Integrand size = 21, antiderivative size = 136 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {3}{16 d (a+a \sin (c+d x))}-\frac {a^5}{24 d \left (a^2+a^2 \sin (c+d x)\right )^3}+\frac {a^5}{32 d \left (a^3-a^3 \sin (c+d x)\right )^2}-\frac {3 a^5}{32 d \left (a^3+a^3 \sin (c+d x)\right )^2} \] Output:
5/16*arctanh(sin(d*x+c))/a/d+1/8/d/(a-a*sin(d*x+c))-3/16/d/(a+a*sin(d*x+c) )-1/24*a^5/d/(a^2+a^2*sin(d*x+c))^3+1/32*a^5/d/(a^3-a^3*sin(d*x+c))^2-3/32 *a^5/d/(a^3+a^3*sin(d*x+c))^2
Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sec ^4(c+d x) \left (-8+25 \sin (c+d x)+25 \sin ^2(c+d x)-15 \sin ^3(c+d x)-15 \sin ^4(c+d x)+15 \text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x))^2 (1+\sin (c+d x))^3\right )}{48 a d (1+\sin (c+d x))} \] Input:
Integrate[Sec[c + d*x]^5/(a + a*Sin[c + d*x]),x]
Output:
(Sec[c + d*x]^4*(-8 + 25*Sin[c + d*x] + 25*Sin[c + d*x]^2 - 15*Sin[c + d*x ]^3 - 15*Sin[c + d*x]^4 + 15*ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x])^2*( 1 + Sin[c + d*x])^3))/(48*a*d*(1 + Sin[c + d*x]))
Time = 0.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 54, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^5 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a^5 \int \frac {1}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^5 \int \left (\frac {1}{8 a^5 (a-a \sin (c+d x))^2}+\frac {3}{16 a^5 (\sin (c+d x) a+a)^2}+\frac {1}{16 a^4 (a-a \sin (c+d x))^3}+\frac {3}{16 a^4 (\sin (c+d x) a+a)^3}+\frac {1}{8 a^3 (\sin (c+d x) a+a)^4}+\frac {5}{16 a^5 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^5 \left (\frac {5 \text {arctanh}(\sin (c+d x))}{16 a^6}+\frac {1}{8 a^5 (a-a \sin (c+d x))}-\frac {3}{16 a^5 (a \sin (c+d x)+a)}+\frac {1}{32 a^4 (a-a \sin (c+d x))^2}-\frac {3}{32 a^4 (a \sin (c+d x)+a)^2}-\frac {1}{24 a^3 (a \sin (c+d x)+a)^3}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^5/(a + a*Sin[c + d*x]),x]
Output:
(a^5*((5*ArcTanh[Sin[c + d*x]])/(16*a^6) + 1/(32*a^4*(a - a*Sin[c + d*x])^ 2) + 1/(8*a^5*(a - a*Sin[c + d*x])) - 1/(24*a^3*(a + a*Sin[c + d*x])^3) - 3/(32*a^4*(a + a*Sin[c + d*x])^2) - 3/(16*a^5*(a + a*Sin[c + d*x]))))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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 + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.89 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.67
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}}{d a}\) | \(91\) |
| default | \(\frac {\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32}}{d a}\) | \(91\) |
| risch | \(-\frac {i \left (30 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}+110 i {\mathrm e}^{6 i \left (d x +c \right )}+40 \,{\mathrm e}^{7 i \left (d x +c \right )}-110 i {\mathrm e}^{4 i \left (d x +c \right )}+18 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}+40 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{16 a d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 a d}\) | \(185\) |
| norman | \(\frac {\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4 d a}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{12 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{12 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}+\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 a d}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 a d}\) | \(239\) |
| parallelrisch | \(\frac {\left (-120 \cos \left (2 d x +2 c \right )-30 \cos \left (4 d x +4 c \right )-30 \sin \left (d x +c \right )-45 \sin \left (3 d x +3 c \right )-15 \sin \left (5 d x +5 c \right )-90\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (120 \cos \left (2 d x +2 c \right )+30 \cos \left (4 d x +4 c \right )+30 \sin \left (d x +c \right )+45 \sin \left (3 d x +3 c \right )+15 \sin \left (5 d x +5 c \right )+90\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-16 \cos \left (2 d x +2 c \right )-14 \cos \left (4 d x +4 c \right )+236 \sin \left (d x +c \right )+84 \sin \left (3 d x +3 c \right )+8 \sin \left (5 d x +5 c \right )+30}{48 a d \left (\sin \left (5 d x +5 c \right )+3 \sin \left (3 d x +3 c \right )+2 \sin \left (d x +c \right )+2 \cos \left (4 d x +4 c \right )+8 \cos \left (2 d x +2 c \right )+6\right )}\) | \(251\) |
Input:
int(sec(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/32/(sin(d*x+c)-1)^2-1/8/(sin(d*x+c)-1)-5/32*ln(sin(d*x+c)-1)-1/24 /(1+sin(d*x+c))^3-3/32/(1+sin(d*x+c))^2-3/16/(1+sin(d*x+c))+5/32*ln(1+sin( d*x+c)))
Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 4}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \] Input:
integrate(sec(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/96*(30*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(sin(d*x + c) + 1) + 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 10*(3*cos(d*x + c)^2 + 2)* sin(d*x + c) - 4)/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)
\[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**5/(a+a*sin(d*x+c)),x)
Output:
Integral(sec(c + d*x)**5/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 15 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 25 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \] Input:
integrate(sec(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/96*(2*(15*sin(d*x + c)^4 + 15*sin(d*x + c)^3 - 25*sin(d*x + c)^2 - 25*s in(d*x + c) + 8)/(a*sin(d*x + c)^5 + a*sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) - 15*log(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)/a)/d
Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.77 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{32 \, a d} - \frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{32 \, a d} - \frac {15 \, \sin \left (d x + c\right )^{4} + 15 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 25 \, \sin \left (d x + c\right ) + 8}{48 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{3} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
5/32*log(abs(sin(d*x + c) + 1))/(a*d) - 5/32*log(abs(sin(d*x + c) - 1))/(a *d) - 1/48*(15*sin(d*x + c)^4 + 15*sin(d*x + c)^3 - 25*sin(d*x + c)^2 - 25 *sin(d*x + c) + 8)/(a*d*(sin(d*x + c) + 1)^3*(sin(d*x + c) - 1)^2)
Time = 25.44 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{16\,a\,d}-\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{16}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{16}-\frac {25\,{\sin \left (c+d\,x\right )}^2}{48}-\frac {25\,\sin \left (c+d\,x\right )}{48}+\frac {1}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \] Input:
int(1/(cos(c + d*x)^5*(a + a*sin(c + d*x))),x)
Output:
(5*atanh(sin(c + d*x)))/(16*a*d) - ((5*sin(c + d*x)^3)/16 - (25*sin(c + d* x)^2)/48 - (25*sin(c + d*x))/48 + (5*sin(c + d*x)^4)/16 + 1/6)/(d*(a + a*s in(c + d*x) - 2*a*sin(c + d*x)^2 - 2*a*sin(c + d*x)^3 + a*sin(c + d*x)^4 + a*sin(c + d*x)^5))
Time = 0.19 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.50 \[ \int \frac {\sec ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{5}-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{5}+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-25 \sin \left (d x +c \right )^{5}-40 \sin \left (d x +c \right )^{4}+35 \sin \left (d x +c \right )^{3}+75 \sin \left (d x +c \right )^{2}-33}{48 a d \left (\sin \left (d x +c \right )^{5}+\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )^{2}+\sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^5/(a+a*sin(d*x+c)),x)
Output:
( - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 30* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 15*log(tan((c + d*x)/2) - 1)*s in(c + d*x) - 15*log(tan((c + d*x)/2) - 1) + 15*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**5 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 30*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**3 - 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 15*log(tan((c + d*x )/2) + 1) - 25*sin(c + d*x)**5 - 40*sin(c + d*x)**4 + 35*sin(c + d*x)**3 + 75*sin(c + d*x)**2 - 33)/(48*a*d*(sin(c + d*x)**5 + sin(c + d*x)**4 - 2*s in(c + d*x)**3 - 2*sin(c + d*x)**2 + sin(c + d*x) + 1))