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

Optimal. Leaf size=60 \[ \frac{a \sin ^5(c+d x)}{5 d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^6(c+d x)}{6 d} \]

[Out]

-(b*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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Rubi [A]  time = 0.0422384, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2668, 641, 194} \[ \frac{a \sin ^5(c+d x)}{5 d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*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)

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0232151, size = 60, normalized size = 1. \[ \frac{a \sin ^5(c+d x)}{5 d}-\frac{2 a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}-\frac{b \cos ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*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 = 0.027, size = 46, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{\frac{b \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+b*sin(d*x+c)),x)

[Out]

1/d*(-1/6*b*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 = 0.948505, size = 95, normalized size = 1.58 \begin{align*} \frac{5 \, b \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} - 15 \, b \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} + 15 \, b \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+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.3126, size = 128, normalized size = 2.13 \begin{align*} -\frac{5 \, b \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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/30*(5*b*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 = 3.76841, size = 83, normalized size = 1.38 \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{b \cos ^{6}{\left (c + d x \right )}}{6 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\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+b*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 - b*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a + b*sin(c))*cos(c)**5, True))

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Giac [A]  time = 1.0811, size = 119, normalized size = 1.98 \begin{align*} -\frac{b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{5 \, b \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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/192*b*cos(6*d*x + 6*c)/d - 1/32*b*cos(4*d*x + 4*c)/d - 5/64*b*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