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

Optimal. Leaf size=73 \[ \frac {\cos ^5(c+d x)}{5 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}+\frac {3 x}{8 a} \]

[Out]

3/8*x/a+1/5*cos(d*x+c)^5/a/d+3/8*cos(d*x+c)*sin(d*x+c)/a/d+1/4*cos(d*x+c)^3*sin(d*x+c)/a/d

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Rubi [A]  time = 0.07, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2682, 2635, 8} \[ \frac {\cos ^5(c+d x)}{5 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}+\frac {3 x}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.80, size = 141, normalized size = 1.93 \[ -\frac {\left (30 \sqrt {1-\sin (c+d x)} \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {\sin (c+d x)+1} \left (8 \sin ^5(c+d x)-18 \sin ^4(c+d x)-6 \sin ^3(c+d x)+41 \sin ^2(c+d x)-17 \sin (c+d x)-8\right )\right ) \cos ^7(c+d x)}{40 a d (\sin (c+d x)-1)^4 (\sin (c+d x)+1)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/40*(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
]]*(-8 - 17*Sin[c + d*x] + 41*Sin[c + d*x]^2 - 6*Sin[c + d*x]^3 - 18*Sin[c + d*x]^4 + 8*Sin[c + d*x]^5)))/(a*d
*(-1 + Sin[c + d*x])^4*(1 + Sin[c + d*x])^(7/2))

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fricas [A]  time = 1.11, size = 50, normalized size = 0.68 \[ \frac {8 \, \cos \left (d x + c\right )^{5} + 15 \, d x + 5 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(8*cos(d*x + c)^5 + 15*d*x + 5*(2*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c))/(a*d)

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giac [A]  time = 0.61, size = 114, normalized size = 1.56 \[ \frac {\frac {15 \, {\left (d x + c\right )}}{a} - \frac {2 \, {\left (25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a}}{40 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(15*(d*x + c)/a - 2*(25*tan(1/2*d*x + 1/2*c)^9 - 40*tan(1/2*d*x + 1/2*c)^8 + 10*tan(1/2*d*x + 1/2*c)^7 -
80*tan(1/2*d*x + 1/2*c)^4 - 10*tan(1/2*d*x + 1/2*c)^3 - 25*tan(1/2*d*x + 1/2*c) - 8)/((tan(1/2*d*x + 1/2*c)^2
+ 1)^5*a))/d

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maple [B]  time = 0.15, size = 245, normalized size = 3.36 \[ -\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {2 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {2}{5 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.67, size = 258, normalized size = 3.53 \[ \frac {\frac {\frac {25 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {80 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {10 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {40 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {25 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 8}{a + \frac {5 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*((25*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 80*sin(d*x + c)^4/(cos(d*
x + c) + 1)^4 - 10*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 40*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 25*sin(d*x +
c)^9/(cos(d*x + c) + 1)^9 + 8)/(a + 5*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 + 10*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a*sin(d*x + c
)^10/(cos(d*x + c) + 1)^10) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 8.15, size = 107, normalized size = 1.47 \[ \frac {3\,x}{8\,a}+\frac {-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {2}{5}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x)/(8*a) + ((5*tan(c/2 + (d*x)/2))/4 + tan(c/2 + (d*x)/2)^3/2 + 4*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)
^7/2 + 2*tan(c/2 + (d*x)/2)^8 - (5*tan(c/2 + (d*x)/2)^9)/4 + 2/5)/(a*d*(tan(c/2 + (d*x)/2)^2 + 1)^5)

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sympy [A]  time = 34.04, size = 1355, normalized size = 18.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((15*d*x*tan(c/2 + d*x/2)**10/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*ta
n(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 75*d*x*tan(c/2 + d*x
/2)**8/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(
c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 150*d*x*tan(c/2 + d*x/2)**6/(40*a*d*tan(c/2 + d*x/2)
**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c
/2 + d*x/2)**2 + 40*a*d) + 150*d*x*tan(c/2 + d*x/2)**4/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)
**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 75*d
*x*tan(c/2 + d*x/2)**2/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**
6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 15*d*x/(40*a*d*tan(c/2 + d*x/2)**10
+ 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 +
d*x/2)**2 + 40*a*d) - 50*tan(c/2 + d*x/2)**9/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*
a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 80*tan(c/2 + d
*x/2)**8/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*ta
n(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) - 20*tan(c/2 + d*x/2)**7/(40*a*d*tan(c/2 + d*x/2)**1
0 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2
+ d*x/2)**2 + 40*a*d) + 160*tan(c/2 + d*x/2)**4/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 4
00*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 20*tan(c/2
+ d*x/2)**3/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d
*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 50*tan(c/2 + d*x/2)/(40*a*d*tan(c/2 + d*x/2)**1
0 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2
+ d*x/2)**2 + 40*a*d) + 16/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/
2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d), Ne(d, 0)), (x*cos(c)**6/(a*sin(c)
+ a), True))

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