3.1.83 \(\int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [83]

Optimal. Leaf size=167 \[ \frac {11 x}{128 a^2}+\frac {11 \cos (c+d x) \sin (c+d x)}{128 a^2 d}-\frac {7 \cos ^3(c+d x) \sin (c+d x)}{64 a^2 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {2 \sin ^7(c+d x)}{7 a^2 d} \]

[Out]

11/128*x/a^2+11/128*cos(d*x+c)*sin(d*x+c)/a^2/d-7/64*cos(d*x+c)^3*sin(d*x+c)/a^2/d-1/16*cos(d*x+c)^5*sin(d*x+c
)/a^2/d-1/6*cos(d*x+c)^3*sin(d*x+c)^3/a^2/d-1/8*cos(d*x+c)^5*sin(d*x+c)^3/a^2/d-2/5*sin(d*x+c)^5/a^2/d+2/7*sin
(d*x+c)^7/a^2/d

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Rubi [A]
time = 0.32, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3957, 2954, 2952, 2648, 2715, 8, 2644, 14} \begin {gather*} \frac {2 \sin ^7(c+d x)}{7 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac {\sin (c+d x) \cos ^5(c+d x)}{16 a^2 d}-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 a^2 d}-\frac {7 \sin (c+d x) \cos ^3(c+d x)}{64 a^2 d}+\frac {11 \sin (c+d x) \cos (c+d x)}{128 a^2 d}+\frac {11 x}{128 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]

[Out]

(11*x)/(128*a^2) + (11*Cos[c + d*x]*Sin[c + d*x])/(128*a^2*d) - (7*Cos[c + d*x]^3*Sin[c + d*x])/(64*a^2*d) - (
Cos[c + d*x]^5*Sin[c + d*x])/(16*a^2*d) - (Cos[c + d*x]^3*Sin[c + d*x]^3)/(6*a^2*d) - (Cos[c + d*x]^5*Sin[c +
d*x]^3)/(8*a^2*d) - (2*Sin[c + d*x]^5)/(5*a^2*d) + (2*Sin[c + d*x]^7)/(7*a^2*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*n]

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2954

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e +
f*x])^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin ^8(c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {\int \cos ^2(c+d x) (-a+a \cos (c+d x))^2 \sin ^4(c+d x) \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cos ^2(c+d x) \sin ^4(c+d x)-2 a^2 \cos ^3(c+d x) \sin ^4(c+d x)+a^2 \cos ^4(c+d x) \sin ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cos ^2(c+d x) \sin ^4(c+d x) \, dx}{a^2}+\frac {\int \cos ^4(c+d x) \sin ^4(c+d x) \, dx}{a^2}-\frac {2 \int \cos ^3(c+d x) \sin ^4(c+d x) \, dx}{a^2}\\ &=-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}+\frac {3 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx}{8 a^2}+\frac {\int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{2 a^2}-\frac {2 \text {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a^2 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}+\frac {\int \cos ^4(c+d x) \, dx}{16 a^2}+\frac {\int \cos ^2(c+d x) \, dx}{8 a^2}-\frac {2 \text {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=\frac {\cos (c+d x) \sin (c+d x)}{16 a^2 d}-\frac {7 \cos ^3(c+d x) \sin (c+d x)}{64 a^2 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {2 \sin ^7(c+d x)}{7 a^2 d}+\frac {3 \int \cos ^2(c+d x) \, dx}{64 a^2}+\frac {\int 1 \, dx}{16 a^2}\\ &=\frac {x}{16 a^2}+\frac {11 \cos (c+d x) \sin (c+d x)}{128 a^2 d}-\frac {7 \cos ^3(c+d x) \sin (c+d x)}{64 a^2 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {2 \sin ^7(c+d x)}{7 a^2 d}+\frac {3 \int 1 \, dx}{128 a^2}\\ &=\frac {11 x}{128 a^2}+\frac {11 \cos (c+d x) \sin (c+d x)}{128 a^2 d}-\frac {7 \cos ^3(c+d x) \sin (c+d x)}{64 a^2 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {2 \sin ^7(c+d x)}{7 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 1.77, size = 131, normalized size = 0.78 \begin {gather*} \frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (9240 d x-10080 \sin (c+d x)-1680 \sin (2 (c+d x))+3360 \sin (3 (c+d x))-2520 \sin (4 (c+d x))+672 \sin (5 (c+d x))+560 \sin (6 (c+d x))-480 \sin (7 (c+d x))+105 \sin (8 (c+d x))+980 \tan \left (\frac {c}{2}\right )\right )}{26880 a^2 d (1+\sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^8/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(9240*d*x - 10080*Sin[c + d*x] - 1680*Sin[2*(c + d*x)] + 3360*Sin[3*(c + d*
x)] - 2520*Sin[4*(c + d*x)] + 672*Sin[5*(c + d*x)] + 560*Sin[6*(c + d*x)] - 480*Sin[7*(c + d*x)] + 105*Sin[8*(
c + d*x)] + 980*Tan[c/2]))/(26880*a^2*d*(1 + Sec[c + d*x])^2)

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Maple [A]
time = 0.15, size = 141, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {\frac {128 \left (-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8192}-\frac {253 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24576}-\frac {4213 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{122880}-\frac {55583 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{860160}+\frac {31007 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{860160}-\frac {20363 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{122880}+\frac {253 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24576}+\frac {11 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{a^{2} d}\)	\(141\)
default	\(\frac {\frac {128 \left (-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8192}-\frac {253 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24576}-\frac {4213 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{122880}-\frac {55583 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{860160}+\frac {31007 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{860160}-\frac {20363 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{122880}+\frac {253 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24576}+\frac {11 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{a^{2} d}\)	\(141\)
risch	\(\frac {11 x}{128 a^{2}}-\frac {3 \sin \left (d x +c \right )}{32 a^{2} d}+\frac {\sin \left (8 d x +8 c \right )}{1024 a^{2} d}-\frac {\sin \left (7 d x +7 c \right )}{224 a^{2} d}+\frac {\sin \left (6 d x +6 c \right )}{192 a^{2} d}+\frac {\sin \left (5 d x +5 c \right )}{160 a^{2} d}-\frac {3 \sin \left (4 d x +4 c \right )}{128 a^{2} d}+\frac {\sin \left (3 d x +3 c \right )}{32 a^{2} d}-\frac {\sin \left (2 d x +2 c \right )}{64 a^{2} d}\)	\(141\)
norman	\(\frac {\frac {11 x}{128 a}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a d}-\frac {253 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}-\frac {4213 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{960 a d}-\frac {55583 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6720 a d}+\frac {31007 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6720 a d}-\frac {20363 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{960 a d}+\frac {253 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 a d}+\frac {11 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a d}+\frac {11 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}+\frac {77 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a}+\frac {77 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}+\frac {385 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a}+\frac {77 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}+\frac {77 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a}+\frac {11 x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}+\frac {11 x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8} a}\)	\(313\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^8/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

128/d/a^2*((-11/8192*tan(1/2*d*x+1/2*c)-253/24576*tan(1/2*d*x+1/2*c)^3-4213/122880*tan(1/2*d*x+1/2*c)^5-55583/
860160*tan(1/2*d*x+1/2*c)^7+31007/860160*tan(1/2*d*x+1/2*c)^9-20363/122880*tan(1/2*d*x+1/2*c)^11+253/24576*tan
(1/2*d*x+1/2*c)^13+11/8192*tan(1/2*d*x+1/2*c)^15)/(1+tan(1/2*d*x+1/2*c)^2)^8+11/8192*arctan(tan(1/2*d*x+1/2*c)
))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (151) = 302\).
time = 0.49, size = 378, normalized size = 2.26 \begin {gather*} -\frac {\frac {\frac {1155 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {8855 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {29491 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {55583 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {31007 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {142541 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {8855 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac {1155 \, \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}}}{a^{2} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {28 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {56 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {70 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {56 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {28 \, a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac {a^{2} \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}} - \frac {1155 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6720 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6720*((1155*sin(d*x + c)/(cos(d*x + c) + 1) + 8855*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 29491*sin(d*x + c)
^5/(cos(d*x + c) + 1)^5 + 55583*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 31007*sin(d*x + c)^9/(cos(d*x + c) + 1)^
9 + 142541*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 8855*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 1155*sin(d*x +
c)^15/(cos(d*x + c) + 1)^15)/(a^2 + 8*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^2*sin(d*x + c)^4/(cos(d*
x + c) + 1)^4 + 56*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 56*a
^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 8*a^2*sin(d*x + c)^1
4/(cos(d*x + c) + 1)^14 + a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 1155*arctan(sin(d*x + c)/(cos(d*x + c)
+ 1))/a^2)/d

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Fricas [A]
time = 3.55, size = 90, normalized size = 0.54 \begin {gather*} \frac {1155 \, d x + {\left (1680 \, \cos \left (d x + c\right )^{7} - 3840 \, \cos \left (d x + c\right )^{6} - 280 \, \cos \left (d x + c\right )^{5} + 6144 \, \cos \left (d x + c\right )^{4} - 3710 \, \cos \left (d x + c\right )^{3} - 768 \, \cos \left (d x + c\right )^{2} + 1155 \, \cos \left (d x + c\right ) - 1536\right )} \sin \left (d x + c\right )}{13440 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/13440*(1155*d*x + (1680*cos(d*x + c)^7 - 3840*cos(d*x + c)^6 - 280*cos(d*x + c)^5 + 6144*cos(d*x + c)^4 - 37
10*cos(d*x + c)^3 - 768*cos(d*x + c)^2 + 1155*cos(d*x + c) - 1536)*sin(d*x + c))/(a^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**8/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.49, size = 139, normalized size = 0.83 \begin {gather*} \frac {\frac {1155 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (1155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 8855 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 142541 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 31007 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 55583 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 29491 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8855 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{8} a^{2}}}{13440 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^8/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/13440*(1155*(d*x + c)/a^2 + 2*(1155*tan(1/2*d*x + 1/2*c)^15 + 8855*tan(1/2*d*x + 1/2*c)^13 - 142541*tan(1/2*
d*x + 1/2*c)^11 + 31007*tan(1/2*d*x + 1/2*c)^9 - 55583*tan(1/2*d*x + 1/2*c)^7 - 29491*tan(1/2*d*x + 1/2*c)^5 -
8855*tan(1/2*d*x + 1/2*c)^3 - 1155*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^8*a^2))/d

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Mupad [B]
time = 3.93, size = 133, normalized size = 0.80 \begin {gather*} \frac {11\,x}{128\,a^2}-\frac {-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}-\frac {253\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}+\frac {20363\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{960}-\frac {31007\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{6720}+\frac {55583\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6720}+\frac {4213\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{960}+\frac {253\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^8/(a + a/cos(c + d*x))^2,x)

[Out]

(11*x)/(128*a^2) - ((11*tan(c/2 + (d*x)/2))/64 + (253*tan(c/2 + (d*x)/2)^3)/192 + (4213*tan(c/2 + (d*x)/2)^5)/
960 + (55583*tan(c/2 + (d*x)/2)^7)/6720 - (31007*tan(c/2 + (d*x)/2)^9)/6720 + (20363*tan(c/2 + (d*x)/2)^11)/96
0 - (253*tan(c/2 + (d*x)/2)^13)/192 - (11*tan(c/2 + (d*x)/2)^15)/64)/(a^2*d*(tan(c/2 + (d*x)/2)^2 + 1)^8)

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