3.1.47 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx\) [47]

3.1.47.1 Optimal result
3.1.47.2 Mathematica [C] (verified)
3.1.47.3 Rubi [A] (verified)
3.1.47.4 Maple [A] (verified)
3.1.47.5 Fricas [A] (verification not implemented)
3.1.47.6 Sympy [F(-1)]
3.1.47.7 Maxima [A] (verification not implemented)
3.1.47.8 Giac [A] (verification not implemented)
3.1.47.9 Mupad [B] (verification not implemented)

3.1.47.1 Optimal result

Integrand size = 21, antiderivative size = 201 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {3003 a^8 x}{16}+\frac {1001 a^8 \cos ^5(c+d x)}{10 d}-\frac {3003 a^8 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {1001 a^8 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac {2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac {26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}+\frac {286 a^{14} \cos ^9(c+d x)}{3 d \left (a^2-a^2 \sin (c+d x)\right )^3}+\frac {143 a^{16} \cos ^7(c+d x)}{2 d \left (a^8-a^8 \sin (c+d x)\right )} \]

output
-3003/16*a^8*x+1001/10*a^8*cos(d*x+c)^5/d-3003/16*a^8*cos(d*x+c)*sin(d*x+c 
)/d-1001/8*a^8*cos(d*x+c)^3*sin(d*x+c)/d+2*a^15*cos(d*x+c)^13/d/(a-a*sin(d 
*x+c))^7+26*a^13*cos(d*x+c)^11/d/(a-a*sin(d*x+c))^5+286/3*a^14*cos(d*x+c)^ 
9/d/(a^2-a^2*sin(d*x+c))^3+143/2*a^16*cos(d*x+c)^7/d/(a^8-a^8*sin(d*x+c))
 
3.1.47.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.27 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {128 \sqrt {2} a^8 \operatorname {Hypergeometric2F1}\left (-\frac {13}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) \sqrt {1+\sin (c+d x)}}{d} \]

input
Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^8,x]
 
output
(128*Sqrt[2]*a^8*Hypergeometric2F1[-13/2, -1/2, 1/2, (1 - Sin[c + d*x])/2] 
*Sec[c + d*x]*Sqrt[1 + Sin[c + d*x]])/d
 
3.1.47.3 Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.12, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.810, Rules used = {3042, 3149, 3042, 3159, 3042, 3159, 3042, 3159, 3042, 3158, 3042, 3161, 3042, 3115, 3042, 3115, 24}

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.

\(\displaystyle \int \sec ^2(c+d x) (a \sin (c+d x)+a)^8 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^8}{\cos (c+d x)^2}dx\)

\(\Big \downarrow \) 3149

\(\displaystyle a^{16} \int \frac {\cos ^{14}(c+d x)}{(a-a \sin (c+d x))^8}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^{16} \int \frac {\cos (c+d x)^{14}}{(a-a \sin (c+d x))^8}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \int \frac {\cos ^{12}(c+d x)}{(a-a \sin (c+d x))^6}dx}{a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \int \frac {\cos (c+d x)^{12}}{(a-a \sin (c+d x))^6}dx}{a^2}\right )\)

\(\Big \downarrow \) 3159

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \int \frac {\cos ^{10}(c+d x)}{(a-a \sin (c+d x))^4}dx}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \int \frac {\cos (c+d x)^{10}}{(a-a \sin (c+d x))^4}dx}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3159

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \int \frac {\cos ^8(c+d x)}{(a-a \sin (c+d x))^2}dx}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \int \frac {\cos (c+d x)^8}{(a-a \sin (c+d x))^2}dx}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3158

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \int \frac {\cos ^6(c+d x)}{a-a \sin (c+d x)}dx}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \int \frac {\cos (c+d x)^6}{a-a \sin (c+d x)}dx}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3161

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {\int \cos ^4(c+d x)dx}{a}-\frac {\cos ^5(c+d x)}{5 a d}\right )}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx}{a}-\frac {\cos ^5(c+d x)}{5 a d}\right )}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}}{a}-\frac {\cos ^5(c+d x)}{5 a d}\right )}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}}{a}-\frac {\cos ^5(c+d x)}{5 a d}\right )}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}}{a}-\frac {\cos ^5(c+d x)}{5 a d}\right )}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle a^{16} \left (\frac {2 \cos ^{13}(c+d x)}{a d (a-a \sin (c+d x))^7}-\frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )}{a}-\frac {\cos ^5(c+d x)}{5 a d}\right )}{6 a}-\frac {\cos ^7(c+d x)}{6 d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}-\frac {2 \cos ^9(c+d x)}{3 a d (a-a \sin (c+d x))^3}\right )}{a^2}-\frac {2 \cos ^{11}(c+d x)}{a d (a-a \sin (c+d x))^5}\right )}{a^2}\right )\)

input
Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^8,x]
 
output
a^16*((2*Cos[c + d*x]^13)/(a*d*(a - a*Sin[c + d*x])^7) - (13*((-2*Cos[c + 
d*x]^11)/(a*d*(a - a*Sin[c + d*x])^5) + (11*((-2*Cos[c + d*x]^9)/(3*a*d*(a 
 - a*Sin[c + d*x])^3) + (3*(-1/6*Cos[c + d*x]^7/(d*(a^2 - a^2*Sin[c + d*x] 
)) + (7*(-1/5*Cos[c + d*x]^5/(a*d) + ((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) 
+ (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)/a))/(6*a)))/a^2))/a^2)) 
/a^2)
 

3.1.47.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3149
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(a/g)^(2*m)   Int[(g*Cos[e + f*x])^(2*m + p)/( 
a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2 
, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]
 

rule 3158
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, 
f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || 
 EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In 
tegersQ[2*m, 2*p]
 

rule 3159
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

rule 3161
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si 
mp[g^2/a   Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x 
] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
 
3.1.47.4 Maple [A] (verified)

Time = 3.77 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.55

method result size
parallelrisch \(\frac {a^{8} \left (-360360 \cos \left (d x +c \right ) d x +96 \cos \left (6 d x +6 c \right )-5824 \cos \left (4 d x +4 c \right )+160160 \cos \left (2 d x +2 c \right )+566272 \cos \left (d x +c \right )+276705 \sin \left (d x +c \right )-910 \sin \left (5 d x +5 c \right )+5 \sin \left (7 d x +7 c \right )+30030 \sin \left (3 d x +3 c \right )+411840\right )}{1920 d \cos \left (d x +c \right )}\) \(111\)
risch \(-\frac {3003 a^{8} x}{16}+\frac {173 a^{8} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {173 a^{8} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {256 a^{8}}{d \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {a^{8} \sin \left (6 d x +6 c \right )}{192 d}+\frac {a^{8} \cos \left (5 d x +5 c \right )}{10 d}-\frac {61 a^{8} \sin \left (4 d x +4 c \right )}{64 d}-\frac {37 a^{8} \cos \left (3 d x +3 c \right )}{6 d}+\frac {2063 a^{8} \sin \left (2 d x +2 c \right )}{64 d}\) \(149\)
derivativedivides \(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )-\frac {35 d x}{16}-\frac {35 c}{16}\right )+8 a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+56 a^{8} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+56 a^{8} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+28 a^{8} \left (\tan \left (d x +c \right )-d x -c \right )+\frac {8 a^{8}}{\cos \left (d x +c \right )}+a^{8} \tan \left (d x +c \right )}{d}\) \(389\)
default \(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )-\frac {35 d x}{16}-\frac {35 c}{16}\right )+8 a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+56 a^{8} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+56 a^{8} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+28 a^{8} \left (\tan \left (d x +c \right )-d x -c \right )+\frac {8 a^{8}}{\cos \left (d x +c \right )}+a^{8} \tan \left (d x +c \right )}{d}\) \(389\)

input
int(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x,method=_RETURNVERBOSE)
 
output
1/1920*a^8/d*(-360360*cos(d*x+c)*d*x+96*cos(6*d*x+6*c)-5824*cos(4*d*x+4*c) 
+160160*cos(2*d*x+2*c)+566272*cos(d*x+c)+276705*sin(d*x+c)-910*sin(5*d*x+5 
*c)+5*sin(7*d*x+7*c)+30030*sin(3*d*x+3*c)+411840)/cos(d*x+c)
 
3.1.47.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.15 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {40 \, a^{8} \cos \left (d x + c\right )^{7} + 384 \, a^{8} \cos \left (d x + c\right )^{6} - 1526 \, a^{8} \cos \left (d x + c\right )^{5} - 6400 \, a^{8} \cos \left (d x + c\right )^{4} + 11865 \, a^{8} \cos \left (d x + c\right )^{3} - 45045 \, a^{8} d x + 46080 \, a^{8} \cos \left (d x + c\right )^{2} + 30720 \, a^{8} - 15 \, {\left (3003 \, a^{8} d x - 4027 \, a^{8}\right )} \cos \left (d x + c\right ) + {\left (40 \, a^{8} \cos \left (d x + c\right )^{6} - 344 \, a^{8} \cos \left (d x + c\right )^{5} - 1870 \, a^{8} \cos \left (d x + c\right )^{4} + 4530 \, a^{8} \cos \left (d x + c\right )^{3} + 45045 \, a^{8} d x + 16395 \, a^{8} \cos \left (d x + c\right )^{2} - 29685 \, a^{8} \cos \left (d x + c\right ) + 30720 \, a^{8}\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

input
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x, algorithm="fricas")
 
output
1/240*(40*a^8*cos(d*x + c)^7 + 384*a^8*cos(d*x + c)^6 - 1526*a^8*cos(d*x + 
 c)^5 - 6400*a^8*cos(d*x + c)^4 + 11865*a^8*cos(d*x + c)^3 - 45045*a^8*d*x 
 + 46080*a^8*cos(d*x + c)^2 + 30720*a^8 - 15*(3003*a^8*d*x - 4027*a^8)*cos 
(d*x + c) + (40*a^8*cos(d*x + c)^6 - 344*a^8*cos(d*x + c)^5 - 1870*a^8*cos 
(d*x + c)^4 + 4530*a^8*cos(d*x + c)^3 + 45045*a^8*d*x + 16395*a^8*cos(d*x 
+ c)^2 - 29685*a^8*cos(d*x + c) + 30720*a^8)*sin(d*x + c))/(d*cos(d*x + c) 
 - d*sin(d*x + c) + d)
 
3.1.47.6 Sympy [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**2*(a+a*sin(d*x+c))**8,x)
 
output
Timed out
 
3.1.47.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.65 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {384 \, {\left (\cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3} + \frac {5}{\cos \left (d x + c\right )} + 15 \, \cos \left (d x + c\right )\right )} a^{8} - 4480 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{8} - 5 \, {\left (105 \, d x + 105 \, c - \frac {87 \, \tan \left (d x + c\right )^{5} + 136 \, \tan \left (d x + c\right )^{3} + 57 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1} - 48 \, \tan \left (d x + c\right )\right )} a^{8} - 840 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{8} - 8400 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{8} - 6720 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{8} + 13440 \, a^{8} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 240 \, a^{8} \tan \left (d x + c\right ) + \frac {1920 \, a^{8}}{\cos \left (d x + c\right )}}{240 \, d} \]

input
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x, algorithm="maxima")
 
output
1/240*(384*(cos(d*x + c)^5 - 5*cos(d*x + c)^3 + 5/cos(d*x + c) + 15*cos(d* 
x + c))*a^8 - 4480*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^8 
- 5*(105*d*x + 105*c - (87*tan(d*x + c)^5 + 136*tan(d*x + c)^3 + 57*tan(d* 
x + c))/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1) - 48*ta 
n(d*x + c))*a^8 - 840*(15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c)) 
/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c))*a^8 - 8400*(3*d 
*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*a^8 - 6720* 
(d*x + c - tan(d*x + c))*a^8 + 13440*a^8*(1/cos(d*x + c) + cos(d*x + c)) + 
 240*a^8*tan(d*x + c) + 1920*a^8/cos(d*x + c))/d
 
3.1.47.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.15 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {45045 \, {\left (d x + c\right )} a^{8} + \frac {61440 \, a^{8}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (14565 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 28800 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 50855 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 174720 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 36930 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 400640 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 36930 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 426240 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 50855 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 211584 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 14565 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 40064 \, a^{8}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

input
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x, algorithm="giac")
 
output
-1/240*(45045*(d*x + c)*a^8 + 61440*a^8/(tan(1/2*d*x + 1/2*c) - 1) + 2*(14 
565*a^8*tan(1/2*d*x + 1/2*c)^11 - 28800*a^8*tan(1/2*d*x + 1/2*c)^10 + 5085 
5*a^8*tan(1/2*d*x + 1/2*c)^9 - 174720*a^8*tan(1/2*d*x + 1/2*c)^8 + 36930*a 
^8*tan(1/2*d*x + 1/2*c)^7 - 400640*a^8*tan(1/2*d*x + 1/2*c)^6 - 36930*a^8* 
tan(1/2*d*x + 1/2*c)^5 - 426240*a^8*tan(1/2*d*x + 1/2*c)^4 - 50855*a^8*tan 
(1/2*d*x + 1/2*c)^3 - 211584*a^8*tan(1/2*d*x + 1/2*c)^2 - 14565*a^8*tan(1/ 
2*d*x + 1/2*c) - 40064*a^8)/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
 
3.1.47.9 Mupad [B] (verification not implemented)

Time = 10.50 (sec) , antiderivative size = 513, normalized size of antiderivative = 2.55 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {3003\,a^8\,x}{16}-\frac {\frac {3003\,a^8\,\left (c+d\,x\right )}{16}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3003\,a^8\,\left (c+d\,x\right )}{16}-\frac {a^8\,\left (45045\,c+45045\,d\,x-50998\right )}{240}\right )-\frac {a^8\,\left (45045\,c+45045\,d\,x-141568\right )}{240}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {3003\,a^8\,\left (c+d\,x\right )}{16}-\frac {a^8\,\left (45045\,c+45045\,d\,x-90570\right )}{240}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {9009\,a^8\,\left (c+d\,x\right )}{8}-\frac {a^8\,\left (270270\,c+270270\,d\,x-86730\right )}{240}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {9009\,a^8\,\left (c+d\,x\right )}{8}-\frac {a^8\,\left (270270\,c+270270\,d\,x-321458\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {9009\,a^8\,\left (c+d\,x\right )}{8}-\frac {a^8\,\left (270270\,c+270270\,d\,x-527950\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {9009\,a^8\,\left (c+d\,x\right )}{8}-\frac {a^8\,\left (270270\,c+270270\,d\,x-762678\right )}{240}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {45045\,a^8\,\left (c+d\,x\right )}{16}-\frac {a^8\,\left (675675\,c+675675\,d\,x-451150\right )}{240}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {45045\,a^8\,\left (c+d\,x\right )}{16}-\frac {a^8\,\left (675675\,c+675675\,d\,x-778620\right )}{240}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {15015\,a^8\,\left (c+d\,x\right )}{4}-\frac {a^8\,\left (900900\,c+900900\,d\,x-875140\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {45045\,a^8\,\left (c+d\,x\right )}{16}-\frac {a^8\,\left (675675\,c+675675\,d\,x-1344900\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {45045\,a^8\,\left (c+d\,x\right )}{16}-\frac {a^8\,\left (675675\,c+675675\,d\,x-1672370\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {15015\,a^8\,\left (c+d\,x\right )}{4}-\frac {a^8\,\left (900900\,c+900900\,d\,x-1956220\right )}{240}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

input
int((a + a*sin(c + d*x))^8/cos(c + d*x)^2,x)
 
output
- (3003*a^8*x)/16 - ((3003*a^8*(c + d*x))/16 - tan(c/2 + (d*x)/2)*((3003*a 
^8*(c + d*x))/16 - (a^8*(45045*c + 45045*d*x - 50998))/240) - (a^8*(45045* 
c + 45045*d*x - 141568))/240 + tan(c/2 + (d*x)/2)^12*((3003*a^8*(c + d*x)) 
/16 - (a^8*(45045*c + 45045*d*x - 90570))/240) - tan(c/2 + (d*x)/2)^11*((9 
009*a^8*(c + d*x))/8 - (a^8*(270270*c + 270270*d*x - 86730))/240) - tan(c/ 
2 + (d*x)/2)^3*((9009*a^8*(c + d*x))/8 - (a^8*(270270*c + 270270*d*x - 321 
458))/240) + tan(c/2 + (d*x)/2)^10*((9009*a^8*(c + d*x))/8 - (a^8*(270270* 
c + 270270*d*x - 527950))/240) + tan(c/2 + (d*x)/2)^2*((9009*a^8*(c + d*x) 
)/8 - (a^8*(270270*c + 270270*d*x - 762678))/240) - tan(c/2 + (d*x)/2)^9*( 
(45045*a^8*(c + d*x))/16 - (a^8*(675675*c + 675675*d*x - 451150))/240) - t 
an(c/2 + (d*x)/2)^5*((45045*a^8*(c + d*x))/16 - (a^8*(675675*c + 675675*d* 
x - 778620))/240) - tan(c/2 + (d*x)/2)^7*((15015*a^8*(c + d*x))/4 - (a^8*( 
900900*c + 900900*d*x - 875140))/240) + tan(c/2 + (d*x)/2)^8*((45045*a^8*( 
c + d*x))/16 - (a^8*(675675*c + 675675*d*x - 1344900))/240) + tan(c/2 + (d 
*x)/2)^4*((45045*a^8*(c + d*x))/16 - (a^8*(675675*c + 675675*d*x - 1672370 
))/240) + tan(c/2 + (d*x)/2)^6*((15015*a^8*(c + d*x))/4 - (a^8*(900900*c + 
 900900*d*x - 1956220))/240))/(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x) 
/2)^2 + 1)^6)