3.379 \(\int \cos ^4(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=159 \[ -\frac{a^2 \cos ^9(c+d x)}{9 d}+\frac{3 a^2 \cos ^7(c+d x)}{7 d}-\frac{2 a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{4 d}-\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{8 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{32 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{64 d}+\frac{3 a^2 x}{64} \]

[Out]

(3*a^2*x)/64 - (2*a^2*Cos[c + d*x]^5)/(5*d) + (3*a^2*Cos[c + d*x]^7)/(7*d) - (a^2*Cos[c + d*x]^9)/(9*d) + (3*a
^2*Cos[c + d*x]*Sin[c + d*x])/(64*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(32*d) - (a^2*Cos[c + d*x]^5*Sin[c +
d*x])/(8*d) - (a^2*Cos[c + d*x]^5*Sin[c + d*x]^3)/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.254925, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2873, 2565, 14, 2568, 2635, 8, 270} \[ -\frac{a^2 \cos ^9(c+d x)}{9 d}+\frac{3 a^2 \cos ^7(c+d x)}{7 d}-\frac{2 a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{4 d}-\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{8 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{32 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{64 d}+\frac{3 a^2 x}{64} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(3*a^2*x)/64 - (2*a^2*Cos[c + d*x]^5)/(5*d) + (3*a^2*Cos[c + d*x]^7)/(7*d) - (a^2*Cos[c + d*x]^9)/(9*d) + (3*a
^2*Cos[c + d*x]*Sin[c + d*x])/(64*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(32*d) - (a^2*Cos[c + d*x]^5*Sin[c +
d*x])/(8*d) - (a^2*Cos[c + d*x]^5*Sin[c + d*x]^3)/(4*d)

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2568

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

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]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cos ^4(c+d x) \sin ^3(c+d x)+2 a^2 \cos ^4(c+d x) \sin ^4(c+d x)+a^2 \cos ^4(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx+a^2 \int \cos ^4(c+d x) \sin ^5(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^4(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac{a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac{1}{4} \left (3 a^2\right ) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx-\frac{a^2 \operatorname{Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{a^2 \operatorname{Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac{a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac{1}{8} a^2 \int \cos ^4(c+d x) \, dx-\frac{a^2 \operatorname{Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{a^2 \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{2 a^2 \cos ^5(c+d x)}{5 d}+\frac{3 a^2 \cos ^7(c+d x)}{7 d}-\frac{a^2 \cos ^9(c+d x)}{9 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{32 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac{a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac{1}{32} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{2 a^2 \cos ^5(c+d x)}{5 d}+\frac{3 a^2 \cos ^7(c+d x)}{7 d}-\frac{a^2 \cos ^9(c+d x)}{9 d}+\frac{3 a^2 \cos (c+d x) \sin (c+d x)}{64 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{32 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac{a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac{1}{64} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac{3 a^2 x}{64}-\frac{2 a^2 \cos ^5(c+d x)}{5 d}+\frac{3 a^2 \cos ^7(c+d x)}{7 d}-\frac{a^2 \cos ^9(c+d x)}{9 d}+\frac{3 a^2 \cos (c+d x) \sin (c+d x)}{64 d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{32 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac{a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.697071, size = 86, normalized size = 0.54 \[ \frac{a^2 (-2520 \sin (4 (c+d x))+315 \sin (8 (c+d x))-11340 \cos (c+d x)-3360 \cos (3 (c+d x))+1008 \cos (5 (c+d x))+450 \cos (7 (c+d x))-70 \cos (9 (c+d x))+7560 c+7560 d x)}{161280 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(7560*c + 7560*d*x - 11340*Cos[c + d*x] - 3360*Cos[3*(c + d*x)] + 1008*Cos[5*(c + d*x)] + 450*Cos[7*(c +
d*x)] - 70*Cos[9*(c + d*x)] - 2520*Sin[4*(c + d*x)] + 315*Sin[8*(c + d*x)]))/(161280*d)

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 162, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{9}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{63}}-{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{315}} \right ) +2\,{a}^{2} \left ( -1/8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}-1/16\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{ \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) }{64}}+{\frac{3\,dx}{128}}+{\frac{3\,c}{128}} \right ) +{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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/9*sin(d*x+c)^4*cos(d*x+c)^5-4/63*sin(d*x+c)^2*cos(d*x+c)^5-8/315*cos(d*x+c)^5)+2*a^2*(-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)+a^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5))

________________________________________________________________________________________

Maxima [A]  time = 1.09879, size = 136, normalized size = 0.86 \begin{align*} -\frac{512 \,{\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} a^{2} - 4608 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} - 315 \,{\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^{2}}{161280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/161280*(512*(35*cos(d*x + c)^9 - 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5)*a^2 - 4608*(5*cos(d*x + c)^7 - 7*co
s(d*x + c)^5)*a^2 - 315*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*a^2)/d

________________________________________________________________________________________

Fricas [A]  time = 1.60285, size = 290, normalized size = 1.82 \begin{align*} -\frac{2240 \, a^{2} \cos \left (d x + c\right )^{9} - 8640 \, a^{2} \cos \left (d x + c\right )^{7} + 8064 \, a^{2} \cos \left (d x + c\right )^{5} - 945 \, a^{2} d x - 315 \,{\left (16 \, a^{2} \cos \left (d x + c\right )^{7} - 24 \, 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 )}{20160 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/20160*(2240*a^2*cos(d*x + c)^9 - 8640*a^2*cos(d*x + c)^7 + 8064*a^2*cos(d*x + c)^5 - 945*a^2*d*x - 315*(16*
a^2*cos(d*x + c)^7 - 24*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 = 22.0236, size = 335, normalized size = 2.11 \begin{align*} \begin{cases} \frac{3 a^{2} x \sin ^{8}{\left (c + d x \right )}}{64} + \frac{3 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{9 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{32} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \cos ^{8}{\left (c + d x \right )}}{64} + \frac{3 a^{2} \sin ^{7}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{64 d} + \frac{11 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{64 d} - \frac{a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{11 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{64 d} - \frac{4 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{3 a^{2} \sin{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac{8 a^{2} \cos ^{9}{\left (c + d x \right )}}{315 d} - \frac{2 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \sin ^{3}{\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)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((3*a**2*x*sin(c + d*x)**8/64 + 3*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/16 + 9*a**2*x*sin(c + d*x)**
4*cos(c + d*x)**4/32 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**6/16 + 3*a**2*x*cos(c + d*x)**8/64 + 3*a**2*sin(
c + d*x)**7*cos(c + d*x)/(64*d) + 11*a**2*sin(c + d*x)**5*cos(c + d*x)**3/(64*d) - a**2*sin(c + d*x)**4*cos(c
+ d*x)**5/(5*d) - 11*a**2*sin(c + d*x)**3*cos(c + d*x)**5/(64*d) - 4*a**2*sin(c + d*x)**2*cos(c + d*x)**7/(35*
d) - a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 3*a**2*sin(c + d*x)*cos(c + d*x)**7/(64*d) - 8*a**2*cos(c +
d*x)**9/(315*d) - 2*a**2*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**3*cos(c)**4, True))

________________________________________________________________________________________

Giac [A]  time = 1.42578, size = 166, normalized size = 1.04 \begin{align*} \frac{3}{64} \, a^{2} x - \frac{a^{2} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac{5 \, a^{2} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac{a^{2} \cos \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac{a^{2} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{9 \, a^{2} \cos \left (d x + c\right )}{128 \, d} + \frac{a^{2} \sin \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac{a^{2} \sin \left (4 \, d x + 4 \, 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)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

3/64*a^2*x - 1/2304*a^2*cos(9*d*x + 9*c)/d + 5/1792*a^2*cos(7*d*x + 7*c)/d + 1/160*a^2*cos(5*d*x + 5*c)/d - 1/
48*a^2*cos(3*d*x + 3*c)/d - 9/128*a^2*cos(d*x + c)/d + 1/512*a^2*sin(8*d*x + 8*c)/d - 1/64*a^2*sin(4*d*x + 4*c
)/d