3.997 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=159 \[ -\frac{7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}-\frac{7 (2 A+B) \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{40 d}+\frac{7 a^3 (2 A+B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{7}{16} a^3 x (2 A+B)-\frac{a (2 A+B) \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{10 d}-\frac{B \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

[Out]

(7*a^3*(2*A + B)*x)/16 - (7*a^3*(2*A + B)*Cos[c + d*x]^3)/(24*d) + (7*a^3*(2*A + B)*Cos[c + d*x]*Sin[c + d*x])
/(16*d) - (a*(2*A + B)*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(10*d) - (B*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^
3)/(6*d) - (7*(2*A + B)*Cos[c + d*x]^3*(a^3 + a^3*Sin[c + d*x]))/(40*d)

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Rubi [A]  time = 0.21743, antiderivative size = 159, 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} \[ -\frac{7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}-\frac{7 (2 A+B) \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{40 d}+\frac{7 a^3 (2 A+B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{7}{16} a^3 x (2 A+B)-\frac{a (2 A+B) \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{10 d}-\frac{B \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(7*a^3*(2*A + B)*x)/16 - (7*a^3*(2*A + B)*Cos[c + d*x]^3)/(24*d) + (7*a^3*(2*A + B)*Cos[c + d*x]*Sin[c + d*x])
/(16*d) - (a*(2*A + B)*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(10*d) - (B*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^
3)/(6*d) - (7*(2*A + B)*Cos[c + d*x]^3*(a^3 + a^3*Sin[c + d*x]))/(40*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 ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=-\frac{B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac{1}{2} (2 A+B) \int \cos ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=-\frac{a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac{B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac{1}{10} (7 a (2 A+B)) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac{B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac{1}{8} \left (7 a^2 (2 A+B)\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}-\frac{a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac{B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac{1}{8} \left (7 a^3 (2 A+B)\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}+\frac{7 a^3 (2 A+B) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac{B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}+\frac{1}{16} \left (7 a^3 (2 A+B)\right ) \int 1 \, dx\\ &=\frac{7}{16} a^3 (2 A+B) x-\frac{7 a^3 (2 A+B) \cos ^3(c+d x)}{24 d}+\frac{7 a^3 (2 A+B) \cos (c+d x) \sin (c+d x)}{16 d}-\frac{a (2 A+B) \cos ^3(c+d x) (a+a \sin (c+d x))^2}{10 d}-\frac{B \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac{7 (2 A+B) \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{40 d}\\ \end{align*}

Mathematica [A]  time = 1.36457, size = 146, normalized size = 0.92 \[ -\frac{a^3 \cos (c+d x) \left (16 (17 A+11 B) \cos (2 (c+d x))-12 (A+3 B) \cos (4 (c+d x))+\frac{420 (2 A+B) \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )}{\sqrt{\cos ^2(c+d x)}}-330 A \sin (c+d x)+90 A \sin (3 (c+d x))+284 A-95 B \sin (c+d x)+110 B \sin (3 (c+d x))-5 B \sin (5 (c+d x))+212 B\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-(a^3*Cos[c + d*x]*(284*A + 212*B + (420*(2*A + B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]])/Sqrt[Cos[c + d*x]^2
] + 16*(17*A + 11*B)*Cos[2*(c + d*x)] - 12*(A + 3*B)*Cos[4*(c + d*x)] - 330*A*Sin[c + d*x] - 95*B*Sin[c + d*x]
+ 90*A*Sin[3*(c + d*x)] + 110*B*Sin[3*(c + d*x)] - 5*B*Sin[5*(c + d*x)]))/(480*d)

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Maple [A]  time = 0.069, size = 279, normalized size = 1.8 \begin{align*}{\frac{1}{d} \left ({a}^{3}A \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +B{a}^{3} \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 ) +3\,{a}^{3}A \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 ) +3\,B{a}^{3} \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 ) -{a}^{3}A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3\,B{a}^{3} \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 ) +{a}^{3}A \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\frac{B{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}} \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))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+B*a^3*(-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)+3*a^3*A*(-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)+3*B*a^3*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)-a^3*A*cos(d*x+c)^3+3
*B*a^3*(-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)+a^3*A*(1/2*cos(d*x+c)*sin(d*x+c)
+1/2*d*x+1/2*c)-1/3*B*a^3*cos(d*x+c)^3)

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Maxima [A]  time = 1.13797, size = 269, normalized size = 1.69 \begin{align*} -\frac{960 \, A a^{3} \cos \left (d x + c\right )^{3} + 320 \, B a^{3} \cos \left (d x + c\right )^{3} - 64 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} A a^{3} - 90 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} A a^{3} - 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 192 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} B a^{3} + 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 )} B a^{3} - 90 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{3}}{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))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/960*(960*A*a^3*cos(d*x + c)^3 + 320*B*a^3*cos(d*x + c)^3 - 64*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*A*a^3 -
90*(4*d*x + 4*c - sin(4*d*x + 4*c))*A*a^3 - 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 192*(3*cos(d*x + c)^
5 - 5*cos(d*x + c)^3)*B*a^3 + 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*B*a^3 - 90*(4*d*x
+ 4*c - sin(4*d*x + 4*c))*B*a^3)/d

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Fricas [A]  time = 1.83732, size = 285, normalized size = 1.79 \begin{align*} \frac{48 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} - 320 \,{\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{3} + 105 \,{\left (2 \, A + B\right )} a^{3} d x + 5 \,{\left (8 \, B a^{3} \cos \left (d x + c\right )^{5} - 2 \,{\left (18 \, A + 25 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 21 \,{\left (2 \, A + B\right )} a^{3} \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))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(48*(A + 3*B)*a^3*cos(d*x + c)^5 - 320*(A + B)*a^3*cos(d*x + c)^3 + 105*(2*A + B)*a^3*d*x + 5*(8*B*a^3*c
os(d*x + c)^5 - 2*(18*A + 25*B)*a^3*cos(d*x + c)^3 + 21*(2*A + B)*a^3*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 5.75516, size = 588, normalized size = 3.7 \begin{align*} \text{result too large to display} \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))**3*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.3148, size = 223, normalized size = 1.4 \begin{align*} \frac{B a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{7}{16} \,{\left (2 \, A a^{3} + B a^{3}\right )} x + \frac{{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{{\left (13 \, A a^{3} + 7 \, B a^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{{\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (d x + c\right )}{8 \, d} - \frac{{\left (6 \, A a^{3} + 7 \, B a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (16 \, A a^{3} - B a^{3}\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)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*B*a^3*sin(6*d*x + 6*c)/d + 7/16*(2*A*a^3 + B*a^3)*x + 1/80*(A*a^3 + 3*B*a^3)*cos(5*d*x + 5*c)/d - 1/48*(
13*A*a^3 + 7*B*a^3)*cos(3*d*x + 3*c)/d - 1/8*(7*A*a^3 + 5*B*a^3)*cos(d*x + c)/d - 1/64*(6*A*a^3 + 7*B*a^3)*sin
(4*d*x + 4*c)/d + 1/64*(16*A*a^3 - B*a^3)*sin(2*d*x + 2*c)/d