b^4 (a+b)^2 (b-b \sin (c+d x))^2}-\frac {2 a A-a B-3 A b+2 b B}{32 b^4 (a-b)^2 (b \sin (c+d x)+b)^2}-\frac {A-B}{48 b^3 (a-b) (b \sin (c+d x)+b)^3}+\frac {A+B}{48 b^3 (a+b) (b-b \sin (c+d x))^3}\right )}{d}\)

Input:

Int[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
 

Output:

(b^6*(-1/32*((5*a^3*A + a^2*b*(20*A + B) + a*b^2*(29*A + 4*B) + b^3*(16*A 
+ 5*B))*Log[b - b*Sin[c + d*x]])/(b^6*(a + b)^4) + ((A*b - a*B)*Log[a + b* 
Sin[c + d*x]])/(a^2 - b^2)^4 + ((5*a^3*A - b^3*(16*A - 5*B) + a*b^2*(29*A 
- 4*B) - a^2*b*(20*A - B))*Log[b + b*Sin[c + d*x]])/(32*(a - b)^4*b^6) + ( 
A + B)/(48*b^3*(a + b)*(b - b*Sin[c + d*x])^3) + (2*a*A + 3*A*b + a*B + 2* 
b*B)/(32*b^4*(a + b)^2*(b - b*Sin[c + d*x])^2) + (a^2*(5*A + B) + 2*a*b*(7 
*A + 2*B) + b^2*(11*A + 5*B))/(32*b^5*(a + b)^3*(b - b*Sin[c + d*x])) - (A 
 - B)/(48*(a - b)*b^3*(b + b*Sin[c + d*x])^3) - (2*a*A - 3*A*b - a*B + 2*b 
*B)/(32*(a - b)^2*b^4*(b + b*Sin[c + d*x])^2) - (b^2*(11*A - 5*B) + a^2*(5 
*A - B) - a*(14*A*b - 4*b*B))/(32*(a - b)^3*b^5*(b + b*Sin[c + d*x]))))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 663

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ 
)^2)^(p_.), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[1/c^p   Int[ExpandI 
ntegrand[(d + e*x)^m*(f + g*x)^n*(-q + c*x)^p*(q + c*x)^p, x], x], x] /;  ! 
FractionalPowerFactorQ[q]] /; FreeQ[{a, c, d, e, f, g}, x] && ILtQ[p, -1] & 
& IntegersQ[m, n] && NiceSqrtQ[(-a)*c]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 4.95 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {\frac {\left (A b -B a \right ) b^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}-\frac {A -B}{3 \left (16 a -16 b \right ) \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {2 a A -3 A b -B a +2 B b}{32 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5 A \,a^{2}-14 A a b +11 b^{2} A -a^{2} B +4 a b B -5 B \,b^{2}}{32 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (5 a^{3} A -20 A \,a^{2} b +29 A a \,b^{2}-16 A \,b^{3}+B \,a^{2} b -4 B a \,b^{2}+5 B \,b^{3}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{32 \left (a -b \right )^{4}}-\frac {A +B}{3 \left (16 a +16 b \right ) \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {-2 a A -3 A b -B a -2 B b}{32 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5 A \,a^{2}+14 A a b +11 b^{2} A +a^{2} B +4 a b B +5 B \,b^{2}}{32 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-5 a^{3} A -20 A \,a^{2} b -29 A a \,b^{2}-16 A \,b^{3}-B \,a^{2} b -4 B a \,b^{2}-5 B \,b^{3}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{32 \left (a +b \right )^{4}}}{d}\)	\(393\)
default	\(\frac {\frac {\left (A b -B a \right ) b^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}-\frac {A -B}{3 \left (16 a -16 b \right ) \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {2 a A -3 A b -B a +2 B b}{32 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5 A \,a^{2}-14 A a b +11 b^{2} A -a^{2} B +4 a b B -5 B \,b^{2}}{32 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (5 a^{3} A -20 A \,a^{2} b +29 A a \,b^{2}-16 A \,b^{3}+B \,a^{2} b -4 B a \,b^{2}+5 B \,b^{3}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{32 \left (a -b \right )^{4}}-\frac {A +B}{3 \left (16 a +16 b \right ) \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {-2 a A -3 A b -B a -2 B b}{32 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5 A \,a^{2}+14 A a b +11 b^{2} A +a^{2} B +4 a b B +5 B \,b^{2}}{32 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-5 a^{3} A -20 A \,a^{2} b -29 A a \,b^{2}-16 A \,b^{3}-B \,a^{2} b -4 B a \,b^{2}-5 B \,b^{3}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{32 \left (a +b \right )^{4}}}{d}\)	\(393\)
parallelrisch	\(\frac {48 b^{6} \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (A b -B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-15 \left (a -b \right )^{4} \left (\left (\frac {16 A}{5}+B \right ) b^{3}+\frac {29 \left (A +\frac {4 B}{29}\right ) a \,b^{2}}{5}+4 \left (A +\frac {B}{20}\right ) a^{2} b +a^{3} A \right ) \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (a +b \right ) \left (\left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (a +b \right )^{3} \left (\left (-\frac {16 A}{5}+B \right ) b^{3}+\frac {29 \left (A -\frac {4 B}{29}\right ) a \,b^{2}}{5}-4 \left (A -\frac {B}{20}\right ) a^{2} b +a^{3} A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {8 \left (a -b \right ) \left (15 \left (a^{2}-\frac {7 b^{2}}{5}\right ) \left (A b -B a \right ) \left (a^{2}-\frac {b^{2}}{2}\right ) \cos \left (2 d x +2 c \right )+6 \left (a^{4}-\frac {7}{2} a^{2} b^{2}+\frac {7}{2} b^{4}\right ) \left (A b -B a \right ) \cos \left (4 d x +4 c \right )+\left (A b -B a \right ) \left (a^{4}-\frac {7}{2} a^{2} b^{2}+\frac {11}{2} b^{4}\right ) \cos \left (6 d x +6 c \right )+\left (-68 A \,a^{3} b^{2}+\frac {259}{4} A a \,b^{4}+\frac {17}{4} B \,a^{4} b -B \,a^{2} b^{3}+\frac {85}{4} A \,a^{5}-\frac {85}{4} B \,b^{5}\right ) \sin \left (3 d x +3 c \right )+\left (-12 A \,a^{3} b^{2}+\frac {57}{4} A a \,b^{4}+\frac {3}{4} B \,a^{4} b -3 B \,a^{2} b^{3}+\frac {15}{4} A \,a^{5}-\frac {15}{4} B \,b^{5}\right ) \sin \left (5 d x +5 c \right )+\left (-120 A \,a^{3} b^{2}+\frac {165}{2} A a \,b^{4}-\frac {57}{2} B \,a^{4} b +66 B \,a^{2} b^{3}+\frac {99}{2} A \,a^{5}-\frac {99}{2} B \,b^{5}\right ) \sin \left (d x +c \right )-22 \left (a^{4}-\frac {53}{22} a^{2} b^{2}+\frac {37}{22} b^{4}\right ) \left (A b -B a \right )\right )}{15}\right )}{48 \left (a -b \right )^{4} \left (a +b \right )^{4} d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\)	\(617\)
norman	\(\text {Expression too large to display}\)	\(1436\)
risch	\(\text {Expression too large to display}\)	\(3078\)

Input:

int(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE 
)
 

Output:

1/d*((A*b-B*a)*b^6/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))-1/3*(A-B)/(16*a-16*b 
)/(1+sin(d*x+c))^3-1/32*(2*A*a-3*A*b-B*a+2*B*b)/(a-b)^2/(1+sin(d*x+c))^2-1 
/32*(5*A*a^2-14*A*a*b+11*A*b^2-B*a^2+4*B*a*b-5*B*b^2)/(a-b)^3/(1+sin(d*x+c 
))+1/32*(5*A*a^3-20*A*a^2*b+29*A*a*b^2-16*A*b^3+B*a^2*b-4*B*a*b^2+5*B*b^3) 
/(a-b)^4*ln(1+sin(d*x+c))-1/3*(A+B)/(16*a+16*b)/(sin(d*x+c)-1)^3-1/32*(-2* 
A*a-3*A*b-B*a-2*B*b)/(a+b)^2/(sin(d*x+c)-1)^2-1/32*(5*A*a^2+14*A*a*b+11*A* 
b^2+B*a^2+4*B*a*b+5*B*b^2)/(a+b)^3/(sin(d*x+c)-1)+1/32/(a+b)^4*(-5*A*a^3-2 
0*A*a^2*b-29*A*a*b^2-16*A*b^3-B*a^2*b-4*B*a*b^2-5*B*b^3)*ln(sin(d*x+c)-1))
 

Fricas [A] (verification not implemented)

Time = 2.16 (sec) , antiderivative size = 643, normalized size of antiderivative = 1.53 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {16 \, B a^{7} - 16 \, A a^{6} b - 48 \, B a^{5} b^{2} + 48 \, A a^{4} b^{3} + 48 \, B a^{3} b^{4} - 48 \, A a^{2} b^{5} - 16 \, B a b^{6} + 16 \, A b^{7} - 96 \, {\left (B a b^{6} - A b^{7}\right )} \cos \left (d x + c\right )^{6} \log \left (b \sin \left (d x + c\right ) + a\right ) + 3 \, {\left (5 \, A a^{7} + B a^{6} b - 21 \, A a^{5} b^{2} - 5 \, B a^{4} b^{3} + 35 \, A a^{3} b^{4} + 15 \, B a^{2} b^{5} - {\left (35 \, A - 16 \, B\right )} a b^{6} - {\left (16 \, A - 5 \, B\right )} b^{7}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a^{7} + B a^{6} b - 21 \, A a^{5} b^{2} - 5 \, B a^{4} b^{3} + 35 \, A a^{3} b^{4} + 15 \, B a^{2} b^{5} - {\left (35 \, A + 16 \, B\right )} a b^{6} + {\left (16 \, A + 5 \, B\right )} b^{7}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 48 \, {\left (B a^{3} b^{4} - A a^{2} b^{5} - B a b^{6} + A b^{7}\right )} \cos \left (d x + c\right )^{4} - 24 \, {\left (B a^{5} b^{2} - A a^{4} b^{3} - 2 \, B a^{3} b^{4} + 2 \, A a^{2} b^{5} + B a b^{6} - A b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, A a^{7} - 8 \, B a^{6} b - 24 \, A a^{5} b^{2} + 24 \, B a^{4} b^{3} + 24 \, A a^{3} b^{4} - 24 \, B a^{2} b^{5} - 8 \, A a b^{6} + 8 \, B b^{7} + 3 \, {\left (5 \, A a^{7} + B a^{6} b - 21 \, A a^{5} b^{2} - 5 \, B a^{4} b^{3} + 35 \, A a^{3} b^{4} - B a^{2} b^{5} - 19 \, A a b^{6} + 5 \, B b^{7}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, A a^{7} + B a^{6} b - 21 \, A a^{5} b^{2} + 3 \, B a^{4} b^{3} + 27 \, A a^{3} b^{4} - 9 \, B a^{2} b^{5} - 11 \, A a b^{6} + 5 \, B b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{6}} \] Input:

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="fri 
cas")
 

Output:

1/96*(16*B*a^7 - 16*A*a^6*b - 48*B*a^5*b^2 + 48*A*a^4*b^3 + 48*B*a^3*b^4 - 
 48*A*a^2*b^5 - 16*B*a*b^6 + 16*A*b^7 - 96*(B*a*b^6 - A*b^7)*cos(d*x + c)^ 
6*log(b*sin(d*x + c) + a) + 3*(5*A*a^7 + B*a^6*b - 21*A*a^5*b^2 - 5*B*a^4* 
b^3 + 35*A*a^3*b^4 + 15*B*a^2*b^5 - (35*A - 16*B)*a*b^6 - (16*A - 5*B)*b^7 
)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a^7 + B*a^6*b - 21*A*a^5*b 
^2 - 5*B*a^4*b^3 + 35*A*a^3*b^4 + 15*B*a^2*b^5 - (35*A + 16*B)*a*b^6 + (16 
*A + 5*B)*b^7)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 48*(B*a^3*b^4 - A*a 
^2*b^5 - B*a*b^6 + A*b^7)*cos(d*x + c)^4 - 24*(B*a^5*b^2 - A*a^4*b^3 - 2*B 
*a^3*b^4 + 2*A*a^2*b^5 + B*a*b^6 - A*b^7)*cos(d*x + c)^2 + 2*(8*A*a^7 - 8* 
B*a^6*b - 24*A*a^5*b^2 + 24*B*a^4*b^3 + 24*A*a^3*b^4 - 24*B*a^2*b^5 - 8*A* 
a*b^6 + 8*B*b^7 + 3*(5*A*a^7 + B*a^6*b - 21*A*a^5*b^2 - 5*B*a^4*b^3 + 35*A 
*a^3*b^4 - B*a^2*b^5 - 19*A*a*b^6 + 5*B*b^7)*cos(d*x + c)^4 + 2*(5*A*a^7 + 
 B*a^6*b - 21*A*a^5*b^2 + 3*B*a^4*b^3 + 27*A*a^3*b^4 - 9*B*a^2*b^5 - 11*A* 
a*b^6 + 5*B*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^8 - 4*a^6*b^2 + 6*a^4*b 
^4 - 4*a^2*b^6 + b^8)*d*cos(d*x + c)^6)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**7*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 632, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\frac {96 \, {\left (B a b^{6} - A b^{7}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {3 \, {\left (5 \, A a^{3} - {\left (20 \, A - B\right )} a^{2} b + {\left (29 \, A - 4 \, B\right )} a b^{2} - {\left (16 \, A - 5 \, B\right )} b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {3 \, {\left (5 \, A a^{3} + {\left (20 \, A + B\right )} a^{2} b + {\left (29 \, A + 4 \, B\right )} a b^{2} + {\left (16 \, A + 5 \, B\right )} b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, {\left (8 \, B a^{5} - 8 \, A a^{4} b - 28 \, B a^{3} b^{2} + 28 \, A a^{2} b^{3} + 44 \, B a b^{4} - 44 \, A b^{5} + 3 \, {\left (5 \, A a^{5} + B a^{4} b - 16 \, A a^{3} b^{2} - 4 \, B a^{2} b^{3} + 19 \, A a b^{4} - 5 \, B b^{5}\right )} \sin \left (d x + c\right )^{5} + 24 \, {\left (B a b^{4} - A b^{5}\right )} \sin \left (d x + c\right )^{4} - 8 \, {\left (5 \, A a^{5} + B a^{4} b - 16 \, A a^{3} b^{2} - 2 \, B a^{2} b^{3} + 17 \, A a b^{4} - 5 \, B b^{5}\right )} \sin \left (d x + c\right )^{3} + 12 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - 5 \, B a b^{4} + 5 \, A b^{5}\right )} \sin \left (d x + c\right )^{2} + 3 \, {\left (11 \, A a^{5} - B a^{4} b - 32 \, A a^{3} b^{2} + 4 \, B a^{2} b^{3} + 29 \, A a b^{4} - 11 \, B b^{5}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sin \left (d x + c\right )^{6} - a^{6} + 3 \, a^{4} b^{2} - 3 \, a^{2} b^{4} + b^{6} - 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sin \left (d x + c\right )^{4} + 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sin \left (d x + c\right )^{2}}}{96 \, d} \] Input:

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="max 
ima")
 

Output:

-1/96*(96*(B*a*b^6 - A*b^7)*log(b*sin(d*x + c) + a)/(a^8 - 4*a^6*b^2 + 6*a 
^4*b^4 - 4*a^2*b^6 + b^8) - 3*(5*A*a^3 - (20*A - B)*a^2*b + (29*A - 4*B)*a 
*b^2 - (16*A - 5*B)*b^3)*log(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 
- 4*a*b^3 + b^4) + 3*(5*A*a^3 + (20*A + B)*a^2*b + (29*A + 4*B)*a*b^2 + (1 
6*A + 5*B)*b^3)*log(sin(d*x + c) - 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 
 + b^4) + 2*(8*B*a^5 - 8*A*a^4*b - 28*B*a^3*b^2 + 28*A*a^2*b^3 + 44*B*a*b^ 
4 - 44*A*b^5 + 3*(5*A*a^5 + B*a^4*b - 16*A*a^3*b^2 - 4*B*a^2*b^3 + 19*A*a* 
b^4 - 5*B*b^5)*sin(d*x + c)^5 + 24*(B*a*b^4 - A*b^5)*sin(d*x + c)^4 - 8*(5 
*A*a^5 + B*a^4*b - 16*A*a^3*b^2 - 2*B*a^2*b^3 + 17*A*a*b^4 - 5*B*b^5)*sin( 
d*x + c)^3 + 12*(B*a^3*b^2 - A*a^2*b^3 - 5*B*a*b^4 + 5*A*b^5)*sin(d*x + c) 
^2 + 3*(11*A*a^5 - B*a^4*b - 32*A*a^3*b^2 + 4*B*a^2*b^3 + 29*A*a*b^4 - 11* 
B*b^5)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sin(d*x + c)^6 - 
 a^6 + 3*a^4*b^2 - 3*a^2*b^4 + b^6 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) 
*sin(d*x + c)^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sin(d*x + c)^2))/d
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 675, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {{\left (B a b^{7} - A b^{8}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b d - 4 \, a^{6} b^{3} d + 6 \, a^{4} b^{5} d - 4 \, a^{2} b^{7} d + b^{9} d} - \frac {{\left (5 \, A a^{3} + 20 \, A a^{2} b + B a^{2} b + 29 \, A a b^{2} + 4 \, B a b^{2} + 16 \, A b^{3} + 5 \, B b^{3}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{32 \, {\left (a^{4} d + 4 \, a^{3} b d + 6 \, a^{2} b^{2} d + 4 \, a b^{3} d + b^{4} d\right )}} + \frac {{\left (5 \, A a^{3} - 20 \, A a^{2} b + B a^{2} b + 29 \, A a b^{2} - 4 \, B a b^{2} - 16 \, A b^{3} + 5 \, B b^{3}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{32 \, {\left (a^{4} d - 4 \, a^{3} b d + 6 \, a^{2} b^{2} d - 4 \, a b^{3} d + b^{4} d\right )}} - \frac {8 \, B a^{7} - 8 \, A a^{6} b - 36 \, B a^{5} b^{2} + 36 \, A a^{4} b^{3} + 72 \, B a^{3} b^{4} - 72 \, A a^{2} b^{5} - 44 \, B a b^{6} + 44 \, A b^{7} + 3 \, {\left (5 \, A a^{7} + B a^{6} b - 21 \, A a^{5} b^{2} - 5 \, B a^{4} b^{3} + 35 \, A a^{3} b^{4} - B a^{2} b^{5} - 19 \, A a b^{6} + 5 \, B b^{7}\right )} \sin \left (d x + c\right )^{5} + 24 \, {\left (B a^{3} b^{4} - A a^{2} b^{5} - B a b^{6} + A b^{7}\right )} \sin \left (d x + c\right )^{4} - 8 \, {\left (5 \, A a^{7} + B a^{6} b - 21 \, A a^{5} b^{2} - 3 \, B a^{4} b^{3} + 33 \, A a^{3} b^{4} - 3 \, B a^{2} b^{5} - 17 \, A a b^{6} + 5 \, B b^{7}\right )} \sin \left (d x + c\right )^{3} + 12 \, {\left (B a^{5} b^{2} - A a^{4} b^{3} - 6 \, B a^{3} b^{4} + 6 \, A a^{2} b^{5} + 5 \, B a b^{6} - 5 \, A b^{7}\right )} \sin \left (d x + c\right )^{2} + 3 \, {\left (11 \, A a^{7} - B a^{6} b - 43 \, A a^{5} b^{2} + 5 \, B a^{4} b^{3} + 61 \, A a^{3} b^{4} - 15 \, B a^{2} b^{5} - 29 \, A a b^{6} + 11 \, B b^{7}\right )} \sin \left (d x + c\right )}{48 \, {\left (a + b\right )}^{4} {\left (a - b\right )}^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:

integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="gia 
c")
 

Output:

-(B*a*b^7 - A*b^8)*log(abs(b*sin(d*x + c) + a))/(a^8*b*d - 4*a^6*b^3*d + 6 
*a^4*b^5*d - 4*a^2*b^7*d + b^9*d) - 1/32*(5*A*a^3 + 20*A*a^2*b + B*a^2*b + 
 29*A*a*b^2 + 4*B*a*b^2 + 16*A*b^3 + 5*B*b^3)*log(abs(-sin(d*x + c) + 1))/ 
(a^4*d + 4*a^3*b*d + 6*a^2*b^2*d + 4*a*b^3*d + b^4*d) + 1/32*(5*A*a^3 - 20 
*A*a^2*b + B*a^2*b + 29*A*a*b^2 - 4*B*a*b^2 - 16*A*b^3 + 5*B*b^3)*log(abs( 
-sin(d*x + c) - 1))/(a^4*d - 4*a^3*b*d + 6*a^2*b^2*d - 4*a*b^3*d + b^4*d) 
- 1/48*(8*B*a^7 - 8*A*a^6*b - 36*B*a^5*b^2 + 36*A*a^4*b^3 + 72*B*a^3*b^4 - 
 72*A*a^2*b^5 - 44*B*a*b^6 + 44*A*b^7 + 3*(5*A*a^7 + B*a^6*b - 21*A*a^5*b^ 
2 - 5*B*a^4*b^3 + 35*A*a^3*b^4 - B*a^2*b^5 - 19*A*a*b^6 + 5*B*b^7)*sin(d*x 
 + c)^5 + 24*(B*a^3*b^4 - A*a^2*b^5 - B*a*b^6 + A*b^7)*sin(d*x + c)^4 - 8* 
(5*A*a^7 + B*a^6*b - 21*A*a^5*b^2 - 3*B*a^4*b^3 + 33*A*a^3*b^4 - 3*B*a^2*b 
^5 - 17*A*a*b^6 + 5*B*b^7)*sin(d*x + c)^3 + 12*(B*a^5*b^2 - A*a^4*b^3 - 6* 
B*a^3*b^4 + 6*A*a^2*b^5 + 5*B*a*b^6 - 5*A*b^7)*sin(d*x + c)^2 + 3*(11*A*a^ 
7 - B*a^6*b - 43*A*a^5*b^2 + 5*B*a^4*b^3 + 61*A*a^3*b^4 - 15*B*a^2*b^5 - 2 
9*A*a*b^6 + 11*B*b^7)*sin(d*x + c))/((a + b)^4*(a - b)^4*d*(sin(d*x + c) + 
 1)^3*(sin(d*x + c) - 1)^3)
 

Mupad [B] (verification not implemented)

Time = 36.95 (sec) , antiderivative size = 729, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (5\,A\,a^3+\left (B-20\,A\right )\,a^2\,b+\left (29\,A-4\,B\right )\,a\,b^2+\left (5\,B-16\,A\right )\,b^3\right )}{d\,\left (32\,a^4-128\,a^3\,b+192\,a^2\,b^2-128\,a\,b^3+32\,b^4\right )}-\frac {\frac {-2\,B\,a^5+2\,A\,a^4\,b+7\,B\,a^3\,b^2-7\,A\,a^2\,b^3-11\,B\,a\,b^4+11\,A\,b^5}{12\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {{\sin \left (c+d\,x\right )}^4\,\left (A\,b^5-B\,a\,b^4\right )}{2\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\sin \left (c+d\,x\right )\,\left (11\,A\,a^5-B\,a^4\,b-32\,A\,a^3\,b^2+4\,B\,a^2\,b^3+29\,A\,a\,b^4-11\,B\,b^5\right )}{16\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (B\,a^3\,b^2-A\,a^2\,b^3-5\,B\,a\,b^4+5\,A\,b^5\right )}{4\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {{\sin \left (c+d\,x\right )}^3\,\left (5\,A\,a^5+B\,a^4\,b-16\,A\,a^3\,b^2-2\,B\,a^2\,b^3+17\,A\,a\,b^4-5\,B\,b^5\right )}{6\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {{\sin \left (c+d\,x\right )}^5\,\left (5\,A\,a^5+B\,a^4\,b-16\,A\,a^3\,b^2-4\,B\,a^2\,b^3+19\,A\,a\,b^4-5\,B\,b^5\right )}{16\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2-{\sin \left (c+d\,x\right )}^6+3\,{\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (5\,A\,a^3+\left (20\,A+B\right )\,a^2\,b+\left (29\,A+4\,B\right )\,a\,b^2+\left (16\,A+5\,B\right )\,b^3\right )}{d\,\left (32\,a^4+128\,a^3\,b+192\,a^2\,b^2+128\,a\,b^3+32\,b^4\right )}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b^7-B\,a\,b^6\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \] Input:

int((A + B*sin(c + d*x))/(cos(c + d*x)^7*(a + b*sin(c + d*x))),x)
 

Output:

(log(sin(c + d*x) + 1)*(5*A*a^3 - b^3*(16*A - 5*B) - a^2*b*(20*A - B) + a* 
b^2*(29*A - 4*B)))/(d*(32*a^4 - 128*a^3*b - 128*a*b^3 + 32*b^4 + 192*a^2*b 
^2)) - ((11*A*b^5 - 2*B*a^5 - 7*A*a^2*b^3 + 7*B*a^3*b^2 + 2*A*a^4*b - 11*B 
*a*b^4)/(12*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (sin(c + d*x)^4*(A*b^5 
- B*a*b^4))/(2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (sin(c + d*x)*(11*A* 
a^5 - 11*B*b^5 - 32*A*a^3*b^2 + 4*B*a^2*b^3 + 29*A*a*b^4 - B*a^4*b))/(16*( 
a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (sin(c + d*x)^2*(5*A*b^5 - A*a^2*b^3 
 + B*a^3*b^2 - 5*B*a*b^4))/(4*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (sin( 
c + d*x)^3*(5*A*a^5 - 5*B*b^5 - 16*A*a^3*b^2 - 2*B*a^2*b^3 + 17*A*a*b^4 + 
B*a^4*b))/(6*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (sin(c + d*x)^5*(5*A*a 
^5 - 5*B*b^5 - 16*A*a^3*b^2 - 4*B*a^2*b^3 + 19*A*a*b^4 + B*a^4*b))/(16*(a^ 
6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/(d*(cos(c + d*x)^2 - 2*sin(c + d*x)^2 + 
 3*sin(c + d*x)^4 - sin(c + d*x)^6)) - (log(sin(c + d*x) - 1)*(5*A*a^3 + b 
^3*(16*A + 5*B) + a*b^2*(29*A + 4*B) + a^2*b*(20*A + B)))/(d*(128*a*b^3 + 
128*a^3*b + 32*a^4 + 32*b^4 + 192*a^2*b^2)) + (log(a + b*sin(c + d*x))*(A* 
b^7 - B*a*b^6))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{6}+45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}-45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{6}-45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}+45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-15 \sin \left (d x +c \right )^{5}+40 \sin \left (d x +c \right )^{3}-33 \sin \left (d x +c \right )}{48 d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x)
 

Output:

( - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 + 45*log(tan((c + d*x)/2) 
 - 1)*sin(c + d*x)**4 - 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 15* 
log(tan((c + d*x)/2) - 1) + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 - 
 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 45*log(tan((c + d*x)/2) + 
1)*sin(c + d*x)**2 - 15*log(tan((c + d*x)/2) + 1) - 15*sin(c + d*x)**5 + 4 
0*sin(c + d*x)**3 - 33*sin(c + d*x))/(48*d*(sin(c + d*x)**6 - 3*sin(c + d* 
x)**4 + 3*sin(c + d*x)**2 - 1))