\(\int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx\) [218]

Optimal result

Integrand size = 27, antiderivative size = 88 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^4 \sin ^5(c+d x)}{5 d}+\frac {a^4 \sin ^6(c+d x)}{d}+\frac {4 a^4 \sin ^7(c+d x)}{7 d}+\frac {a^4 \sin ^8(c+d x)}{8 d} \]

[Out]

1/4*a^4*sin(d*x+c)^4/d+4/5*a^4*sin(d*x+c)^5/d+a^4*sin(d*x+c)^6/d+4/7*a^4*sin(d*x+c)^7/d+1/8*a^4*sin(d*x+c)^8/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 45} \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^8(c+d x)}{8 d}+\frac {4 a^4 \sin ^7(c+d x)}{7 d}+\frac {a^4 \sin ^6(c+d x)}{d}+\frac {4 a^4 \sin ^5(c+d x)}{5 d}+\frac {a^4 \sin ^4(c+d x)}{4 d} \]

[In]

Int[Cos[c + d*x]*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*Sin[c + d*x]^4)/(4*d) + (4*a^4*Sin[c + d*x]^5)/(5*d) + (a^4*Sin[c + d*x]^6)/d + (4*a^4*Sin[c + d*x]^7)/(7
*d) + (a^4*Sin[c + d*x]^8)/(8*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^3 (a+x)^4}{a^3} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int x^3 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (a^4 x^3+4 a^3 x^4+6 a^2 x^5+4 a x^6+x^7\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d} \\ & = \frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^4 \sin ^5(c+d x)}{5 d}+\frac {a^4 \sin ^6(c+d x)}{d}+\frac {4 a^4 \sin ^7(c+d x)}{7 d}+\frac {a^4 \sin ^8(c+d x)}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 (36400-69720 \cos (2 (c+d x))+26460 \cos (4 (c+d x))-4200 \cos (6 (c+d x))+105 \cos (8 (c+d x))+87360 \sin (c+d x)-47040 \sin (3 (c+d x))+12096 \sin (5 (c+d x))-960 \sin (7 (c+d x)))}{107520 d} \]

[In]

Integrate[Cos[c + d*x]*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(36400 - 69720*Cos[2*(c + d*x)] + 26460*Cos[4*(c + d*x)] - 4200*Cos[6*(c + d*x)] + 105*Cos[8*(c + d*x)] +
87360*Sin[c + d*x] - 47040*Sin[3*(c + d*x)] + 12096*Sin[5*(c + d*x)] - 960*Sin[7*(c + d*x)]))/(107520*d)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {\frac {a^{4} \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {4 a^{4} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+a^{4} \left (\sin ^{6}\left (d x +c \right )\right )+\frac {4 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d}\)	\(70\)
default	\(\frac {\frac {a^{4} \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {4 a^{4} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+a^{4} \left (\sin ^{6}\left (d x +c \right )\right )+\frac {4 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d}\)	\(70\)
parallelrisch	\(-\frac {a^{4} \left (23240 \cos \left (2 d x +2 c \right )-35 \cos \left (8 d x +8 c \right )+320 \sin \left (7 d x +7 c \right )-4032 \sin \left (5 d x +5 c \right )+1400 \cos \left (6 d x +6 c \right )-29120 \sin \left (d x +c \right )+15680 \sin \left (3 d x +3 c \right )-8820 \cos \left (4 d x +4 c \right )-15785\right )}{35840 d}\)	\(96\)
risch	\(\frac {13 a^{4} \sin \left (d x +c \right )}{16 d}+\frac {a^{4} \cos \left (8 d x +8 c \right )}{1024 d}-\frac {a^{4} \sin \left (7 d x +7 c \right )}{112 d}-\frac {5 a^{4} \cos \left (6 d x +6 c \right )}{128 d}+\frac {9 a^{4} \sin \left (5 d x +5 c \right )}{80 d}+\frac {63 a^{4} \cos \left (4 d x +4 c \right )}{256 d}-\frac {7 a^{4} \sin \left (3 d x +3 c \right )}{16 d}-\frac {83 a^{4} \cos \left (2 d x +2 c \right )}{128 d}\)	\(135\)
norman	\(\frac {\frac {128 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {5248 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {5248 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {128 a^{4} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {4 a^{4} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{4} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {80 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {80 a^{4} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {184 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}\)	\(189\)

[In]

int(cos(d*x+c)*sin(d*x+c)^3*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/8*a^4*sin(d*x+c)^8+4/7*a^4*sin(d*x+c)^7+a^4*sin(d*x+c)^6+4/5*a^4*sin(d*x+c)^5+1/4*a^4*sin(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.26 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {35 \, a^{4} \cos \left (d x + c\right )^{8} - 420 \, a^{4} \cos \left (d x + c\right )^{6} + 1120 \, a^{4} \cos \left (d x + c\right )^{4} - 1120 \, a^{4} \cos \left (d x + c\right )^{2} - 32 \, {\left (5 \, a^{4} \cos \left (d x + c\right )^{6} - 22 \, a^{4} \cos \left (d x + c\right )^{4} + 29 \, a^{4} \cos \left (d x + c\right )^{2} - 12 \, a^{4}\right )} \sin \left (d x + c\right )}{280 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^3*(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/280*(35*a^4*cos(d*x + c)^8 - 420*a^4*cos(d*x + c)^6 + 1120*a^4*cos(d*x + c)^4 - 1120*a^4*cos(d*x + c)^2 - 32
*(5*a^4*cos(d*x + c)^6 - 22*a^4*cos(d*x + c)^4 + 29*a^4*cos(d*x + c)^2 - 12*a^4)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\begin {cases} \frac {a^{4} \sin ^{8}{\left (c + d x \right )}}{8 d} + \frac {4 a^{4} \sin ^{7}{\left (c + d x \right )}}{7 d} + \frac {a^{4} \sin ^{6}{\left (c + d x \right )}}{d} + \frac {4 a^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a^{4} \sin ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{4} \sin ^{3}{\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**3*(a+a*sin(d*x+c))**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {35 \, a^{4} \sin \left (d x + c\right )^{8} + 160 \, a^{4} \sin \left (d x + c\right )^{7} + 280 \, a^{4} \sin \left (d x + c\right )^{6} + 224 \, a^{4} \sin \left (d x + c\right )^{5} + 70 \, a^{4} \sin \left (d x + c\right )^{4}}{280 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^3*(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/280*(35*a^4*sin(d*x + c)^8 + 160*a^4*sin(d*x + c)^7 + 280*a^4*sin(d*x + c)^6 + 224*a^4*sin(d*x + c)^5 + 70*a
^4*sin(d*x + c)^4)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {35 \, a^{4} \sin \left (d x + c\right )^{8} + 160 \, a^{4} \sin \left (d x + c\right )^{7} + 280 \, a^{4} \sin \left (d x + c\right )^{6} + 224 \, a^{4} \sin \left (d x + c\right )^{5} + 70 \, a^{4} \sin \left (d x + c\right )^{4}}{280 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^3*(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/280*(35*a^4*sin(d*x + c)^8 + 160*a^4*sin(d*x + c)^7 + 280*a^4*sin(d*x + c)^6 + 224*a^4*sin(d*x + c)^5 + 70*a
^4*sin(d*x + c)^4)/d

Mupad [B] (verification not implemented)

Time = 9.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^4\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {4\,a^4\,{\sin \left (c+d\,x\right )}^7}{7}+a^4\,{\sin \left (c+d\,x\right )}^6+\frac {4\,a^4\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a^4\,{\sin \left (c+d\,x\right )}^4}{4}}{d} \]

[In]

int(cos(c + d*x)*sin(c + d*x)^3*(a + a*sin(c + d*x))^4,x)

[Out]

((a^4*sin(c + d*x)^4)/4 + (4*a^4*sin(c + d*x)^5)/5 + a^4*sin(c + d*x)^6 + (4*a^4*sin(c + d*x)^7)/7 + (a^4*sin(
c + d*x)^8)/8)/d