Integrand size = 21, antiderivative size = 217 \[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {b^5 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {1}{16 (a+b) d (1-\sin (c+d x))^2}+\frac {3 a+5 b}{16 (a+b)^2 d (1-\sin (c+d x))}-\frac {1}{16 (a-b) d (1+\sin (c+d x))^2}-\frac {3 a-5 b}{16 (a-b)^2 d (1+\sin (c+d x))} \] Output:
-1/16*(3*a^2+9*a*b+8*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*(3*a^2-9*a*b+8*b ^2)*ln(1+sin(d*x+c))/(a-b)^3/d-b^5*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/16/( a+b)/d/(1-sin(d*x+c))^2+1/16*(3*a+5*b)/(a+b)^2/d/(1-sin(d*x+c))-1/16/(a-b) /d/(1+sin(d*x+c))^2-1/16*(3*a-5*b)/(a-b)^2/d/(1+sin(d*x+c))
Time = 5.66 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {(a-b)^3 \left (3 a^2+9 a b+8 b^2\right ) \log (1-\sin (c+d x))-3 a^5 \log (1+\sin (c+d x))+10 a^3 b^2 \log (1+\sin (c+d x))-15 a b^4 \log (1+\sin (c+d x))-8 b^5 \log (1+\sin (c+d x))+16 b^5 \log (a+b \sin (c+d x))+8 b^3 \left (-a^2+b^2\right ) \sec ^2(c+d x)+4 b \left (a^2-b^2\right )^2 \sec ^4(c+d x)-2 a \left (3 a^4-10 a^2 b^2+7 b^4\right ) \sec (c+d x) \tan (c+d x)-4 a \left (a^2-b^2\right )^2 \sec ^3(c+d x) \tan (c+d x)}{16 (-a+b)^3 (a+b)^3 d} \] Input:
Integrate[Sec[c + d*x]^5/(a + b*Sin[c + d*x]),x]
Output:
((a - b)^3*(3*a^2 + 9*a*b + 8*b^2)*Log[1 - Sin[c + d*x]] - 3*a^5*Log[1 + S in[c + d*x]] + 10*a^3*b^2*Log[1 + Sin[c + d*x]] - 15*a*b^4*Log[1 + Sin[c + d*x]] - 8*b^5*Log[1 + Sin[c + d*x]] + 16*b^5*Log[a + b*Sin[c + d*x]] + 8* b^3*(-a^2 + b^2)*Sec[c + d*x]^2 + 4*b*(a^2 - b^2)^2*Sec[c + d*x]^4 - 2*a*( 3*a^4 - 10*a^2*b^2 + 7*b^4)*Sec[c + d*x]*Tan[c + d*x] - 4*a*(a^2 - b^2)^2* Sec[c + d*x]^3*Tan[c + d*x])/(16*(-a + b)^3*(a + b)^3*d)
Time = 0.53 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 477, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^5 \int \frac {1}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle \frac {\int \left (-\frac {b^6}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {b^3}{8 (a+b) (b-b \sin (c+d x))^3}+\frac {b^3}{8 (a-b) (\sin (c+d x) b+b)^3}+\frac {(3 a+5 b) b^2}{16 (a+b)^2 (b-b \sin (c+d x))^2}+\frac {(3 a-5 b) b^2}{16 (a-b)^2 (\sin (c+d x) b+b)^2}+\frac {\left (3 a^2+9 b a+8 b^2\right ) b}{16 (a+b)^3 (b-b \sin (c+d x))}+\frac {\left (3 a^2-9 b a+8 b^2\right ) b}{16 (a-b)^3 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b \left (3 a^2+9 a b+8 b^2\right ) \log (b-b \sin (c+d x))}{16 (a+b)^3}+\frac {b \left (3 a^2-9 a b+8 b^2\right ) \log (b \sin (c+d x)+b)}{16 (a-b)^3}-\frac {b^6 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}+\frac {b^3}{16 (a+b) (b-b \sin (c+d x))^2}-\frac {b^3}{16 (a-b) (b \sin (c+d x)+b)^2}+\frac {b^2 (3 a+5 b)}{16 (a+b)^2 (b-b \sin (c+d x))}-\frac {b^2 (3 a-5 b)}{16 (a-b)^2 (b \sin (c+d x)+b)}}{b d}\) |
Input:
Int[Sec[c + d*x]^5/(a + b*Sin[c + d*x]),x]
Output:
(-1/16*(b*(3*a^2 + 9*a*b + 8*b^2)*Log[b - b*Sin[c + d*x]])/(a + b)^3 - (b^ 6*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^3 + (b*(3*a^2 - 9*a*b + 8*b^2)*Log[ b + b*Sin[c + d*x]])/(16*(a - b)^3) + b^3/(16*(a + b)*(b - b*Sin[c + d*x]) ^2) + (b^2*(3*a + 5*b))/(16*(a + b)^2*(b - b*Sin[c + d*x])) - b^3/(16*(a - b)*(b + b*Sin[c + d*x])^2) - ((3*a - 5*b)*b^2)/(16*(a - b)^2*(b + b*Sin[c + d*x])))/(b*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 1.13 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -5 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a^{2}-9 a b +8 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3 a +5 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a^{2}-9 a b -8 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(190\) |
| default | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -5 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a^{2}-9 a b +8 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3 a +5 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a^{2}-9 a b -8 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(190\) |
| parallelrisch | \(\frac {-8 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) b^{5} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-3 \left (a -b \right )^{3} \left (a^{2}+3 a b +\frac {8}{3} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\left (a^{2}-3 a b +\frac {8}{3} b^{2}\right ) \left (a +b \right )^{2} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {11 \left (a -b \right ) \left (\frac {4 \left (a^{2} b -b^{3}\right ) \cos \left (2 d x +2 c \right )}{11}+\frac {b \left (a^{2}-3 b^{2}\right ) \cos \left (4 d x +4 c \right )}{11}+\frac {\left (3 a^{3}-7 a \,b^{2}\right ) \sin \left (3 d x +3 c \right )}{11}+\left (a^{3}-\frac {15}{11} a \,b^{2}\right ) \sin \left (d x +c \right )-\frac {5 a^{2} b}{11}+\frac {7 b^{3}}{11}\right )}{6}\right ) \left (a +b \right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} d \left (3+4 \cos \left (2 d x +2 c \right )+\cos \left (4 d x +4 c \right )\right )}\) | \(314\) |
| norman | \(\frac {-\frac {4 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {2 \left (a^{2} b -2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {2 \left (a^{2} b -2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {a \left (5 a^{2}-9 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {a \left (5 a^{2}-9 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {\left (3 a^{2}+b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {\left (3 a^{2}+b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {b^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {\left (3 a^{2}-9 a b +8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\left (3 a^{2}+9 a b +8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}\) | \(488\) |
| risch | \(\frac {2 i b^{5} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {9 i a b c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {2 i b^{5} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {9 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i a^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i b^{2} x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {3 i a^{2} c}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {9 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 i a^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i b^{2} c}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {3 i a^{2} c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {i b^{2} c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {9 i a b c}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {i b^{2} x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {-3 i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+7 i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-11 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+15 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+11 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-15 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+32 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-7 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{4 \left (-a^{2}+b^{2}\right )^{2} 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^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(964\) |
Input:
int(sec(d*x+c)^5/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(3*a-5*b)/(a-b)^2/(1+sin(d*x+c)) +1/16*(3*a^2-9*a*b+8*b^2)/(a-b)^3*ln(1+sin(d*x+c))-b^5/(a+b)^3/(a-b)^3*ln( a+b*sin(d*x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(3*a+5*b)/(a+b)^2/(sin (d*x+c)-1)+1/16/(a+b)^3*(-3*a^2-9*a*b-8*b^2)*ln(sin(d*x+c)-1))
Time = 0.21 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {16 \, b^{5} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 15 \, a b^{4} - 8 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{4} b - 8 \, a^{2} b^{3} + 4 \, b^{5} - 8 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} + {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/16*(16*b^5*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (3*a^5 - 10*a^ 3*b^2 + 15*a*b^4 - 8*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^4*b - 8*a^2*b^3 + 4*b^5 - 8*(a^2*b^3 - b^5)*cos(d*x + c)^2 - 2*(2*a^5 - 4*a^3* b^2 + 2*a*b^4 + (3*a^5 - 10*a^3*b^2 + 7*a*b^4)*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)
\[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**5/(a+b*sin(d*x+c)),x)
Output:
Integral(sec(c + d*x)**5/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.28 \[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, b^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, b^{3} \sin \left (d x + c\right )^{2} + {\left (3 \, a^{3} - 7 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{2} b - 6 \, b^{3} - {\left (5 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \] Input:
integrate(sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/16*(16*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^2 - 9*a*b + 8*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 + 9*a*b + 8*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a *b^2 + b^3) + 2*(4*b^3*sin(d*x + c)^2 + (3*a^3 - 7*a*b^2)*sin(d*x + c)^3 + 2*a^2*b - 6*b^3 - (5*a^3 - 9*a*b^2)*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 = 0.14 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} - \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} - \frac {2 \, a^{4} b - 8 \, a^{2} b^{3} + 6 \, b^{5} + {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} + 4 \, {\left (a^{2} b^{3} - b^{5}\right )} \sin \left (d x + c\right )^{2} - {\left (5 \, a^{5} - 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-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^2 + 9*a*b + 8*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^2 - 9*a*b + 8*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*a^4*b - 8 *a^2*b^3 + 6*b^5 + (3*a^5 - 10*a^3*b^2 + 7*a*b^4)*sin(d*x + c)^3 + 4*(a^2* b^3 - b^5)*sin(d*x + c)^2 - (5*a^5 - 14*a^3*b^2 + 9*a*b^4)*sin(d*x + c))/( (a + b)^3*(a - b)^3*d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 0.35 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.48 \[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b^2}{8\,{\left (a-b\right )}^3}-\frac {3\,b}{16\,{\left (a-b\right )}^2}+\frac {3}{16\,\left (a-b\right )}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {3\,b}{16\,{\left (a+b\right )}^2}+\frac {3}{16\,\left (a+b\right )}+\frac {b^2}{8\,{\left (a+b\right )}^3}\right )}{d}-\frac {\frac {a^2\,b-3\,b^3}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b^3\,{\sin \left (c+d\,x\right )}^2}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (7\,a\,b^2-3\,a^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\sin \left (c+d\,x\right )\,\left (9\,a\,b^2-5\,a^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2+{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^2\right )}-\frac {b^5\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \] Input:
int(1/(cos(c + d*x)^5*(a + b*sin(c + d*x))),x)
Output:
(log(sin(c + d*x) + 1)*(b^2/(8*(a - b)^3) - (3*b)/(16*(a - b)^2) + 3/(16*( a - b))))/d - (log(sin(c + d*x) - 1)*((3*b)/(16*(a + b)^2) + 3/(16*(a + b) ) + b^2/(8*(a + b)^3)))/d - ((a^2*b - 3*b^3)/(4*(a^4 + b^4 - 2*a^2*b^2)) + (b^3*sin(c + d*x)^2)/(2*(a^4 + b^4 - 2*a^2*b^2)) - (sin(c + d*x)^3*(7*a*b ^2 - 3*a^3))/(8*(a^4 + b^4 - 2*a^2*b^2)) + (sin(c + d*x)*(9*a*b^2 - 5*a^3) )/(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)) - (b^5*log(a + b*sin(c + d*x)))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3* a^4*b^2))
Time = 0.19 (sec) , antiderivative size = 1008, normalized size of antiderivative = 4.65 \[ \int \frac {\sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^5/(a+b*sin(d*x+c)),x)
Output:
( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**5 + 10*log(tan((c + d*x )/2) - 1)*sin(c + d*x)**4*a**3*b**2 - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**4 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**5 + 6*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5 - 20*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**2*a**3*b**2 + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2* a*b**4 - 16*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**5 - 3*log(tan((c + d*x)/2) - 1)*a**5 + 10*log(tan((c + d*x)/2) - 1)*a**3*b**2 - 15*log(tan( (c + d*x)/2) - 1)*a*b**4 + 8*log(tan((c + d*x)/2) - 1)*b**5 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**5 - 10*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3*b**2 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**4 + 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**5 - 6*log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**2*a**5 + 20*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 *a**3*b**2 - 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**4 - 16*log( tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**5 + 3*log(tan((c + d*x)/2) + 1)*a **5 - 10*log(tan((c + d*x)/2) + 1)*a**3*b**2 + 15*log(tan((c + d*x)/2) + 1 )*a*b**4 + 8*log(tan((c + d*x)/2) + 1)*b**5 - 8*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**4*b**5 + 16*log(tan((c + d*x)/2) **2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*b**5 - 8*log(tan((c + d* x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b**5 - 2*sin(c + d*x)**4*a**4*b + 4 *sin(c + d*x)**4*a**2*b**3 - 2*sin(c + d*x)**4*b**5 - 3*sin(c + d*x)**3...