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

Optimal. Leaf size=44 \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0318842, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2668, 641} \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2668

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]

Rule 641

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]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+b \sin (c+d x)) \, dx &=\frac{\operatorname{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 \operatorname{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]  time = 0.0126988, size = 44, normalized size = 1. \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 36, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}}+{\frac{a \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.95303, size = 65, normalized size = 1.48 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.12835, size = 97, normalized size = 2.2 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 1.08351, size = 82, normalized size = 1.86 \begin{align*} \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 \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac{b \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right ) \cos ^{3}{\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)**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*sin(c + d*x)**4/(4*d) + b*sin(c +
d*x)**2*cos(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*sin(c))*cos(c)**3, True))

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Giac [A]  time = 1.07937, size = 65, normalized size = 1.48 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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