Integrand size = 31, antiderivative size = 221 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {5 (7 A+B) \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {a^2 (A+B)}{96 d (a-a \sin (c+d x))^3}+\frac {5 (3 A+B)}{128 d (a-a \sin (c+d x))}-\frac {a^2 (2 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {5 A}{32 d (a+a \sin (c+d x))}-\frac {a^7 (A-B)}{64 d \left (a^2+a^2 \sin (c+d x)\right )^4}+\frac {a^7 (5 A+3 B)}{128 d \left (a^4-a^4 \sin (c+d x)\right )^2}-\frac {a^7 (5 A-B)}{64 d \left (a^4+a^4 \sin (c+d x)\right )^2} \] Output:
5/128*(7*A+B)*arctanh(sin(d*x+c))/a/d+1/96*a^2*(A+B)/d/(a-a*sin(d*x+c))^3+ 5/128*(3*A+B)/d/(a-a*sin(d*x+c))-1/48*a^2*(2*A-B)/d/(a+a*sin(d*x+c))^3-5/3 2*A/d/(a+a*sin(d*x+c))-1/64*a^7*(A-B)/d/(a^2+a^2*sin(d*x+c))^4+1/128*a^7*( 5*A+3*B)/d/(a^4-a^4*sin(d*x+c))^2-1/64*a^7*(5*A-B)/d/(a^4+a^4*sin(d*x+c))^ 2
Time = 0.96 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {15 (7 A+B) \text {arctanh}(\sin (c+d x))+\frac {48 (A-B)-33 (7 A+B) \sin (c+d x)-33 (7 A+B) \sin ^2(c+d x)+40 (7 A+B) \sin ^3(c+d x)+40 (7 A+B) \sin ^4(c+d x)-15 (7 A+B) \sin ^5(c+d x)-15 (7 A+B) \sin ^6(c+d x)}{(-1+\sin (c+d x))^3 (1+\sin (c+d x))^4}}{384 a d} \] Input:
Integrate[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(15*(7*A + B)*ArcTanh[Sin[c + d*x]] + (48*(A - B) - 33*(7*A + B)*Sin[c + d *x] - 33*(7*A + B)*Sin[c + d*x]^2 + 40*(7*A + B)*Sin[c + d*x]^3 + 40*(7*A + B)*Sin[c + d*x]^4 - 15*(7*A + B)*Sin[c + d*x]^5 - 15*(7*A + B)*Sin[c + d *x]^6)/((-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^4))/(384*a*d)
Time = 0.45 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 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.
| \(\displaystyle \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x)^7 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^7 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^6 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^6 \int \left (\frac {5 A}{32 a^6 (\sin (c+d x) a+a)^2}+\frac {5 (7 A+B)}{128 a^6 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {5 (3 A+B)}{128 a^6 (a-a \sin (c+d x))^2}+\frac {5 A+3 B}{64 a^5 (a-a \sin (c+d x))^3}+\frac {5 A-B}{32 a^5 (\sin (c+d x) a+a)^3}+\frac {A+B}{32 a^4 (a-a \sin (c+d x))^4}+\frac {2 A-B}{16 a^4 (\sin (c+d x) a+a)^4}+\frac {A-B}{16 a^3 (\sin (c+d x) a+a)^5}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^6 \left (\frac {5 (7 A+B) \text {arctanh}(\sin (c+d x))}{128 a^7}+\frac {5 (3 A+B)}{128 a^6 (a-a \sin (c+d x))}-\frac {5 A}{32 a^6 (a \sin (c+d x)+a)}+\frac {5 A+3 B}{128 a^5 (a-a \sin (c+d x))^2}-\frac {5 A-B}{64 a^5 (a \sin (c+d x)+a)^2}+\frac {A+B}{96 a^4 (a-a \sin (c+d x))^3}-\frac {2 A-B}{48 a^4 (a \sin (c+d x)+a)^3}-\frac {A-B}{64 a^3 (a \sin (c+d x)+a)^4}\right )}{d}\) |
Input:
Int[(Sec[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(a^6*((5*(7*A + B)*ArcTanh[Sin[c + d*x]])/(128*a^7) + (A + B)/(96*a^4*(a - a*Sin[c + d*x])^3) + (5*A + 3*B)/(128*a^5*(a - a*Sin[c + d*x])^2) + (5*(3 *A + B))/(128*a^6*(a - a*Sin[c + d*x])) - (A - B)/(64*a^3*(a + a*Sin[c + d *x])^4) - (2*A - B)/(48*a^4*(a + a*Sin[c + d*x])^3) - (5*A - B)/(64*a^5*(a + a*Sin[c + d*x])^2) - (5*A)/(32*a^6*(a + a*Sin[c + d*x]))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 2.43 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.77
| method | result | size |
| derivativedivides | \(\frac {-\frac {5 A}{32 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{16}-\frac {B}{16}}{4 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {\frac {A}{8}-\frac {B}{16}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {5 A}{32}-\frac {B}{32}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {35 A}{256}+\frac {5 B}{256}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {35 A}{256}-\frac {5 B}{256}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {5 A}{64}-\frac {3 B}{64}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{32}+\frac {B}{32}}{3 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {\frac {15 A}{128}+\frac {5 B}{128}}{\sin \left (d x +c \right )-1}}{d a}\) | \(170\) |
| default | \(\frac {-\frac {5 A}{32 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{16}-\frac {B}{16}}{4 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {\frac {A}{8}-\frac {B}{16}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {5 A}{32}-\frac {B}{32}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {35 A}{256}+\frac {5 B}{256}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {35 A}{256}-\frac {5 B}{256}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {5 A}{64}-\frac {3 B}{64}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{32}+\frac {B}{32}}{3 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {\frac {15 A}{128}+\frac {5 B}{128}}{\sin \left (d x +c \right )-1}}{d a}\) | \(170\) |
| parallelrisch | \(\frac {-525 \left (A +\frac {B}{7}\right ) \left (4+\frac {\sin \left (7 d x +7 c \right )}{5}+\sin \left (5 d x +5 c \right )+\frac {9 \sin \left (3 d x +3 c \right )}{5}+\sin \left (d x +c \right )+\frac {2 \cos \left (6 d x +6 c \right )}{5}+\frac {12 \cos \left (4 d x +4 c \right )}{5}+6 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+525 \left (A +\frac {B}{7}\right ) \left (4+\frac {\sin \left (7 d x +7 c \right )}{5}+\sin \left (5 d x +5 c \right )+\frac {9 \sin \left (3 d x +3 c \right )}{5}+\sin \left (d x +c \right )+\frac {2 \cos \left (6 d x +6 c \right )}{5}+\frac {12 \cos \left (4 d x +4 c \right )}{5}+6 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-8512 A -1216 B \right ) \cos \left (2 d x +2 c \right )+\left (-3752 A -536 B \right ) \cos \left (4 d x +4 c \right )+\left (-672 A -96 B \right ) \cos \left (6 d x +6 c \right )+\left (301 A +43 B \right ) \sin \left (3 d x +3 c \right )+\left (-735 A -105 B \right ) \sin \left (5 d x +5 c \right )+\left (-231 A -33 B \right ) \sin \left (7 d x +7 c \right )+\left (4389 A +627 B \right ) \sin \left (d x +c \right )-4920 A +2808 B}{384 a d \left (20+\sin \left (7 d x +7 c \right )+5 \sin \left (5 d x +5 c \right )+9 \sin \left (3 d x +3 c \right )+5 \sin \left (d x +c \right )+2 \cos \left (6 d x +6 c \right )+12 \cos \left (4 d x +4 c \right )+30 \cos \left (2 d x +2 c \right )\right )}\) | \(390\) |
| norman | \(\frac {\frac {\left (257 A +311 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{192 d a}+\frac {\left (257 A +311 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{192 d a}+\frac {5 \left (337 A +103 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{192 d a}+\frac {5 \left (337 A +103 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{192 d a}-\frac {\left (11 A -163 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{48 d a}-\frac {\left (11 A -163 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{48 d a}+\frac {\left (29 A +59 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{32 d a}+\frac {\left (29 A +59 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{32 d a}+\frac {5 \left (5 A +19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 d a}+\frac {\left (93 A -5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {\left (93 A -5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{64 d a}+\frac {\left (41 A +719 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{96 d a}+\frac {\left (41 A +719 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{96 d a}+\frac {\left (1363 A -299 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{192 d a}+\frac {\left (1363 A -299 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{192 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}-\frac {5 \left (7 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 d a}+\frac {5 \left (7 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 d a}\) | \(483\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (105 A +15 B +791 A \,{\mathrm e}^{4 i \left (d x +c \right )}+113 B \,{\mathrm e}^{4 i \left (d x +c \right )}+490 A \,{\mathrm e}^{2 i \left (d x +c \right )}+30 i B \,{\mathrm e}^{11 i \left (d x +c \right )}+1190 i A \,{\mathrm e}^{9 i \left (d x +c \right )}+210 i A \,{\mathrm e}^{11 i \left (d x +c \right )}+170 i B \,{\mathrm e}^{9 i \left (d x +c \right )}-210 i A \,{\mathrm e}^{i \left (d x +c \right )}-2772 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+70 B \,{\mathrm e}^{2 i \left (d x +c \right )}-1190 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-170 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-30 i B \,{\mathrm e}^{i \left (d x +c \right )}+70 B \,{\mathrm e}^{10 i \left (d x +c \right )}+105 A \,{\mathrm e}^{12 i \left (d x +c \right )}+15 B \,{\mathrm e}^{12 i \left (d x +c \right )}+490 A \,{\mathrm e}^{10 i \left (d x +c \right )}+791 A \,{\mathrm e}^{8 i \left (d x +c \right )}+300 A \,{\mathrm e}^{6 i \left (d x +c \right )}+113 B \,{\mathrm e}^{8 i \left (d x +c \right )}-3468 B \,{\mathrm e}^{6 i \left (d x +c \right )}+396 i B \,{\mathrm e}^{7 i \left (d x +c \right )}-396 i B \,{\mathrm e}^{5 i \left (d x +c \right )}+2772 i A \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{128 d a}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{128 d a}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{128 d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{128 d a}\) | \(487\) |
Input:
int(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d/a*(-5/32*A/(1+sin(d*x+c))-1/4*(1/16*A-1/16*B)/(1+sin(d*x+c))^4-1/3*(1/ 8*A-1/16*B)/(1+sin(d*x+c))^3-1/2*(5/32*A-1/32*B)/(1+sin(d*x+c))^2+(35/256* A+5/256*B)*ln(1+sin(d*x+c))+(-35/256*A-5/256*B)*ln(sin(d*x+c)-1)-1/2*(-5/6 4*A-3/64*B)/(sin(d*x+c)-1)^2-1/3*(1/32*A+1/32*B)/(sin(d*x+c)-1)^3-(15/128* A+5/128*B)/(sin(d*x+c)-1))
Time = 0.10 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.01 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {30 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} - 10 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left ({\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left ({\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (7 \, A + B\right )} \cos \left (d x + c\right )^{2} + 56 \, A + 8 \, B\right )} \sin \left (d x + c\right ) - 16 \, A - 112 \, B}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \] Input:
integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/768*(30*(7*A + B)*cos(d*x + c)^6 - 10*(7*A + B)*cos(d*x + c)^4 - 4*(7*A + B)*cos(d*x + c)^2 - 15*((7*A + B)*cos(d*x + c)^6*sin(d*x + c) + (7*A + B)*cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 15*((7*A + B)*cos(d*x + c)^6*si n(d*x + c) + (7*A + B)*cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(15*(7*A + B)*cos(d*x + c)^4 + 10*(7*A + B)*cos(d*x + c)^2 + 56*A + 8*B)*sin(d*x + c) - 16*A - 112*B)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*x + c)^6)
Timed out. \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {15 \, {\left (7 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {15 \, {\left (7 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{6} + 15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{5} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{4} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{3} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{2} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right ) - 48 \, A + 48 \, B\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a}}{768 \, d} \] Input:
integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="max ima")
Output:
1/768*(15*(7*A + B)*log(sin(d*x + c) + 1)/a - 15*(7*A + B)*log(sin(d*x + c ) - 1)/a - 2*(15*(7*A + B)*sin(d*x + c)^6 + 15*(7*A + B)*sin(d*x + c)^5 - 40*(7*A + B)*sin(d*x + c)^4 - 40*(7*A + B)*sin(d*x + c)^3 + 33*(7*A + B)*s in(d*x + c)^2 + 33*(7*A + B)*sin(d*x + c) - 48*A + 48*B)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)^4 + 3*a*sin(d* x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a))/d
Time = 0.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.77 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {5 \, {\left (7 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a d} - \frac {5 \, {\left (7 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a d} - \frac {15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{6} + 15 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{5} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{4} - 40 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{3} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right )^{2} + 33 \, {\left (7 \, A + B\right )} \sin \left (d x + c\right ) - 48 \, A + 48 \, B}{384 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{4} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="gia c")
Output:
5/256*(7*A + B)*log(abs(sin(d*x + c) + 1))/(a*d) - 5/256*(7*A + B)*log(abs (sin(d*x + c) - 1))/(a*d) - 1/384*(15*(7*A + B)*sin(d*x + c)^6 + 15*(7*A + B)*sin(d*x + c)^5 - 40*(7*A + B)*sin(d*x + c)^4 - 40*(7*A + B)*sin(d*x + c)^3 + 33*(7*A + B)*sin(d*x + c)^2 + 33*(7*A + B)*sin(d*x + c) - 48*A + 48 *B)/(a*d*(sin(d*x + c) + 1)^4*(sin(d*x + c) - 1)^3)
Time = 34.76 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\left (\frac {35\,A}{128}+\frac {5\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^6+\left (\frac {35\,A}{128}+\frac {5\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {35\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^4+\left (-\frac {35\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {77\,A}{128}+\frac {11\,B}{128}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {77\,A}{128}+\frac {11\,B}{128}\right )\,\sin \left (c+d\,x\right )-\frac {A}{8}+\frac {B}{8}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^7-a\,{\sin \left (c+d\,x\right )}^6+3\,a\,{\sin \left (c+d\,x\right )}^5+3\,a\,{\sin \left (c+d\,x\right )}^4-3\,a\,{\sin \left (c+d\,x\right )}^3-3\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )}+\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (7\,A+B\right )}{128\,a\,d} \] Input:
int((A + B*sin(c + d*x))/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)
Output:
(B/8 - A/8 + sin(c + d*x)*((77*A)/128 + (11*B)/128) - sin(c + d*x)^3*((35* A)/48 + (5*B)/48) - sin(c + d*x)^4*((35*A)/48 + (5*B)/48) + sin(c + d*x)^5 *((35*A)/128 + (5*B)/128) + sin(c + d*x)^6*((35*A)/128 + (5*B)/128) + sin( c + d*x)^2*((77*A)/128 + (11*B)/128))/(d*(a + a*sin(c + d*x) - 3*a*sin(c + d*x)^2 - 3*a*sin(c + d*x)^3 + 3*a*sin(c + d*x)^4 + 3*a*sin(c + d*x)^5 - a *sin(c + d*x)^6 - a*sin(c + d*x)^7)) + (5*atanh(sin(c + d*x))*(7*A + B))/( 128*a*d)
Time = 0.18 (sec) , antiderivative size = 911, normalized size of antiderivative = 4.12 \[ \int \frac {\sec ^7(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
( - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7*a - 15*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**7*b - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 *a - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*b + 315*log(tan((c + d*x )/2) - 1)*sin(c + d*x)**5*a + 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 *b + 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a + 45*log(tan((c + d*x )/2) - 1)*sin(c + d*x)**4*b - 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 3*a - 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*b - 315*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**2*a - 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 2*b + 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a + 15*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)*b + 105*log(tan((c + d*x)/2) - 1)*a + 15*log(tan((c + d*x)/2) - 1)*b + 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**7*a + 15*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**7*b + 105*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**6*a + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*b - 315*lo g(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*a - 45*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**5*b - 315*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a - 45*lo g(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 315*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**3*a + 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*b + 315*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 45*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**2*b - 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a - 15*log( tan((c + d*x)/2) + 1)*sin(c + d*x)*b - 105*log(tan((c + d*x)/2) + 1)*a ...