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]

49/16*a^4*x+8*a^4*sin(d*x+c)/d+49/16*a^4*cos(d*x+c)*sin(d*x+c)/d+41/24*a^4*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a^4*c
os(d*x+c)^5*sin(d*x+c)/d-4*a^4*sin(d*x+c)^3/d+4/5*a^4*sin(d*x+c)^5/d

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Rubi [A]  time = 0.16, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.20, 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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fricas [A]  time = 0.91, size = 89, normalized size = 0.70 \[ \frac {735 \, a^{4} d x + {\left (40 \, a^{4} \cos \left (d x + c\right )^{5} + 192 \, a^{4} \cos \left (d x + c\right )^{4} + 410 \, a^{4} \cos \left (d x + c\right )^{3} + 576 \, a^{4} \cos \left (d x + c\right )^{2} + 735 \, a^{4} \cos \left (d x + c\right ) + 1152 \, a^{4}\right )} \sin \left (d x + c\right )}{240 \, d} \]

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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giac [A]  time = 0.56, size = 106, normalized size = 0.83 \[ \frac {49}{16} \, a^{4} x + \frac {a^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a^{4} \sin \left (5 \, d x + 5 \, c\right )}{20 \, d} + \frac {15 \, a^{4} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {3 \, a^{4} \sin \left (3 \, d x + 3 \, c\right )}{4 \, d} + \frac {127 \, a^{4} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {11 \, a^{4} \sin \left (d x + c\right )}{2 \, d} \]

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

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maple [A]  time = 0.05, size = 169, normalized size = 1.33 \[ \frac {a^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {4 a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+6 a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*cos(d*x+c))^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 = 1.75, size = 165, normalized size = 1.30 \[ \frac {256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} + 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4}}{960 \, d} \]

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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mupad [B]  time = 2.85, size = 121, normalized size = 0.95 \[ \frac {49\,a^4\,x}{16}+\frac {\frac {49\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {833\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {1617\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}+\frac {1967\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {1471\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {207\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(49*a^4*x)/16 + ((1471*a^4*tan(c/2 + (d*x)/2)^3)/24 + (1967*a^4*tan(c/2 + (d*x)/2)^5)/20 + (1617*a^4*tan(c/2 +
(d*x)/2)^7)/20 + (833*a^4*tan(c/2 + (d*x)/2)^9)/24 + (49*a^4*tan(c/2 + (d*x)/2)^11)/8 + (207*a^4*tan(c/2 + (d
*x)/2))/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)

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sympy [A]  time = 3.87, size = 434, normalized size = 3.42 \[ \begin {cases} \frac {5 a^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{4} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {15 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {9 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{4} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {5 a^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{4} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {a^{4} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {5 a^{4} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {32 a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {5 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {16 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {8 a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {11 a^{4} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {4 a^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {4 a^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{4} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

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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