3.83 \(\int \frac{\sin ^8(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

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

[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)

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Rubi [A]  time = 0.440065, antiderivative size = 167, normalized size of antiderivative = 1., 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 = {3872, 2875, 2873, 2568, 2635, 8, 2564, 14} \[ \frac{2 \sin ^7(c+d x)}{7 a^2 d}-\frac{2 \sin ^5(c+d x)}{5 a^2 d}-\frac{\sin ^3(c+d x) \cos ^5(c+d x)}{8 a^2 d}-\frac{\sin ^3(c+d x) \cos ^3(c+d x)}{6 a^2 d}-\frac{\sin (c+d x) \cos ^5(c+d x)}{16 a^2 d}-\frac{7 \sin (c+d x) \cos ^3(c+d x)}{64 a^2 d}+\frac{11 \sin (c+d x) \cos (c+d x)}{128 a^2 d}+\frac{11 x}{128 a^2} \]

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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2875

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 2873

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 2568

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 2635

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] && Integer
Q[2*n]

Rule 8

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

Rule 2564

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 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]]

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 \operatorname{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 \operatorname{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*}

Mathematica [A]  time = 2.75518, size = 131, normalized size = 0.78 \[ \frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-10080 \sin (c+d x)-1680 \sin (2 (c+d x))+3360 \sin (3 (c+d x))-2520 \sin (4 (c+d x))+672 \sin (5 (c+d x))+560 \sin (6 (c+d x))-480 \sin (7 (c+d x))+105 \sin (8 (c+d x))+980 \tan \left (\frac{c}{2}\right )+9240 d x\right )}{26880 a^2 d (\sec (c+d x)+1)^2} \]

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 = 0.102, size = 290, normalized size = 1.7 \begin{align*} -{\frac{11}{64\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}-{\frac{253}{192\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}-{\frac{4213}{960\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}-{\frac{55583}{6720\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{31007}{6720\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}-{\frac{20363}{960\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{11} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{253}{192\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{13} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{11}{64\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{15} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-8}}+{\frac{11}{64\,d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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)

[Out]

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

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Maxima [B]  time = 1.53106, size = 510, normalized size = 3.05 \begin{align*} -\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{align*}

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 = 1.74874, size = 270, normalized size = 1.62 \begin{align*} \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{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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 = 1.33326, size = 188, normalized size = 1.13 \begin{align*} \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{align*}

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