Integrand size = 21, antiderivative size = 181 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {35 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}+\frac {15}{128 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {5}{32 d (a+a \sin (c+d x))}-\frac {a^7}{64 d \left (a^2+a^2 \sin (c+d x)\right )^4}+\frac {5 a^7}{128 d \left (a^4-a^4 \sin (c+d x)\right )^2}-\frac {5 a^7}{64 d \left (a^4+a^4 \sin (c+d x)\right )^2} \] Output:
35/128*arctanh(sin(d*x+c))/a/d+1/96*a^2/d/(a-a*sin(d*x+c))^3+15/128/d/(a-a *sin(d*x+c))-1/24*a^2/d/(a+a*sin(d*x+c))^3-5/32/d/(a+a*sin(d*x+c))-1/64*a^ 7/d/(a^2+a^2*sin(d*x+c))^4+5/128*a^7/d/(a^4-a^4*sin(d*x+c))^2-5/64*a^7/d/( a^4+a^4*sin(d*x+c))^2
Time = 0.52 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^6(c+d x) \left (48-105 \text {arctanh}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^8-231 \sin (c+d x)-231 \sin ^2(c+d x)+280 \sin ^3(c+d x)+280 \sin ^4(c+d x)-105 \sin ^5(c+d x)-105 \sin ^6(c+d x)\right )}{384 a d (1+\sin (c+d x))} \] Input:
Integrate[Sec[c + d*x]^7/(a + a*Sin[c + d*x]),x]
Output:
-1/384*(Sec[c + d*x]^6*(48 - 105*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^6*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8 - 231*Sin[c + d*x] - 231*Sin[c + d*x]^2 + 280*Sin[c + d*x]^3 + 280*Sin[c + d*x]^4 - 105 *Sin[c + d*x]^5 - 105*Sin[c + d*x]^6))/(a*d*(1 + Sin[c + d*x]))
Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.87, 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.
| \(\displaystyle \int \frac {\sec ^7(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^7 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a^7 \int \frac {1}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^7 \int \left (\frac {15}{128 a^7 (a-a \sin (c+d x))^2}+\frac {5}{32 a^7 (\sin (c+d x) a+a)^2}+\frac {5}{64 a^6 (a-a \sin (c+d x))^3}+\frac {5}{32 a^6 (\sin (c+d x) a+a)^3}+\frac {1}{32 a^5 (a-a \sin (c+d x))^4}+\frac {1}{8 a^5 (\sin (c+d x) a+a)^4}+\frac {1}{16 a^4 (\sin (c+d x) a+a)^5}+\frac {35}{128 a^7 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^7 \left (\frac {35 \text {arctanh}(\sin (c+d x))}{128 a^8}+\frac {15}{128 a^7 (a-a \sin (c+d x))}-\frac {5}{32 a^7 (a \sin (c+d x)+a)}+\frac {5}{128 a^6 (a-a \sin (c+d x))^2}-\frac {5}{64 a^6 (a \sin (c+d x)+a)^2}+\frac {1}{96 a^5 (a-a \sin (c+d x))^3}-\frac {1}{24 a^5 (a \sin (c+d x)+a)^3}-\frac {1}{64 a^4 (a \sin (c+d x)+a)^4}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^7/(a + a*Sin[c + d*x]),x]
Output:
(a^7*((35*ArcTanh[Sin[c + d*x]])/(128*a^8) + 1/(96*a^5*(a - a*Sin[c + d*x] )^3) + 5/(128*a^6*(a - a*Sin[c + d*x])^2) + 15/(128*a^7*(a - a*Sin[c + d*x ])) - 1/(64*a^4*(a + a*Sin[c + d*x])^4) - 1/(24*a^5*(a + a*Sin[c + d*x])^3 ) - 5/(64*a^6*(a + a*Sin[c + d*x])^2) - 5/(32*a^7*(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 = 1.93 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.64
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {35 \ln \left (1+\sin \left (d x +c \right )\right )}{256}-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {15}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {35 \ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d a}\) | \(115\) |
| default | \(\frac {-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {35 \ln \left (1+\sin \left (d x +c \right )\right )}{256}-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {15}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {35 \ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d a}\) | \(115\) |
| risch | \(-\frac {i \left (210 i {\mathrm e}^{12 i \left (d x +c \right )}+105 \,{\mathrm e}^{13 i \left (d x +c \right )}+1190 i {\mathrm e}^{10 i \left (d x +c \right )}+490 \,{\mathrm e}^{11 i \left (d x +c \right )}+2772 i {\mathrm e}^{8 i \left (d x +c \right )}+791 \,{\mathrm e}^{9 i \left (d x +c \right )}-2772 i {\mathrm e}^{6 i \left (d x +c \right )}+300 \,{\mathrm e}^{7 i \left (d x +c \right )}-1190 i {\mathrm e}^{4 i \left (d x +c \right )}+791 \,{\mathrm e}^{5 i \left (d x +c \right )}-210 i {\mathrm e}^{2 i \left (d x +c \right )}+490 \,{\mathrm e}^{3 i \left (d x +c \right )}+105 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 a d}\) | \(231\) |
| norman | \(\frac {\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{16 d a}+\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16 d a}+\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16 d a}+\frac {93 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {93 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{64 a d}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{32 d a}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{32 a d}-\frac {109 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{96 d a}-\frac {109 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{96 a d}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{96 d a}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{96 a d}+\frac {1385 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{192 d a}+\frac {1385 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{192 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}-\frac {35 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 a d}+\frac {35 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 a d}\) | \(315\) |
| parallelrisch | \(\frac {\left (-525 \sin \left (5 d x +5 c \right )-105 \sin \left (7 d x +7 c \right )-3150 \cos \left (2 d x +2 c \right )-1260 \cos \left (4 d x +4 c \right )-210 \cos \left (6 d x +6 c \right )-525 \sin \left (d x +c \right )-945 \sin \left (3 d x +3 c \right )-2100\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (525 \sin \left (5 d x +5 c \right )+105 \sin \left (7 d x +7 c \right )+3150 \cos \left (2 d x +2 c \right )+1260 \cos \left (4 d x +4 c \right )+210 \cos \left (6 d x +6 c \right )+525 \sin \left (d x +c \right )+945 \sin \left (3 d x +3 c \right )+2100\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-735 \sin \left (5 d x +5 c \right )-231 \sin \left (7 d x +7 c \right )-8512 \cos \left (2 d x +2 c \right )-3752 \cos \left (4 d x +4 c \right )-672 \cos \left (6 d x +6 c \right )+4389 \sin \left (d x +c \right )+301 \sin \left (3 d x +3 c \right )-4920}{384 a d \left (20+\sin \left (7 d x +7 c \right )+5 \sin \left (5 d x +5 c \right )+9 \sin \left (3 d x +3 c \right )+5 \sin \left (d x +c \right )+2 \cos \left (6 d x +6 c \right )+12 \cos \left (4 d x +4 c \right )+30 \cos \left (2 d x +2 c \right )\right )}\) | \(339\) |
Input:
int(sec(d*x+c)^7/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/64/(1+sin(d*x+c))^4-1/24/(1+sin(d*x+c))^3-5/64/(1+sin(d*x+c))^2- 5/32/(1+sin(d*x+c))+35/256*ln(1+sin(d*x+c))-1/96/(sin(d*x+c)-1)^3+5/128/(s in(d*x+c)-1)^2-15/128/(sin(d*x+c)-1)-35/256*ln(sin(d*x+c)-1))
Time = 0.09 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.92 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {210 \, \cos \left (d x + c\right )^{6} - 70 \, \cos \left (d x + c\right )^{4} - 28 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 14 \, {\left (15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 16}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \] Input:
integrate(sec(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/768*(210*cos(d*x + c)^6 - 70*cos(d*x + c)^4 - 28*cos(d*x + c)^2 - 105*( cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 105* (cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 14 *(15*cos(d*x + c)^4 + 10*cos(d*x + c)^2 + 8)*sin(d*x + c) - 16)/(a*d*cos(d *x + c)^6*sin(d*x + c) + a*d*cos(d*x + c)^6)
\[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sec ^{7}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**7/(a+a*sin(d*x+c)),x)
Output:
Integral(sec(c + d*x)**7/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{6} + 105 \, \sin \left (d x + c\right )^{5} - 280 \, \sin \left (d x + c\right )^{4} - 280 \, \sin \left (d x + c\right )^{3} + 231 \, \sin \left (d x + c\right )^{2} + 231 \, \sin \left (d x + c\right ) - 48\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \] Input:
integrate(sec(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/768*(2*(105*sin(d*x + c)^6 + 105*sin(d*x + c)^5 - 280*sin(d*x + c)^4 - 280*sin(d*x + c)^3 + 231*sin(d*x + c)^2 + 231*sin(d*x + c) - 48)/(a*sin(d* x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)^4 + 3* a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a) - 105*log(sin( d*x + c) + 1)/a + 105*log(sin(d*x + c) - 1)/a)/d
Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {35 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a d} - \frac {35 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a d} - \frac {105 \, \sin \left (d x + c\right )^{6} + 105 \, \sin \left (d x + c\right )^{5} - 280 \, \sin \left (d x + c\right )^{4} - 280 \, \sin \left (d x + c\right )^{3} + 231 \, \sin \left (d x + c\right )^{2} + 231 \, \sin \left (d x + c\right ) - 48}{384 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{4} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(sec(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
35/256*log(abs(sin(d*x + c) + 1))/(a*d) - 35/256*log(abs(sin(d*x + c) - 1) )/(a*d) - 1/384*(105*sin(d*x + c)^6 + 105*sin(d*x + c)^5 - 280*sin(d*x + c )^4 - 280*sin(d*x + c)^3 + 231*sin(d*x + c)^2 + 231*sin(d*x + c) - 48)/(a* d*(sin(d*x + c) + 1)^4*(sin(d*x + c) - 1)^3)
Time = 0.24 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {35\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{128\,a\,d}+\frac {\frac {35\,{\sin \left (c+d\,x\right )}^6}{128}+\frac {35\,{\sin \left (c+d\,x\right )}^5}{128}-\frac {35\,{\sin \left (c+d\,x\right )}^4}{48}-\frac {35\,{\sin \left (c+d\,x\right )}^3}{48}+\frac {77\,{\sin \left (c+d\,x\right )}^2}{128}+\frac {77\,\sin \left (c+d\,x\right )}{128}-\frac {1}{8}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^7-a\,{\sin \left (c+d\,x\right )}^6+3\,a\,{\sin \left (c+d\,x\right )}^5+3\,a\,{\sin \left (c+d\,x\right )}^4-3\,a\,{\sin \left (c+d\,x\right )}^3-3\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \] Input:
int(1/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)
Output:
(35*atanh(sin(c + d*x)))/(128*a*d) + ((77*sin(c + d*x))/128 + (77*sin(c + d*x)^2)/128 - (35*sin(c + d*x)^3)/48 - (35*sin(c + d*x)^4)/48 + (35*sin(c + d*x)^5)/128 + (35*sin(c + d*x)^6)/128 - 1/8)/(d*(a + a*sin(c + d*x) - 3* a*sin(c + d*x)^2 - 3*a*sin(c + d*x)^3 + 3*a*sin(c + d*x)^4 + 3*a*sin(c + d *x)^5 - a*sin(c + d*x)^6 - a*sin(c + d*x)^7))
Time = 0.21 (sec) , antiderivative size = 470, normalized size of antiderivative = 2.60 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^7/(a+a*sin(d*x+c)),x)
Output:
( - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 - 105*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)**6 + 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 + 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 - 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 - 315*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 105* log(tan((c + d*x)/2) - 1)*sin(c + d*x) + 105*log(tan((c + d*x)/2) - 1) + 1 05*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**7 + 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 - 315*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 - 315*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 315*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**3 + 315*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - 105*log(t an((c + d*x)/2) + 1)*sin(c + d*x) - 105*log(tan((c + d*x)/2) + 1) - 231*si n(c + d*x)**7 - 336*sin(c + d*x)**6 + 588*sin(c + d*x)**5 + 973*sin(c + d* x)**4 - 413*sin(c + d*x)**3 - 924*sin(c + d*x)**2 + 279)/(384*a*d*(sin(c + d*x)**7 + sin(c + d*x)**6 - 3*sin(c + d*x)**5 - 3*sin(c + d*x)**4 + 3*sin (c + d*x)**3 + 3*sin(c + d*x)**2 - sin(c + d*x) - 1))