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

Optimal. Leaf size=103 \[ \frac{a \cos ^7(c+d x)}{7 d}-\frac{a \cos ^5(c+d x)}{5 d}-\frac{a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{a \sin (c+d x) \cos (c+d x)}{16 d}+\frac{a x}{16} \]

[Out]

(a*x)/16 - (a*Cos[c + d*x]^5)/(5*d) + (a*Cos[c + d*x]^7)/(7*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a*Cos
[c + d*x]^3*Sin[c + d*x])/(24*d) - (a*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)

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Rubi [A]  time = 0.132512, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2838, 2568, 2635, 8, 2565, 14} \[ \frac{a \cos ^7(c+d x)}{7 d}-\frac{a \cos ^5(c+d x)}{5 d}-\frac{a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{a \sin (c+d x) \cos (c+d x)}{16 d}+\frac{a x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/16 - (a*Cos[c + d*x]^5)/(5*d) + (a*Cos[c + d*x]^7)/(7*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a*Cos
[c + d*x]^3*Sin[c + d*x])/(24*d) - (a*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

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 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]]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx+a \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx\\ &=-\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{6} a \int \cos ^4(c+d x) \, dx-\frac{a \operatorname{Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{a \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{8} a \int \cos ^2(c+d x) \, dx-\frac{a \operatorname{Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a \cos ^5(c+d x)}{5 d}+\frac{a \cos ^7(c+d x)}{7 d}+\frac{a \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{16} a \int 1 \, dx\\ &=\frac{a x}{16}-\frac{a \cos ^5(c+d x)}{5 d}+\frac{a \cos ^7(c+d x)}{7 d}+\frac{a \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.201471, size = 81, normalized size = 0.79 \[ \frac{a (105 \sin (2 (c+d x))-105 \sin (4 (c+d x))-35 \sin (6 (c+d x))-315 \cos (c+d x)-105 \cos (3 (c+d x))+21 \cos (5 (c+d x))+15 \cos (7 (c+d x))+420 d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(420*d*x - 315*Cos[c + d*x] - 105*Cos[3*(c + d*x)] + 21*Cos[5*(c + d*x)] + 15*Cos[7*(c + d*x)] + 105*Sin[2*
(c + d*x)] - 105*Sin[4*(c + d*x)] - 35*Sin[6*(c + d*x)]))/(6720*d)

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Maple [A]  time = 0.03, size = 88, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( a \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 ) +a \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{\sin \left ( dx+c \right ) }{24} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{dx}{16}}+{\frac{c}{16}} \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)^2*(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.00037, size = 88, normalized size = 0.85 \begin{align*} \frac{192 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a + 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}{6720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6720*(192*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a + 35*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x +
4*c))*a)/d

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Fricas [A]  time = 1.53086, size = 198, normalized size = 1.92 \begin{align*} \frac{240 \, a \cos \left (d x + c\right )^{7} - 336 \, a \cos \left (d x + c\right )^{5} + 105 \, a d x - 35 \,{\left (8 \, a \cos \left (d x + c\right )^{5} - 2 \, a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1680*(240*a*cos(d*x + c)^7 - 336*a*cos(d*x + c)^5 + 105*a*d*x - 35*(8*a*cos(d*x + c)^5 - 2*a*cos(d*x + c)^3
- 3*a*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 7.72064, size = 192, normalized size = 1.86 \begin{align*} \begin{cases} \frac{a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{a \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{a \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{a \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{2 a \cos ^{7}{\left (c + d x \right )}}{35 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \sin ^{2}{\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)**2*(a+a*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.34899, size = 144, normalized size = 1.4 \begin{align*} \frac{1}{16} \, a x + \frac{a \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{a \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{a \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac{3 \, a \cos \left (d x + c\right )}{64 \, d} - \frac{a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{a \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)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*a*x + 1/448*a*cos(7*d*x + 7*c)/d + 1/320*a*cos(5*d*x + 5*c)/d - 1/64*a*cos(3*d*x + 3*c)/d - 3/64*a*cos(d*
x + c)/d - 1/192*a*sin(6*d*x + 6*c)/d - 1/64*a*sin(4*d*x + 4*c)/d + 1/64*a*sin(2*d*x + 2*c)/d