3.77 \(\int \frac{\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=77 \[ \frac{5 \cos ^3(c+d x)}{3 a^3 d}+\frac{5 \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{5 x}{2 a^3}+\frac{2 \cos ^5(c+d x)}{a d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.0999435, antiderivative size = 77, 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 = {2680, 2682, 2635, 8} \[ \frac{5 \cos ^3(c+d x)}{3 a^3 d}+\frac{5 \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac{5 x}{2 a^3}+\frac{2 \cos ^5(c+d x)}{a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 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))^3} \, dx &=\frac{2 \cos ^5(c+d x)}{a d (a+a \sin (c+d x))^2}+\frac{5 \int \frac{\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac{5 \cos ^3(c+d x)}{3 a^3 d}+\frac{2 \cos ^5(c+d x)}{a d (a+a \sin (c+d x))^2}+\frac{5 \int \cos ^2(c+d x) \, dx}{a^3}\\ &=\frac{5 \cos ^3(c+d x)}{3 a^3 d}+\frac{5 \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac{2 \cos ^5(c+d x)}{a d (a+a \sin (c+d x))^2}+\frac{5 \int 1 \, dx}{2 a^3}\\ &=\frac{5 x}{2 a^3}+\frac{5 \cos ^3(c+d x)}{3 a^3 d}+\frac{5 \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac{2 \cos ^5(c+d x)}{a d (a+a \sin (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.485727, size = 121, normalized size = 1.57 \[ -\frac{\left (30 \sqrt{1-\sin (c+d x)} \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )+\sqrt{\sin (c+d x)+1} \left (2 \sin ^3(c+d x)-11 \sin ^2(c+d x)+31 \sin (c+d x)-22\right )\right ) \cos ^7(c+d x)}{6 a^3 d (\sin (c+d x)-1)^4 (\sin (c+d x)+1)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^3,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]]*(-
22 + 31*Sin[c + d*x] - 11*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3)))/(6*a^3*d*(-1 + Sin[c + d*x])^4*(1 + Sin[c + d*x
])^(7/2))

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Maple [B]  time = 0.084, size = 177, normalized size = 2.3 \begin{align*} 3\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+6\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+16\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-3\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{22}{3\,d{a}^{3}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+5\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \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))^3,x)

[Out]

3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^
4+16/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*
c)+22/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3+5/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.44955, size = 248, normalized size = 3.22 \begin{align*} -\frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{9 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 22}{a^{3} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, 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))^3,x, algorithm="maxima")

[Out]

-1/3*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 18*sin(d*x + c)^4/(cos(d*x
+ c) + 1)^4 - 9*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 22)/(a^3 + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) - 15*arctan(sin(d*x + c)/
(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.98291, size = 122, normalized size = 1.58 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{3} - 15 \, d x + 9 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 24 \, \cos \left (d x + c\right )}{6 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

-1/6*(2*cos(d*x + c)^3 - 15*d*x + 9*cos(d*x + c)*sin(d*x + c) - 24*cos(d*x + c))/(a^3*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15625, size = 119, normalized size = 1.55 \begin{align*} \frac{\frac{15 \,{\left (d x + c\right )}}{a^{3}} + \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 48 \, \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 ) + 22\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, 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))^3,x, algorithm="giac")

[Out]

1/6*(15*(d*x + c)/a^3 + 2*(9*tan(1/2*d*x + 1/2*c)^5 + 18*tan(1/2*d*x + 1/2*c)^4 + 48*tan(1/2*d*x + 1/2*c)^2 -
9*tan(1/2*d*x + 1/2*c) + 22)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d