3.1.46 \(\int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx\) [46]

3.1.46.1 Optimal result
3.1.46.2 Mathematica [A] (verified)
3.1.46.3 Rubi [A] (verified)
3.1.46.4 Maple [A] (verified)
3.1.46.5 Fricas [A] (verification not implemented)
3.1.46.6 Sympy [F(-1)]
3.1.46.7 Maxima [A] (verification not implemented)
3.1.46.8 Giac [A] (verification not implemented)
3.1.46.9 Mupad [B] (verification not implemented)

3.1.46.1 Optimal result

Integrand size = 19, antiderivative size = 162 \[ \int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {128 a^8 \log (1-\sin (c+d x))}{d}-\frac {64 a^8 \sin (c+d x)}{d}-\frac {16 a^5 (a+a \sin (c+d x))^3}{3 d}-\frac {4 a^3 (a+a \sin (c+d x))^5}{5 d}-\frac {a^2 (a+a \sin (c+d x))^6}{3 d}-\frac {a (a+a \sin (c+d x))^7}{7 d}-\frac {2 \left (a^2+a^2 \sin (c+d x)\right )^4}{d}-\frac {16 \left (a^4+a^4 \sin (c+d x)\right )^2}{d} \]

output
-128*a^8*ln(1-sin(d*x+c))/d-64*a^8*sin(d*x+c)/d-16/3*a^5*(a+a*sin(d*x+c))^ 
3/d-4/5*a^3*(a+a*sin(d*x+c))^5/d-1/3*a^2*(a+a*sin(d*x+c))^6/d-1/7*a*(a+a*s 
in(d*x+c))^7/d-2*(a^2+a^2*sin(d*x+c))^4/d-16*(a^4+a^4*sin(d*x+c))^2/d
 
3.1.46.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.59 \[ \int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {a^8 \left (-128 \log (1-\sin (c+d x))-127 \sin (c+d x)-60 \sin ^2(c+d x)-33 \sin ^3(c+d x)-16 \sin ^4(c+d x)-\frac {29}{5} \sin ^5(c+d x)-\frac {4}{3} \sin ^6(c+d x)-\frac {1}{7} \sin ^7(c+d x)\right )}{d} \]

input
Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^8,x]
 
output
(a^8*(-128*Log[1 - Sin[c + d*x]] - 127*Sin[c + d*x] - 60*Sin[c + d*x]^2 - 
33*Sin[c + d*x]^3 - 16*Sin[c + d*x]^4 - (29*Sin[c + d*x]^5)/5 - (4*Sin[c + 
 d*x]^6)/3 - Sin[c + d*x]^7/7))/d
 
3.1.46.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3146, 49, 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.

\(\displaystyle \int \sec (c+d x) (a \sin (c+d x)+a)^8 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^8}{\cos (c+d x)}dx\)

\(\Big \downarrow \) 3146

\(\displaystyle \frac {a \int \frac {(\sin (c+d x) a+a)^7}{a-a \sin (c+d x)}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {a \int \left (\frac {128 a^7}{a-a \sin (c+d x)}-64 a^6-32 (\sin (c+d x) a+a) a^5-16 (\sin (c+d x) a+a)^2 a^4-8 (\sin (c+d x) a+a)^3 a^3-4 (\sin (c+d x) a+a)^4 a^2-2 (\sin (c+d x) a+a)^5 a-(\sin (c+d x) a+a)^6\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \left (-64 a^7 \sin (c+d x)-128 a^7 \log (a-a \sin (c+d x))-16 a^5 (a \sin (c+d x)+a)^2-\frac {16}{3} a^4 (a \sin (c+d x)+a)^3-2 a^3 (a \sin (c+d x)+a)^4-\frac {4}{5} a^2 (a \sin (c+d x)+a)^5-\frac {1}{3} a (a \sin (c+d x)+a)^6-\frac {1}{7} (a \sin (c+d x)+a)^7\right )}{d}\)

input
Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^8,x]
 
output
(a*(-128*a^7*Log[a - a*Sin[c + d*x]] - 64*a^7*Sin[c + d*x] - 16*a^5*(a + a 
*Sin[c + d*x])^2 - (16*a^4*(a + a*Sin[c + d*x])^3)/3 - 2*a^3*(a + a*Sin[c 
+ d*x])^4 - (4*a^2*(a + a*Sin[c + d*x])^5)/5 - (a*(a + a*Sin[c + d*x])^6)/ 
3 - (a + a*Sin[c + d*x])^7/7))/d
 

3.1.46.3.1 Defintions of rubi rules used

rule 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[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])
 
3.1.46.4 Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.70

method result size
parallelrisch \(-\frac {a^{8} \left (1720320 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-860160 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+244720-259560 \cos \left (2 d x +2 c \right )-67935 \sin \left (3 d x +3 c \right )+15120 \cos \left (4 d x +4 c \right )-280 \cos \left (6 d x +6 c \right )+1044645 \sin \left (d x +c \right )+2541 \sin \left (5 d x +5 c \right )-15 \sin \left (7 d x +7 c \right )\right )}{6720 d}\) \(113\)
risch \(\frac {9949 i a^{8} {\mathrm e}^{i \left (d x +c \right )}}{128 d}+128 i a^{8} x +\frac {256 i a^{8} c}{d}-\frac {9949 i a^{8} {\mathrm e}^{-i \left (d x +c \right )}}{128 d}-\frac {256 a^{8} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}+\frac {a^{8} \sin \left (7 d x +7 c \right )}{448 d}+\frac {a^{8} \cos \left (6 d x +6 c \right )}{24 d}-\frac {121 a^{8} \sin \left (5 d x +5 c \right )}{320 d}-\frac {9 a^{8} \cos \left (4 d x +4 c \right )}{4 d}+\frac {647 a^{8} \sin \left (3 d x +3 c \right )}{64 d}+\frac {309 a^{8} \cos \left (2 d x +2 c \right )}{8 d}\) \(178\)
derivativedivides \(\frac {a^{8} \left (-\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+8 a^{8} \left (-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+28 a^{8} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+56 a^{8} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+56 a^{8} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+28 a^{8} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-8 a^{8} \ln \left (\cos \left (d x +c \right )\right )+a^{8} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(312\)
default \(\frac {a^{8} \left (-\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+8 a^{8} \left (-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+28 a^{8} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+56 a^{8} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+56 a^{8} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+28 a^{8} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-8 a^{8} \ln \left (\cos \left (d x +c \right )\right )+a^{8} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(312\)
norman \(\frac {-\frac {254 a^{8} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2042 a^{8} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {34198 a^{8} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {423678 a^{8} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}-\frac {423678 a^{8} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}-\frac {34198 a^{8} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {2042 a^{8} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {254 a^{8} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {240 a^{8} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {240 a^{8} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {1696 a^{8} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {1696 a^{8} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {14128 a^{8} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {14128 a^{8} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {19520 a^{8} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}-\frac {256 a^{8} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {128 a^{8} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) \(344\)

input
int(sec(d*x+c)*(a+a*sin(d*x+c))^8,x,method=_RETURNVERBOSE)
 
output
-1/6720*a^8*(1720320*ln(tan(1/2*d*x+1/2*c)-1)-860160*ln(sec(1/2*d*x+1/2*c) 
^2)+244720-259560*cos(2*d*x+2*c)-67935*sin(3*d*x+3*c)+15120*cos(4*d*x+4*c) 
-280*cos(6*d*x+6*c)+1044645*sin(d*x+c)+2541*sin(5*d*x+5*c)-15*sin(7*d*x+7* 
c))/d
 
3.1.46.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.70 \[ \int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {140 \, a^{8} \cos \left (d x + c\right )^{6} - 2100 \, a^{8} \cos \left (d x + c\right )^{4} + 10080 \, a^{8} \cos \left (d x + c\right )^{2} - 13440 \, a^{8} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (5 \, a^{8} \cos \left (d x + c\right )^{6} - 218 \, a^{8} \cos \left (d x + c\right )^{4} + 1576 \, a^{8} \cos \left (d x + c\right )^{2} - 5808 \, a^{8}\right )} \sin \left (d x + c\right )}{105 \, d} \]

input
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^8,x, algorithm="fricas")
 
output
1/105*(140*a^8*cos(d*x + c)^6 - 2100*a^8*cos(d*x + c)^4 + 10080*a^8*cos(d* 
x + c)^2 - 13440*a^8*log(-sin(d*x + c) + 1) + 3*(5*a^8*cos(d*x + c)^6 - 21 
8*a^8*cos(d*x + c)^4 + 1576*a^8*cos(d*x + c)^2 - 5808*a^8)*sin(d*x + c))/d
 
3.1.46.6 Sympy [F(-1)]

Timed out. \[ \int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)*(a+a*sin(d*x+c))**8,x)
 
output
Timed out
 
3.1.46.7 Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.67 \[ \int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {15 \, a^{8} \sin \left (d x + c\right )^{7} + 140 \, a^{8} \sin \left (d x + c\right )^{6} + 609 \, a^{8} \sin \left (d x + c\right )^{5} + 1680 \, a^{8} \sin \left (d x + c\right )^{4} + 3465 \, a^{8} \sin \left (d x + c\right )^{3} + 6300 \, a^{8} \sin \left (d x + c\right )^{2} + 13440 \, a^{8} \log \left (\sin \left (d x + c\right ) - 1\right ) + 13335 \, a^{8} \sin \left (d x + c\right )}{105 \, d} \]

input
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^8,x, algorithm="maxima")
 
output
-1/105*(15*a^8*sin(d*x + c)^7 + 140*a^8*sin(d*x + c)^6 + 609*a^8*sin(d*x + 
 c)^5 + 1680*a^8*sin(d*x + c)^4 + 3465*a^8*sin(d*x + c)^3 + 6300*a^8*sin(d 
*x + c)^2 + 13440*a^8*log(sin(d*x + c) - 1) + 13335*a^8*sin(d*x + c))/d
 
3.1.46.8 Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.78 \[ \int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {2 \, {\left (6720 \, a^{8} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 13440 \, a^{8} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {17424 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{14} + 13335 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 134568 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 93870 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 442344 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 265209 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 780640 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 370308 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 780640 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 265209 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 442344 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 93870 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 134568 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13335 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17424 \, a^{8}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7}}\right )}}{105 \, d} \]

input
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^8,x, algorithm="giac")
 
output
2/105*(6720*a^8*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 13440*a^8*log(abs(tan(1/ 
2*d*x + 1/2*c) - 1)) - (17424*a^8*tan(1/2*d*x + 1/2*c)^14 + 13335*a^8*tan( 
1/2*d*x + 1/2*c)^13 + 134568*a^8*tan(1/2*d*x + 1/2*c)^12 + 93870*a^8*tan(1 
/2*d*x + 1/2*c)^11 + 442344*a^8*tan(1/2*d*x + 1/2*c)^10 + 265209*a^8*tan(1 
/2*d*x + 1/2*c)^9 + 780640*a^8*tan(1/2*d*x + 1/2*c)^8 + 370308*a^8*tan(1/2 
*d*x + 1/2*c)^7 + 780640*a^8*tan(1/2*d*x + 1/2*c)^6 + 265209*a^8*tan(1/2*d 
*x + 1/2*c)^5 + 442344*a^8*tan(1/2*d*x + 1/2*c)^4 + 93870*a^8*tan(1/2*d*x 
+ 1/2*c)^3 + 134568*a^8*tan(1/2*d*x + 1/2*c)^2 + 13335*a^8*tan(1/2*d*x + 1 
/2*c) + 17424*a^8)/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d
 
3.1.46.9 Mupad [B] (verification not implemented)

Time = 5.46 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.67 \[ \int \sec (c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {128\,a^8\,\ln \left (\sin \left (c+d\,x\right )-1\right )+127\,a^8\,\sin \left (c+d\,x\right )+60\,a^8\,{\sin \left (c+d\,x\right )}^2+33\,a^8\,{\sin \left (c+d\,x\right )}^3+16\,a^8\,{\sin \left (c+d\,x\right )}^4+\frac {29\,a^8\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {4\,a^8\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {a^8\,{\sin \left (c+d\,x\right )}^7}{7}}{d} \]

input
int((a + a*sin(c + d*x))^8/cos(c + d*x),x)
 
output
-(128*a^8*log(sin(c + d*x) - 1) + 127*a^8*sin(c + d*x) + 60*a^8*sin(c + d* 
x)^2 + 33*a^8*sin(c + d*x)^3 + 16*a^8*sin(c + d*x)^4 + (29*a^8*sin(c + d*x 
)^5)/5 + (4*a^8*sin(c + d*x)^6)/3 + (a^8*sin(c + d*x)^7)/7)/d