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

Optimal. Leaf size=137 \[ -\frac{7 a^4 \cos ^3(c+d x)}{8 d}-\frac{3 \cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )^2}{10 d}-\frac{21 \cos ^3(c+d x) \left (a^4 \sin (c+d x)+a^4\right )}{40 d}+\frac{21 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{21 a^4 x}{16}-\frac{a \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

[Out]

(21*a^4*x)/16 - (7*a^4*Cos[c + d*x]^3)/(8*d) + (21*a^4*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a*Cos[c + d*x]^3*(
a + a*Sin[c + d*x])^3)/(6*d) - (3*Cos[c + d*x]^3*(a^2 + a^2*Sin[c + d*x])^2)/(10*d) - (21*Cos[c + d*x]^3*(a^4
+ a^4*Sin[c + d*x]))/(40*d)

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Rubi [A]  time = 0.157458, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2678, 2669, 2635, 8} \[ -\frac{7 a^4 \cos ^3(c+d x)}{8 d}-\frac{3 \cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )^2}{10 d}-\frac{21 \cos ^3(c+d x) \left (a^4 \sin (c+d x)+a^4\right )}{40 d}+\frac{21 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{21 a^4 x}{16}-\frac{a \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*a^4*x)/16 - (7*a^4*Cos[c + d*x]^3)/(8*d) + (21*a^4*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a*Cos[c + d*x]^3*(
a + a*Sin[c + d*x])^3)/(6*d) - (3*Cos[c + d*x]^3*(a^2 + a^2*Sin[c + d*x])^2)/(10*d) - (21*Cos[c + d*x]^3*(a^4
+ a^4*Sin[c + d*x]))/(40*d)

Rule 2678

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^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 8

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

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \sin (c+d x))^4 \, dx &=-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac{1}{2} (3 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}+\frac{1}{10} \left (21 a^2\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac{21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}+\frac{1}{8} \left (21 a^3\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{7 a^4 \cos ^3(c+d x)}{8 d}-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac{21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}+\frac{1}{8} \left (21 a^4\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{7 a^4 \cos ^3(c+d x)}{8 d}+\frac{21 a^4 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac{21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}+\frac{1}{16} \left (21 a^4\right ) \int 1 \, dx\\ &=\frac{21 a^4 x}{16}-\frac{7 a^4 \cos ^3(c+d x)}{8 d}+\frac{21 a^4 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac{21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}\\ \end{align*}

Mathematica [A]  time = 0.432605, size = 151, normalized size = 1.1 \[ -\frac{a^4 \left (630 \sqrt{1-\sin (c+d x)} \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )+\sqrt{\sin (c+d x)+1} \left (40 \sin ^6(c+d x)+152 \sin ^5(c+d x)+158 \sin ^4(c+d x)-94 \sin ^3(c+d x)-331 \sin ^2(c+d x)-373 \sin (c+d x)+448\right )\right ) \cos ^3(c+d x)}{240 d (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^4*Cos[c + d*x]^3*(630*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x
]]*(448 - 373*Sin[c + d*x] - 331*Sin[c + d*x]^2 - 94*Sin[c + d*x]^3 + 158*Sin[c + d*x]^4 + 152*Sin[c + d*x]^5
+ 40*Sin[c + d*x]^6)))/(240*d*(-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.055, size = 182, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{8}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16}}+{\frac{dx}{16}}+{\frac{c}{16}} \right ) +4\,{a}^{4} \left ( -1/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-2/15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3} \right ) +6\,{a}^{4} \left ( -1/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1/8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/8\,dx+c/8 \right ) -{\frac{4\,{a}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{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+a*sin(d*x+c))^4,x)

[Out]

1/d*(a^4*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*
c)+4*a^4*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+6*a^4*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)
*sin(d*x+c)+1/8*d*x+1/8*c)-4/3*a^4*cos(d*x+c)^3+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 = 1.12893, size = 173, normalized size = 1.26 \begin{align*} -\frac{1280 \, a^{4} \cos \left (d x + c\right )^{3} - 256 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{4} + 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} - 180 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, 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*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/960*(1280*a^4*cos(d*x + c)^3 - 256*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^4 + 5*(4*sin(2*d*x + 2*c)^3 - 12
*d*x - 12*c + 3*sin(4*d*x + 4*c))*a^4 - 180*(4*d*x + 4*c - sin(4*d*x + 4*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 = 1.64891, size = 215, normalized size = 1.57 \begin{align*} \frac{192 \, a^{4} \cos \left (d x + c\right )^{5} - 640 \, a^{4} \cos \left (d x + c\right )^{3} + 315 \, a^{4} d x + 5 \,{\left (8 \, a^{4} \cos \left (d x + c\right )^{5} - 86 \, a^{4} \cos \left (d x + c\right )^{3} + 63 \, a^{4} \cos \left (d x + c\right )\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*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/240*(192*a^4*cos(d*x + c)^5 - 640*a^4*cos(d*x + c)^3 + 315*a^4*d*x + 5*(8*a^4*cos(d*x + c)^5 - 86*a^4*cos(d*
x + c)^3 + 63*a^4*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 6.28719, size = 381, normalized size = 2.78 \begin{align*} \begin{cases} \frac{a^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 a^{4} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{3 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{3 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{a^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{4} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{a^{4} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{4} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} - \frac{a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{3 a^{4} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} - \frac{4 a^{4} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{a^{4} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{3 a^{4} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{a^{4} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{8 a^{4} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac{4 a^{4} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\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*sin(d*x+c))**4,x)

[Out]

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

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Giac [A]  time = 1.39885, size = 143, normalized size = 1.04 \begin{align*} \frac{21}{16} \, a^{4} x + \frac{a^{4} \cos \left (5 \, d x + 5 \, c\right )}{20 \, d} - \frac{5 \, a^{4} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac{3 \, a^{4} \cos \left (d x + c\right )}{2 \, d} + \frac{a^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{13 \, a^{4} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{15 \, a^{4} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

21/16*a^4*x + 1/20*a^4*cos(5*d*x + 5*c)/d - 5/12*a^4*cos(3*d*x + 3*c)/d - 3/2*a^4*cos(d*x + c)/d + 1/192*a^4*s
in(6*d*x + 6*c)/d - 13/64*a^4*sin(4*d*x + 4*c)/d + 15/64*a^4*sin(2*d*x + 2*c)/d