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

Optimal. Leaf size=65 \[ -\frac {a \cos ^3(c+d x)}{3 d}-\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a x}{8} \]

[Out]

1/8*a*x-1/3*a*cos(d*x+c)^3/d+1/8*a*cos(d*x+c)*sin(d*x+c)/d-1/4*a*cos(d*x+c)^3*sin(d*x+c)/d

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Rubi [A]  time = 0.09, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2838, 2565, 30, 2568, 2635, 8} \[ -\frac {a \cos ^3(c+d x)}{3 d}-\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/8 - (a*Cos[c + d*x]^3)/(3*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*Cos[c + d*x]^3*Sin[c + d*x])/(4*
d)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

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Mathematica [A]  time = 0.10, size = 42, normalized size = 0.65 \[ -\frac {a (3 (\sin (4 (c+d x))-4 d x)+24 \cos (c+d x)+8 \cos (3 (c+d x)))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(a*(24*Cos[c + d*x] + 8*Cos[3*(c + d*x)] + 3*(-4*d*x + Sin[4*(c + d*x)])))/d

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fricas [A]  time = 0.53, size = 51, normalized size = 0.78 \[ -\frac {8 \, a \cos \left (d x + c\right )^{3} - 3 \, a d x + 3 \, {\left (2 \, a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(8*a*cos(d*x + c)^3 - 3*a*d*x + 3*(2*a*cos(d*x + c)^3 - a*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.16, size = 47, normalized size = 0.72 \[ \frac {1}{8} \, a x - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {a \cos \left (d x + c\right )}{4 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*x - 1/12*a*cos(3*d*x + 3*c)/d - 1/4*a*cos(d*x + c)/d - 1/32*a*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.08, size = 57, normalized size = 0.88 \[ \frac {a \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) a}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1/8*c)-1/3*cos(d*x+c)^3*a)

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maxima [A]  time = 0.31, size = 39, normalized size = 0.60 \[ -\frac {32 \, a \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*a*cos(d*x + c)^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a)/d

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mupad [B]  time = 12.33, size = 198, normalized size = 3.05 \[ \frac {a\,x}{8}+\frac {\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\left (\frac {a\,\left (12\,c+12\,d\,x-48\right )}{24}-\frac {a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\left (\frac {a\,\left (18\,c+18\,d\,x-48\right )}{24}-\frac {3\,a\,\left (c+d\,x\right )}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\left (\frac {a\,\left (12\,c+12\,d\,x-16\right )}{24}-\frac {a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a\,\left (3\,c+3\,d\,x-16\right )}{24}-\frac {a\,\left (c+d\,x\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x)/8 + ((a*(3*c + 3*d*x - 16))/24 - (a*tan(c/2 + (d*x)/2))/4 - (a*(c + d*x))/8 + tan(c/2 + (d*x)/2)^2*((a*(
12*c + 12*d*x - 16))/24 - (a*(c + d*x))/2) + tan(c/2 + (d*x)/2)^6*((a*(12*c + 12*d*x - 48))/24 - (a*(c + d*x))
/2) + tan(c/2 + (d*x)/2)^4*((a*(18*c + 18*d*x - 48))/24 - (3*a*(c + d*x))/4) + (7*a*tan(c/2 + (d*x)/2)^3)/4 -
(7*a*tan(c/2 + (d*x)/2)^5)/4 + (a*tan(c/2 + (d*x)/2)^7)/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)

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sympy [A]  time = 1.23, size = 119, normalized size = 1.83 \[ \begin {cases} \frac {a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {a \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right ) \sin {\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x*sin(c + d*x)**4/8 + a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a*x*cos(c + d*x)**4/8 + a*sin(c + d
*x)**3*cos(c + d*x)/(8*d) - a*sin(c + d*x)*cos(c + d*x)**3/(8*d) - a*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*s
in(c) + a)*sin(c)*cos(c)**2, True))

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