3.16 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=45 \[ \frac{(a \sin (c+d x)+a)^4}{2 a^2 d}-\frac{(a \sin (c+d x)+a)^5}{5 a^3 d} \]

[Out]

(a + a*Sin[c + d*x])^4/(2*a^2*d) - (a + a*Sin[c + d*x])^5/(5*a^3*d)

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Rubi [A]  time = 0.046204, 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} \[ \frac{(a \sin (c+d x)+a)^4}{2 a^2 d}-\frac{(a \sin (c+d x)+a)^5}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Mathematica [A]  time = 0.081595, size = 46, normalized size = 1.02 \[ -\frac{a^2 \sin (c+d x) \left (2 \sin ^4(c+d x)+5 \sin ^3(c+d x)-10 \sin (c+d x)-10\right )}{10 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*Sin[c + d*x]*(-10 - 10*Sin[c + d*x] + 5*Sin[c + d*x]^3 + 2*Sin[c + d*x]^4))/(10*d)

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Maple [A]  time = 0.036, size = 79, normalized size = 1.8 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2}}+{\frac{{a}^{2} \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))^2,x)

[Out]

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

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Maxima [A]  time = 0.94727, size = 76, normalized size = 1.69 \begin{align*} -\frac{2 \, a^{2} \sin \left (d x + c\right )^{5} + 5 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{2} - 10 \, a^{2} \sin \left (d x + c\right )}{10 \, 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))^2,x, algorithm="maxima")

[Out]

-1/10*(2*a^2*sin(d*x + c)^5 + 5*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^2 - 10*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.67835, size = 136, normalized size = 3.02 \begin{align*} -\frac{5 \, a^{2} \cos \left (d x + c\right )^{4} + 2 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2}\right )} \sin \left (d x + c\right )}{10 \, 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))^2,x, algorithm="fricas")

[Out]

-1/10*(5*a^2*cos(d*x + c)^4 + 2*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 - 4*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 2.52839, size = 129, normalized size = 2.87 \begin{align*} \begin{cases} \frac{2 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{a^{2} \sin ^{4}{\left (c + d x \right )}}{2 d} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{a^{2} \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 )^{2} \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))**2,x)

[Out]

Piecewise((2*a**2*sin(c + d*x)**5/(15*d) + a**2*sin(c + d*x)**4/(2*d) + a**2*sin(c + d*x)**3*cos(c + d*x)**2/(
3*d) + 2*a**2*sin(c + d*x)**3/(3*d) + a**2*sin(c + d*x)**2*cos(c + d*x)**2/d + a**2*sin(c + d*x)*cos(c + d*x)*
*2/d, Ne(d, 0)), (x*(a*sin(c) + a)**2*cos(c)**3, True))

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Giac [A]  time = 1.13963, size = 76, normalized size = 1.69 \begin{align*} -\frac{2 \, a^{2} \sin \left (d x + c\right )^{5} + 5 \, a^{2} \sin \left (d x + c\right )^{4} - 10 \, a^{2} \sin \left (d x + c\right )^{2} - 10 \, a^{2} \sin \left (d x + c\right )}{10 \, 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))^2,x, algorithm="giac")

[Out]

-1/10*(2*a^2*sin(d*x + c)^5 + 5*a^2*sin(d*x + c)^4 - 10*a^2*sin(d*x + c)^2 - 10*a^2*sin(d*x + c))/d