\(\int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx\) [50]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {80 a^8 \log (1-\sin (c+d x))}{d}-\frac {31 a^8 \sin (c+d x)}{d}-\frac {4 a^8 \sin ^2(c+d x)}{d}-\frac {a^8 \sin ^3(c+d x)}{3 d}+\frac {16 a^{10}}{d (a-a \sin (c+d x))^2}-\frac {80 a^9}{d (a-a \sin (c+d x))} \]

[Out]

-80*a^8*ln(1-sin(d*x+c))/d-31*a^8*sin(d*x+c)/d-4*a^8*sin(d*x+c)^2/d-1/3*a^8*sin(d*x+c)^3/d+16*a^10/d/(a-a*sin(
d*x+c))^2-80*a^9/d/(a-a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 45} \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {16 a^{10}}{d (a-a \sin (c+d x))^2}-\frac {80 a^9}{d (a-a \sin (c+d x))}-\frac {a^8 \sin ^3(c+d x)}{3 d}-\frac {4 a^8 \sin ^2(c+d x)}{d}-\frac {31 a^8 \sin (c+d x)}{d}-\frac {80 a^8 \log (1-\sin (c+d x))}{d} \]

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^8,x]

[Out]

(-80*a^8*Log[1 - Sin[c + d*x]])/d - (31*a^8*Sin[c + d*x])/d - (4*a^8*Sin[c + d*x]^2)/d - (a^8*Sin[c + d*x]^3)/
(3*d) + (16*a^10)/(d*(a - a*Sin[c + d*x])^2) - (80*a^9)/(d*(a - a*Sin[c + d*x]))

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[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]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {(a+x)^5}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (-31 a^2+\frac {32 a^5}{(a-x)^3}-\frac {80 a^4}{(a-x)^2}+\frac {80 a^3}{a-x}-8 a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {80 a^8 \log (1-\sin (c+d x))}{d}-\frac {31 a^8 \sin (c+d x)}{d}-\frac {4 a^8 \sin ^2(c+d x)}{d}-\frac {a^8 \sin ^3(c+d x)}{3 d}+\frac {16 a^{10}}{d (a-a \sin (c+d x))^2}-\frac {80 a^9}{d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {a^8 \left (-80 \log (1-\sin (c+d x))-31 \sin (c+d x)-4 \sin ^2(c+d x)-\frac {1}{3} \sin ^3(c+d x)+\frac {16 (-4+5 \sin (c+d x))}{(-1+\sin (c+d x))^2}\right )}{d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^8,x]

[Out]

(a^8*(-80*Log[1 - Sin[c + d*x]] - 31*Sin[c + d*x] - 4*Sin[c + d*x]^2 - Sin[c + d*x]^3/3 + (16*(-4 + 5*Sin[c +
d*x]))/(-1 + Sin[c + d*x])^2))/d

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\frac {80 \left (\left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 \cos \left (2 d x +2 c \right )+6-8 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {89 \cos \left (2 d x +2 c \right )}{60}+\frac {\cos \left (4 d x +4 c \right )}{96}+\frac {2321 \sin \left (d x +c \right )}{960}-\frac {19 \sin \left (3 d x +3 c \right )}{128}+\frac {\sin \left (5 d x +5 c \right )}{1920}-\frac {239}{160}\right ) a^{8}}{d \left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right )}\)	\(150\)
risch	\(80 i a^{8} x -\frac {i a^{8} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {a^{8} {\mathrm e}^{2 i \left (d x +c \right )}}{d}+\frac {125 i a^{8} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {125 i a^{8} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a^{8} {\mathrm e}^{-2 i \left (d x +c \right )}}{d}+\frac {i a^{8} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {160 i a^{8} c}{d}+\frac {32 i a^{8} \left (-8 i {\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{3 i \left (d x +c \right )}-5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4}}-\frac {160 a^{8} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}\)	\(202\)
derivativedivides	\(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+8 a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+56 a^{8} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {14 a^{8} \left (\sin ^{4}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{4}}+28 a^{8} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {2 a^{8}}{\cos \left (d x +c \right )^{4}}+a^{8} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(527\)
default	\(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+8 a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+56 a^{8} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {14 a^{8} \left (\sin ^{4}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{4}}+28 a^{8} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {2 a^{8}}{\cos \left (d x +c \right )^{4}}+a^{8} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(527\)

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x,method=_RETURNVERBOSE)

[Out]

80*((-3+cos(2*d*x+2*c)+4*sin(d*x+c))*ln(sec(1/2*d*x+1/2*c)^2)+(-2*cos(2*d*x+2*c)+6-8*sin(d*x+c))*ln(tan(1/2*d*
x+1/2*c)-1)+89/60*cos(2*d*x+2*c)+1/96*cos(4*d*x+4*c)+2321/960*sin(d*x+c)-19/128*sin(3*d*x+3*c)+1/1920*sin(5*d*
x+5*c)-239/160)*a^8/d/(-3+cos(2*d*x+2*c)+4*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.26 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {10 \, a^{8} \cos \left (d x + c\right )^{4} + 160 \, a^{8} \cos \left (d x + c\right )^{2} + 16 \, a^{8} - 240 \, {\left (a^{8} \cos \left (d x + c\right )^{2} + 2 \, a^{8} \sin \left (d x + c\right ) - 2 \, a^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{8} \cos \left (d x + c\right )^{4} - 72 \, a^{8} \cos \left (d x + c\right )^{2} - 64 \, a^{8}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

1/3*(10*a^8*cos(d*x + c)^4 + 160*a^8*cos(d*x + c)^2 + 16*a^8 - 240*(a^8*cos(d*x + c)^2 + 2*a^8*sin(d*x + c) -
2*a^8)*log(-sin(d*x + c) + 1) + (a^8*cos(d*x + c)^4 - 72*a^8*cos(d*x + c)^2 - 64*a^8)*sin(d*x + c))/(d*cos(d*x
+ c)^2 + 2*d*sin(d*x + c) - 2*d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**8,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {a^{8} \sin \left (d x + c\right )^{3} + 12 \, a^{8} \sin \left (d x + c\right )^{2} + 240 \, a^{8} \log \left (\sin \left (d x + c\right ) - 1\right ) + 93 \, a^{8} \sin \left (d x + c\right ) - \frac {48 \, {\left (5 \, a^{8} \sin \left (d x + c\right ) - 4 \, a^{8}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

-1/3*(a^8*sin(d*x + c)^3 + 12*a^8*sin(d*x + c)^2 + 240*a^8*log(sin(d*x + c) - 1) + 93*a^8*sin(d*x + c) - 48*(5
*a^8*sin(d*x + c) - 4*a^8)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (110) = 220\).

Time = 0.42 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.21 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=\frac {2 \, {\left (120 \, a^{8} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 240 \, a^{8} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {220 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 93 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 684 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 190 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 684 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 93 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 220 \, a^{8}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}} + \frac {4 \, {\left (125 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 536 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 846 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 536 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 125 \, a^{8}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}\right )}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^8,x, algorithm="giac")

[Out]

2/3*(120*a^8*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 240*a^8*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (220*a^8*tan(1/2*d
*x + 1/2*c)^6 + 93*a^8*tan(1/2*d*x + 1/2*c)^5 + 684*a^8*tan(1/2*d*x + 1/2*c)^4 + 190*a^8*tan(1/2*d*x + 1/2*c)^
3 + 684*a^8*tan(1/2*d*x + 1/2*c)^2 + 93*a^8*tan(1/2*d*x + 1/2*c) + 220*a^8)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3 + 4
*(125*a^8*tan(1/2*d*x + 1/2*c)^4 - 536*a^8*tan(1/2*d*x + 1/2*c)^3 + 846*a^8*tan(1/2*d*x + 1/2*c)^2 - 536*a^8*t
an(1/2*d*x + 1/2*c) + 125*a^8)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^8 \, dx=-\frac {80\,a^8\,\ln \left (\sin \left (c+d\,x\right )-1\right )+31\,a^8\,\sin \left (c+d\,x\right )-\frac {80\,a^8\,\sin \left (c+d\,x\right )-64\,a^8}{{\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1}+4\,a^8\,{\sin \left (c+d\,x\right )}^2+\frac {a^8\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]

[In]

int((a + a*sin(c + d*x))^8/cos(c + d*x)^5,x)

[Out]

-(80*a^8*log(sin(c + d*x) - 1) + 31*a^8*sin(c + d*x) - (80*a^8*sin(c + d*x) - 64*a^8)/(sin(c + d*x)^2 - 2*sin(
c + d*x) + 1) + 4*a^8*sin(c + d*x)^2 + (a^8*sin(c + d*x)^3)/3)/d