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

Optimal. Leaf size=165 \[ -\frac{a^2 (7 A+2 B) \cos ^5(c+d x)}{30 d}-\frac{(7 A+2 B) \cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{42 d}+\frac{a^2 (7 A+2 B) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{a^2 (7 A+2 B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a^2 x (7 A+2 B)-\frac{B \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

[Out]

(a^2*(7*A + 2*B)*x)/16 - (a^2*(7*A + 2*B)*Cos[c + d*x]^5)/(30*d) + (a^2*(7*A + 2*B)*Cos[c + d*x]*Sin[c + d*x])
/(16*d) + (a^2*(7*A + 2*B)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (B*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(7*
d) - ((7*A + 2*B)*Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(42*d)

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Rubi [A]  time = 0.189213, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2860, 2678, 2669, 2635, 8} \[ -\frac{a^2 (7 A+2 B) \cos ^5(c+d x)}{30 d}-\frac{(7 A+2 B) \cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{42 d}+\frac{a^2 (7 A+2 B) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{a^2 (7 A+2 B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a^2 x (7 A+2 B)-\frac{B \cos ^5(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*A + 2*B)*x)/16 - (a^2*(7*A + 2*B)*Cos[c + d*x]^5)/(30*d) + (a^2*(7*A + 2*B)*Cos[c + d*x]*Sin[c + d*x])
/(16*d) + (a^2*(7*A + 2*B)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (B*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(7*
d) - ((7*A + 2*B)*Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(42*d)

Rule 2860

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

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 ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac{1}{7} (7 A+2 B) \int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{(7 A+2 B) \cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{42 d}+\frac{1}{6} (a (7 A+2 B)) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{a^2 (7 A+2 B) \cos ^5(c+d x)}{30 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{(7 A+2 B) \cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{42 d}+\frac{1}{6} \left (a^2 (7 A+2 B)\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac{a^2 (7 A+2 B) \cos ^5(c+d x)}{30 d}+\frac{a^2 (7 A+2 B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{(7 A+2 B) \cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{42 d}+\frac{1}{8} \left (a^2 (7 A+2 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{a^2 (7 A+2 B) \cos ^5(c+d x)}{30 d}+\frac{a^2 (7 A+2 B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 (7 A+2 B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{(7 A+2 B) \cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{42 d}+\frac{1}{16} \left (a^2 (7 A+2 B)\right ) \int 1 \, dx\\ &=\frac{1}{16} a^2 (7 A+2 B) x-\frac{a^2 (7 A+2 B) \cos ^5(c+d x)}{30 d}+\frac{a^2 (7 A+2 B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 (7 A+2 B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{(7 A+2 B) \cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{42 d}\\ \end{align*}

Mathematica [A]  time = 1.70164, size = 171, normalized size = 1.04 \[ -\frac{a^2 \cos (c+d x) \left ((672 A+447 B) \cos (2 (c+d x))+6 (28 A+13 B) \cos (4 (c+d x))+\frac{420 (7 A+2 B) \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )}{\sqrt{\cos ^2(c+d x)}}-1645 A \sin (c+d x)-140 A \sin (3 (c+d x))+35 A \sin (5 (c+d x))+504 A-350 B \sin (c+d x)+140 B \sin (3 (c+d x))+70 B \sin (5 (c+d x))-15 B \cos (6 (c+d x))+354 B\right )}{3360 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Cos[c + d*x]*(504*A + 354*B + (420*(7*A + 2*B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]])/Sqrt[Cos[c + d*x]
^2] + (672*A + 447*B)*Cos[2*(c + d*x)] + 6*(28*A + 13*B)*Cos[4*(c + d*x)] - 15*B*Cos[6*(c + d*x)] - 1645*A*Sin
[c + d*x] - 350*B*Sin[c + d*x] - 140*A*Sin[3*(c + d*x)] + 140*B*Sin[3*(c + d*x)] + 35*A*Sin[5*(c + d*x)] + 70*
B*Sin[5*(c + d*x)]))/(3360*d)

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Maple [A]  time = 0.073, size = 215, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{\sin \left ( dx+c \right ) }{24} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{dx}{16}}+{\frac{c}{16}} \right ) +B{a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{7}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) -{\frac{2\,{a}^{2}A \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+2\,B{a}^{2} \left ( -1/6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) +{a}^{2}A \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) -{\frac{B{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*A*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)+B*a^2*
(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)-2/5*a^2*A*cos(d*x+c)^5+2*B*a^2*(-1/6*sin(d*x+c)*cos(d*x+c)^
5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)+a^2*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(
d*x+c)+3/8*d*x+3/8*c)-1/5*B*a^2*cos(d*x+c)^5)

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Maxima [A]  time = 1.05968, size = 231, normalized size = 1.4 \begin{align*} -\frac{2688 \, A a^{2} \cos \left (d x + c\right )^{5} + 1344 \, B a^{2} \cos \left (d x + c\right )^{5} - 35 \,{\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 a^{2} - 210 \,{\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 a^{2} - 192 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} B a^{2} - 70 \,{\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 )} B a^{2}}{6720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6720*(2688*A*a^2*cos(d*x + c)^5 + 1344*B*a^2*cos(d*x + c)^5 - 35*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*
sin(4*d*x + 4*c))*A*a^2 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 - 192*(5*cos(d*x +
c)^7 - 7*cos(d*x + c)^5)*B*a^2 - 70*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*B*a^2)/d

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Fricas [A]  time = 2.03827, size = 290, normalized size = 1.76 \begin{align*} \frac{240 \, B a^{2} \cos \left (d x + c\right )^{7} - 672 \,{\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{5} + 105 \,{\left (7 \, A + 2 \, B\right )} a^{2} d x - 35 \,{\left (8 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} - 2 \,{\left (7 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} - 3 \,{\left (7 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1680*(240*B*a^2*cos(d*x + c)^7 - 672*(A + B)*a^2*cos(d*x + c)^5 + 105*(7*A + 2*B)*a^2*d*x - 35*(8*(A + 2*B)*
a^2*cos(d*x + c)^5 - 2*(7*A + 2*B)*a^2*cos(d*x + c)^3 - 3*(7*A + 2*B)*a^2*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 9.31177, size = 539, normalized size = 3.27 \begin{align*} \begin{cases} \frac{A a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 A a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 A a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 A a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{3 A a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{A a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 A a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{A a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{3 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{A a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{5 A a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{2 A a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{B a^{2} x \sin ^{6}{\left (c + d x \right )}}{8} + \frac{3 B a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac{3 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac{B a^{2} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac{B a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{B a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{B a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac{2 B a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac{B a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a \sin{\left (c \right )} + a\right )^{2} \cos ^{4}{\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)**4*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((A*a**2*x*sin(c + d*x)**6/16 + 3*A*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*A*a**2*x*sin(c + d*
x)**4/8 + 3*A*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*A*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + A*a**
2*x*cos(c + d*x)**6/16 + 3*A*a**2*x*cos(c + d*x)**4/8 + A*a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + A*a**2*si
n(c + d*x)**3*cos(c + d*x)**3/(6*d) + 3*A*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - A*a**2*sin(c + d*x)*cos(c
+ d*x)**5/(16*d) + 5*A*a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*A*a**2*cos(c + d*x)**5/(5*d) + B*a**2*x*sin
(c + d*x)**6/8 + 3*B*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 3*B*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/8 +
B*a**2*x*cos(c + d*x)**6/8 + B*a**2*sin(c + d*x)**5*cos(c + d*x)/(8*d) + B*a**2*sin(c + d*x)**3*cos(c + d*x)*
*3/(3*d) - B*a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - B*a**2*sin(c + d*x)*cos(c + d*x)**5/(8*d) - 2*B*a**2
*cos(c + d*x)**7/(35*d) - B*a**2*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos(c)*
*4, True))

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Giac [A]  time = 1.30595, size = 259, normalized size = 1.57 \begin{align*} \frac{B a^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{1}{16} \,{\left (7 \, A a^{2} + 2 \, B a^{2}\right )} x - \frac{{\left (8 \, A a^{2} + 3 \, B a^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{{\left (8 \, A a^{2} + 5 \, B a^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac{{\left (16 \, A a^{2} + 11 \, B a^{2}\right )} \cos \left (d x + c\right )}{64 \, d} - \frac{{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{{\left (A a^{2} - 2 \, B a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (17 \, A a^{2} + 2 \, B a^{2}\right )} \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)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/448*B*a^2*cos(7*d*x + 7*c)/d + 1/16*(7*A*a^2 + 2*B*a^2)*x - 1/320*(8*A*a^2 + 3*B*a^2)*cos(5*d*x + 5*c)/d - 1
/64*(8*A*a^2 + 5*B*a^2)*cos(3*d*x + 3*c)/d - 1/64*(16*A*a^2 + 11*B*a^2)*cos(d*x + c)/d - 1/192*(A*a^2 + 2*B*a^
2)*sin(6*d*x + 6*c)/d + 1/64*(A*a^2 - 2*B*a^2)*sin(4*d*x + 4*c)/d + 1/64*(17*A*a^2 + 2*B*a^2)*sin(2*d*x + 2*c)
/d