3.53 \(\int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=49 \[ \frac {\cos ^3(c+d x)}{3 a d}+\frac {\sin (c+d x) \cos (c+d x)}{2 a d}+\frac {x}{2 a} \]

[Out]

1/2*x/a+1/3*cos(d*x+c)^3/a/d+1/2*cos(d*x+c)*sin(d*x+c)/a/d

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Rubi [A]  time = 0.06, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2682, 2635, 8} \[ \frac {\cos ^3(c+d x)}{3 a d}+\frac {\sin (c+d x) \cos (c+d x)}{2 a d}+\frac {x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x]),x]

[Out]

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

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Mathematica [B]  time = 0.32, size = 119, normalized size = 2.43 \[ -\frac {\left (\sqrt {\sin (c+d x)+1} \left (2 \sin ^3(c+d x)-5 \sin ^2(c+d x)+\sin (c+d x)+2\right )-6 \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}\right ) \cos ^5(c+d x)}{6 a d (\sin (c+d x)-1)^3 (\sin (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x]),x]

[Out]

-1/6*(Cos[c + d*x]^5*(-6*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]
]*(2 + Sin[c + d*x] - 5*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3)))/(a*d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/
2))

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fricas [A]  time = 0.65, size = 37, normalized size = 0.76 \[ \frac {2 \, \cos \left (d x + c\right )^{3} + 3 \, d x + 3 \, \cos \left (d x + c\right ) \sin \left (d x + c\right )}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(2*cos(d*x + c)^3 + 3*d*x + 3*cos(d*x + c)*sin(d*x + c))/(a*d)

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giac [A]  time = 0.86, size = 75, normalized size = 1.53 \[ \frac {\frac {3 \, {\left (d x + c\right )}}{a} - \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*(d*x + c)/a - 2*(3*tan(1/2*d*x + 1/2*c)^5 - 6*tan(1/2*d*x + 1/2*c)^4 - 3*tan(1/2*d*x + 1/2*c) - 2)/((ta
n(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d

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maple [B]  time = 0.14, size = 141, normalized size = 2.88 \[ -\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.60, size = 156, normalized size = 3.18 \[ \frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 2}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.87, size = 66, normalized size = 1.35 \[ \frac {x}{2\,a}+\frac {-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {2}{3}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + a*sin(c + d*x)),x)

[Out]

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

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sympy [A]  time = 11.38, size = 558, normalized size = 11.39 \[ \begin {cases} \frac {3 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {9 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {9 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {3 d x}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} - \frac {6 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {12 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} + \frac {4}{6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((3*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2
+ d*x/2)**2 + 6*a*d) + 9*d*x*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*
a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 9*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x
/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 3*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 +
18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 6*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x
/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 12*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(
c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 6*tan(c/2 + d*x/2)/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d
*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d
*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d), Ne(d, 0)), (x*cos(c)**4/(a*sin(c) + a), True))

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