3.9.89 \(\int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx\) [889]

3.9.89.1 Optimal result
3.9.89.2 Mathematica [A] (verified)
3.9.89.3 Rubi [A] (verified)
3.9.89.4 Maple [A] (verified)
3.9.89.5 Fricas [A] (verification not implemented)
3.9.89.6 Sympy [F]
3.9.89.7 Maxima [A] (verification not implemented)
3.9.89.8 Giac [A] (verification not implemented)
3.9.89.9 Mupad [B] (verification not implemented)

3.9.89.1 Optimal result

Integrand size = 21, antiderivative size = 165 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {35 \text {arctanh}(\sin (c+d x))}{128 a d}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}+\frac {5 a}{128 d (a-a \sin (c+d x))^2}+\frac {15}{128 d (a-a \sin (c+d x))}-\frac {a^3}{64 d (a+a \sin (c+d x))^4}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {5 a}{64 d (a+a \sin (c+d x))^2}-\frac {5}{32 d (a+a \sin (c+d x))} \]

output
35/128*arctanh(sin(d*x+c))/a/d+1/96*a^2/d/(a-a*sin(d*x+c))^3+5/128*a/d/(a- 
a*sin(d*x+c))^2+15/128/d/(a-a*sin(d*x+c))-1/64*a^3/d/(a+a*sin(d*x+c))^4-1/ 
24*a^2/d/(a+a*sin(d*x+c))^3-5/64*a/d/(a+a*sin(d*x+c))^2-5/32/d/(a+a*sin(d* 
x+c))
 
3.9.89.2 Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.88 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^6(c+d x) \left (48-105 \text {arctanh}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^8-231 \sin (c+d x)-231 \sin ^2(c+d x)+280 \sin ^3(c+d x)+280 \sin ^4(c+d x)-105 \sin ^5(c+d x)-105 \sin ^6(c+d x)\right )}{384 a d (1+\sin (c+d x))} \]

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]))
 
3.9.89.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96, 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
 

3.9.89.3.1 Defintions of rubi rules used

rule 54
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])
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3146
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])
 
3.9.89.4 Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.70

method result size
derivativedivides \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {15}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {35 \ln \left (\sin \left (d x +c \right )-1\right )}{256}-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {35 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) \(115\)
default \(\frac {-\frac {1}{96 \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {5}{128 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {15}{128 \left (\sin \left (d x +c \right )-1\right )}-\frac {35 \ln \left (\sin \left (d x +c \right )-1\right )}{256}-\frac {1}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{24 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {35 \ln \left (1+\sin \left (d x +c \right )\right )}{256}}{d a}\) \(115\)
risch \(-\frac {i \left (210 i {\mathrm e}^{12 i \left (d x +c \right )}+105 \,{\mathrm e}^{13 i \left (d x +c \right )}+1190 i {\mathrm e}^{10 i \left (d x +c \right )}+490 \,{\mathrm e}^{11 i \left (d x +c \right )}+2772 i {\mathrm e}^{8 i \left (d x +c \right )}+791 \,{\mathrm e}^{9 i \left (d x +c \right )}-2772 i {\mathrm e}^{6 i \left (d x +c \right )}+300 \,{\mathrm e}^{7 i \left (d x +c \right )}-1190 i {\mathrm e}^{4 i \left (d x +c \right )}+791 \,{\mathrm e}^{5 i \left (d x +c \right )}-210 i {\mathrm e}^{2 i \left (d x +c \right )}+490 \,{\mathrm e}^{3 i \left (d x +c \right )}+105 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{192 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d a}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 d a}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a d}\) \(231\)
norman \(\frac {\frac {25 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {25 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {25 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {93 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}+\frac {93 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d a}+\frac {29 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {29 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}-\frac {109 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {109 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}-\frac {11 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d a}+\frac {1385 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}+\frac {1385 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}-\frac {35 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{128 a d}+\frac {35 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{128 a d}\) \(315\)
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*(-1/96/(sin(d*x+c)-1)^3+5/128/(sin(d*x+c)-1)^2-15/128/(sin(d*x+c)-1) 
-35/256*ln(sin(d*x+c)-1)-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)))
 
3.9.89.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.01 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {210 \, \cos \left (d x + c\right )^{6} - 70 \, \cos \left (d x + c\right )^{4} - 28 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 14 \, {\left (15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 16}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

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)
 
3.9.89.6 Sympy [F]

\[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sec ^{7}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

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
 
3.9.89.7 Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{6} + 105 \, \sin \left (d x + c\right )^{5} - 280 \, \sin \left (d x + c\right )^{4} - 280 \, \sin \left (d x + c\right )^{3} + 231 \, \sin \left (d x + c\right )^{2} + 231 \, \sin \left (d x + c\right ) - 48\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]

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
 
3.9.89.8 Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (385 \, \sin \left (d x + c\right )^{3} - 1335 \, \sin \left (d x + c\right )^{2} + 1575 \, \sin \left (d x + c\right ) - 641\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {875 \, \sin \left (d x + c\right )^{4} + 3980 \, \sin \left (d x + c\right )^{3} + 6930 \, \sin \left (d x + c\right )^{2} + 5548 \, \sin \left (d x + c\right ) + 1771}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

input
integrate(sec(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="giac")
 
output
1/3072*(420*log(abs(sin(d*x + c) + 1))/a - 420*log(abs(sin(d*x + c) - 1))/ 
a + 2*(385*sin(d*x + c)^3 - 1335*sin(d*x + c)^2 + 1575*sin(d*x + c) - 641) 
/(a*(sin(d*x + c) - 1)^3) - (875*sin(d*x + c)^4 + 3980*sin(d*x + c)^3 + 69 
30*sin(d*x + c)^2 + 5548*sin(d*x + c) + 1771)/(a*(sin(d*x + c) + 1)^4))/d
 
3.9.89.9 Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {35\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{128\,a\,d}+\frac {\frac {35\,{\sin \left (c+d\,x\right )}^6}{128}+\frac {35\,{\sin \left (c+d\,x\right )}^5}{128}-\frac {35\,{\sin \left (c+d\,x\right )}^4}{48}-\frac {35\,{\sin \left (c+d\,x\right )}^3}{48}+\frac {77\,{\sin \left (c+d\,x\right )}^2}{128}+\frac {77\,\sin \left (c+d\,x\right )}{128}-\frac {1}{8}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^7-a\,{\sin \left (c+d\,x\right )}^6+3\,a\,{\sin \left (c+d\,x\right )}^5+3\,a\,{\sin \left (c+d\,x\right )}^4-3\,a\,{\sin \left (c+d\,x\right )}^3-3\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]

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))