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

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

[Out]

(a + a*Sin[c + d*x])^4/(4*a*d)

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Rubi [A]  time = 0.0253023, 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 = {2667, 32} \[ \frac{(a \sin (c+d x)+a)^4}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + a*Sin[c + d*x])^4/(4*a*d)

Rule 2667

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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

Mathematica [B]  time = 0.027618, size = 65, normalized size = 2.95 \[ \frac{a^3 \sin ^4(c+d x)}{4 d}+\frac{a^3 \sin ^3(c+d x)}{d}+\frac{3 a^3 \sin ^2(c+d x)}{2 d}+\frac{a^3 \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 21, normalized size = 1. \begin{align*}{\frac{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{4}}{4\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.94555, size = 27, normalized size = 1.23 \begin{align*} \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*sin(d*x + c) + a)^4/(a*d)

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Fricas [B]  time = 1.68935, size = 131, normalized size = 5.95 \begin{align*} \frac{a^{3} \cos \left (d x + c\right )^{4} - 8 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \,{\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\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+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(a^3*cos(d*x + c)^4 - 8*a^3*cos(d*x + c)^2 - 4*(a^3*cos(d*x + c)^2 - 2*a^3)*sin(d*x + c))/d

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Sympy [A]  time = 1.3255, size = 94, normalized size = 4.27 \begin{align*} \begin{cases} \frac{a^{3} \sin ^{3}{\left (c + d x \right )}}{d} - \frac{a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{3 a^{3} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac{a^{3} \sin{\left (c + d x \right )}}{d} - \frac{a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\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+a*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.15015, size = 27, normalized size = 1.23 \begin{align*} \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*sin(d*x + c) + a)^4/(a*d)