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

Optimal. Leaf size=105 \[ \frac {a \cos ^5(c+d x)}{5 d}-\frac {a \cos ^3(c+d x)}{3 d}-\frac {a \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac {a \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {a x}{16} \]

[Out]

1/16*a*x-1/3*a*cos(d*x+c)^3/d+1/5*a*cos(d*x+c)^5/d+1/16*a*cos(d*x+c)*sin(d*x+c)/d-1/8*a*cos(d*x+c)^3*sin(d*x+c
)/d-1/6*a*cos(d*x+c)^3*sin(d*x+c)^3/d

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Rubi [A]  time = 0.17, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2838, 2565, 14, 2568, 2635, 8} \[ \frac {a \cos ^5(c+d x)}{5 d}-\frac {a \cos ^3(c+d x)}{3 d}-\frac {a \sin ^3(c+d x) \cos ^3(c+d x)}{6 d}-\frac {a \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {a x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/16 - (a*Cos[c + d*x]^3)/(3*d) + (a*Cos[c + d*x]^5)/(5*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a*Cos
[c + d*x]^3*Sin[c + d*x])/(8*d) - (a*Cos[c + d*x]^3*Sin[c + d*x]^3)/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

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Mathematica [A]  time = 0.20, size = 71, normalized size = 0.68 \[ \frac {a (-15 \sin (2 (c+d x))-15 \sin (4 (c+d x))+5 \sin (6 (c+d x))-120 \cos (c+d x)-20 \cos (3 (c+d x))+12 \cos (5 (c+d x))+60 d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(60*d*x - 120*Cos[c + d*x] - 20*Cos[3*(c + d*x)] + 12*Cos[5*(c + d*x)] - 15*Sin[2*(c + d*x)] - 15*Sin[4*(c
+ d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.49, size = 73, normalized size = 0.70 \[ \frac {48 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 15 \, a d x + 5 \, {\left (8 \, a \cos \left (d x + c\right )^{5} - 14 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(48*a*cos(d*x + c)^5 - 80*a*cos(d*x + c)^3 + 15*a*d*x + 5*(8*a*cos(d*x + c)^5 - 14*a*cos(d*x + c)^3 + 3*
a*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.17, size = 92, normalized size = 0.88 \[ \frac {1}{16} \, a x + \frac {a \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {a \cos \left (d x + c\right )}{8 \, d} + \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a*x + 1/80*a*cos(5*d*x + 5*c)/d - 1/48*a*cos(3*d*x + 3*c)/d - 1/8*a*cos(d*x + c)/d + 1/192*a*sin(6*d*x +
6*c)/d - 1/64*a*sin(4*d*x + 4*c)/d - 1/64*a*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.13, size = 95, normalized size = 0.90 \[ \frac {a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*c)
+a*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3))

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maxima [A]  time = 0.33, size = 65, normalized size = 0.62 \[ \frac {64 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(64*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a - 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*
c))*a)/d

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mupad [B]  time = 12.13, size = 226, normalized size = 2.15 \[ \frac {a\,x}{16}+\frac {\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {17\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\left (\frac {a\,\left (225\,c+225\,d\,x-960\right )}{240}-\frac {15\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\left (\frac {a\,\left (300\,c+300\,d\,x-640\right )}{240}-\frac {5\,a\,\left (c+d\,x\right )}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {17\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\left (\frac {a\,\left (90\,c+90\,d\,x-384\right )}{240}-\frac {3\,a\,\left (c+d\,x\right )}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {a\,\left (15\,c+15\,d\,x-64\right )}{240}-\frac {a\,\left (c+d\,x\right )}{16}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x)/16 + ((a*(15*c + 15*d*x - 64))/240 - (a*tan(c/2 + (d*x)/2))/8 - (a*(c + d*x))/16 + tan(c/2 + (d*x)/2)^2*
((a*(90*c + 90*d*x - 384))/240 - (3*a*(c + d*x))/8) + tan(c/2 + (d*x)/2)^6*((a*(300*c + 300*d*x - 640))/240 -
(5*a*(c + d*x))/4) + tan(c/2 + (d*x)/2)^8*((a*(225*c + 225*d*x - 960))/240 - (15*a*(c + d*x))/16) - (17*a*tan(
c/2 + (d*x)/2)^3)/24 + (19*a*tan(c/2 + (d*x)/2)^5)/4 - (19*a*tan(c/2 + (d*x)/2)^7)/4 + (17*a*tan(c/2 + (d*x)/2
)^9)/24 + (a*tan(c/2 + (d*x)/2)^11)/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)

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sympy [A]  time = 4.34, size = 192, normalized size = 1.83 \[ \begin {cases} \frac {a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {2 a \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right ) \sin ^{3}{\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**3*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((a*x*sin(c + d*x)**6/16 + 3*a*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a*x*sin(c + d*x)**2*cos(c + d
*x)**4/16 + a*x*cos(c + d*x)**6/16 + a*sin(c + d*x)**5*cos(c + d*x)/(16*d) - a*sin(c + d*x)**3*cos(c + d*x)**3
/(6*d) - a*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - a*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 2*a*cos(c + d*x)**5
/(15*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**3*cos(c)**2, True))

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