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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 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]

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 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]

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 = 0.02, size = 60, normalized size = 1.00 \[ \frac {a \sin ^5(c+d x)}{5 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {b \cos ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*Cos[c + d*x]^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)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 51, normalized size = 0.85 \[ -\frac {5 \, b \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{30 \, d} \]

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

________________________________________________________________________________________

giac [A]  time = 0.92, size = 88, normalized size = 1.47 \[ -\frac {b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, b \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.15, size = 46, normalized size = 0.77 \[ \frac {-\frac {b \left (\cos ^{6}\left (d x +c \right )\right )}{6}+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

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))

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 70, normalized size = 1.17 \[ \frac {5 \, b \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} - 15 \, b \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} + 15 \, b \sin \left (d x + c\right )^{2} + 30 \, a \sin \left (d x + c\right )}{30 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 68, normalized size = 1.13 \[ \frac {\frac {b\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {b\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 4.42, size = 83, normalized size = 1.38 \[ \begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {b \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right ) \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________