3.1.66 \(\int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [66]

Optimal. Leaf size=56 \[ \frac {3 x}{2 a^2}+\frac {3 \cos (c+d x)}{2 a^2 d}+\frac {\cos ^3(c+d x)}{2 d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 56, 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 = {2758, 2761, 8} \begin {gather*} \frac {3 \cos (c+d x)}{2 a^2 d}+\frac {\cos ^3(c+d x)}{2 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {3 x}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 8

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

Rule 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[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 2761

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

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Mathematica [A]
time = 0.12, size = 109, normalized size = 1.95 \begin {gather*} -\frac {\cos ^5(c+d x) \left (-6 \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (4-5 \sin (c+d x)+\sin ^2(c+d x)\right )\right )}{2 a^2 d (-1+\sin (c+d x))^3 (1+\sin (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]
time = 0.20, size = 77, normalized size = 1.38

method	result	size
risch	\(\frac {3 x}{2 a^{2}}+\frac {2 \cos \left (d x +c \right )}{a^{2} d}-\frac {\sin \left (2 d x +2 c \right )}{4 a^{2} d}\)	\(39\)
derivativedivides	\(\frac {\frac {2 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}\)	\(77\)
default	\(\frac {\frac {2 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+2\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}\)	\(77\)
norman	\(\frac {\frac {4}{a d}+\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {16 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {36 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {3 x}{2 a}+\frac {9 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {21 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {39 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {27 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {33 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {33 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {27 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {39 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {21 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {9 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {3 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {7 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {21 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {28 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {44 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {42 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {46 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(420\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (50) = 100\).
time = 0.50, size = 140, normalized size = 2.50 \begin {gather*} -\frac {\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 4}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 35, normalized size = 0.62 \begin {gather*} \frac {3 \, d x - \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, \cos \left (d x + c\right )}{2 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 403 vs. \(2 (48) = 96\).
time = 9.44, size = 403, normalized size = 7.20 \begin {gather*} \begin {cases} \frac {3 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {6 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {3 d x}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {2 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {8 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} - \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} + \frac {8}{2 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((3*d*x*tan(c/2 + d*x/2)**4/(2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d)
+ 6*d*x*tan(c/2 + d*x/2)**2/(2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) + 3*d*x/(
2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) + 2*tan(c/2 + d*x/2)**3/(2*a**2*d*tan(
c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) + 8*tan(c/2 + d*x/2)**2/(2*a**2*d*tan(c/2 + d*x/2)*
*4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d) - 2*tan(c/2 + d*x/2)/(2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*ta
n(c/2 + d*x/2)**2 + 2*a**2*d) + 8/(2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d), Ne
(d, 0)), (x*cos(c)**4/(a*sin(c) + a)**2, True))

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Giac [A]
time = 6.03, size = 73, normalized size = 1.30 \begin {gather*} \frac {\frac {3 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.65, size = 32, normalized size = 0.57 \begin {gather*} \frac {4\,\cos \left (c+d\,x\right )-\frac {\sin \left (2\,c+2\,d\,x\right )}{2}+3\,d\,x}{2\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + a*sin(c + d*x))^2,x)

[Out]

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

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