3.17 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=78 \[ -\frac{5 a^2 \cos ^3(c+d x)}{12 d}-\frac{\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}+\frac{5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5 a^2 x}{8} \]

[Out]

(5*a^2*x)/8 - (5*a^2*Cos[c + d*x]^3)/(12*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (Cos[c + d*x]^3*(a^2 +
a^2*Sin[c + d*x]))/(4*d)

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Rubi [A]  time = 0.0885705, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2678, 2669, 2635, 8} \[ -\frac{5 a^2 \cos ^3(c+d x)}{12 d}-\frac{\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}+\frac{5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5 a^2 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*x)/8 - (5*a^2*Cos[c + d*x]^3)/(12*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (Cos[c + d*x]^3*(a^2 +
a^2*Sin[c + d*x]))/(4*d)

Rule 2678

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

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 \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac{\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}+\frac{1}{4} (5 a) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{5 a^2 \cos ^3(c+d x)}{12 d}-\frac{\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}+\frac{1}{4} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{5 a^2 \cos ^3(c+d x)}{12 d}+\frac{5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac{\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}+\frac{1}{8} \left (5 a^2\right ) \int 1 \, dx\\ &=\frac{5 a^2 x}{8}-\frac{5 a^2 \cos ^3(c+d x)}{12 d}+\frac{5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac{\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.302669, size = 131, normalized size = 1.68 \[ -\frac{a^2 \left (30 \sqrt{1-\sin (c+d x)} \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )+\sqrt{\sin (c+d x)+1} \left (6 \sin ^4(c+d x)+10 \sin ^3(c+d x)-7 \sin ^2(c+d x)-25 \sin (c+d x)+16\right )\right ) \cos ^3(c+d x)}{24 d (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Cos[c + d*x]^3*(30*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]
]*(16 - 25*Sin[c + d*x] - 7*Sin[c + d*x]^2 + 10*Sin[c + d*x]^3 + 6*Sin[c + d*x]^4)))/(24*d*(-1 + Sin[c + d*x])
^2*(1 + Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.036, size = 87, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8}}+{\frac{dx}{8}}+{\frac{c}{8}} \right ) -{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}}+{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\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)^2*(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.942792, size = 88, normalized size = 1.13 \begin{align*} -\frac{64 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(64*a^2*cos(d*x + c)^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^
2)/d

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Fricas [A]  time = 1.67521, size = 144, normalized size = 1.85 \begin{align*} -\frac{16 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} d x + 3 \,{\left (2 \, a^{2} \cos \left (d x + c\right )^{3} - 5 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(16*a^2*cos(d*x + c)^3 - 15*a^2*d*x + 3*(2*a^2*cos(d*x + c)^3 - 5*a^2*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 1.37272, size = 180, normalized size = 2.31 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{2 a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \cos ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13125, size = 97, normalized size = 1.24 \begin{align*} \frac{5}{8} \, a^{2} x - \frac{a^{2} \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac{a^{2} \cos \left (d x + c\right )}{2 \, d} - \frac{a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*a^2*x - 1/6*a^2*cos(3*d*x + 3*c)/d - 1/2*a^2*cos(d*x + c)/d - 1/32*a^2*sin(4*d*x + 4*c)/d + 1/4*a^2*sin(2*
d*x + 2*c)/d