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

Optimal. Leaf size=22 \[ \frac{(a+b \sin (c+d x))^4}{4 b d} \]

[Out]

(a + b*Sin[c + d*x])^4/(4*b*d)

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Rubi [A]  time = 0.0263097, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2668, 32} \[ \frac{(a+b \sin (c+d x))^4}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*Sin[c + d*x])^4/(4*b*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 0.0693703, size = 57, normalized size = 2.59 \[ \frac{\sin (c+d x) \left (6 a^2 b \sin (c+d x)+4 a^3+4 a b^2 \sin ^2(c+d x)+b^3 \sin ^3(c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]*(4*a^3 + 6*a^2*b*Sin[c + d*x] + 4*a*b^2*Sin[c + d*x]^2 + b^3*Sin[c + d*x]^3))/(4*d)

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Maple [A]  time = 0.019, size = 21, normalized size = 1. \begin{align*}{\frac{ \left ( a+b\sin \left ( dx+c \right ) \right ) ^{4}}{4\,bd}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.954898, size = 27, normalized size = 1.23 \begin{align*} \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{4 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*sin(d*x + c) + a)^4/(b*d)

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Fricas [B]  time = 2.24263, size = 158, normalized size = 7.18 \begin{align*} \frac{b^{3} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.22593, size = 97, normalized size = 4.41 \begin{align*} \begin{cases} \frac{a^{3} \sin{\left (c + d x \right )}}{d} + \frac{3 a^{2} b \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac{a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} - \frac{b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} - \frac{b^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{3} \cos{\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)*(a+b*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.09528, size = 27, normalized size = 1.23 \begin{align*} \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{4 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*sin(d*x + c) + a)^4/(b*d)