Integrand size = 29, antiderivative size = 143 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(3 a A-b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(a+b) (A+B)}{16 d (1-\sin (c+d x))^2}+\frac {b (A-B)+a (3 A+B)}{16 d (1-\sin (c+d x))}-\frac {(a-b) (A-B)}{16 d (1+\sin (c+d x))^2}-\frac {a (3 A-B)-b (A+B)}{16 d (1+\sin (c+d x))} \] Output:
1/8*(3*A*a-B*b)*arctanh(sin(d*x+c))/d+1/16*(a+b)*(A+B)/d/(1-sin(d*x+c))^2+ 1/16*(b*(A-B)+a*(3*A+B))/d/(1-sin(d*x+c))-1/16*(a-b)*(A-B)/d/(1+sin(d*x+c) )^2-1/16*(a*(3*A-B)-b*(A+B))/d/(1+sin(d*x+c))
Time = 0.59 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.57 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\sec ^4(c+d x) \left (2 (A b+a B)+(3 a A-b B) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+(5 a A+b B) \sin (c+d x)+(-3 a A+b B) \sin ^3(c+d x)\right )}{8 d} \] Input:
Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(Sec[c + d*x]^4*(2*(A*b + a*B) + (3*a*A - b*B)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (5*a*A + b*B)*Sin[c + d*x] + (-3*a*A + b*B)*Sin[c + d*x]^3))/( 8*d)
Time = 0.45 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.10, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x)) (A+B \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b)}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^4 \int \frac {(a+b \sin (c+d x)) (A b+B \sin (c+d x) b)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle -\frac {b^4 \int \left (-\frac {(a-b) (A-B)}{8 b^2 (\sin (c+d x) b+b)^3}-\frac {3 a A-b B}{8 b^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {b (A-B)+a (3 A+B)}{16 b^3 (b-b \sin (c+d x))^2}-\frac {a (3 A-B)-b (A+B)}{16 b^3 (\sin (c+d x) b+b)^2}-\frac {(a+b) (A+B)}{8 b^2 (b-b \sin (c+d x))^3}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^4 \left (-\frac {(3 a A-b B) \text {arctanh}(\sin (c+d x))}{8 b^4}-\frac {a (3 A+B)+b (A-B)}{16 b^3 (b-b \sin (c+d x))}+\frac {a (3 A-B)-b (A+B)}{16 b^3 (b \sin (c+d x)+b)}+\frac {(a-b) (A-B)}{16 b^2 (b \sin (c+d x)+b)^2}-\frac {(a+b) (A+B)}{16 b^2 (b-b \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
-((b^4*(-1/8*((3*a*A - b*B)*ArcTanh[Sin[c + d*x]])/b^4 - ((a + b)*(A + B)) /(16*b^2*(b - b*Sin[c + d*x])^2) - (b*(A - B) + a*(3*A + B))/(16*b^3*(b - b*Sin[c + d*x])) + ((a - b)*(A - B))/(16*b^2*(b + b*Sin[c + d*x])^2) + (a* (3*A - B) - b*(A + B))/(16*b^3*(b + b*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[((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 = 1.75 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B a}{4 \cos \left (d x +c \right )^{4}}+\frac {A b}{4 \cos \left (d x +c \right )^{4}}+B b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(141\) |
| default | \(\frac {a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B a}{4 \cos \left (d x +c \right )^{4}}+\frac {A b}{4 \cos \left (d x +c \right )^{4}}+B b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(141\) |
| parallelrisch | \(\frac {-12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a A -\frac {B b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a A -\frac {B b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-8 A b -8 B a \right ) \cos \left (2 d x +2 c \right )+\left (-2 A b -2 B a \right ) \cos \left (4 d x +4 c \right )+\left (6 a A -2 B b \right ) \sin \left (3 d x +3 c \right )+\left (22 a A +14 B b \right ) \sin \left (d x +c \right )+10 A b +10 B a}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(200\) |
| risch | \(\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (-3 A a \,{\mathrm e}^{6 i \left (d x +c \right )}+B b \,{\mathrm e}^{6 i \left (d x +c \right )}-11 A a \,{\mathrm e}^{4 i \left (d x +c \right )}-7 B b \,{\mathrm e}^{4 i \left (d x +c \right )}+11 A a \,{\mathrm e}^{2 i \left (d x +c \right )}-16 i A b \,{\mathrm e}^{3 i \left (d x +c \right )}+7 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-16 i B a \,{\mathrm e}^{3 i \left (d x +c \right )}+3 a A -B b \right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{8 d}\) | \(242\) |
| norman | \(\frac {\frac {\left (4 A b +4 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (4 A b +4 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {\left (5 a A +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (5 a A +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {\left (7 a A +11 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {\left (7 a A +11 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}+\frac {\left (13 a A +9 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {\left (13 a A +9 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {2 \left (A b +B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 \left (A b +B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {4 \left (A b +B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {\left (3 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(349\) |
Input:
int(sec(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d*(a*A*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c) +tan(d*x+c)))+1/4*B*a/cos(d*x+c)^4+1/4*A*b/cos(d*x+c)^4+B*b*(1/4*sin(d*x+c )^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d *x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.80 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, B a + 4 \, A b + 2 \, {\left ({\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{2} + 2 \, A a + 2 \, B b\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fri cas")
Output:
1/16*((3*A*a - B*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*A*a - B*b)*c os(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*B*a + 4*A*b + 2*((3*A*a - B*b)*co s(d*x + c)^2 + 2*A*a + 2*B*b)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\int \left (A + B \sin {\left (c + d x \right )}\right ) \left (a + b \sin {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
Integral((A + B*sin(c + d*x))*(a + b*sin(c + d*x))*sec(c + d*x)**5, x)
Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (3 \, A a - B b\right )} \sin \left (d x + c\right )^{3} - 2 \, B a - 2 \, A b - {\left (5 \, A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="max ima")
Output:
1/16*((3*A*a - B*b)*log(sin(d*x + c) + 1) - (3*A*a - B*b)*log(sin(d*x + c) - 1) - 2*((3*A*a - B*b)*sin(d*x + c)^3 - 2*B*a - 2*A*b - (5*A*a + B*b)*si n(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (3 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {3 \, A a \sin \left (d x + c\right )^{3} - B b \sin \left (d x + c\right )^{3} - 5 \, A a \sin \left (d x + c\right ) - B b \sin \left (d x + c\right ) - 2 \, B a - 2 \, A b}{8 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2} d} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="gia c")
Output:
1/16*(3*A*a - B*b)*log(abs(sin(d*x + c) + 1))/d - 1/16*(3*A*a - B*b)*log(a bs(sin(d*x + c) - 1))/d - 1/8*(3*A*a*sin(d*x + c)^3 - B*b*sin(d*x + c)^3 - 5*A*a*sin(d*x + c) - B*b*sin(d*x + c) - 2*B*a - 2*A*b)/((sin(d*x + c)^2 - 1)^2*d)
Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.64 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\left (\frac {B\,b}{8}-\frac {3\,A\,a}{8}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,A\,a}{8}+\frac {B\,b}{8}\right )\,\sin \left (c+d\,x\right )+\frac {A\,b}{4}+\frac {B\,a}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {3\,A\,a}{8}-\frac {B\,b}{8}\right )}{d} \] Input:
int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x)))/cos(c + d*x)^5,x)
Output:
((A*b)/4 + (B*a)/4 + sin(c + d*x)*((5*A*a)/8 + (B*b)/8) - sin(c + d*x)^3*( (3*A*a)/8 - (B*b)/8))/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) + (atanh (sin(c + d*x))*((3*A*a)/8 - (B*b)/8))/d
Time = 0.18 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.55 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-4 \sin \left (d x +c \right )^{4} a b -3 \sin \left (d x +c \right )^{3} a^{2}+\sin \left (d x +c \right )^{3} b^{2}+8 \sin \left (d x +c \right )^{2} a b +5 \sin \left (d x +c \right ) a^{2}+\sin \left (d x +c \right ) b^{2}}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 + log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**4*b**2 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2* a**2 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 - 3*log(tan((c + d *x)/2) - 1)*a**2 + log(tan((c + d*x)/2) - 1)*b**2 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2 - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b** 2 - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 + 2*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**2*b**2 + 3*log(tan((c + d*x)/2) + 1)*a**2 - log(tan ((c + d*x)/2) + 1)*b**2 - 4*sin(c + d*x)**4*a*b - 3*sin(c + d*x)**3*a**2 + sin(c + d*x)**3*b**2 + 8*sin(c + d*x)**2*a*b + 5*sin(c + d*x)*a**2 + sin( c + d*x)*b**2)/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))