\(\int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx\) [219]

Optimal result

Integrand size = 27, antiderivative size = 67 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {(a+a \sin (c+d x))^5}{5 a d}-\frac {(a+a \sin (c+d x))^6}{3 a^2 d}+\frac {(a+a \sin (c+d x))^7}{7 a^3 d} \]

[Out]

1/5*(a+a*sin(d*x+c))^5/a/d-1/3*(a+a*sin(d*x+c))^6/a^2/d+1/7*(a+a*sin(d*x+c))^7/a^3/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, 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} \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {(a \sin (c+d x)+a)^7}{7 a^3 d}-\frac {(a \sin (c+d x)+a)^6}{3 a^2 d}+\frac {(a \sin (c+d x)+a)^5}{5 a d} \]

[In]

Int[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^4,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {a^4 (-630+5460 \cos (2 (c+d x))-1680 \cos (4 (c+d x))+140 \cos (6 (c+d x))-7245 \sin (c+d x)+3395 \sin (3 (c+d x))-609 \sin (5 (c+d x))+15 \sin (7 (c+d x)))}{6720 d} \]

[In]

Integrate[Cos[c + d*x]*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^4,x]

[Out]

-1/6720*(a^4*(-630 + 5460*Cos[2*(c + d*x)] - 1680*Cos[4*(c + d*x)] + 140*Cos[6*(c + d*x)] - 7245*Sin[c + d*x]
+ 3395*Sin[3*(c + d*x)] - 609*Sin[5*(c + d*x)] + 15*Sin[7*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\frac {a^{4} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {2 a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {6 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+a^{4} \left (\sin ^{4}\left (d x +c \right )\right )+\frac {a^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(70\)
default	\(\frac {\frac {a^{4} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {2 a^{4} \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {6 a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+a^{4} \left (\sin ^{4}\left (d x +c \right )\right )+\frac {a^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(70\)
parallelrisch	\(\frac {a^{4} \left (-\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-564 \cos \left (2 d x +2 c \right )+1260 \sin \left (d x +c \right )-140 \sin \left (3 d x +3 c \right )+15 \cos \left (4 d x +4 c \right )+829\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1680 d}\)	\(96\)
risch	\(\frac {69 a^{4} \sin \left (d x +c \right )}{64 d}-\frac {a^{4} \sin \left (7 d x +7 c \right )}{448 d}-\frac {a^{4} \cos \left (6 d x +6 c \right )}{48 d}+\frac {29 a^{4} \sin \left (5 d x +5 c \right )}{320 d}+\frac {a^{4} \cos \left (4 d x +4 c \right )}{4 d}-\frac {97 a^{4} \sin \left (3 d x +3 c \right )}{192 d}-\frac {13 a^{4} \cos \left (2 d x +2 c \right )}{16 d}\)	\(118\)
norman	\(\frac {\frac {16 a^{4} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{4} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {736 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {3888 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {736 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 a^{4} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {272 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {272 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\)	\(189\)

[In]

int(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/7*a^4*sin(d*x+c)^7+2/3*a^4*sin(d*x+c)^6+6/5*a^4*sin(d*x+c)^5+a^4*sin(d*x+c)^4+1/3*a^4*sin(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.45 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {70 \, a^{4} \cos \left (d x + c\right )^{6} - 315 \, a^{4} \cos \left (d x + c\right )^{4} + 420 \, a^{4} \cos \left (d x + c\right )^{2} + {\left (15 \, a^{4} \cos \left (d x + c\right )^{6} - 171 \, a^{4} \cos \left (d x + c\right )^{4} + 332 \, a^{4} \cos \left (d x + c\right )^{2} - 176 \, a^{4}\right )} \sin \left (d x + c\right )}{105 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/105*(70*a^4*cos(d*x + c)^6 - 315*a^4*cos(d*x + c)^4 + 420*a^4*cos(d*x + c)^2 + (15*a^4*cos(d*x + c)^6 - 171
*a^4*cos(d*x + c)^4 + 332*a^4*cos(d*x + c)^2 - 176*a^4)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.42 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\begin {cases} \frac {a^{4} \sin ^{7}{\left (c + d x \right )}}{7 d} + \frac {2 a^{4} \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac {6 a^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {a^{4} \sin ^{4}{\left (c + d x \right )}}{d} + \frac {a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{4} \sin ^{2}{\left (c \right )} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**2*(a+a*sin(d*x+c))**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {15 \, a^{4} \sin \left (d x + c\right )^{7} + 70 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5} + 105 \, a^{4} \sin \left (d x + c\right )^{4} + 35 \, a^{4} \sin \left (d x + c\right )^{3}}{105 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/105*(15*a^4*sin(d*x + c)^7 + 70*a^4*sin(d*x + c)^6 + 126*a^4*sin(d*x + c)^5 + 105*a^4*sin(d*x + c)^4 + 35*a^
4*sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {15 \, a^{4} \sin \left (d x + c\right )^{7} + 70 \, a^{4} \sin \left (d x + c\right )^{6} + 126 \, a^{4} \sin \left (d x + c\right )^{5} + 105 \, a^{4} \sin \left (d x + c\right )^{4} + 35 \, a^{4} \sin \left (d x + c\right )^{3}}{105 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/105*(15*a^4*sin(d*x + c)^7 + 70*a^4*sin(d*x + c)^6 + 126*a^4*sin(d*x + c)^5 + 105*a^4*sin(d*x + c)^4 + 35*a^
4*sin(d*x + c)^3)/d

Mupad [B] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {\frac {a^4\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {2\,a^4\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {6\,a^4\,{\sin \left (c+d\,x\right )}^5}{5}+a^4\,{\sin \left (c+d\,x\right )}^4+\frac {a^4\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]

[In]

int(cos(c + d*x)*sin(c + d*x)^2*(a + a*sin(c + d*x))^4,x)

[Out]

((a^4*sin(c + d*x)^3)/3 + a^4*sin(c + d*x)^4 + (6*a^4*sin(c + d*x)^5)/5 + (2*a^4*sin(c + d*x)^6)/3 + (a^4*sin(
c + d*x)^7)/7)/d