3.521 \(\int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=111 \[ \frac{(a \sin (c+d x)+a)^{10}}{10 a^7 d}-\frac{2 (a \sin (c+d x)+a)^9}{3 a^6 d}+\frac{13 (a \sin (c+d x)+a)^8}{8 a^5 d}-\frac{12 (a \sin (c+d x)+a)^7}{7 a^4 d}+\frac{2 (a \sin (c+d x)+a)^6}{3 a^3 d} \]

[Out]

(2*(a + a*Sin[c + d*x])^6)/(3*a^3*d) - (12*(a + a*Sin[c + d*x])^7)/(7*a^4*d) + (13*(a + a*Sin[c + d*x])^8)/(8*
a^5*d) - (2*(a + a*Sin[c + d*x])^9)/(3*a^6*d) + (a + a*Sin[c + d*x])^10/(10*a^7*d)

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Rubi [A]  time = 0.128079, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac{(a \sin (c+d x)+a)^{10}}{10 a^7 d}-\frac{2 (a \sin (c+d x)+a)^9}{3 a^6 d}+\frac{13 (a \sin (c+d x)+a)^8}{8 a^5 d}-\frac{12 (a \sin (c+d x)+a)^7}{7 a^4 d}+\frac{2 (a \sin (c+d x)+a)^6}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(2*(a + a*Sin[c + d*x])^6)/(3*a^3*d) - (12*(a + a*Sin[c + d*x])^7)/(7*a^4*d) + (13*(a + a*Sin[c + d*x])^8)/(8*
a^5*d) - (2*(a + a*Sin[c + d*x])^9)/(3*a^6*d) + (a + a*Sin[c + d*x])^10/(10*a^7*d)

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^2 (a+x)^5}{a^2} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int (a-x)^2 x^2 (a+x)^5 \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (4 a^4 (a+x)^5-12 a^3 (a+x)^6+13 a^2 (a+x)^7-6 a (a+x)^8+(a+x)^9\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{2 (a+a \sin (c+d x))^6}{3 a^3 d}-\frac{12 (a+a \sin (c+d x))^7}{7 a^4 d}+\frac{13 (a+a \sin (c+d x))^8}{8 a^5 d}-\frac{2 (a+a \sin (c+d x))^9}{3 a^6 d}+\frac{(a+a \sin (c+d x))^{10}}{10 a^7 d}\\ \end{align*}

Mathematica [A]  time = 0.909289, size = 110, normalized size = 0.99 \[ -\frac{a^3 (-63840 \sin (c+d x)+8960 \sin (3 (c+d x))+8064 \sin (5 (c+d x))+240 \sin (7 (c+d x))-560 \sin (9 (c+d x))+34440 \cos (2 (c+d x))+5040 \cos (4 (c+d x))-4060 \cos (6 (c+d x))-1260 \cos (8 (c+d x))+84 \cos (10 (c+d x))-2835)}{430080 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

-(a^3*(-2835 + 34440*Cos[2*(c + d*x)] + 5040*Cos[4*(c + d*x)] - 4060*Cos[6*(c + d*x)] - 1260*Cos[8*(c + d*x)]
+ 84*Cos[10*(c + d*x)] - 63840*Sin[c + d*x] + 8960*Sin[3*(c + d*x)] + 8064*Sin[5*(c + d*x)] + 240*Sin[7*(c + d
*x)] - 560*Sin[9*(c + d*x)]))/(430080*d)

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Maple [B]  time = 0.043, size = 208, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{10}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{20}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{60}} \right ) +3\,{a}^{3} \left ( -1/9\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}-1/21\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{ \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{105}} \right ) +3\,{a}^{3} \left ( -1/8\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}-1/24\, \left ( \cos \left ( dx+c \right ) \right ) ^{6} \right ) +{a}^{3} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(-1/10*sin(d*x+c)^4*cos(d*x+c)^6-1/20*sin(d*x+c)^2*cos(d*x+c)^6-1/60*cos(d*x+c)^6)+3*a^3*(-1/9*sin(d*
x+c)^3*cos(d*x+c)^6-1/21*sin(d*x+c)*cos(d*x+c)^6+1/105*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+3*a^3*(
-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6)+a^3*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3
*cos(d*x+c)^2)*sin(d*x+c)))

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Maxima [A]  time = 1.15809, size = 149, normalized size = 1.34 \begin{align*} \frac{84 \, a^{3} \sin \left (d x + c\right )^{10} + 280 \, a^{3} \sin \left (d x + c\right )^{9} + 105 \, a^{3} \sin \left (d x + c\right )^{8} - 600 \, a^{3} \sin \left (d x + c\right )^{7} - 700 \, a^{3} \sin \left (d x + c\right )^{6} + 168 \, a^{3} \sin \left (d x + c\right )^{5} + 630 \, a^{3} \sin \left (d x + c\right )^{4} + 280 \, a^{3} \sin \left (d x + c\right )^{3}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(84*a^3*sin(d*x + c)^10 + 280*a^3*sin(d*x + c)^9 + 105*a^3*sin(d*x + c)^8 - 600*a^3*sin(d*x + c)^7 - 700
*a^3*sin(d*x + c)^6 + 168*a^3*sin(d*x + c)^5 + 630*a^3*sin(d*x + c)^4 + 280*a^3*sin(d*x + c)^3)/d

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Fricas [A]  time = 1.4803, size = 277, normalized size = 2.5 \begin{align*} -\frac{84 \, a^{3} \cos \left (d x + c\right )^{10} - 525 \, a^{3} \cos \left (d x + c\right )^{8} + 560 \, a^{3} \cos \left (d x + c\right )^{6} - 8 \,{\left (35 \, a^{3} \cos \left (d x + c\right )^{8} - 65 \, a^{3} \cos \left (d x + c\right )^{6} + 6 \, a^{3} \cos \left (d x + c\right )^{4} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 16 \, a^{3}\right )} \sin \left (d x + c\right )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/840*(84*a^3*cos(d*x + c)^10 - 525*a^3*cos(d*x + c)^8 + 560*a^3*cos(d*x + c)^6 - 8*(35*a^3*cos(d*x + c)^8 -
65*a^3*cos(d*x + c)^6 + 6*a^3*cos(d*x + c)^4 + 8*a^3*cos(d*x + c)^2 + 16*a^3)*sin(d*x + c))/d

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Sympy [A]  time = 33.7642, size = 255, normalized size = 2.3 \begin{align*} \begin{cases} \frac{8 a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac{12 a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{8 a^{3} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac{4 a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} - \frac{a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac{a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac{a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{12 d} - \frac{a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{2 d} - \frac{a^{3} \cos ^{10}{\left (c + d x \right )}}{60 d} - \frac{a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{3} \sin ^{2}{\left (c \right )} \cos ^{5}{\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)**5*sin(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((8*a**3*sin(c + d*x)**9/(105*d) + 12*a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 8*a**3*sin(c + d*
x)**7/(105*d) + 3*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + 4*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) -
a**3*sin(c + d*x)**4*cos(c + d*x)**6/(6*d) + a**3*sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - a**3*sin(c + d*x)**
2*cos(c + d*x)**8/(12*d) - a**3*sin(c + d*x)**2*cos(c + d*x)**6/(2*d) - a**3*cos(c + d*x)**10/(60*d) - a**3*co
s(c + d*x)**8/(8*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**2*cos(c)**5, True))

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Giac [A]  time = 1.30594, size = 227, normalized size = 2.05 \begin{align*} -\frac{a^{3} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac{3 \, a^{3} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac{29 \, a^{3} \cos \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac{3 \, a^{3} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{41 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} + \frac{a^{3} \sin \left (9 \, d x + 9 \, c\right )}{768 \, d} - \frac{a^{3} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac{3 \, a^{3} \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac{a^{3} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{19 \, a^{3} \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)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/5120*a^3*cos(10*d*x + 10*c)/d + 3/1024*a^3*cos(8*d*x + 8*c)/d + 29/3072*a^3*cos(6*d*x + 6*c)/d - 3/256*a^3*
cos(4*d*x + 4*c)/d - 41/512*a^3*cos(2*d*x + 2*c)/d + 1/768*a^3*sin(9*d*x + 9*c)/d - 1/1792*a^3*sin(7*d*x + 7*c
)/d - 3/160*a^3*sin(5*d*x + 5*c)/d - 1/48*a^3*sin(3*d*x + 3*c)/d + 19/128*a^3*sin(d*x + c)/d