3.498 \(\int \cos ^5(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=65 \[ \frac{a \sin ^7(c+d x)}{7 d}-\frac{2 a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \cos ^6(c+d x)}{6 d} \]

[Out]

-(a*Cos[c + d*x]^6)/(6*d) + (a*Sin[c + d*x]^3)/(3*d) - (2*a*Sin[c + d*x]^5)/(5*d) + (a*Sin[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.0908986, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2834, 2565, 30, 2564, 270} \[ \frac{a \sin ^7(c+d x)}{7 d}-\frac{2 a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \cos ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

-(a*Cos[c + d*x]^6)/(6*d) + (a*Sin[c + d*x]^3)/(3*d) - (2*a*Sin[c + d*x]^5)/(5*d) + (a*Sin[c + d*x]^7)/(7*d)

Rule 2834

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

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

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

Mathematica [A]  time = 0.156388, size = 78, normalized size = 1.2 \[ -\frac{a (-525 \sin (c+d x)+35 \sin (3 (c+d x))+63 \sin (5 (c+d x))+15 \sin (7 (c+d x))+525 \cos (2 (c+d x))+210 \cos (4 (c+d x))+35 \cos (6 (c+d x))+350)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

-(a*(350 + 525*Cos[2*(c + d*x)] + 210*Cos[4*(c + d*x)] + 35*Cos[6*(c + d*x)] - 525*Sin[c + d*x] + 35*Sin[3*(c
+ d*x)] + 63*Sin[5*(c + d*x)] + 15*Sin[7*(c + d*x)]))/(6720*d)

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Maple [A]  time = 0.025, size = 64, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

1/d*(a*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-1/6*a*cos(d*x+c)^6)

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Maxima [A]  time = 1.02143, size = 97, normalized size = 1.49 \begin{align*} \frac{30 \, a \sin \left (d x + c\right )^{7} + 35 \, a \sin \left (d x + c\right )^{6} - 84 \, a \sin \left (d x + c\right )^{5} - 105 \, a \sin \left (d x + c\right )^{4} + 70 \, a \sin \left (d x + c\right )^{3} + 105 \, a \sin \left (d x + c\right )^{2}}{210 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/210*(30*a*sin(d*x + c)^7 + 35*a*sin(d*x + c)^6 - 84*a*sin(d*x + c)^5 - 105*a*sin(d*x + c)^4 + 70*a*sin(d*x +
c)^3 + 105*a*sin(d*x + c)^2)/d

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Fricas [A]  time = 1.12755, size = 161, normalized size = 2.48 \begin{align*} -\frac{35 \, a \cos \left (d x + c\right )^{6} + 2 \,{\left (15 \, a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} - 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right )}{210 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/210*(35*a*cos(d*x + c)^6 + 2*(15*a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^4 - 4*a*cos(d*x + c)^2 - 8*a)*sin(d*x
+ c))/d

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Sympy [A]  time = 6.95768, size = 90, normalized size = 1.38 \begin{align*} \begin{cases} \frac{8 a \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{4 a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{a \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac{a \cos ^{6}{\left (c + d x \right )}}{6 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \sin{\left (c \right )} \cos ^{5}{\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)**5*sin(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((8*a*sin(c + d*x)**7/(105*d) + 4*a*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + a*sin(c + d*x)**3*cos(c
+ d*x)**4/(3*d) - a*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)*cos(c)**5, True))

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Giac [A]  time = 1.26566, size = 139, normalized size = 2.14 \begin{align*} -\frac{a \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{a \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{5 \, a \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} - \frac{a \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{3 \, a \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{a \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{5 \, a \sin \left (d x + c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/192*a*cos(6*d*x + 6*c)/d - 1/32*a*cos(4*d*x + 4*c)/d - 5/64*a*cos(2*d*x + 2*c)/d - 1/448*a*sin(7*d*x + 7*c)
/d - 3/320*a*sin(5*d*x + 5*c)/d - 1/192*a*sin(3*d*x + 3*c)/d + 5/64*a*sin(d*x + c)/d