3.31 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=106 \[ -\frac{7 a^3 \cos ^3(c+d x)}{12 d}-\frac{7 \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{20 d}+\frac{7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{7 a^3 x}{8}-\frac{a \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

[Out]

(7*a^3*x)/8 - (7*a^3*Cos[c + d*x]^3)/(12*d) + (7*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*Cos[c + d*x]^3*(a +
a*Sin[c + d*x])^2)/(5*d) - (7*Cos[c + d*x]^3*(a^3 + a^3*Sin[c + d*x]))/(20*d)

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Rubi [A]  time = 0.12388, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2678, 2669, 2635, 8} \[ -\frac{7 a^3 \cos ^3(c+d x)}{12 d}-\frac{7 \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{20 d}+\frac{7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{7 a^3 x}{8}-\frac{a \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^3*x)/8 - (7*a^3*Cos[c + d*x]^3)/(12*d) + (7*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*Cos[c + d*x]^3*(a +
a*Sin[c + d*x])^2)/(5*d) - (7*Cos[c + d*x]^3*(a^3 + a^3*Sin[c + d*x]))/(20*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))^3 \, dx &=-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac{1}{5} (7 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac{7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}+\frac{1}{4} \left (7 a^2\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{7 a^3 \cos ^3(c+d x)}{12 d}-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac{7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}+\frac{1}{4} \left (7 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{7 a^3 \cos ^3(c+d x)}{12 d}+\frac{7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac{7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}+\frac{1}{8} \left (7 a^3\right ) \int 1 \, dx\\ &=\frac{7 a^3 x}{8}-\frac{7 a^3 \cos ^3(c+d x)}{12 d}+\frac{7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac{7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}\\ \end{align*}

Mathematica [A]  time = 0.42379, size = 141, normalized size = 1.33 \[ -\frac{a^3 \left (210 \sqrt{1-\sin (c+d x)} \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )+\sqrt{\sin (c+d x)+1} \left (24 \sin ^5(c+d x)+66 \sin ^4(c+d x)+22 \sin ^3(c+d x)-97 \sin ^2(c+d x)-151 \sin (c+d x)+136\right )\right ) \cos ^3(c+d x)}{120 d (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Cos[c + d*x]^3*(210*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x
]]*(136 - 151*Sin[c + d*x] - 97*Sin[c + d*x]^2 + 22*Sin[c + d*x]^3 + 66*Sin[c + d*x]^4 + 24*Sin[c + d*x]^5)))/
(120*d*(-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.039, size = 121, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +3\,{a}^{3} \left ( -1/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1/8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/8\,dx+c/8 \right ) -{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{a}^{3} \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))^3,x)

[Out]

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

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Maxima [A]  time = 0.962166, size = 123, normalized size = 1.16 \begin{align*} -\frac{480 \, a^{3} \cos \left (d x + c\right )^{3} - 32 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 45 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{480 \, 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))^3,x, algorithm="maxima")

[Out]

-1/480*(480*a^3*cos(d*x + c)^3 - 32*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^3 - 45*(4*d*x + 4*c - sin(4*d*x +
4*c))*a^3 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3)/d

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Fricas [A]  time = 1.76386, size = 181, normalized size = 1.71 \begin{align*} \frac{24 \, a^{3} \cos \left (d x + c\right )^{5} - 160 \, a^{3} \cos \left (d x + c\right )^{3} + 105 \, a^{3} d x - 15 \,{\left (6 \, a^{3} \cos \left (d x + c\right )^{3} - 7 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, 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))^3,x, algorithm="fricas")

[Out]

1/120*(24*a^3*cos(d*x + c)^5 - 160*a^3*cos(d*x + c)^3 + 105*a^3*d*x - 15*(6*a^3*cos(d*x + c)^3 - 7*a^3*cos(d*x
+ c))*sin(d*x + c))/d

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Sympy [A]  time = 2.6959, size = 226, normalized size = 2.13 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{3 a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{2 a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{a^{3} \cos ^{3}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{3} \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))**3,x)

[Out]

Piecewise((3*a**3*x*sin(c + d*x)**4/8 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**3*x*sin(c + d*x)**2/2
+ 3*a**3*x*cos(c + d*x)**4/8 + a**3*x*cos(c + d*x)**2/2 + 3*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) - a**3*sin
(c + d*x)**2*cos(c + d*x)**3/(3*d) - 3*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + a**3*sin(c + d*x)*cos(c + d*x
)/(2*d) - 2*a**3*cos(c + d*x)**5/(15*d) - a**3*cos(c + d*x)**3/d, Ne(d, 0)), (x*(a*sin(c) + a)**3*cos(c)**2, T
rue))

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Giac [A]  time = 1.14277, size = 120, normalized size = 1.13 \begin{align*} \frac{7}{8} \, a^{3} x + \frac{a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{13 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{7 \, a^{3} \cos \left (d x + c\right )}{8 \, d} - \frac{3 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a^{3} \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))^3,x, algorithm="giac")

[Out]

7/8*a^3*x + 1/80*a^3*cos(5*d*x + 5*c)/d - 13/48*a^3*cos(3*d*x + 3*c)/d - 7/8*a^3*cos(d*x + c)/d - 3/32*a^3*sin
(4*d*x + 4*c)/d + 1/4*a^3*sin(2*d*x + 2*c)/d