\(\int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx\) [377]

Optimal result

Integrand size = 19, antiderivative size = 44 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos ^4(c+d x)}{4 d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2747, 655} \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^4(c+d x)}{4 d} \]

[In]

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

[Out]

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

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos ^4(c+d x)}{4 d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(-\frac {\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}-a \sin \left (d x +c \right )}{d}\)	\(49\)
default	\(-\frac {\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}-a \sin \left (d x +c \right )}{d}\)	\(49\)
risch	\(\frac {3 a \sin \left (d x +c \right )}{4 d}-\frac {b \cos \left (4 d x +4 c \right )}{32 d}+\frac {a \sin \left (3 d x +3 c \right )}{12 d}-\frac {b \cos \left (2 d x +2 c \right )}{8 d}\)	\(59\)
parallelrisch	\(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\frac {5 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {5 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(97\)
norman	\(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {10 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {10 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(118\)

[In]

int(cos(d*x+c)^3*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/d*(1/4*b*sin(d*x+c)^4+1/3*a*sin(d*x+c)^3-1/2*sin(d*x+c)^2*b-a*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {3 \, b \cos \left (d x + c\right )^{4} - 4 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.36 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {2 a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {b \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 6 \, b \sin \left (d x + c\right )^{2} - 12 \, a \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

-1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 6*b*sin(d*x + c)^2 - 12*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {3 \, b \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 6 \, b \sin \left (d x + c\right )^{2} - 12 \, a \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

-1/12*(3*b*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 6*b*sin(d*x + c)^2 - 12*a*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx=\frac {-\frac {b\,{\sin \left (c+d\,x\right )}^4}{4}-\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

(a*sin(c + d*x) - (a*sin(c + d*x)^3)/3 + (b*sin(c + d*x)^2)/2 - (b*sin(c + d*x)^4)/4)/d