3.659 \(\int \cos ^7(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=97 \[ -\frac {a \sin ^{11}(c+d x)}{11 d}+\frac {a \sin ^9(c+d x)}{3 d}-\frac {3 a \sin ^7(c+d x)}{7 d}+\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \cos ^{10}(c+d x)}{10 d}-\frac {a \cos ^8(c+d x)}{8 d} \]

[Out]

-1/8*a*cos(d*x+c)^8/d+1/10*a*cos(d*x+c)^10/d+1/5*a*sin(d*x+c)^5/d-3/7*a*sin(d*x+c)^7/d+1/3*a*sin(d*x+c)^9/d-1/
11*a*sin(d*x+c)^11/d

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Rubi [A]  time = 0.13, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2834, 2565, 14, 2564, 270} \[ -\frac {a \sin ^{11}(c+d x)}{11 d}+\frac {a \sin ^9(c+d x)}{3 d}-\frac {3 a \sin ^7(c+d x)}{7 d}+\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \cos ^{10}(c+d x)}{10 d}-\frac {a \cos ^8(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*Sin[c + d*x]^3*(a + a*Sin[c + d*x]),x]

[Out]

-(a*Cos[c + d*x]^8)/(8*d) + (a*Cos[c + d*x]^10)/(10*d) + (a*Sin[c + d*x]^5)/(5*d) - (3*a*Sin[c + d*x]^7)/(7*d)
+ (a*Sin[c + d*x]^9)/(3*d) - (a*Sin[c + d*x]^11)/(11*d)

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 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]

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 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 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])

Rubi steps

\begin {align*} \int \cos ^7(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^7(c+d x) \sin ^3(c+d x) \, dx+a \int \cos ^7(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac {a \operatorname {Subst}\left (\int x^7 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {a \operatorname {Subst}\left (\int x^4 \left (1-x^2\right )^3 \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a \operatorname {Subst}\left (\int \left (x^7-x^9\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {a \operatorname {Subst}\left (\int \left (x^4-3 x^6+3 x^8-x^{10}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a \cos ^8(c+d x)}{8 d}+\frac {a \cos ^{10}(c+d x)}{10 d}+\frac {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)}{3 d}-\frac {a \sin ^{11}(c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 117, normalized size = 1.21 \[ \frac {a (16170 \sin (c+d x)-2310 \sin (3 (c+d x))-2541 \sin (5 (c+d x))-165 \sin (7 (c+d x))+385 \sin (9 (c+d x))+105 \sin (11 (c+d x))-16170 \cos (2 (c+d x))-4620 \cos (4 (c+d x))+1155 \cos (6 (c+d x))+1155 \cos (8 (c+d x))+231 \cos (10 (c+d x)))}{1182720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*Sin[c + d*x]^3*(a + a*Sin[c + d*x]),x]

[Out]

(a*(-16170*Cos[2*(c + d*x)] - 4620*Cos[4*(c + d*x)] + 1155*Cos[6*(c + d*x)] + 1155*Cos[8*(c + d*x)] + 231*Cos[
10*(c + d*x)] + 16170*Sin[c + d*x] - 2310*Sin[3*(c + d*x)] - 2541*Sin[5*(c + d*x)] - 165*Sin[7*(c + d*x)] + 38
5*Sin[9*(c + d*x)] + 105*Sin[11*(c + d*x)]))/(1182720*d)

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fricas [A]  time = 0.89, size = 95, normalized size = 0.98 \[ \frac {924 \, a \cos \left (d x + c\right )^{10} - 1155 \, a \cos \left (d x + c\right )^{8} + 8 \, {\left (105 \, a \cos \left (d x + c\right )^{10} - 140 \, a \cos \left (d x + c\right )^{8} + 5 \, a \cos \left (d x + c\right )^{6} + 6 \, a \cos \left (d x + c\right )^{4} + 8 \, a \cos \left (d x + c\right )^{2} + 16 \, a\right )} \sin \left (d x + c\right )}{9240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/9240*(924*a*cos(d*x + c)^10 - 1155*a*cos(d*x + c)^8 + 8*(105*a*cos(d*x + c)^10 - 140*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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giac [A]  time = 0.29, size = 163, normalized size = 1.68 \[ \frac {a \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {a \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {a \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {a \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {7 \, a \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} + \frac {a \sin \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac {a \sin \left (9 \, d x + 9 \, c\right )}{3072 \, d} - \frac {a \sin \left (7 \, d x + 7 \, c\right )}{7168 \, d} - \frac {11 \, a \sin \left (5 \, d x + 5 \, c\right )}{5120 \, d} - \frac {a \sin \left (3 \, d x + 3 \, c\right )}{512 \, d} + \frac {7 \, a \sin \left (d x + c\right )}{512 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/5120*a*cos(10*d*x + 10*c)/d + 1/1024*a*cos(8*d*x + 8*c)/d + 1/1024*a*cos(6*d*x + 6*c)/d - 1/256*a*cos(4*d*x
+ 4*c)/d - 7/512*a*cos(2*d*x + 2*c)/d + 1/11264*a*sin(11*d*x + 11*c)/d + 1/3072*a*sin(9*d*x + 9*c)/d - 1/7168*
a*sin(7*d*x + 7*c)/d - 11/5120*a*sin(5*d*x + 5*c)/d - 1/512*a*sin(3*d*x + 3*c)/d + 7/512*a*sin(d*x + c)/d

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maple [A]  time = 0.24, size = 112, normalized size = 1.15 \[ \frac {a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{8}\left (d x +c \right )\right )}{11}-\frac {\sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )}{33}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{231}\right )+a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{8}\left (d x +c \right )\right )}{10}-\frac {\left (\cos ^{8}\left (d x +c \right )\right )}{40}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*sin(d*x+c)^3*(a+a*sin(d*x+c)),x)

[Out]

1/d*(a*(-1/11*sin(d*x+c)^3*cos(d*x+c)^8-1/33*sin(d*x+c)*cos(d*x+c)^8+1/231*(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))+a*(-1/10*sin(d*x+c)^2*cos(d*x+c)^8-1/40*cos(d*x+c)^8))

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maxima [A]  time = 0.31, size = 94, normalized size = 0.97 \[ -\frac {840 \, a \sin \left (d x + c\right )^{11} + 924 \, a \sin \left (d x + c\right )^{10} - 3080 \, a \sin \left (d x + c\right )^{9} - 3465 \, a \sin \left (d x + c\right )^{8} + 3960 \, a \sin \left (d x + c\right )^{7} + 4620 \, a \sin \left (d x + c\right )^{6} - 1848 \, a \sin \left (d x + c\right )^{5} - 2310 \, a \sin \left (d x + c\right )^{4}}{9240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/9240*(840*a*sin(d*x + c)^11 + 924*a*sin(d*x + c)^10 - 3080*a*sin(d*x + c)^9 - 3465*a*sin(d*x + c)^8 + 3960*
a*sin(d*x + c)^7 + 4620*a*sin(d*x + c)^6 - 1848*a*sin(d*x + c)^5 - 2310*a*sin(d*x + c)^4)/d

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mupad [B]  time = 0.08, size = 93, normalized size = 0.96 \[ \frac {-\frac {a\,{\sin \left (c+d\,x\right )}^{11}}{11}-\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}+\frac {a\,{\sin \left (c+d\,x\right )}^9}{3}+\frac {3\,a\,{\sin \left (c+d\,x\right )}^8}{8}-\frac {3\,a\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a\,{\sin \left (c+d\,x\right )}^6}{2}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*sin(c + d*x)^3*(a + a*sin(c + d*x)),x)

[Out]

((a*sin(c + d*x)^4)/4 + (a*sin(c + d*x)^5)/5 - (a*sin(c + d*x)^6)/2 - (3*a*sin(c + d*x)^7)/7 + (3*a*sin(c + d*
x)^8)/8 + (a*sin(c + d*x)^9)/3 - (a*sin(c + d*x)^10)/10 - (a*sin(c + d*x)^11)/11)/d

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sympy [A]  time = 37.47, size = 138, normalized size = 1.42 \[ \begin {cases} \frac {16 a \sin ^{11}{\left (c + d x \right )}}{1155 d} + \frac {8 a \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{105 d} + \frac {6 a \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{35 d} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{5 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac {a \cos ^{10}{\left (c + d x \right )}}{40 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right ) \sin ^{3}{\relax (c )} \cos ^{7}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*sin(d*x+c)**3*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((16*a*sin(c + d*x)**11/(1155*d) + 8*a*sin(c + d*x)**9*cos(c + d*x)**2/(105*d) + 6*a*sin(c + d*x)**7*
cos(c + d*x)**4/(35*d) + a*sin(c + d*x)**5*cos(c + d*x)**6/(5*d) - a*sin(c + d*x)**2*cos(c + d*x)**8/(8*d) - a
*cos(c + d*x)**10/(40*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**3*cos(c)**7, True))

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