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]

5/2*x/a^3+5/3*cos(d*x+c)^3/a^3/d+5/2*cos(d*x+c)*sin(d*x+c)/a^3/d+2*cos(d*x+c)^5/a/d/(a+a*sin(d*x+c))^2

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Rubi [A]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 8

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

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 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]

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*}

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Mathematica [A]  time = 0.47, 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]

-1/6*(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)))/(a^3*d*(-1 + Sin[c + d*x])^4*(1 + Sin[c + d
*x])^(7/2))

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fricas [A]  time = 0.61, size = 45, normalized size = 0.58 \[ -\frac {2 \, \cos \left (d x + c\right )^{3} - 15 \, d x + 9 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 24 \, \cos \left (d x + c\right )}{6 \, a^{3} d} \]

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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giac [A]  time = 0.42, size = 88, normalized size = 1.14 \[ \frac {\frac {15 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]

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

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maple [B]  time = 0.20, size = 177, normalized size = 2.30 \[ \frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {16 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a^{3} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {22}{3 a^{3} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d} \]

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

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maxima [B]  time = 0.47, size = 184, normalized size = 2.39 \[ -\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {9 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 22}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, d} \]

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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mupad [B]  time = 4.64, size = 57, normalized size = 0.74 \[ \frac {5\,x}{2\,a^3}+\frac {4\,\cos \left (c+d\,x\right )}{a^3\,d}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^3\,d}-\frac {3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^3,x)

[Out]

(5*x)/(2*a^3) + (4*cos(c + d*x))/(a^3*d) - cos(c + d*x)^3/(3*a^3*d) - (3*cos(c + d*x)*sin(c + d*x))/(2*a^3*d)

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sympy [A]  time = 108.27, size = 690, normalized size = 8.96 \[ \begin {cases} \frac {15 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {45 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {45 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {15 d x}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {18 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {36 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {96 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} - \frac {18 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {44}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{6}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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]

Piecewise((15*d*x*tan(c/2 + d*x/2)**6/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*
d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 45*d*x*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c
/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 45*d*x*tan(c/2 + d*x/2)**2/(6*a**3*d*tan(c/2 + d*
x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 15*d*x/(6*a**3*d*tan(c/2
+ d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 18*tan(c/2 + d*x/2)
**5/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d)
+ 36*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d
*x/2)**2 + 6*a**3*d) + 96*tan(c/2 + d*x/2)**2/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 +
18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) - 18*tan(c/2 + d*x/2)/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(
c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 44/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan
(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d), Ne(d, 0)), (x*cos(c)**6/(a*sin(c) + a)**3, True)
)

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