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

Optimal. Leaf size=125 \[ \frac{\sin ^7(c+d x)}{7 a d}+\frac{\sin ^5(c+d x) \cos ^3(c+d x)}{8 a d}+\frac{5 \sin ^3(c+d x) \cos ^3(c+d x)}{48 a d}+\frac{5 \sin (c+d x) \cos ^3(c+d x)}{64 a d}-\frac{5 \sin (c+d x) \cos (c+d x)}{128 a d}-\frac{5 x}{128 a} \]

[Out]

(-5*x)/(128*a) - (5*Cos[c + d*x]*Sin[c + d*x])/(128*a*d) + (5*Cos[c + d*x]^3*Sin[c + d*x])/(64*a*d) + (5*Cos[c
+ d*x]^3*Sin[c + d*x]^3)/(48*a*d) + (Cos[c + d*x]^3*Sin[c + d*x]^5)/(8*a*d) + Sin[c + d*x]^7/(7*a*d)

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Rubi [A]  time = 0.210315, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2839, 2564, 30, 2568, 2635, 8} \[ \frac{\sin ^7(c+d x)}{7 a d}+\frac{\sin ^5(c+d x) \cos ^3(c+d x)}{8 a d}+\frac{5 \sin ^3(c+d x) \cos ^3(c+d x)}{48 a d}+\frac{5 \sin (c+d x) \cos ^3(c+d x)}{64 a d}-\frac{5 \sin (c+d x) \cos (c+d x)}{128 a d}-\frac{5 x}{128 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x)/(128*a) - (5*Cos[c + d*x]*Sin[c + d*x])/(128*a*d) + (5*Cos[c + d*x]^3*Sin[c + d*x])/(64*a*d) + (5*Cos[c
+ d*x]^3*Sin[c + d*x]^3)/(48*a*d) + (Cos[c + d*x]^3*Sin[c + d*x]^5)/(8*a*d) + Sin[c + d*x]^7/(7*a*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 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

Rubi steps

\begin{align*} \int \frac{\sin ^8(c+d x)}{a+a \sec (c+d x)} \, dx &=-\int \frac{\cos (c+d x) \sin ^8(c+d x)}{-a-a \cos (c+d x)} \, dx\\ &=\frac{\int \cos (c+d x) \sin ^6(c+d x) \, dx}{a}-\frac{\int \cos ^2(c+d x) \sin ^6(c+d x) \, dx}{a}\\ &=\frac{\cos ^3(c+d x) \sin ^5(c+d x)}{8 a d}-\frac{5 \int \cos ^2(c+d x) \sin ^4(c+d x) \, dx}{8 a}+\frac{\operatorname{Subst}\left (\int x^6 \, dx,x,\sin (c+d x)\right )}{a d}\\ &=\frac{5 \cos ^3(c+d x) \sin ^3(c+d x)}{48 a d}+\frac{\cos ^3(c+d x) \sin ^5(c+d x)}{8 a d}+\frac{\sin ^7(c+d x)}{7 a d}-\frac{5 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{16 a}\\ &=\frac{5 \cos ^3(c+d x) \sin (c+d x)}{64 a d}+\frac{5 \cos ^3(c+d x) \sin ^3(c+d x)}{48 a d}+\frac{\cos ^3(c+d x) \sin ^5(c+d x)}{8 a d}+\frac{\sin ^7(c+d x)}{7 a d}-\frac{5 \int \cos ^2(c+d x) \, dx}{64 a}\\ &=-\frac{5 \cos (c+d x) \sin (c+d x)}{128 a d}+\frac{5 \cos ^3(c+d x) \sin (c+d x)}{64 a d}+\frac{5 \cos ^3(c+d x) \sin ^3(c+d x)}{48 a d}+\frac{\cos ^3(c+d x) \sin ^5(c+d x)}{8 a d}+\frac{\sin ^7(c+d x)}{7 a d}-\frac{5 \int 1 \, dx}{128 a}\\ &=-\frac{5 x}{128 a}-\frac{5 \cos (c+d x) \sin (c+d x)}{128 a d}+\frac{5 \cos ^3(c+d x) \sin (c+d x)}{64 a d}+\frac{5 \cos ^3(c+d x) \sin ^3(c+d x)}{48 a d}+\frac{\cos ^3(c+d x) \sin ^5(c+d x)}{8 a d}+\frac{\sin ^7(c+d x)}{7 a d}\\ \end{align*}

Mathematica [A]  time = 1.19528, size = 132, normalized size = 1.06 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (1680 \sin (c+d x)+336 \sin (2 (c+d x))-1008 \sin (3 (c+d x))+168 \sin (4 (c+d x))+336 \sin (5 (c+d x))-112 \sin (6 (c+d x))-48 \sin (7 (c+d x))+21 \sin (8 (c+d x))+1176 c-1176 \tan \left (\frac{c}{2}\right )-840 d x\right )}{10752 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(1176*c - 840*d*x + 1680*Sin[c + d*x] + 336*Sin[2*(c + d*x)] - 1008*Sin[3*(c
+ d*x)] + 168*Sin[4*(c + d*x)] + 336*Sin[5*(c + d*x)] - 112*Sin[6*(c + d*x)] - 48*Sin[7*(c + d*x)] + 21*Sin[8*
(c + d*x)] - 1176*Tan[c/2]))/(10752*a*d*(1 + Sec[c + d*x]))

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Maple [B]  time = 0.099, size = 290, normalized size = 2.3 \begin{align*}{\frac{5}{64\,da}\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{115}{192\,da} \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{383}{192\,da} \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{5053}{1344\,da} \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{44099}{1344\,da} \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{383}{192\,da} \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{115}{192\,da} \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{5}{64\,da} \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{5}{64\,da}\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)),x)

[Out]

5/64/a/d/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)+115/192/a/d/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*
c)^3+383/192/a/d/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^5+5053/1344/a/d/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(
1/2*d*x+1/2*c)^7+44099/1344/a/d/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^9-383/192/a/d/(1+tan(1/2*d*x+1/2
*c)^2)^8*tan(1/2*d*x+1/2*c)^11-115/192/a/d/(1+tan(1/2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^13-5/64/a/d/(1+tan(1/
2*d*x+1/2*c)^2)^8*tan(1/2*d*x+1/2*c)^15-5/64/d/a*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.52456, size = 486, normalized size = 3.89 \begin{align*} \frac{\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{2681 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{5053 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{44099 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac{2681 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac{805 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac{105 \, \sin \left (d x + c\right )^{15}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{15}}}{a + \frac{8 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{28 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{56 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{70 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{56 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac{28 \, a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac{8 \, a \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}} + \frac{a \sin \left (d x + c\right )^{16}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{16}}} - \frac{105 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{1344 \, 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)),x, algorithm="maxima")

[Out]

1/1344*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2681*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5 + 5053*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 44099*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 2
681*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 805*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 105*sin(d*x + c)^15/(c
os(d*x + c) + 1)^15)/(a + 8*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 +
56*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 56*a*sin(d*x + c)^10/(c
os(d*x + c) + 1)^10 + 28*a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 8*a*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 +
a*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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Fricas [A]  time = 1.79057, size = 261, normalized size = 2.09 \begin{align*} -\frac{105 \, d x -{\left (336 \, \cos \left (d x + c\right )^{7} - 384 \, \cos \left (d x + c\right )^{6} - 952 \, \cos \left (d x + c\right )^{5} + 1152 \, \cos \left (d x + c\right )^{4} + 826 \, \cos \left (d x + c\right )^{3} - 1152 \, \cos \left (d x + c\right )^{2} - 105 \, \cos \left (d x + c\right ) + 384\right )} \sin \left (d x + c\right )}{2688 \, a 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)),x, algorithm="fricas")

[Out]

-1/2688*(105*d*x - (336*cos(d*x + c)^7 - 384*cos(d*x + c)^6 - 952*cos(d*x + c)^5 + 1152*cos(d*x + c)^4 + 826*c
os(d*x + c)^3 - 1152*cos(d*x + c)^2 - 105*cos(d*x + c) + 384)*sin(d*x + c))/(a*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)),x)

[Out]

Timed out

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Giac [A]  time = 1.31033, size = 188, normalized size = 1.5 \begin{align*} -\frac{\frac{105 \,{\left (d x + c\right )}}{a} + \frac{2 \,{\left (105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{15} + 805 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 2681 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 44099 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 5053 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 2681 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 805 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, \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}}{2688 \, 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)),x, algorithm="giac")

[Out]

-1/2688*(105*(d*x + c)/a + 2*(105*tan(1/2*d*x + 1/2*c)^15 + 805*tan(1/2*d*x + 1/2*c)^13 + 2681*tan(1/2*d*x + 1
/2*c)^11 - 44099*tan(1/2*d*x + 1/2*c)^9 - 5053*tan(1/2*d*x + 1/2*c)^7 - 2681*tan(1/2*d*x + 1/2*c)^5 - 805*tan(
1/2*d*x + 1/2*c)^3 - 105*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^8*a))/d