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

Optimal. Leaf size=45 \[ \frac{2 (a \sin (c+d x)+a)^5}{5 a^2 d}-\frac{(a \sin (c+d x)+a)^6}{6 a^3 d} \]

[Out]

(2*(a + a*Sin[c + d*x])^5)/(5*a^2*d) - (a + a*Sin[c + d*x])^6/(6*a^3*d)

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Rubi [A]  time = 0.0471518, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2667, 43} \[ \frac{2 (a \sin (c+d x)+a)^5}{5 a^2 d}-\frac{(a \sin (c+d x)+a)^6}{6 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + a*Sin[c + d*x])^5)/(5*a^2*d) - (a + a*Sin[c + d*x])^6/(6*a^3*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int (a-x) (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a (a+x)^4-(a+x)^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac{2 (a+a \sin (c+d x))^5}{5 a^2 d}-\frac{(a+a \sin (c+d x))^6}{6 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.1451, size = 43, normalized size = 0.96 \[ -\frac{a^3 (5 \sin (c+d x)-7) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^{10}}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^10*(-7 + 5*Sin[c + d*x]))/(30*d)

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Maple [B]  time = 0.042, size = 113, normalized size = 2.5 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{12}} \right ) +3\,{a}^{3} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}}+{\frac{{a}^{3} \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+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+3*a^3*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*
x+c)^2)*sin(d*x+c))-3/4*a^3*cos(d*x+c)^4+1/3*a^3*(2+cos(d*x+c)^2)*sin(d*x+c))

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Maxima [B]  time = 0.947696, size = 111, normalized size = 2.47 \begin{align*} -\frac{5 \, a^{3} \sin \left (d x + c\right )^{6} + 18 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 20 \, a^{3} \sin \left (d x + c\right )^{3} - 45 \, a^{3} \sin \left (d x + c\right )^{2} - 30 \, a^{3} \sin \left (d x + c\right )}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(5*a^3*sin(d*x + c)^6 + 18*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 - 20*a^3*sin(d*x + c)^3 - 45*a^3*s
in(d*x + c)^2 - 30*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.72699, size = 171, normalized size = 3.8 \begin{align*} \frac{5 \, a^{3} \cos \left (d x + c\right )^{6} - 30 \, a^{3} \cos \left (d x + c\right )^{4} - 2 \,{\left (9 \, a^{3} \cos \left (d x + c\right )^{4} - 8 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3}\right )} \sin \left (d x + c\right )}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(5*a^3*cos(d*x + c)^6 - 30*a^3*cos(d*x + c)^4 - 2*(9*a^3*cos(d*x + c)^4 - 8*a^3*cos(d*x + c)^2 - 16*a^3)*
sin(d*x + c))/d

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Sympy [A]  time = 4.83883, size = 172, normalized size = 3.82 \begin{align*} \begin{cases} \frac{a^{3} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac{2 a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac{a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac{3 a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac{a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{2 a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{3 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{a^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{3} \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+a*sin(d*x+c))**3,x)

[Out]

Piecewise((a**3*sin(c + d*x)**6/(12*d) + 2*a**3*sin(c + d*x)**5/(5*d) + a**3*sin(c + d*x)**4*cos(c + d*x)**2/(
4*d) + 3*a**3*sin(c + d*x)**4/(4*d) + a**3*sin(c + d*x)**3*cos(c + d*x)**2/d + 2*a**3*sin(c + d*x)**3/(3*d) +
3*a**3*sin(c + d*x)**2*cos(c + d*x)**2/(2*d) + a**3*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(a*sin(c) +
a)**3*cos(c)**3, True))

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Giac [B]  time = 1.1651, size = 111, normalized size = 2.47 \begin{align*} -\frac{5 \, a^{3} \sin \left (d x + c\right )^{6} + 18 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 20 \, a^{3} \sin \left (d x + c\right )^{3} - 45 \, a^{3} \sin \left (d x + c\right )^{2} - 30 \, a^{3} \sin \left (d x + c\right )}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(5*a^3*sin(d*x + c)^6 + 18*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 - 20*a^3*sin(d*x + c)^3 - 45*a^3*s
in(d*x + c)^2 - 30*a^3*sin(d*x + c))/d