3.27 \(\int \cos ^6(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=154 \[ -\frac{11 a^3 \cos ^7(c+d x)}{56 d}-\frac{11 \cos ^7(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{72 d}+\frac{11 a^3 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac{55 a^3 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac{55 a^3 \sin (c+d x) \cos (c+d x)}{128 d}+\frac{55 a^3 x}{128}-\frac{a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d} \]

[Out]

(55*a^3*x)/128 - (11*a^3*Cos[c + d*x]^7)/(56*d) + (55*a^3*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (55*a^3*Cos[c +
d*x]^3*Sin[c + d*x])/(192*d) + (11*a^3*Cos[c + d*x]^5*Sin[c + d*x])/(48*d) - (a*Cos[c + d*x]^7*(a + a*Sin[c +
d*x])^2)/(9*d) - (11*Cos[c + d*x]^7*(a^3 + a^3*Sin[c + d*x]))/(72*d)

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Rubi [A]  time = 0.153738, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2678, 2669, 2635, 8} \[ -\frac{11 a^3 \cos ^7(c+d x)}{56 d}-\frac{11 \cos ^7(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{72 d}+\frac{11 a^3 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac{55 a^3 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac{55 a^3 \sin (c+d x) \cos (c+d x)}{128 d}+\frac{55 a^3 x}{128}-\frac{a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + a*Sin[c + d*x])^3,x]

[Out]

(55*a^3*x)/128 - (11*a^3*Cos[c + d*x]^7)/(56*d) + (55*a^3*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (55*a^3*Cos[c +
d*x]^3*Sin[c + d*x])/(192*d) + (11*a^3*Cos[c + d*x]^5*Sin[c + d*x])/(48*d) - (a*Cos[c + d*x]^7*(a + a*Sin[c +
d*x])^2)/(9*d) - (11*Cos[c + d*x]^7*(a^3 + a^3*Sin[c + d*x]))/(72*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 ^6(c+d x) (a+a \sin (c+d x))^3 \, dx &=-\frac{a \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}+\frac{1}{9} (11 a) \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{a \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac{11 \cos ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{72 d}+\frac{1}{8} \left (11 a^2\right ) \int \cos ^6(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{11 a^3 \cos ^7(c+d x)}{56 d}-\frac{a \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac{11 \cos ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{72 d}+\frac{1}{8} \left (11 a^3\right ) \int \cos ^6(c+d x) \, dx\\ &=-\frac{11 a^3 \cos ^7(c+d x)}{56 d}+\frac{11 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac{a \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac{11 \cos ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{72 d}+\frac{1}{48} \left (55 a^3\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac{11 a^3 \cos ^7(c+d x)}{56 d}+\frac{55 a^3 \cos ^3(c+d x) \sin (c+d x)}{192 d}+\frac{11 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac{a \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac{11 \cos ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{72 d}+\frac{1}{64} \left (55 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{11 a^3 \cos ^7(c+d x)}{56 d}+\frac{55 a^3 \cos (c+d x) \sin (c+d x)}{128 d}+\frac{55 a^3 \cos ^3(c+d x) \sin (c+d x)}{192 d}+\frac{11 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac{a \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac{11 \cos ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{72 d}+\frac{1}{128} \left (55 a^3\right ) \int 1 \, dx\\ &=\frac{55 a^3 x}{128}-\frac{11 a^3 \cos ^7(c+d x)}{56 d}+\frac{55 a^3 \cos (c+d x) \sin (c+d x)}{128 d}+\frac{55 a^3 \cos ^3(c+d x) \sin (c+d x)}{192 d}+\frac{11 a^3 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac{a \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac{11 \cos ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{72 d}\\ \end{align*}

Mathematica [A]  time = 1.99511, size = 181, normalized size = 1.18 \[ -\frac{a^3 \left (6930 \sqrt{1-\sin (c+d x)} \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )+\sqrt{\sin (c+d x)+1} \left (896 \sin ^9(c+d x)+2128 \sin ^8(c+d x)-2000 \sin ^7(c+d x)-8248 \sin ^6(c+d x)-1224 \sin ^5(c+d x)+11514 \sin ^4(c+d x)+7174 \sin ^3(c+d x)-5641 \sin ^2(c+d x)-8311 \sin (c+d x)+3712\right )\right ) \cos ^7(c+d x)}{8064 d (\sin (c+d x)-1)^4 (\sin (c+d x)+1)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sin[c + d*x])^3,x]

[Out]

-(a^3*Cos[c + d*x]^7*(6930*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*
x]]*(3712 - 8311*Sin[c + d*x] - 5641*Sin[c + d*x]^2 + 7174*Sin[c + d*x]^3 + 11514*Sin[c + d*x]^4 - 1224*Sin[c
+ d*x]^5 - 8248*Sin[c + d*x]^6 - 2000*Sin[c + d*x]^7 + 2128*Sin[c + d*x]^8 + 896*Sin[c + d*x]^9)))/(8064*d*(-1
+ Sin[c + d*x])^4*(1 + Sin[c + d*x])^(7/2))

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Maple [A]  time = 0.043, size = 163, 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 ) ^{7}}{9}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{63}} \right ) +3\,{a}^{3} \left ( -1/8\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+1/48\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) \sin \left ( dx+c \right ) +{\frac{5\,dx}{128}}+{\frac{5\,c}{128}} \right ) -{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{7}}+{a}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7)+3*a^3*(-1/8*sin(d*x+c)*cos(d*x+c)^7+1/48*(cos(d*x+
c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/128*d*x+5/128*c)-3/7*a^3*cos(d*x+c)^7+a^3*(1/6*(cos(d*x+c)
^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))

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Maxima [A]  time = 0.98203, size = 190, normalized size = 1.23 \begin{align*} -\frac{27648 \, a^{3} \cos \left (d x + c\right )^{7} - 1024 \,{\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{3} - 63 \,{\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} + 336 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{64512 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/64512*(27648*a^3*cos(d*x + c)^7 - 1024*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^3 - 63*(64*sin(2*d*x + 2*c)^
3 + 120*d*x + 120*c - 3*sin(8*d*x + 8*c) - 24*sin(4*d*x + 4*c))*a^3 + 336*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*
c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^3)/d

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Fricas [A]  time = 1.87002, size = 258, normalized size = 1.68 \begin{align*} \frac{896 \, a^{3} \cos \left (d x + c\right )^{9} - 4608 \, a^{3} \cos \left (d x + c\right )^{7} + 3465 \, a^{3} d x - 21 \,{\left (144 \, a^{3} \cos \left (d x + c\right )^{7} - 88 \, a^{3} \cos \left (d x + c\right )^{5} - 110 \, a^{3} \cos \left (d x + c\right )^{3} - 165 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8064 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/8064*(896*a^3*cos(d*x + c)^9 - 4608*a^3*cos(d*x + c)^7 + 3465*a^3*d*x - 21*(144*a^3*cos(d*x + c)^7 - 88*a^3*
cos(d*x + c)^5 - 110*a^3*cos(d*x + c)^3 - 165*a^3*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 24.3576, size = 439, normalized size = 2.85 \begin{align*} \begin{cases} \frac{15 a^{3} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac{15 a^{3} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac{5 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{45 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac{15 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac{15 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{15 a^{3} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac{5 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{15 a^{3} \sin ^{7}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{128 d} + \frac{55 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac{5 a^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{73 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} + \frac{5 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac{15 a^{3} \sin{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} + \frac{11 a^{3} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{2 a^{3} \cos ^{9}{\left (c + d x \right )}}{63 d} - \frac{3 a^{3} \cos ^{7}{\left (c + d x \right )}}{7 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{3} \cos ^{6}{\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)**6*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((15*a**3*x*sin(c + d*x)**8/128 + 15*a**3*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 5*a**3*x*sin(c + d*x
)**6/16 + 45*a**3*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 15*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a**
3*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 15*a**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 15*a**3*x*cos(c + d*x)
**8/128 + 5*a**3*x*cos(c + d*x)**6/16 + 15*a**3*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 55*a**3*sin(c + d*x)**5
*cos(c + d*x)**3/(128*d) + 5*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 73*a**3*sin(c + d*x)**3*cos(c + d*x)**
5/(128*d) + 5*a**3*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - a**3*sin(c + d*x)**2*cos(c + d*x)**7/(7*d) - 15*a**
3*sin(c + d*x)*cos(c + d*x)**7/(128*d) + 11*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 2*a**3*cos(c + d*x)**9/
(63*d) - 3*a**3*cos(c + d*x)**7/(7*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*cos(c)**6, True))

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Giac [A]  time = 1.18716, size = 212, normalized size = 1.38 \begin{align*} \frac{55}{128} \, a^{3} x + \frac{a^{3} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac{9 \, a^{3} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac{3 \, a^{3} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac{29 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac{33 \, a^{3} \cos \left (d x + c\right )}{128 \, d} - \frac{3 \, a^{3} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{a^{3} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} + \frac{3 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac{9 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

55/128*a^3*x + 1/2304*a^3*cos(9*d*x + 9*c)/d - 9/1792*a^3*cos(7*d*x + 7*c)/d - 3/64*a^3*cos(5*d*x + 5*c)/d - 2
9/192*a^3*cos(3*d*x + 3*c)/d - 33/128*a^3*cos(d*x + c)/d - 3/1024*a^3*sin(8*d*x + 8*c)/d - 1/96*a^3*sin(6*d*x
+ 6*c)/d + 3/128*a^3*sin(4*d*x + 4*c)/d + 9/32*a^3*sin(2*d*x + 2*c)/d