Integrand size = 31, antiderivative size = 277 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\left (3 a^2 A+a b (9 A+B)+b^2 (8 A+3 B)\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (3 a^2 A+b^2 (8 A-3 B)-a b (9 A-B)\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {b^4 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {A+B}{16 (a+b) d (1-\sin (c+d x))^2}+\frac {3 a A+5 A b+a B+3 b B}{16 (a+b)^2 d (1-\sin (c+d x))}-\frac {A-B}{16 (a-b) d (1+\sin (c+d x))^2}-\frac {3 a A-5 A b-a B+3 b B}{16 (a-b)^2 d (1+\sin (c+d x))} \] Output:
-1/16*(3*A*a^2+a*b*(9*A+B)+b^2*(8*A+3*B))*ln(1-sin(d*x+c))/(a+b)^3/d+1/16* (3*A*a^2+b^2*(8*A-3*B)-a*b*(9*A-B))*ln(1+sin(d*x+c))/(a-b)^3/d-b^4*(A*b-B* a)*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/16*(A+B)/(a+b)/d/(1-sin(d*x+c))^2+1/ 16*(3*A*a+5*A*b+B*a+3*B*b)/(a+b)^2/d/(1-sin(d*x+c))-1/16*(A-B)/(a-b)/d/(1+ sin(d*x+c))^2-1/16*(3*A*a-5*A*b-B*a+3*B*b)/(a-b)^2/d/(1+sin(d*x+c))
Time = 6.26 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {b^5 \left (-\frac {\sec ^4(c+d x) \left (-A b^2+a b B-b (-a A+b B) \sin (c+d x)\right )}{4 b^6 \left (-a^2+b^2\right )}+\frac {\frac {-\frac {(a-b)^2 \left (3 a^2 A+a b (9 A+B)+b^2 (8 A+3 B)\right ) \log (1-\sin (c+d x))}{2 b (a+b)}+\frac {(a+b)^2 \left (3 a^2 A+b^2 (8 A-3 B)-a b (9 A-B)\right ) \log (1+\sin (c+d x))}{2 (a-b) b}-\frac {8 b^3 (A b-a B) \log (a+b \sin (c+d x))}{(a-b) (a+b)}}{2 b^2 \left (-a^2+b^2\right )}-\frac {\sec ^2(c+d x) \left (-3 a b^2 (a A-b B)-b^2 \left (-3 a^2 A+4 A b^2-a b B\right )-b \left (-3 b^2 (a A-b B)-a \left (-3 a^2 A+4 A b^2-a b B\right )\right ) \sin (c+d x)\right )}{2 b^4 \left (-a^2+b^2\right )}}{4 b^2 \left (-a^2+b^2\right )}\right )}{d} \] Input:
Integrate[(Sec[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
Output:
(b^5*(-1/4*(Sec[c + d*x]^4*(-(A*b^2) + a*b*B - b*(-(a*A) + b*B)*Sin[c + d* x]))/(b^6*(-a^2 + b^2)) + ((-1/2*((a - b)^2*(3*a^2*A + a*b*(9*A + B) + b^2 *(8*A + 3*B))*Log[1 - Sin[c + d*x]])/(b*(a + b)) + ((a + b)^2*(3*a^2*A + b ^2*(8*A - 3*B) - a*b*(9*A - B))*Log[1 + Sin[c + d*x]])/(2*(a - b)*b) - (8* b^3*(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*(a*A - b*B) - b^2*(-3*a^2*A + 4*A*b^2 - a*b*B) - b*(-3*b^2*(a*A - b*B) - a*(-3*a^2*A + 4*A*b^2 - a*b*B))*Sin[c + d*x]))/(2*b^4*(-a^2 + b^2)))/(4*b^2*(-a^2 + b^2))))/d
Time = 0.74 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.04, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {A b+B \sin (c+d x) b}{b (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^4 \int \frac {A b+B \sin (c+d x) b}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle -\frac {b^4 \int \left (-\frac {3 A a^2-b (9 A-B) a+b^2 (8 A-3 B)}{16 (a-b)^3 b^4 (\sin (c+d x) b+b)}-\frac {3 A a^2+b (9 A+B) a+b^2 (8 A+3 B)}{16 b^4 (a+b)^3 (b-b \sin (c+d x))}+\frac {A b-a B}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {3 a A+5 b A+a B+3 b B}{16 b^3 (a+b)^2 (b-b \sin (c+d x))^2}-\frac {3 a A-5 b A-a B+3 b B}{16 (a-b)^2 b^3 (\sin (c+d x) b+b)^2}-\frac {A+B}{8 b^2 (a+b) (b-b \sin (c+d x))^3}-\frac {A-B}{8 (a-b) b^2 (\sin (c+d x) b+b)^3}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^4 \left (\frac {(A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}+\frac {\left (3 a^2 A+a b (9 A+B)+b^2 (8 A+3 B)\right ) \log (b-b \sin (c+d x))}{16 b^4 (a+b)^3}-\frac {\left (3 a^2 A-a b (9 A-B)+b^2 (8 A-3 B)\right ) \log (b \sin (c+d x)+b)}{16 b^4 (a-b)^3}-\frac {3 a A+a B+5 A b+3 b B}{16 b^3 (a+b)^2 (b-b \sin (c+d x))}+\frac {3 a A-a B-5 A b+3 b B}{16 b^3 (a-b)^2 (b \sin (c+d x)+b)}+\frac {A-B}{16 b^2 (a-b) (b \sin (c+d x)+b)^2}-\frac {A+B}{16 b^2 (a+b) (b-b \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[(Sec[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
Output:
-((b^4*(((3*a^2*A + a*b*(9*A + B) + b^2*(8*A + 3*B))*Log[b - b*Sin[c + d*x ]])/(16*b^4*(a + b)^3) + ((A*b - a*B)*Log[a + b*Sin[c + d*x]])/(a^2 - b^2) ^3 - ((3*a^2*A + b^2*(8*A - 3*B) - a*b*(9*A - B))*Log[b + b*Sin[c + d*x]]) /(16*(a - b)^3*b^4) - (A + B)/(16*b^2*(a + b)*(b - b*Sin[c + d*x])^2) - (3 *a*A + 5*A*b + a*B + 3*b*B)/(16*b^3*(a + b)^2*(b - b*Sin[c + d*x])) + (A - B)/(16*(a - b)*b^2*(b + b*Sin[c + d*x])^2) + (3*a*A - 5*A*b - a*B + 3*b*B )/(16*(a - b)^2*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 = 4.43 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {-\frac {-A -B}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3 a A +5 A b +B a +3 B b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 A \,a^{2}-9 A a b -8 b^{2} A -a b B -3 B \,b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {\left (A b -B a \right ) b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {A -B}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a A -5 A b -B a +3 B b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 A \,a^{2}-9 A a b +8 b^{2} A +a b B -3 B \,b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}}{d}\) | \(256\) |
| default | \(\frac {-\frac {-A -B}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3 a A +5 A b +B a +3 B b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 A \,a^{2}-9 A a b -8 b^{2} A -a b B -3 B \,b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {\left (A b -B a \right ) b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {A -B}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a A -5 A b -B a +3 B b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 A \,a^{2}-9 A a b +8 b^{2} A +a b B -3 B \,b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}}{d}\) | \(256\) |
| parallelrisch | \(\frac {-16 b^{4} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A b -B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-6 \left (a -b \right )^{3} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (\left (\frac {8 A}{3}+B \right ) b^{2}+3 \left (A +\frac {B}{9}\right ) a b +A \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+11 \left (\frac {6 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right )^{2} \left (\left (\frac {8 A}{3}-B \right ) b^{2}-3 \left (A -\frac {B}{9}\right ) a b +A \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{11}+\left (a -b \right ) \left (\frac {4 \left (a -b \right ) \left (a +b \right ) \left (A b -B a \right ) \cos \left (2 d x +2 c \right )}{11}+\frac {\left (a^{2}-3 b^{2}\right ) \left (A b -B a \right ) \cos \left (4 d x +4 c \right )}{11}+\frac {\left (3 a^{3} A -7 A a \,b^{2}+B \,a^{2} b +3 B \,b^{3}\right ) \sin \left (3 d x +3 c \right )}{11}+\left (a^{3} A -\frac {15}{11} A a \,b^{2}-\frac {7}{11} B \,a^{2} b +B \,b^{3}\right ) \sin \left (d x +c \right )-\frac {5 \left (a^{2}-\frac {7 b^{2}}{5}\right ) \left (A b -B a \right )}{11}\right )\right ) \left (a +b \right )}{4 \left (a -b \right )^{3} \left (a +b \right )^{3} d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(395\) |
| norman | \(\frac {\frac {2 \left (a A -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d \left (a^{2}-b^{2}\right )}+\frac {2 \left (a A -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d \left (a^{2}-b^{2}\right )}+\frac {\left (5 a^{3} A -9 A a \,b^{2}-B \,a^{2} b +5 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (5 a^{3} A -9 A a \,b^{2}-B \,a^{2} b +5 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (2 A \,a^{2} b -4 A \,b^{3}-2 a^{3} B +4 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (2 A \,a^{2} b -4 A \,b^{3}-2 a^{3} B +4 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (3 a^{3} A +A a \,b^{2}-7 B \,a^{2} b +3 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 \left (A \,a^{2} b -a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 \left (A \,a^{2} b -a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}}{\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 )}+\frac {\left (3 A \,a^{2}-9 A a b +8 b^{2} A +a b B -3 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\left (3 A \,a^{2}+9 A a b +8 b^{2} A +a b B +3 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {b^{4} \left (A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(698\) |
| risch | \(\text {Expression too large to display}\) | \(1751\) |
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*(-1/2*(-A-B)/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(3*A*a+5*A*b+B*a+3*B*b)/( a+b)^2/(sin(d*x+c)-1)+1/16/(a+b)^3*(-3*A*a^2-9*A*a*b-8*A*b^2-B*a*b-3*B*b^2 )*ln(sin(d*x+c)-1)-(A*b-B*a)*b^4/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))-1/2*(A -B)/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(3*A*a-5*A*b-B*a+3*B*b)/(a-b)^2/(1+sin (d*x+c))+1/16*(3*A*a^2-9*A*a*b+8*A*b^2+B*a*b-3*B*b^2)/(a-b)^3*ln(1+sin(d*x +c)))
Time = 0.92 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {4 \, B a^{5} - 4 \, A a^{4} b - 8 \, B a^{3} b^{2} + 8 \, A a^{2} b^{3} + 4 \, B a b^{4} - 4 \, A b^{5} + 16 \, {\left (B a b^{4} - A b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (3 \, A a^{5} + B a^{4} b - 10 \, A a^{3} b^{2} - 6 \, B a^{2} b^{3} + {\left (15 \, A - 8 \, B\right )} a b^{4} + {\left (8 \, A - 3 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a^{5} + B a^{4} b - 10 \, A a^{3} b^{2} - 6 \, B a^{2} b^{3} + {\left (15 \, A + 8 \, B\right )} a b^{4} - {\left (8 \, A + 3 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A a^{5} - 2 \, B a^{4} b - 4 \, A a^{3} b^{2} + 4 \, B a^{2} b^{3} + 2 \, A a b^{4} - 2 \, B b^{5} + {\left (3 \, A a^{5} + B a^{4} b - 10 \, A a^{3} b^{2} + 2 \, B a^{2} b^{3} + 7 \, A a b^{4} - 3 \, B b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \] 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*(4*B*a^5 - 4*A*a^4*b - 8*B*a^3*b^2 + 8*A*a^2*b^3 + 4*B*a*b^4 - 4*A*b^ 5 + 16*(B*a*b^4 - A*b^5)*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + (3*A*a^5 + B*a^4*b - 10*A*a^3*b^2 - 6*B*a^2*b^3 + (15*A - 8*B)*a*b^4 + (8*A - 3*B) *b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*A*a^5 + B*a^4*b - 10*A*a^3 *b^2 - 6*B*a^2*b^3 + (15*A + 8*B)*a*b^4 - (8*A + 3*B)*b^5)*cos(d*x + c)^4* log(-sin(d*x + c) + 1) - 8*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d *x + c)^2 + 2*(2*A*a^5 - 2*B*a^4*b - 4*A*a^3*b^2 + 4*B*a^2*b^3 + 2*A*a*b^4 - 2*B*b^5 + (3*A*a^5 + B*a^4*b - 10*A*a^3*b^2 + 2*B*a^2*b^3 + 7*A*a*b^4 - 3*B*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^ 6)*d*cos(d*x + c)^4)
\[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\int \frac {\left (A + B \sin {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] 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))*sec(c + d*x)**5/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, {\left (B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, A a^{2} - {\left (9 \, A - B\right )} a b + {\left (8 \, A - 3 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (3 \, A a^{2} + {\left (9 \, A + B\right )} a b + {\left (8 \, A + 3 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (2 \, B a^{3} - 2 \, A a^{2} b - 6 \, B a b^{2} + 6 \, A b^{3} - {\left (3 \, A a^{3} + B a^{2} b - 7 \, A a b^{2} + 3 \, B b^{3}\right )} \sin \left (d x + c\right )^{3} + 4 \, {\left (B a b^{2} - A b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (5 \, A a^{3} - B a^{2} b - 9 \, A a b^{2} + 5 \, B b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \] 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*(16*(B*a*b^4 - A*b^5)*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^ 2*b^4 - b^6) + (3*A*a^2 - (9*A - B)*a*b + (8*A - 3*B)*b^2)*log(sin(d*x + c ) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*A*a^2 + (9*A + B)*a*b + (8*A + 3*B)*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(2*B* a^3 - 2*A*a^2*b - 6*B*a*b^2 + 6*A*b^3 - (3*A*a^3 + B*a^2*b - 7*A*a*b^2 + 3 *B*b^3)*sin(d*x + c)^3 + 4*(B*a*b^2 - A*b^3)*sin(d*x + c)^2 + (5*A*a^3 - B *a^2*b - 9*A*a*b^2 + 5*B*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d *x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c) ^2))/d
Time = 0.37 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.54 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {{\left (B a b^{5} - A b^{6}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} - \frac {{\left (3 \, A a^{2} + 9 \, A a b + B a b + 8 \, A b^{2} + 3 \, B b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {{\left (3 \, A a^{2} - 9 \, A a b + B a b + 8 \, A b^{2} - 3 \, B b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} + \frac {2 \, B a^{5} - 2 \, A a^{4} b - 8 \, B a^{3} b^{2} + 8 \, A a^{2} b^{3} + 6 \, B a b^{4} - 6 \, A b^{5} - {\left (3 \, A a^{5} + B a^{4} b - 10 \, A a^{3} b^{2} + 2 \, B a^{2} b^{3} + 7 \, A a b^{4} - 3 \, B b^{5}\right )} \sin \left (d x + c\right )^{3} + 4 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \sin \left (d x + c\right )^{2} + {\left (5 \, A a^{5} - B a^{4} b - 14 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} + 9 \, A a b^{4} - 5 \, B b^{5}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="gia c")
Output:
(B*a*b^5 - A*b^6)*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 3*a^4*b^3*d + 3* a^2*b^5*d - b^7*d) - 1/16*(3*A*a^2 + 9*A*a*b + B*a*b + 8*A*b^2 + 3*B*b^2)* log(abs(-sin(d*x + c) + 1))/(a^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) + 1/16 *(3*A*a^2 - 9*A*a*b + B*a*b + 8*A*b^2 - 3*B*b^2)*log(abs(-sin(d*x + c) - 1 ))/(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) + 1/8*(2*B*a^5 - 2*A*a^4*b - 8* B*a^3*b^2 + 8*A*a^2*b^3 + 6*B*a*b^4 - 6*A*b^5 - (3*A*a^5 + B*a^4*b - 10*A* a^3*b^2 + 2*B*a^2*b^3 + 7*A*a*b^4 - 3*B*b^5)*sin(d*x + c)^3 + 4*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5)*sin(d*x + c)^2 + (5*A*a^5 - B*a^4*b - 14*A *a^3*b^2 + 6*B*a^2*b^3 + 9*A*a*b^4 - 5*B*b^5)*sin(d*x + c))/((a + b)^3*(a - b)^3*d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 36.71 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.54 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {B\,a^3-A\,a^2\,b-3\,B\,a\,b^2+3\,A\,b^3}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\sin \left (c+d\,x\right )\,\left (5\,A\,a^3-B\,a^2\,b-9\,A\,a\,b^2+5\,B\,b^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (A\,b^3-B\,a\,b^2\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (3\,A\,a^3+B\,a^2\,b-7\,A\,a\,b^2+3\,B\,b^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2+{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (3\,A\,a^2+\left (9\,A+B\right )\,a\,b+\left (8\,A+3\,B\right )\,b^2\right )}{d\,\left (16\,a^3+48\,a^2\,b+48\,a\,b^2+16\,b^3\right )}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b^5-B\,a\,b^4\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (3\,A\,a^2+\left (B-9\,A\right )\,a\,b+\left (8\,A-3\,B\right )\,b^2\right )}{d\,\left (16\,a^3-48\,a^2\,b+48\,a\,b^2-16\,b^3\right )} \] Input:
int((A + B*sin(c + d*x))/(cos(c + d*x)^5*(a + b*sin(c + d*x))),x)
Output:
((3*A*b^3 + B*a^3 - A*a^2*b - 3*B*a*b^2)/(4*(a^4 + b^4 - 2*a^2*b^2)) + (si n(c + d*x)*(5*A*a^3 + 5*B*b^3 - 9*A*a*b^2 - B*a^2*b))/(8*(a^4 + b^4 - 2*a^ 2*b^2)) - (sin(c + d*x)^2*(A*b^3 - B*a*b^2))/(2*(a^4 + b^4 - 2*a^2*b^2)) - (sin(c + d*x)^3*(3*A*a^3 + 3*B*b^3 - 7*A*a*b^2 + B*a^2*b))/(8*(a^4 + b^4 - 2*a^2*b^2)))/(d*(cos(c + d*x)^2 - sin(c + d*x)^2 + sin(c + d*x)^4)) - (l og(sin(c + d*x) - 1)*(3*A*a^2 + b^2*(8*A + 3*B) + a*b*(9*A + B)))/(d*(48*a *b^2 + 48*a^2*b + 16*a^3 + 16*b^3)) - (log(a + b*sin(c + d*x))*(A*b^5 - B* a*b^4))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (log(sin(c + d*x) + 1)*( 3*A*a^2 + b^2*(8*A - 3*B) - a*b*(9*A - B)))/(d*(48*a*b^2 - 48*a^2*b + 16*a ^3 - 16*b^3))
Time = 0.18 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.58 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \sin \left (d x +c \right )^{3}+5 \sin \left (d x +c \right )}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] 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 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 3*log(tan((c + d*x)/2) - 1) + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 3*lo g(tan((c + d*x)/2) + 1) - 3*sin(c + d*x)**3 + 5*sin(c + d*x))/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))