3.440 \(\int \cos (c+d x) (a+b \cos (c+d x))^4 \, dx\)

Optimal. Leaf size=170 \[ \frac{2 \left (28 a^2 b^2+3 a^4+4 b^4\right ) \sin (c+d x)}{15 d}+\frac{\left (3 a^2+4 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d}+\frac{a b \left (6 a^2+29 b^2\right ) \sin (c+d x) \cos (c+d x)}{30 d}+\frac{1}{2} a b x \left (4 a^2+3 b^2\right )+\frac{\sin (c+d x) (a+b \cos (c+d x))^4}{5 d}+\frac{a \sin (c+d x) (a+b \cos (c+d x))^3}{5 d} \]

[Out]

(a*b*(4*a^2 + 3*b^2)*x)/2 + (2*(3*a^4 + 28*a^2*b^2 + 4*b^4)*Sin[c + d*x])/(15*d) + (a*b*(6*a^2 + 29*b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(30*d) + ((3*a^2 + 4*b^2)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(15*d) + (a*(a + b*Cos[c
+ d*x])^3*Sin[c + d*x])/(5*d) + ((a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.203751, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2753, 2734} \[ \frac{2 \left (28 a^2 b^2+3 a^4+4 b^4\right ) \sin (c+d x)}{15 d}+\frac{\left (3 a^2+4 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d}+\frac{a b \left (6 a^2+29 b^2\right ) \sin (c+d x) \cos (c+d x)}{30 d}+\frac{1}{2} a b x \left (4 a^2+3 b^2\right )+\frac{\sin (c+d x) (a+b \cos (c+d x))^4}{5 d}+\frac{a \sin (c+d x) (a+b \cos (c+d x))^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + b*Cos[c + d*x])^4,x]

[Out]

(a*b*(4*a^2 + 3*b^2)*x)/2 + (2*(3*a^4 + 28*a^2*b^2 + 4*b^4)*Sin[c + d*x])/(15*d) + (a*b*(6*a^2 + 29*b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(30*d) + ((3*a^2 + 4*b^2)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(15*d) + (a*(a + b*Cos[c
+ d*x])^3*Sin[c + d*x])/(5*d) + ((a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*d)

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+b \cos (c+d x))^4 \, dx &=\frac{(a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{5} \int (4 b+4 a \cos (c+d x)) (a+b \cos (c+d x))^3 \, dx\\ &=\frac{a (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{(a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{20} \int (a+b \cos (c+d x))^2 \left (28 a b+4 \left (3 a^2+4 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{\left (3 a^2+4 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac{a (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{(a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{60} \int (a+b \cos (c+d x)) \left (4 b \left (27 a^2+8 b^2\right )+4 a \left (6 a^2+29 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{1}{2} a b \left (4 a^2+3 b^2\right ) x+\frac{2 \left (3 a^4+28 a^2 b^2+4 b^4\right ) \sin (c+d x)}{15 d}+\frac{a b \left (6 a^2+29 b^2\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 a^2+4 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac{a (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{(a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.489407, size = 133, normalized size = 0.78 \[ \frac{30 \left (36 a^2 b^2+8 a^4+5 b^4\right ) \sin (c+d x)+b \left (240 a \left (a^2+b^2\right ) \sin (2 (c+d x))+5 \left (24 a^2 b+5 b^3\right ) \sin (3 (c+d x))+480 a^3 c+480 a^3 d x+30 a b^2 \sin (4 (c+d x))+360 a b^2 c+360 a b^2 d x+3 b^3 \sin (5 (c+d x))\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + b*Cos[c + d*x])^4,x]

[Out]

(30*(8*a^4 + 36*a^2*b^2 + 5*b^4)*Sin[c + d*x] + b*(480*a^3*c + 360*a*b^2*c + 480*a^3*d*x + 360*a*b^2*d*x + 240
*a*(a^2 + b^2)*Sin[2*(c + d*x)] + 5*(24*a^2*b + 5*b^3)*Sin[3*(c + d*x)] + 30*a*b^2*Sin[4*(c + d*x)] + 3*b^3*Si
n[5*(c + d*x)]))/(240*d)

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Maple [A]  time = 0.034, size = 138, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{4}\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 ) }+4\,a{b}^{3} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +2\,{a}^{2}{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +4\,{a}^{3}b \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{4}\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+b*cos(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 1.00177, size = 180, normalized size = 1.06 \begin{align*} \frac{120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b - 240 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b^{2} + 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{3} + 8 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} b^{4} + 120 \, a^{4} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/120*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3*b - 240*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2*b^2 + 15*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b^3 + 8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x
+ c))*b^4 + 120*a^4*sin(d*x + c))/d

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Fricas [A]  time = 2.01594, size = 285, normalized size = 1.68 \begin{align*} \frac{15 \,{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} d x +{\left (6 \, b^{4} \cos \left (d x + c\right )^{4} + 30 \, a b^{3} \cos \left (d x + c\right )^{3} + 30 \, a^{4} + 120 \, a^{2} b^{2} + 16 \, b^{4} + 4 \,{\left (15 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )\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)*(a+b*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/30*(15*(4*a^3*b + 3*a*b^3)*d*x + (6*b^4*cos(d*x + c)^4 + 30*a*b^3*cos(d*x + c)^3 + 30*a^4 + 120*a^2*b^2 + 16
*b^4 + 4*(15*a^2*b^2 + 2*b^4)*cos(d*x + c)^2 + 15*(4*a^3*b + 3*a*b^3)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 2.54356, size = 301, normalized size = 1.77 \begin{align*} \begin{cases} \frac{a^{4} \sin{\left (c + d x \right )}}{d} + 2 a^{3} b x \sin ^{2}{\left (c + d x \right )} + 2 a^{3} b x \cos ^{2}{\left (c + d x \right )} + \frac{2 a^{3} b \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{4 a^{2} b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac{6 a^{2} b^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 a b^{3} x \sin ^{4}{\left (c + d x \right )}}{2} + 3 a b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + \frac{3 a b^{3} x \cos ^{4}{\left (c + d x \right )}}{2} + \frac{3 a b^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{5 a b^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac{8 b^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 b^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{b^{4} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \cos{\left (c \right )}\right )^{4} \cos{\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)*(a+b*cos(d*x+c))**4,x)

[Out]

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

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Giac [A]  time = 1.37816, size = 181, normalized size = 1.06 \begin{align*} \frac{b^{4} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{a b^{3} \sin \left (4 \, d x + 4 \, c\right )}{8 \, d} + \frac{1}{2} \,{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} x + \frac{{\left (24 \, a^{2} b^{2} + 5 \, b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (a^{3} b + a b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac{{\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \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)*(a+b*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/80*b^4*sin(5*d*x + 5*c)/d + 1/8*a*b^3*sin(4*d*x + 4*c)/d + 1/2*(4*a^3*b + 3*a*b^3)*x + 1/48*(24*a^2*b^2 + 5*
b^4)*sin(3*d*x + 3*c)/d + (a^3*b + a*b^3)*sin(2*d*x + 2*c)/d + 1/8*(8*a^4 + 36*a^2*b^2 + 5*b^4)*sin(d*x + c)/d