3.64 \(\int \frac{\cos ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=80 \[ \frac{5 \cos ^3(c+d x)}{12 a^2 d}+\frac{\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{5 \sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac{5 x}{8 a^2} \]

[Out]

(5*x)/(8*a^2) + (5*Cos[c + d*x]^3)/(12*a^2*d) + (5*Cos[c + d*x]*Sin[c + d*x])/(8*a^2*d) + Cos[c + d*x]^5/(4*d*
(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.100451, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2679, 2682, 2635, 8} \[ \frac{5 \cos ^3(c+d x)}{12 a^2 d}+\frac{\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{5 \sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac{5 x}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^2,x]

[Out]

(5*x)/(8*a^2) + (5*Cos[c + d*x]^3)/(12*a^2*d) + (5*Cos[c + d*x]*Sin[c + d*x])/(8*a^2*d) + Cos[c + d*x]^5/(4*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 ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\cos ^5(c+d x)}{4 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{5 \int \frac{\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx}{4 a}\\ &=\frac{5 \cos ^3(c+d x)}{12 a^2 d}+\frac{\cos ^5(c+d x)}{4 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{5 \int \cos ^2(c+d x) \, dx}{4 a^2}\\ &=\frac{5 \cos ^3(c+d x)}{12 a^2 d}+\frac{5 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac{\cos ^5(c+d x)}{4 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{5 \int 1 \, dx}{8 a^2}\\ &=\frac{5 x}{8 a^2}+\frac{5 \cos ^3(c+d x)}{12 a^2 d}+\frac{5 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac{\cos ^5(c+d x)}{4 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.721368, size = 131, normalized size = 1.64 \[ -\frac{\left (30 \sqrt{1-\sin (c+d x)} \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )+\sqrt{\sin (c+d x)+1} \left (6 \sin ^4(c+d x)-22 \sin ^3(c+d x)+25 \sin ^2(c+d x)+7 \sin (c+d x)-16\right )\right ) \cos ^7(c+d x)}{24 a^2 d (\sin (c+d x)-1)^4 (\sin (c+d x)+1)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^2,x]

[Out]

-(Cos[c + d*x]^7*(30*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]]*(-
16 + 7*Sin[c + d*x] + 25*Sin[c + d*x]^2 - 22*Sin[c + d*x]^3 + 6*Sin[c + d*x]^4)))/(24*a^2*d*(-1 + Sin[c + d*x]
)^4*(1 + Sin[c + d*x])^(7/2))

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Maple [B]  time = 0.067, size = 279, normalized size = 3.5 \begin{align*} -{\frac{3}{4\,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 ) ^{-4}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{11}{4\,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 ) ^{-4}}+4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{11}{4\,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 ) ^{-4}}+{\frac{4}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+{\frac{3}{4\,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 ) ^{-4}}+{\frac{4}{3\,d{a}^{2}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+{\frac{5}{4\,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(cos(d*x+c)^6/(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 1.43447, size = 360, normalized size = 4.5 \begin{align*} \frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{16 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{33 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{48 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{33 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{48 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{9 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 16}{a^{2} + \frac{4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 33*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 48*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 33*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 48*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 - 9*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 16)/(a^2 + 4*a^2*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 6*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d
*x + c)^8/(cos(d*x + c) + 1)^8) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.84829, size = 130, normalized size = 1.62 \begin{align*} \frac{16 \, \cos \left (d x + c\right )^{3} + 15 \, d x - 3 \,{\left (2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(16*cos(d*x + c)^3 + 15*d*x - 3*(2*cos(d*x + c)^3 - 5*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)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.12186, size = 171, normalized size = 2.14 \begin{align*} \frac{\frac{15 \,{\left (d x + c\right )}}{a^{2}} - \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 33 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 33 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 16\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(15*(d*x + c)/a^2 - 2*(9*tan(1/2*d*x + 1/2*c)^7 - 48*tan(1/2*d*x + 1/2*c)^6 + 33*tan(1/2*d*x + 1/2*c)^5 -
48*tan(1/2*d*x + 1/2*c)^4 - 33*tan(1/2*d*x + 1/2*c)^3 - 16*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) -
16)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^2))/d