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]

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

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Rubi [A]  time = 0.0550841, antiderivative size = 49, normalized size of antiderivative = 1., 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 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 ^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*}

Mathematica [B]  time = 0.32849, 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]

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

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Maple [B]  time = 0.048, size = 141, normalized size = 2.9 \begin{align*} -{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+{\frac{2}{3\,da} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+{\frac{1}{da}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+1
/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)+2/3/d/a/(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 = 1.42872, size = 211, normalized size = 4.31 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.52181, size = 92, normalized size = 1.88 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 20.8105, size = 697, normalized size = 14.22 \begin{align*} \text{result too large to display} \end{align*}

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((6*d*x*tan(c/2 + d*x/2)**6/(12*a*d*tan(c/2 + d*x/2)**6 + 36*a*d*tan(c/2 + d*x/2)**4 + 36*a*d*tan(c/2
+ d*x/2)**2 + 12*a*d) + 18*d*x*tan(c/2 + d*x/2)**4/(12*a*d*tan(c/2 + d*x/2)**6 + 36*a*d*tan(c/2 + d*x/2)**4 +
36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) + 18*d*x*tan(c/2 + d*x/2)**2/(12*a*d*tan(c/2 + d*x/2)**6 + 36*a*d*tan(c/
2 + d*x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) + 6*d*x/(12*a*d*tan(c/2 + d*x/2)**6 + 36*a*d*tan(c/2 + d*
x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) - 3*tan(c/2 + d*x/2)**6/(12*a*d*tan(c/2 + d*x/2)**6 + 36*a*d*ta
n(c/2 + d*x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) - 12*tan(c/2 + d*x/2)**5/(12*a*d*tan(c/2 + d*x/2)**6
+ 36*a*d*tan(c/2 + d*x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) + 15*tan(c/2 + d*x/2)**4/(12*a*d*tan(c/2 +
d*x/2)**6 + 36*a*d*tan(c/2 + d*x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) - 9*tan(c/2 + d*x/2)**2/(12*a*d
*tan(c/2 + d*x/2)**6 + 36*a*d*tan(c/2 + d*x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) + 12*tan(c/2 + d*x/2)
/(12*a*d*tan(c/2 + d*x/2)**6 + 36*a*d*tan(c/2 + d*x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d) + 5/(12*a*d*t
an(c/2 + d*x/2)**6 + 36*a*d*tan(c/2 + d*x/2)**4 + 36*a*d*tan(c/2 + d*x/2)**2 + 12*a*d), Ne(d, 0)), (x*cos(c)**
4/(a*sin(c) + a), True))

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Giac [A]  time = 1.16654, size = 101, normalized size = 2.06 \begin{align*} \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} \end{align*}

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