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

Optimal. Leaf size=104 \[ \frac{7 \cos ^5(c+d x)}{30 a^2 d}+\frac{\cos ^7(c+d x)}{6 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{7 \sin (c+d x) \cos ^3(c+d x)}{24 a^2 d}+\frac{7 \sin (c+d x) \cos (c+d x)}{16 a^2 d}+\frac{7 x}{16 a^2} \]

[Out]

(7*x)/(16*a^2) + (7*Cos[c + d*x]^5)/(30*a^2*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(16*a^2*d) + (7*Cos[c + d*x]^3*
Sin[c + d*x])/(24*a^2*d) + Cos[c + d*x]^7/(6*d*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.11249, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2679, 2682, 2635, 8} \[ \frac{7 \cos ^5(c+d x)}{30 a^2 d}+\frac{\cos ^7(c+d x)}{6 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{7 \sin (c+d x) \cos ^3(c+d x)}{24 a^2 d}+\frac{7 \sin (c+d x) \cos (c+d x)}{16 a^2 d}+\frac{7 x}{16 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*x)/(16*a^2) + (7*Cos[c + d*x]^5)/(30*a^2*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(16*a^2*d) + (7*Cos[c + d*x]^3*
Sin[c + d*x])/(24*a^2*d) + Cos[c + d*x]^7/(6*d*(a^2 + a^2*Sin[c + d*x]))

Rule 2679

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] + Dist[(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] && IntegersQ[2*m, 2*p]

Rule 2682

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] + Dist[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]

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{\cos ^8(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\cos ^7(c+d x)}{6 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{7 \int \frac{\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx}{6 a}\\ &=\frac{7 \cos ^5(c+d x)}{30 a^2 d}+\frac{\cos ^7(c+d x)}{6 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{7 \int \cos ^4(c+d x) \, dx}{6 a^2}\\ &=\frac{7 \cos ^5(c+d x)}{30 a^2 d}+\frac{7 \cos ^3(c+d x) \sin (c+d x)}{24 a^2 d}+\frac{\cos ^7(c+d x)}{6 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{7 \int \cos ^2(c+d x) \, dx}{8 a^2}\\ &=\frac{7 \cos ^5(c+d x)}{30 a^2 d}+\frac{7 \cos (c+d x) \sin (c+d x)}{16 a^2 d}+\frac{7 \cos ^3(c+d x) \sin (c+d x)}{24 a^2 d}+\frac{\cos ^7(c+d x)}{6 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{7 \int 1 \, dx}{16 a^2}\\ &=\frac{7 x}{16 a^2}+\frac{7 \cos ^5(c+d x)}{30 a^2 d}+\frac{7 \cos (c+d x) \sin (c+d x)}{16 a^2 d}+\frac{7 \cos ^3(c+d x) \sin (c+d x)}{24 a^2 d}+\frac{\cos ^7(c+d x)}{6 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.12024, size = 151, normalized size = 1.45 \[ -\frac{\left (\sqrt{\sin (c+d x)+1} \left (40 \sin ^6(c+d x)-136 \sin ^5(c+d x)+86 \sin ^4(c+d x)+202 \sin ^3(c+d x)-327 \sin ^2(c+d x)+39 \sin (c+d x)+96\right )-210 \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right ) \sqrt{1-\sin (c+d x)}\right ) \cos ^9(c+d x)}{240 a^2 d (\sin (c+d x)-1)^5 (\sin (c+d x)+1)^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[c + d*x]^9*(-210*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]]*
(96 + 39*Sin[c + d*x] - 327*Sin[c + d*x]^2 + 202*Sin[c + d*x]^3 + 86*Sin[c + d*x]^4 - 136*Sin[c + d*x]^5 + 40*
Sin[c + d*x]^6)))/(240*a^2*d*(-1 + Sin[c + d*x])^5*(1 + Sin[c + d*x])^(9/2))

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Maple [B]  time = 0.082, size = 415, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2
*c)^10-89/24/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*
d*x+1/2*c)^8+11/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7+8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan
(1/2*d*x+1/2*c)^6-11/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5+8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^
6*tan(1/2*d*x+1/2*c)^4+89/24/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3+4/5/d/a^2/(1+tan(1/2*d*x+1/
2*c)^2)^6*tan(1/2*d*x+1/2*c)^2+9/8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)+4/5/d/a^2/(1+tan(1/2*d*
x+1/2*c)^2)^6+7/8/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.46637, size = 531, normalized size = 5.11 \begin{align*} \frac{\frac{\frac{135 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{96 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{445 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{960 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{330 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{960 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{330 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{480 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{445 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac{480 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac{135 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + 96}{a^{2} + \frac{6 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{15 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{20 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{15 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{6 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac{a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} + \frac{105 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*((135*sin(d*x + c)/(cos(d*x + c) + 1) + 96*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 445*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 960*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 330*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 960*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6 + 330*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 480*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 - 445*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 480*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 135*sin(d*x + c
)^11/(cos(d*x + c) + 1)^11 + 96)/(a^2 + 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^2*sin(d*x + c)^4/(cos
(d*x + c) + 1)^4 + 20*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 6
*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 105*arctan(sin(d*x +
c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 2.00226, size = 161, normalized size = 1.55 \begin{align*} \frac{96 \, \cos \left (d x + c\right )^{5} + 105 \, d x - 5 \,{\left (8 \, \cos \left (d x + c\right )^{5} - 14 \, \cos \left (d x + c\right )^{3} - 21 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(96*cos(d*x + c)^5 + 105*d*x - 5*(8*cos(d*x + c)^5 - 14*cos(d*x + c)^3 - 21*cos(d*x + c))*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(cos(d*x+c)**8/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.14215, size = 242, normalized size = 2.33 \begin{align*} \frac{\frac{105 \,{\left (d x + c\right )}}{a^{2}} - \frac{2 \,{\left (135 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 480 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 445 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 480 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 330 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 960 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 330 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 960 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 445 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 96 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 135 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 96\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{6} a^{2}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(105*(d*x + c)/a^2 - 2*(135*tan(1/2*d*x + 1/2*c)^11 - 480*tan(1/2*d*x + 1/2*c)^10 + 445*tan(1/2*d*x + 1/
2*c)^9 - 480*tan(1/2*d*x + 1/2*c)^8 - 330*tan(1/2*d*x + 1/2*c)^7 - 960*tan(1/2*d*x + 1/2*c)^6 + 330*tan(1/2*d*
x + 1/2*c)^5 - 960*tan(1/2*d*x + 1/2*c)^4 - 445*tan(1/2*d*x + 1/2*c)^3 - 96*tan(1/2*d*x + 1/2*c)^2 - 135*tan(1
/2*d*x + 1/2*c) - 96)/((tan(1/2*d*x + 1/2*c)^2 + 1)^6*a^2))/d