3.381 \(\int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=129 \[ -\frac{a^2 \cos ^5(c+d x)}{15 d}-\frac{\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{21 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a^2 x}{8}-\frac{\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

[Out]

(a^2*x)/8 - (a^2*Cos[c + d*x]^5)/(15*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*Cos[c + d*x]^3*Sin[c +
d*x])/(12*d) - (Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(7*d) - (Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(21*d
)

________________________________________________________________________________________

Rubi [A]  time = 0.138963, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{a^2 \cos ^5(c+d x)}{15 d}-\frac{\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{21 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a^2 x}{8}-\frac{\cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/8 - (a^2*Cos[c + d*x]^5)/(15*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*Cos[c + d*x]^3*Sin[c +
d*x])/(12*d) - (Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(7*d) - (Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(21*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))^2 \, dx &=-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac{2}{7} \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac{1}{3} a \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{a^2 \cos ^5(c+d x)}{15 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac{1}{3} a^2 \int \cos ^4(c+d x) \, dx\\ &=-\frac{a^2 \cos ^5(c+d x)}{15 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac{1}{4} a^2 \int \cos ^2(c+d x) \, dx\\ &=-\frac{a^2 \cos ^5(c+d x)}{15 d}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}+\frac{1}{8} a^2 \int 1 \, dx\\ &=\frac{a^2 x}{8}-\frac{a^2 \cos ^5(c+d x)}{15 d}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}-\frac{\cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{21 d}\\ \end{align*}

Mathematica [A]  time = 0.324188, size = 86, normalized size = 0.67 \[ \frac{a^2 (210 \sin (2 (c+d x))-210 \sin (4 (c+d x))-70 \sin (6 (c+d x))-1155 \cos (c+d x)-525 \cos (3 (c+d x))-63 \cos (5 (c+d x))+15 \cos (7 (c+d x))+840 c+840 d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(840*c + 840*d*x - 1155*Cos[c + d*x] - 525*Cos[3*(c + d*x)] - 63*Cos[5*(c + d*x)] + 15*Cos[7*(c + d*x)] +
210*Sin[2*(c + d*x)] - 210*Sin[4*(c + d*x)] - 70*Sin[6*(c + d*x)]))/(6720*d)

________________________________________________________________________________________

Maple [A]  time = 0.033, size = 106, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{7}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) +2\,{a}^{2} \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}^{2} \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))^2,x)

[Out]

1/d*(a^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+2*a^2*(-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^2*cos(d*x+c)^5)

________________________________________________________________________________________

Maxima [A]  time = 1.11102, size = 111, normalized size = 0.86 \begin{align*} -\frac{672 \, a^{2} \cos \left (d x + c\right )^{5} - 96 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} - 35 \,{\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^{2}}{3360 \, 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))^2,x, algorithm="maxima")

[Out]

-1/3360*(672*a^2*cos(d*x + c)^5 - 96*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^2 - 35*(4*sin(2*d*x + 2*c)^3 + 12
*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2)/d

________________________________________________________________________________________

Fricas [A]  time = 1.53621, size = 213, normalized size = 1.65 \begin{align*} \frac{120 \, a^{2} \cos \left (d x + c\right )^{7} - 336 \, a^{2} \cos \left (d x + c\right )^{5} + 105 \, a^{2} d x - 35 \,{\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 2 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, 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))^2,x, algorithm="fricas")

[Out]

1/840*(120*a^2*cos(d*x + c)^7 - 336*a^2*cos(d*x + c)^5 + 105*a^2*d*x - 35*(8*a^2*cos(d*x + c)^5 - 2*a^2*cos(d*
x + c)^3 - 3*a^2*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [A]  time = 7.51552, size = 223, normalized size = 1.73 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{6}{\left (c + d x \right )}}{8} + \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a^{2} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac{a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac{2 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac{a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \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))**2,x)

[Out]

Piecewise((a**2*x*sin(c + d*x)**6/8 + 3*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 3*a**2*x*sin(c + d*x)**2*co
s(c + d*x)**4/8 + a**2*x*cos(c + d*x)**6/8 + a**2*sin(c + d*x)**5*cos(c + d*x)/(8*d) + a**2*sin(c + d*x)**3*co
s(c + d*x)**3/(3*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - a**2*sin(c + d*x)*cos(c + d*x)**5/(8*d) - 2
*a**2*cos(c + d*x)**7/(35*d) - a**2*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)*cos(c)**4, T
rue))

________________________________________________________________________________________

Giac [A]  time = 1.33229, size = 166, normalized size = 1.29 \begin{align*} \frac{1}{8} \, a^{2} x + \frac{a^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{3 \, a^{2} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{5 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac{11 \, a^{2} \cos \left (d x + c\right )}{64 \, d} - \frac{a^{2} \sin \left (6 \, d x + 6 \, c\right )}{96 \, 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 )}{32 \, 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))^2,x, algorithm="giac")

[Out]

1/8*a^2*x + 1/448*a^2*cos(7*d*x + 7*c)/d - 3/320*a^2*cos(5*d*x + 5*c)/d - 5/64*a^2*cos(3*d*x + 3*c)/d - 11/64*
a^2*cos(d*x + c)/d - 1/96*a^2*sin(6*d*x + 6*c)/d - 1/32*a^2*sin(4*d*x + 4*c)/d + 1/32*a^2*sin(2*d*x + 2*c)/d