3.33 \(\int \cos ^2(c+d x) (a+a \cos (c+d x))^4 \, dx\)

Optimal. Leaf size=127 \[ \frac{4 a^4 \sin ^5(c+d x)}{5 d}-\frac{4 a^4 \sin ^3(c+d x)}{d}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{a^4 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{41 a^4 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{49 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{49 a^4 x}{16} \]

[Out]

(49*a^4*x)/16 + (8*a^4*Sin[c + d*x])/d + (49*a^4*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (41*a^4*Cos[c + d*x]^3*Si
n[c + d*x])/(24*d) + (a^4*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (4*a^4*Sin[c + d*x]^3)/d + (4*a^4*Sin[c + d*x]^
5)/(5*d)

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Rubi [A]  time = 0.156639, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2757, 2635, 8, 2633} \[ \frac{4 a^4 \sin ^5(c+d x)}{5 d}-\frac{4 a^4 \sin ^3(c+d x)}{d}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{a^4 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{41 a^4 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{49 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{49 a^4 x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^4,x]

[Out]

(49*a^4*x)/16 + (8*a^4*Sin[c + d*x])/d + (49*a^4*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (41*a^4*Cos[c + d*x]^3*Si
n[c + d*x])/(24*d) + (a^4*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (4*a^4*Sin[c + d*x]^3)/d + (4*a^4*Sin[c + d*x]^
5)/(5*d)

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \cos (c+d x))^4 \, dx &=\int \left (a^4 \cos ^2(c+d x)+4 a^4 \cos ^3(c+d x)+6 a^4 \cos ^4(c+d x)+4 a^4 \cos ^5(c+d x)+a^4 \cos ^6(c+d x)\right ) \, dx\\ &=a^4 \int \cos ^2(c+d x) \, dx+a^4 \int \cos ^6(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^3(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^5(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^4(c+d x) \, dx\\ &=\frac{a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a^4 \cos ^3(c+d x) \sin (c+d x)}{2 d}+\frac{a^4 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{2} a^4 \int 1 \, dx+\frac{1}{6} \left (5 a^4\right ) \int \cos ^4(c+d x) \, dx+\frac{1}{2} \left (9 a^4\right ) \int \cos ^2(c+d x) \, dx-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{a^4 x}{2}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{11 a^4 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{41 a^4 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a^4 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{4 a^4 \sin ^3(c+d x)}{d}+\frac{4 a^4 \sin ^5(c+d x)}{5 d}+\frac{1}{8} \left (5 a^4\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{4} \left (9 a^4\right ) \int 1 \, dx\\ &=\frac{11 a^4 x}{4}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{49 a^4 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{41 a^4 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a^4 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{4 a^4 \sin ^3(c+d x)}{d}+\frac{4 a^4 \sin ^5(c+d x)}{5 d}+\frac{1}{16} \left (5 a^4\right ) \int 1 \, dx\\ &=\frac{49 a^4 x}{16}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{49 a^4 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{41 a^4 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a^4 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{4 a^4 \sin ^3(c+d x)}{d}+\frac{4 a^4 \sin ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.174498, size = 73, normalized size = 0.57 \[ \frac{a^4 (5280 \sin (c+d x)+1905 \sin (2 (c+d x))+720 \sin (3 (c+d x))+225 \sin (4 (c+d x))+48 \sin (5 (c+d x))+5 \sin (6 (c+d x))+2940 d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^4,x]

[Out]

(a^4*(2940*d*x + 5280*Sin[c + d*x] + 1905*Sin[2*(c + d*x)] + 720*Sin[3*(c + d*x)] + 225*Sin[4*(c + d*x)] + 48*
Sin[5*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.046, size = 169, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{4\,{a}^{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 ) }+6\,{a}^{4} \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 ) +{\frac{4\,{a}^{4} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{4} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+cos(d*x+c)*a)^4,x)

[Out]

1/d*(a^4*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+4/5*a^4*(8/3+cos(d*x
+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+6*a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+4/3*a^4*
(2+cos(d*x+c)^2)*sin(d*x+c)+a^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 1.11564, size = 223, normalized size = 1.76 \begin{align*} \frac{256 \,{\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 )} a^{4} - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 1280 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} + 180 \,{\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^{4} + 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/960*(256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^4 - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 6
0*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^4 - 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 + 180*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^4 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^4)/d

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Fricas [A]  time = 1.65537, size = 231, normalized size = 1.82 \begin{align*} \frac{735 \, a^{4} d x +{\left (40 \, a^{4} \cos \left (d x + c\right )^{5} + 192 \, a^{4} \cos \left (d x + c\right )^{4} + 410 \, a^{4} \cos \left (d x + c\right )^{3} + 576 \, a^{4} \cos \left (d x + c\right )^{2} + 735 \, a^{4} \cos \left (d x + c\right ) + 1152 \, a^{4}\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

1/240*(735*a^4*d*x + (40*a^4*cos(d*x + c)^5 + 192*a^4*cos(d*x + c)^4 + 410*a^4*cos(d*x + c)^3 + 576*a^4*cos(d*
x + c)^2 + 735*a^4*cos(d*x + c) + 1152*a^4)*sin(d*x + c))/d

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Sympy [A]  time = 4.89644, size = 434, normalized size = 3.42 \begin{align*} \begin{cases} \frac{5 a^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 a^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{9 a^{4} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{15 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{9 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{4} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{5 a^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{9 a^{4} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{a^{4} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{5 a^{4} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{32 a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{5 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{16 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{9 a^{4} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{8 a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{11 a^{4} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{4 a^{4} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{15 a^{4} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{4 a^{4} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{a^{4} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + a\right )^{4} \cos ^{2}{\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)**2*(a+a*cos(d*x+c))**4,x)

[Out]

Piecewise((5*a**4*x*sin(c + d*x)**6/16 + 15*a**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a**4*x*sin(c + d*x)*
*4/4 + 15*a**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 9*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + a**4*x*sin(
c + d*x)**2/2 + 5*a**4*x*cos(c + d*x)**6/16 + 9*a**4*x*cos(c + d*x)**4/4 + a**4*x*cos(c + d*x)**2/2 + 5*a**4*s
in(c + d*x)**5*cos(c + d*x)/(16*d) + 32*a**4*sin(c + d*x)**5/(15*d) + 5*a**4*sin(c + d*x)**3*cos(c + d*x)**3/(
6*d) + 16*a**4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 9*a**4*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 8*a**4*sin(
c + d*x)**3/(3*d) + 11*a**4*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 4*a**4*sin(c + d*x)*cos(c + d*x)**4/d + 15*a
**4*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 4*a**4*sin(c + d*x)*cos(c + d*x)**2/d + a**4*sin(c + d*x)*cos(c + d*x
)/(2*d), Ne(d, 0)), (x*(a*cos(c) + a)**4*cos(c)**2, True))

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Giac [A]  time = 1.38786, size = 143, normalized size = 1.13 \begin{align*} \frac{49}{16} \, a^{4} x + \frac{a^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{a^{4} \sin \left (5 \, d x + 5 \, c\right )}{20 \, d} + \frac{15 \, a^{4} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{3 \, a^{4} \sin \left (3 \, d x + 3 \, c\right )}{4 \, d} + \frac{127 \, a^{4} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{11 \, a^{4} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

49/16*a^4*x + 1/192*a^4*sin(6*d*x + 6*c)/d + 1/20*a^4*sin(5*d*x + 5*c)/d + 15/64*a^4*sin(4*d*x + 4*c)/d + 3/4*
a^4*sin(3*d*x + 3*c)/d + 127/64*a^4*sin(2*d*x + 2*c)/d + 11/2*a^4*sin(d*x + c)/d