3.3 \(\int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=64 \[ \frac{(a \sin (c+d x)+a)^6}{6 a^5 d}-\frac{4 (a \sin (c+d x)+a)^5}{5 a^4 d}+\frac{(a \sin (c+d x)+a)^4}{a^3 d} \]

[Out]

(a + a*Sin[c + d*x])^4/(a^3*d) - (4*(a + a*Sin[c + d*x])^5)/(5*a^4*d) + (a + a*Sin[c + d*x])^6/(6*a^5*d)

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Rubi [A]  time = 0.0444546, antiderivative size = 64, 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 = {2667, 43} \[ \frac{(a \sin (c+d x)+a)^6}{6 a^5 d}-\frac{4 (a \sin (c+d x)+a)^5}{5 a^4 d}+\frac{(a \sin (c+d x)+a)^4}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

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

Mathematica [A]  time = 0.0226898, size = 60, normalized size = 0.94 \[ \frac{a \sin ^5(c+d x)}{5 d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{a \cos ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.025, size = 46, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6}}+{\frac{a\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.952699, size = 95, normalized size = 1.48 \begin{align*} \frac{5 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} + 15 \, a \sin \left (d x + c\right )^{2} + 30 \, a \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)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.62355, size = 128, normalized size = 2. \begin{align*} -\frac{5 \, a \cos \left (d x + c\right )^{6} - 2 \,{\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\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)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.31667, size = 83, normalized size = 1.3 \begin{align*} \begin{cases} \frac{8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{a \cos ^{6}{\left (c + d x \right )}}{6 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \cos ^{5}{\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)**5*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((8*a*sin(c + d*x)**5/(15*d) + 4*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a*sin(c + d*x)*cos(c + d*x
)**4/d - a*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*sin(c) + a)*cos(c)**5, True))

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Giac [A]  time = 1.18751, size = 119, normalized size = 1.86 \begin{align*} -\frac{a \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{a \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{5 \, a \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{5 \, a \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/192*a*cos(6*d*x + 6*c)/d - 1/32*a*cos(4*d*x + 4*c)/d - 5/64*a*cos(2*d*x + 2*c)/d + 1/80*a*sin(5*d*x + 5*c)/
d + 5/48*a*sin(3*d*x + 3*c)/d + 5/8*a*sin(d*x + c)/d