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]

5/8*x/a^2+5/12*cos(d*x+c)^3/a^2/d+5/8*cos(d*x+c)*sin(d*x+c)/a^2/d+1/4*cos(d*x+c)^5/d/(a^2+a^2*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 80, 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 = {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 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 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]

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 = 0.51, 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]

-1/24*(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)))/(a^2*d*(-1 + Sin[c + d*
x])^4*(1 + Sin[c + d*x])^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 50, normalized size = 0.62 \[ \frac {16 \, \cos \left (d x + c\right )^{3} + 15 \, d x - 3 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a^{2} d} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.65, size = 127, normalized size = 1.59 \[ \frac {\frac {15 \, {\left (d x + c\right )}}{a^{2}} - \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 16\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \]

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

________________________________________________________________________________________

maple [B]  time = 0.18, size = 279, normalized size = 3.49 \[ -\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {11 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {4}{3 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.73, size = 267, normalized size = 3.34 \[ \frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {33 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {48 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {33 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {48 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {9 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 16}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{12 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 4.75, size = 65, normalized size = 0.81 \[ \frac {5\,x}{8\,a^2}+\frac {2\,{\cos \left (c+d\,x\right )}^3}{3\,a^2\,d}-\frac {{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,a^2\,d}+\frac {5\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^2,x)

[Out]

(5*x)/(8*a^2) + (2*cos(c + d*x)^3)/(3*a^2*d) - (cos(c + d*x)^3*sin(c + d*x))/(4*a^2*d) + (5*cos(c + d*x)*sin(c
+ d*x))/(8*a^2*d)

________________________________________________________________________________________

sympy [A]  time = 58.83, size = 1243, normalized size = 15.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((15*d*x*tan(c/2 + d*x/2)**8/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**
2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 60*d*x*tan(c/2 + d*x/2)**6/(24*a**2*d*t
an(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/
2)**2 + 24*a**2*d) + 90*d*x*tan(c/2 + d*x/2)**4/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6
+ 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 60*d*x*tan(c/2 + d*x/2)**2/(2
4*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(
c/2 + d*x/2)**2 + 24*a**2*d) + 15*d*x/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**
2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 18*tan(c/2 + d*x/2)**7/(24*a**2*d*tan(c
/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**
2 + 24*a**2*d) + 96*tan(c/2 + d*x/2)**6/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a
**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) - 66*tan(c/2 + d*x/2)**5/(24*a**2*d*tan
(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)
**2 + 24*a**2*d) + 96*tan(c/2 + d*x/2)**4/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144
*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 66*tan(c/2 + d*x/2)**3/(24*a**2*d*t
an(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/
2)**2 + 24*a**2*d) + 32*tan(c/2 + d*x/2)**2/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 1
44*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 18*tan(c/2 + d*x/2)/(24*a**2*d*ta
n(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2
)**2 + 24*a**2*d) + 32/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d
*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d), Ne(d, 0)), (x*cos(c)**6/(a*sin(c) + a)**2, True))

________________________________________________________________________________________