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

Optimal. Leaf size=143 \[ -\frac {a \cos ^9(c+d x)}{9 d}+\frac {2 a \cos ^7(c+d x)}{7 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {a \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {a \sin (c+d x) \cos ^5(c+d x)}{16 d}+\frac {a \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3 a x}{128} \]

[Out]

3/128*a*x-1/5*a*cos(d*x+c)^5/d+2/7*a*cos(d*x+c)^7/d-1/9*a*cos(d*x+c)^9/d+3/128*a*cos(d*x+c)*sin(d*x+c)/d+1/64*
a*cos(d*x+c)^3*sin(d*x+c)/d-1/16*a*cos(d*x+c)^5*sin(d*x+c)/d-1/8*a*cos(d*x+c)^5*sin(d*x+c)^3/d

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Rubi [A]  time = 0.17, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 9, 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, 270} \[ -\frac {a \cos ^9(c+d x)}{9 d}+\frac {2 a \cos ^7(c+d x)}{7 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {a \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {a \sin (c+d x) \cos ^5(c+d x)}{16 d}+\frac {a \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 a \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3 a x}{128} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x)/128 - (a*Cos[c + d*x]^5)/(5*d) + (2*a*Cos[c + d*x]^7)/(7*d) - (a*Cos[c + d*x]^9)/(9*d) + (3*a*Cos[c +
d*x]*Sin[c + d*x])/(128*d) + (a*Cos[c + d*x]^3*Sin[c + d*x])/(64*d) - (a*Cos[c + d*x]^5*Sin[c + d*x])/(16*d) -
(a*Cos[c + d*x]^5*Sin[c + d*x]^3)/(8*d)

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]

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 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 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]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) \sin ^4(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^4(c+d x) \sin ^4(c+d x) \, dx+a \int \cos ^4(c+d x) \sin ^5(c+d x) \, dx\\ &=-\frac {a \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{8} (3 a) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx-\frac {a \operatorname {Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{16} a \int \cos ^4(c+d x) \, dx-\frac {a \operatorname {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a \cos ^5(c+d x)}{5 d}+\frac {2 a \cos ^7(c+d x)}{7 d}-\frac {a \cos ^9(c+d x)}{9 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {a \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{64} (3 a) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a \cos ^5(c+d x)}{5 d}+\frac {2 a \cos ^7(c+d x)}{7 d}-\frac {a \cos ^9(c+d x)}{9 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{128 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {a \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{128} (3 a) \int 1 \, dx\\ &=\frac {3 a x}{128}-\frac {a \cos ^5(c+d x)}{5 d}+\frac {2 a \cos ^7(c+d x)}{7 d}-\frac {a \cos ^9(c+d x)}{9 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{128 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{64 d}-\frac {a \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 84, normalized size = 0.59 \[ \frac {a (-2520 \sin (4 (c+d x))+315 \sin (8 (c+d x))-7560 \cos (c+d x)-1680 \cos (3 (c+d x))+1008 \cos (5 (c+d x))+180 \cos (7 (c+d x))-140 \cos (9 (c+d x))+7560 c+7560 d x)}{322560 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(7560*c + 7560*d*x - 7560*Cos[c + d*x] - 1680*Cos[3*(c + d*x)] + 1008*Cos[5*(c + d*x)] + 180*Cos[7*(c + d*x
)] - 140*Cos[9*(c + d*x)] - 2520*Sin[4*(c + d*x)] + 315*Sin[8*(c + d*x)]))/(322560*d)

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fricas [A]  time = 0.50, size = 95, normalized size = 0.66 \[ -\frac {4480 \, a \cos \left (d x + c\right )^{9} - 11520 \, a \cos \left (d x + c\right )^{7} + 8064 \, a \cos \left (d x + c\right )^{5} - 945 \, a d x - 315 \, {\left (16 \, a \cos \left (d x + c\right )^{7} - 24 \, a \cos \left (d x + c\right )^{5} + 2 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40320 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40320*(4480*a*cos(d*x + c)^9 - 11520*a*cos(d*x + c)^7 + 8064*a*cos(d*x + c)^5 - 945*a*d*x - 315*(16*a*cos(d
*x + c)^7 - 24*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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giac [A]  time = 0.24, size = 107, normalized size = 0.75 \[ \frac {3}{128} \, a x - \frac {a \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {a \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {a \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {3 \, a \cos \left (d x + c\right )}{128 \, d} + \frac {a \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*a*x - 1/2304*a*cos(9*d*x + 9*c)/d + 1/1792*a*cos(7*d*x + 7*c)/d + 1/320*a*cos(5*d*x + 5*c)/d - 1/192*a*c
os(3*d*x + 3*c)/d - 3/128*a*cos(d*x + c)/d + 1/1024*a*sin(8*d*x + 8*c)/d - 1/128*a*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.25, size = 124, normalized size = 0.87 \[ \frac {a \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{9}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{63}-\frac {8 \left (\cos ^{5}\left (d x +c \right )\right )}{315}\right )+a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{8}-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-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)+a*(-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))

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maxima [A]  time = 0.33, size = 71, normalized size = 0.50 \[ -\frac {1024 \, {\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} a - 315 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a}{322560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/322560*(1024*(35*cos(d*x + c)^9 - 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5)*a - 315*(24*d*x + 24*c + sin(8*d*x
+ 8*c) - 8*sin(4*d*x + 4*c))*a)/d

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mupad [B]  time = 12.34, size = 353, normalized size = 2.47 \[ \frac {3\,a\,x}{128}+\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{64}+\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{32}-\frac {155\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{32}+\left (\frac {a\,\left (79380\,c+79380\,d\,x-430080\right )}{40320}-\frac {63\,a\,\left (c+d\,x\right )}{32}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {169\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{32}+\left (\frac {a\,\left (119070\,c+119070\,d\,x+645120\right )}{40320}-\frac {189\,a\,\left (c+d\,x\right )}{64}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {a\,\left (119070\,c+119070\,d\,x-903168\right )}{40320}-\frac {189\,a\,\left (c+d\,x\right )}{64}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {169\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{32}+\left (\frac {a\,\left (79380\,c+79380\,d\,x+258048\right )}{40320}-\frac {63\,a\,\left (c+d\,x\right )}{32}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {155\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{32}+\left (\frac {a\,\left (34020\,c+34020\,d\,x-73728\right )}{40320}-\frac {27\,a\,\left (c+d\,x\right )}{32}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32}+\left (\frac {a\,\left (8505\,c+8505\,d\,x-18432\right )}{40320}-\frac {27\,a\,\left (c+d\,x\right )}{128}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {a\,\left (945\,c+945\,d\,x-2048\right )}{40320}-\frac {3\,a\,\left (c+d\,x\right )}{128}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*x)/128 + ((a*(945*c + 945*d*x - 2048))/40320 - (3*a*tan(c/2 + (d*x)/2))/64 - (3*a*(c + d*x))/128 + tan(c/
2 + (d*x)/2)^2*((a*(8505*c + 8505*d*x - 18432))/40320 - (27*a*(c + d*x))/128) + tan(c/2 + (d*x)/2)^4*((a*(3402
0*c + 34020*d*x - 73728))/40320 - (27*a*(c + d*x))/32) + tan(c/2 + (d*x)/2)^6*((a*(79380*c + 79380*d*x + 25804
8))/40320 - (63*a*(c + d*x))/32) + tan(c/2 + (d*x)/2)^12*((a*(79380*c + 79380*d*x - 430080))/40320 - (63*a*(c
+ d*x))/32) + tan(c/2 + (d*x)/2)^10*((a*(119070*c + 119070*d*x + 645120))/40320 - (189*a*(c + d*x))/64) + tan(
c/2 + (d*x)/2)^8*((a*(119070*c + 119070*d*x - 903168))/40320 - (189*a*(c + d*x))/64) - (13*a*tan(c/2 + (d*x)/2
)^3)/32 + (155*a*tan(c/2 + (d*x)/2)^5)/32 - (169*a*tan(c/2 + (d*x)/2)^7)/32 + (169*a*tan(c/2 + (d*x)/2)^11)/32
- (155*a*tan(c/2 + (d*x)/2)^13)/32 + (13*a*tan(c/2 + (d*x)/2)^15)/32 + (3*a*tan(c/2 + (d*x)/2)^17)/64)/(d*(ta
n(c/2 + (d*x)/2)^2 + 1)^9)

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sympy [A]  time = 19.38, size = 272, normalized size = 1.90 \[ \begin {cases} \frac {3 a x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {3 a x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {9 a x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 a x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 a \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {11 a \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} - \frac {a \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {11 a \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} - \frac {4 a \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {3 a \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {8 a \cos ^{9}{\left (c + d x \right )}}{315 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right ) \sin ^{4}{\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**4*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((3*a*x*sin(c + d*x)**8/128 + 3*a*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 9*a*x*sin(c + d*x)**4*cos(c
+ d*x)**4/64 + 3*a*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 3*a*x*cos(c + d*x)**8/128 + 3*a*sin(c + d*x)**7*cos(
c + d*x)/(128*d) + 11*a*sin(c + d*x)**5*cos(c + d*x)**3/(128*d) - a*sin(c + d*x)**4*cos(c + d*x)**5/(5*d) - 11
*a*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) - 4*a*sin(c + d*x)**2*cos(c + d*x)**7/(35*d) - 3*a*sin(c + d*x)*cos
(c + d*x)**7/(128*d) - 8*a*cos(c + d*x)**9/(315*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**4*cos(c)**4, True))

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