Integrand size = 29, antiderivative size = 118 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(5 a A-b B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{6 d}+\frac {(5 a A-b B) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(5 a A-b B) \sec ^3(c+d x) \tan (c+d x)}{24 d} \]
1/16*(5*A*a-B*b)*arctanh(sin(d*x+c))/d+1/6*sec(d*x+c)^6*(A*b+B*a+(A*a+B*b) *sin(d*x+c))/d+1/16*(5*A*a-B*b)*sec(d*x+c)*tan(d*x+c)/d+1/24*(5*A*a-B*b)*s ec(d*x+c)^3*tan(d*x+c)/d
Time = 0.67 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.88 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\sec ^6(c+d x) \left (-8 (A b+a B)-3 (5 a A-b B) \text {arctanh}(\sin (c+d x)) \cos ^6(c+d x)-3 (11 a A+b B) \sin (c+d x)+8 (5 a A-b B) \sin ^3(c+d x)+(-15 a A+3 b B) \sin ^5(c+d x)\right )}{48 d} \]
-1/48*(Sec[c + d*x]^6*(-8*(A*b + a*B) - 3*(5*a*A - b*B)*ArcTanh[Sin[c + d* x]]*Cos[c + d*x]^6 - 3*(11*a*A + b*B)*Sin[c + d*x] + 8*(5*a*A - b*B)*Sin[c + d*x]^3 + (-15*a*A + 3*b*B)*Sin[c + d*x]^5))/d
Time = 0.54 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.86, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x)) (A+B \sin (c+d x))}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^7 \int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b)}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^4}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^6 \int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^4}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^6 \int \left (-\frac {A b-a (2 A-B)}{16 b^4 (\sin (c+d x) b+b)^3}+\frac {5 a A-b B}{16 b^5 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {b (A-B)+a (5 A+B)}{32 b^5 (b-b \sin (c+d x))^2}+\frac {a (5 A-B)-b (A+B)}{32 b^5 (\sin (c+d x) b+b)^2}+\frac {A b+a (2 A+B)}{16 b^4 (b-b \sin (c+d x))^3}+\frac {(a+b) (A+B)}{16 b^3 (b-b \sin (c+d x))^4}+\frac {(a-b) (A-B)}{16 b^3 (\sin (c+d x) b+b)^4}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^6 \left (\frac {(5 a A-b B) \text {arctanh}(\sin (c+d x))}{16 b^6}+\frac {a (5 A+B)+b (A-B)}{32 b^5 (b-b \sin (c+d x))}-\frac {a (5 A-B)-b (A+B)}{32 b^5 (b \sin (c+d x)+b)}+\frac {A b-a (2 A-B)}{32 b^4 (b \sin (c+d x)+b)^2}+\frac {a (2 A+B)+A b}{32 b^4 (b-b \sin (c+d x))^2}+\frac {(a+b) (A+B)}{48 b^3 (b-b \sin (c+d x))^3}-\frac {(a-b) (A-B)}{48 b^3 (b \sin (c+d x)+b)^3}\right )}{d}\) |
(b^6*(((5*a*A - b*B)*ArcTanh[Sin[c + d*x]])/(16*b^6) + ((a + b)*(A + B))/( 48*b^3*(b - b*Sin[c + d*x])^3) + (A*b + a*(2*A + B))/(32*b^4*(b - b*Sin[c + d*x])^2) + (b*(A - B) + a*(5*A + B))/(32*b^5*(b - b*Sin[c + d*x])) - ((a - b)*(A - B))/(48*b^3*(b + b*Sin[c + d*x])^3) + (A*b - a*(2*A - B))/(32*b ^4*(b + b*Sin[c + d*x])^2) - (a*(5*A - B) - b*(A + B))/(32*b^5*(b + b*Sin[ c + d*x]))))/d
3.16.35.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Time = 1.15 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.43
| method | result | size |
| derivativedivides | \(\frac {a A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B a}{6 \cos \left (d x +c \right )^{6}}+\frac {A b}{6 \cos \left (d x +c \right )^{6}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(169\) |
| default | \(\frac {a A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B a}{6 \cos \left (d x +c \right )^{6}}+\frac {A b}{6 \cos \left (d x +c \right )^{6}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(169\) |
| parallelrisch | \(\frac {-15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (a A -\frac {B b}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (a A -\frac {B b}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-120 A b -120 B a \right ) \cos \left (2 d x +2 c \right )+\left (-48 A b -48 B a \right ) \cos \left (4 d x +4 c \right )+\left (-8 A b -8 B a \right ) \cos \left (6 d x +6 c \right )+\left (170 a A -34 B b \right ) \sin \left (3 d x +3 c \right )+\left (30 a A -6 B b \right ) \sin \left (5 d x +5 c \right )+\left (396 a A +228 B b \right ) \sin \left (d x +c \right )+176 A b +176 B a}{48 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\) | \(271\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (15 A a \,{\mathrm e}^{10 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{10 i \left (d x +c \right )}+85 A a \,{\mathrm e}^{8 i \left (d x +c \right )}-17 B b \,{\mathrm e}^{8 i \left (d x +c \right )}+198 A a \,{\mathrm e}^{6 i \left (d x +c \right )}+114 B b \,{\mathrm e}^{6 i \left (d x +c \right )}-198 A a \,{\mathrm e}^{4 i \left (d x +c \right )}+256 i A b \,{\mathrm e}^{5 i \left (d x +c \right )}-114 B b \,{\mathrm e}^{4 i \left (d x +c \right )}+256 i B a \,{\mathrm e}^{5 i \left (d x +c \right )}-85 A a \,{\mathrm e}^{2 i \left (d x +c \right )}+17 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-15 a A +3 B b \right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{16 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{16 d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{16 d}\) | \(303\) |
| norman | \(\frac {\frac {\left (11 a A +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (11 a A +B b \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {7 \left (19 a A +25 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {7 \left (19 a A +25 B b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (71 a A +53 B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (71 a A +53 B b \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (275 a A +281 B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (275 a A +281 B b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (A b +B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (A b +B a \right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 \left (4 A b +4 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 A b +13 B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 A b +13 B a \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\left (5 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {\left (5 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\) | \(449\) |
1/d*(a*A*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec(d*x+c))*tan(d*x+c )+5/16*ln(sec(d*x+c)+tan(d*x+c)))+1/6*B*a/cos(d*x+c)^6+1/6*A*b/cos(d*x+c)^ 6+B*b*(1/6*sin(d*x+c)^3/cos(d*x+c)^6+1/8*sin(d*x+c)^3/cos(d*x+c)^4+1/16*si n(d*x+c)^3/cos(d*x+c)^2+1/16*sin(d*x+c)-1/16*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.14 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, B a + 16 \, A b + 2 \, {\left (3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{2} + 8 \, A a + 8 \, B b\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \]
1/96*(3*(5*A*a - B*b)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a - B* b)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 16*B*a + 16*A*b + 2*(3*(5*A*a - B*b)*cos(d*x + c)^4 + 2*(5*A*a - B*b)*cos(d*x + c)^2 + 8*A*a + 8*B*b)*sin (d*x + c))/(d*cos(d*x + c)^6)
Timed out. \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.21 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A a - B b\right )} \sin \left (d x + c\right )^{5} - 8 \, {\left (5 \, A a - B b\right )} \sin \left (d x + c\right )^{3} + 8 \, B a + 8 \, A b + 3 \, {\left (11 \, A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \]
1/96*(3*(5*A*a - B*b)*log(sin(d*x + c) + 1) - 3*(5*A*a - B*b)*log(sin(d*x + c) - 1) - 2*(3*(5*A*a - B*b)*sin(d*x + c)^5 - 8*(5*A*a - B*b)*sin(d*x + c)^3 + 8*B*a + 8*A*b + 3*(11*A*a + B*b)*sin(d*x + c))/(sin(d*x + c)^6 - 3* sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1))/d
Time = 0.36 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.18 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a \sin \left (d x + c\right )^{5} - 3 \, B b \sin \left (d x + c\right )^{5} - 40 \, A a \sin \left (d x + c\right )^{3} + 8 \, B b \sin \left (d x + c\right )^{3} + 33 \, A a \sin \left (d x + c\right ) + 3 \, B b \sin \left (d x + c\right ) + 8 \, B a + 8 \, A b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \]
1/96*(3*(5*A*a - B*b)*log(abs(sin(d*x + c) + 1)) - 3*(5*A*a - B*b)*log(abs (sin(d*x + c) - 1)) - 2*(15*A*a*sin(d*x + c)^5 - 3*B*b*sin(d*x + c)^5 - 40 *A*a*sin(d*x + c)^3 + 8*B*b*sin(d*x + c)^3 + 33*A*a*sin(d*x + c) + 3*B*b*s in(d*x + c) + 8*B*a + 8*A*b)/(sin(d*x + c)^2 - 1)^3)/d
Time = 12.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {5\,A\,a}{16}-\frac {B\,b}{16}\right )}{d}-\frac {\left (\frac {5\,A\,a}{16}-\frac {B\,b}{16}\right )\,{\sin \left (c+d\,x\right )}^5+\left (\frac {B\,b}{6}-\frac {5\,A\,a}{6}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {11\,A\,a}{16}+\frac {B\,b}{16}\right )\,\sin \left (c+d\,x\right )+\frac {A\,b}{6}+\frac {B\,a}{6}}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \]