\(\int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx\) [1052]

Optimal result

Integrand size = 25, antiderivative size = 65 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b x}{8}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {b \cos ^3(c+d x) \sin (c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2917, 2645, 30, 2648, 2715, 8} \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \cos ^3(c+d x)}{3 d}-\frac {b \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {b x}{8} \]

[In]

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

[Out]

(b*x)/8 - (a*Cos[c + d*x]^3)/(3*d) + (b*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (b*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 2645

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 2648

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 2715

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] && Integ
erQ[2*n]

Rule 2917

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b x}{8}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {1}{8} b \left (-\frac {\cos (4 d x) \sin (4 c)}{4 d}-\frac {\cos (4 c) \sin (4 d x)}{4 d}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {b x}{8}-\frac {a \cos \left (d x +c \right )}{4 d}-\frac {b \sin \left (4 d x +4 c \right )}{32 d}-\frac {a \cos \left (3 d x +3 c \right )}{12 d}\)	\(48\)
parallelrisch	\(\frac {12 b x d -24 a \cos \left (d x +c \right )-3 b \sin \left (4 d x +4 c \right )-8 a \cos \left (3 d x +3 c \right )-32 a}{96 d}\)	\(48\)
derivativedivides	\(\frac {-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{3}+b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\)	\(57\)
default	\(\frac {-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{3}+b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\)	\(57\)
norman	\(\frac {\frac {b x}{8}-\frac {2 a}{3 d}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {7 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {7 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(205\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {8 \, a \cos \left (d x + c\right )^{3} - 3 \, b d x + 3 \, {\left (2 \, b \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (56) = 112\).

Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.83 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} - \frac {a \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \sin {\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {1}{8} \, b x - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {a \cos \left (d x + c\right )}{4 \, d} - \frac {b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.92 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b\,x}{8}-\frac {-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {2\,a}{3}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

[In]

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

[Out]

(b*x)/8 - ((2*a)/3 + (b*tan(c/2 + (d*x)/2))/4 + (2*a*tan(c/2 + (d*x)/2)^2)/3 + 2*a*tan(c/2 + (d*x)/2)^4 + 2*a*
tan(c/2 + (d*x)/2)^6 - (7*b*tan(c/2 + (d*x)/2)^3)/4 + (7*b*tan(c/2 + (d*x)/2)^5)/4 - (b*tan(c/2 + (d*x)/2)^7)/
4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)