3.661 \(\int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=81 \[ -\frac{a \sin ^9(c+d x)}{9 d}+\frac{3 a \sin ^7(c+d x)}{7 d}-\frac{3 a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \cos ^8(c+d x)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.08965, antiderivative size = 81, 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} \[ -\frac{a \sin ^9(c+d x)}{9 d}+\frac{3 a \sin ^7(c+d x)}{7 d}-\frac{3 a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \cos ^8(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

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

Mathematica [A]  time = 0.380543, size = 60, normalized size = 0.74 \[ \frac{a \left (\sin ^3(c+d x) (1389 \cos (2 (c+d x))+330 \cos (4 (c+d x))+35 \cos (6 (c+d x))[phone] \cos ^8(c+d x)\right )}{10080 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]

[Out]

(a*(-1260*Cos[c + d*x]^8 + (1606 + 1389*Cos[2*(c + d*x)] + 330*Cos[4*(c + d*x)] + 35*Cos[6*(c + d*x)])*Sin[c +
d*x]^3))/(10080*d)

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Maple [A]  time = 0.025, size = 74, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{9}}+{\frac{\sin \left ( dx+c \right ) }{63} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

1/d*(a*(-1/9*sin(d*x+c)*cos(d*x+c)^8+1/63*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))-1/
8*a*cos(d*x+c)^8)

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Maxima [A]  time = 1.00755, size = 127, normalized size = 1.57 \begin{align*} -\frac{280 \, a \sin \left (d x + c\right )^{9} + 315 \, a \sin \left (d x + c\right )^{8} - 1080 \, a \sin \left (d x + c\right )^{7} - 1260 \, a \sin \left (d x + c\right )^{6} + 1512 \, a \sin \left (d x + c\right )^{5} + 1890 \, a \sin \left (d x + c\right )^{4} - 840 \, a \sin \left (d x + c\right )^{3} - 1260 \, a \sin \left (d x + c\right )^{2}}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/2520*(280*a*sin(d*x + c)^9 + 315*a*sin(d*x + c)^8 - 1080*a*sin(d*x + c)^7 - 1260*a*sin(d*x + c)^6 + 1512*a*
sin(d*x + c)^5 + 1890*a*sin(d*x + c)^4 - 840*a*sin(d*x + c)^3 - 1260*a*sin(d*x + c)^2)/d

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Fricas [A]  time = 1.48095, size = 193, normalized size = 2.38 \begin{align*} -\frac{315 \, a \cos \left (d x + c\right )^{8} + 8 \,{\left (35 \, a \cos \left (d x + c\right )^{8} - 5 \, a \cos \left (d x + c\right )^{6} - 6 \, a \cos \left (d x + c\right )^{4} - 8 \, a \cos \left (d x + c\right )^{2} - 16 \, a\right )} \sin \left (d x + c\right )}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2520*(315*a*cos(d*x + c)^8 + 8*(35*a*cos(d*x + c)^8 - 5*a*cos(d*x + c)^6 - 6*a*cos(d*x + c)^4 - 8*a*cos(d*x
+ c)^2 - 16*a)*sin(d*x + c))/d

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Sympy [A]  time = 21.0276, size = 114, normalized size = 1.41 \begin{align*} \begin{cases} \frac{16 a \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac{8 a \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{2 a \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac{a \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac{a \cos ^{8}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \sin{\left (c \right )} \cos ^{7}{\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)**7*sin(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.19917, size = 159, normalized size = 1.96 \begin{align*} -\frac{a \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{a \cos \left (6 \, d x + 6 \, c\right )}{128 \, d} - \frac{7 \, a \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{7 \, a \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac{a \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac{5 \, a \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac{a \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} + \frac{7 \, a \sin \left (d x + c\right )}{128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/1024*a*cos(8*d*x + 8*c)/d - 1/128*a*cos(6*d*x + 6*c)/d - 7/256*a*cos(4*d*x + 4*c)/d - 7/128*a*cos(2*d*x + 2
*c)/d - 1/2304*a*sin(9*d*x + 9*c)/d - 5/1792*a*sin(7*d*x + 7*c)/d - 1/160*a*sin(5*d*x + 5*c)/d + 7/128*a*sin(d
*x + c)/d