3.395 \(\int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=157 \[ -\frac{9 a^3 \cos ^5(c+d x)}{80 d}-\frac{9 \cos ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{112 d}+\frac{9 a^3 \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac{27 a^3 \sin (c+d x) \cos (c+d x)}{128 d}+\frac{27 a^3 x}{128}-\frac{\cos ^5(c+d x) (a \sin (c+d x)+a)^3}{8 d}-\frac{3 a \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{56 d} \]

[Out]

(27*a^3*x)/128 - (9*a^3*Cos[c + d*x]^5)/(80*d) + (27*a^3*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (9*a^3*Cos[c + d
*x]^3*Sin[c + d*x])/(64*d) - (3*a*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(56*d) - (Cos[c + d*x]^5*(a + a*Sin[c
+ d*x])^3)/(8*d) - (9*Cos[c + d*x]^5*(a^3 + a^3*Sin[c + d*x]))/(112*d)

________________________________________________________________________________________

Rubi [A]  time = 0.193872, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2860, 2678, 2669, 2635, 8} \[ -\frac{9 a^3 \cos ^5(c+d x)}{80 d}-\frac{9 \cos ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{112 d}+\frac{9 a^3 \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac{27 a^3 \sin (c+d x) \cos (c+d x)}{128 d}+\frac{27 a^3 x}{128}-\frac{\cos ^5(c+d x) (a \sin (c+d x)+a)^3}{8 d}-\frac{3 a \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{56 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(27*a^3*x)/128 - (9*a^3*Cos[c + d*x]^5)/(80*d) + (27*a^3*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (9*a^3*Cos[c + d
*x]^3*Sin[c + d*x])/(64*d) - (3*a*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(56*d) - (Cos[c + d*x]^5*(a + a*Sin[c
+ d*x])^3)/(8*d) - (9*Cos[c + d*x]^5*(a^3 + a^3*Sin[c + d*x]))/(112*d)

Rule 2860

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

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 ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^3 \, dx &=-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}+\frac{3}{8} \int \cos ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=-\frac{3 a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}+\frac{1}{56} (27 a) \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{3 a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac{9 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac{1}{16} \left (9 a^2\right ) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{9 a^3 \cos ^5(c+d x)}{80 d}-\frac{3 a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac{9 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac{1}{16} \left (9 a^3\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac{9 a^3 \cos ^5(c+d x)}{80 d}+\frac{9 a^3 \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac{3 a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac{9 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac{1}{64} \left (27 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{9 a^3 \cos ^5(c+d x)}{80 d}+\frac{27 a^3 \cos (c+d x) \sin (c+d x)}{128 d}+\frac{9 a^3 \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac{3 a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac{9 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}+\frac{1}{128} \left (27 a^3\right ) \int 1 \, dx\\ &=\frac{27 a^3 x}{128}-\frac{9 a^3 \cos ^5(c+d x)}{80 d}+\frac{27 a^3 \cos (c+d x) \sin (c+d x)}{128 d}+\frac{9 a^3 \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac{3 a \cos ^5(c+d x) (a+a \sin (c+d x))^2}{56 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^3}{8 d}-\frac{9 \cos ^5(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{112 d}\\ \end{align*}

Mathematica [A]  time = 0.451391, size = 96, normalized size = 0.61 \[ \frac{a^3 (1680 \sin (2 (c+d x))-1960 \sin (4 (c+d x))-560 \sin (6 (c+d x))+35 \sin (8 (c+d x))-9520 \cos (c+d x)-3920 \cos (3 (c+d x))-112 \cos (5 (c+d x))+240 \cos (7 (c+d x))+8400 c+7560 d x)}{35840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(8400*c + 7560*d*x - 9520*Cos[c + d*x] - 3920*Cos[3*(c + d*x)] - 112*Cos[5*(c + d*x)] + 240*Cos[7*(c + d*
x)] + 1680*Sin[2*(c + d*x)] - 1960*Sin[4*(c + d*x)] - 560*Sin[6*(c + d*x)] + 35*Sin[8*(c + d*x)]))/(35840*d)

________________________________________________________________________________________

Maple [A]  time = 0.038, size = 178, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8}}-{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{16}}+{\frac{\sin \left ( dx+c \right ) }{64} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{128}}+{\frac{3\,c}{128}} \right ) +3\,{a}^{3} \left ( -1/7\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) +3\,{a}^{3} \left ( -1/6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(-1/8*sin(d*x+c)^3*cos(d*x+c)^5-1/16*sin(d*x+c)*cos(d*x+c)^5+1/64*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d
*x+c)+3/128*d*x+3/128*c)+3*a^3*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+3*a^3*(-1/6*sin(d*x+c)*cos(d
*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-1/5*a^3*cos(d*x+c)^5)

________________________________________________________________________________________

Maxima [A]  time = 1.12321, size = 155, normalized size = 0.99 \begin{align*} -\frac{7168 \, a^{3} \cos \left (d x + c\right )^{5} - 3072 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{3} - 560 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 35 \,{\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{35840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/35840*(7168*a^3*cos(d*x + c)^5 - 3072*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^3 - 560*(4*sin(2*d*x + 2*c)^3
+ 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^3 - 35*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*a^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.33474, size = 254, normalized size = 1.62 \begin{align*} \frac{1920 \, a^{3} \cos \left (d x + c\right )^{7} - 3584 \, a^{3} \cos \left (d x + c\right )^{5} + 945 \, a^{3} d x + 35 \,{\left (16 \, a^{3} \cos \left (d x + c\right )^{7} - 88 \, a^{3} \cos \left (d x + c\right )^{5} + 18 \, a^{3} \cos \left (d x + c\right )^{3} + 27 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4480*(1920*a^3*cos(d*x + c)^7 - 3584*a^3*cos(d*x + c)^5 + 945*a^3*d*x + 35*(16*a^3*cos(d*x + c)^7 - 88*a^3*c
os(d*x + c)^5 + 18*a^3*cos(d*x + c)^3 + 27*a^3*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [A]  time = 12.8882, size = 440, normalized size = 2.8 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac{3 a^{3} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac{3 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{9 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac{9 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac{9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{3 a^{3} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac{3 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{3} \sin ^{7}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{128 d} + \frac{11 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac{3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{11 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} + \frac{a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac{3 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{3 a^{3} \sin{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac{3 a^{3} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{6 a^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac{a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{3} \sin{\left (c \right )} \cos ^{4}{\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)**4*sin(d*x+c)*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((3*a**3*x*sin(c + d*x)**8/128 + 3*a**3*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 3*a**3*x*sin(c + d*x)*
*6/16 + 9*a**3*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 9*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**3*x*s
in(c + d*x)**2*cos(c + d*x)**6/32 + 9*a**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a**3*x*cos(c + d*x)**8/128
+ 3*a**3*x*cos(c + d*x)**6/16 + 3*a**3*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 11*a**3*sin(c + d*x)**5*cos(c +
d*x)**3/(128*d) + 3*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) - 11*a**3*sin(c + d*x)**3*cos(c + d*x)**5/(128*d
) + a**3*sin(c + d*x)**3*cos(c + d*x)**3/(2*d) - 3*a**3*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 3*a**3*sin(c +
d*x)*cos(c + d*x)**7/(128*d) - 3*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 6*a**3*cos(c + d*x)**7/(35*d) - a
**3*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)*cos(c)**4, True))

________________________________________________________________________________________

Giac [A]  time = 1.25603, size = 189, normalized size = 1.2 \begin{align*} \frac{27}{128} \, a^{3} x + \frac{3 \, a^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{a^{3} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{7 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac{17 \, a^{3} \cos \left (d x + c\right )}{64 \, d} + \frac{a^{3} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{a^{3} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} - \frac{7 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac{3 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

27/128*a^3*x + 3/448*a^3*cos(7*d*x + 7*c)/d - 1/320*a^3*cos(5*d*x + 5*c)/d - 7/64*a^3*cos(3*d*x + 3*c)/d - 17/
64*a^3*cos(d*x + c)/d + 1/1024*a^3*sin(8*d*x + 8*c)/d - 1/64*a^3*sin(6*d*x + 6*c)/d - 7/128*a^3*sin(4*d*x + 4*
c)/d + 3/64*a^3*sin(2*d*x + 2*c)/d