3.32 \(\int (a+b \sin (c+d x)) (A+B \sin (c+d x)+C \sin ^2(c+d x)) \, dx\)

Optimal. Leaf size=81 \[ -\frac{\cos (c+d x) (a B+A b+b C)}{d}+\frac{1}{2} x (a (2 A+C)+b B)-\frac{(a C+b B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{b C \cos ^3(c+d x)}{3 d} \]

[Out]

((b*B + a*(2*A + C))*x)/2 - ((A*b + a*B + b*C)*Cos[c + d*x])/d + (b*C*Cos[c + d*x]^3)/(3*d) - ((b*B + a*C)*Cos
[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.102217, antiderivative size = 113, normalized size of antiderivative = 1.4, number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {3023, 2734} \[ -\frac{\cos (c+d x) \left (a (3 b B-a C)+b^2 (3 A+2 C)\right )}{3 b d}+\frac{1}{2} x (a (2 A+C)+b B)-\frac{(3 b B-a C) \sin (c+d x) \cos (c+d x)}{6 d}-\frac{C \cos (c+d x) (a+b \sin (c+d x))^2}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B + a*(2*A + C))*x)/2 - ((b^2*(3*A + 2*C) + a*(3*b*B - a*C))*Cos[c + d*x])/(3*b*d) - ((3*b*B - a*C)*Cos[c
+ d*x]*Sin[c + d*x])/(6*d) - (C*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(3*b*d)

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int (a+b \sin (c+d x)) \left (A+B \sin (c+d x)+C \sin ^2(c+d x)\right ) \, dx &=-\frac{C \cos (c+d x) (a+b \sin (c+d x))^2}{3 b d}+\frac{\int (a+b \sin (c+d x)) (b (3 A+2 C)+(3 b B-a C) \sin (c+d x)) \, dx}{3 b}\\ &=\frac{1}{2} (b B+a (2 A+C)) x-\frac{\left (b^2 (3 A+2 C)+a (3 b B-a C)\right ) \cos (c+d x)}{3 b d}-\frac{(3 b B-a C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{C \cos (c+d x) (a+b \sin (c+d x))^2}{3 b d}\\ \end{align*}

Mathematica [A]  time = 0.207472, size = 92, normalized size = 1.14 \[ \frac{-3 \cos (c+d x) (4 a B+4 A b+3 b C)+12 a A d x-3 a C \sin (2 (c+d x))+6 a c C+6 a C d x-3 b B \sin (2 (c+d x))+6 b B c+6 b B d x+b C \cos (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*B*c + 6*a*c*C + 12*a*A*d*x + 6*b*B*d*x + 6*a*C*d*x - 3*(4*A*b + 4*a*B + 3*b*C)*Cos[c + d*x] + b*C*Cos[3*(
c + d*x)] - 3*b*B*Sin[2*(c + d*x)] - 3*a*C*Sin[2*(c + d*x)])/(12*d)

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Maple [A]  time = 0.034, size = 104, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{\frac{Cb \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}}+Bb \left ( -{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +aC \left ( -{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) -Ab\cos \left ( dx+c \right ) -Ba\cos \left ( dx+c \right ) +Aa \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.954225, size = 138, normalized size = 1.7 \begin{align*} \frac{12 \,{\left (d x + c\right )} A a + 3 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 3 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} B b + 4 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} C b - 12 \, B a \cos \left (d x + c\right ) - 12 \, A b \cos \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(d*x + c)*A*a + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*C*a + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*B*b + 4*(
cos(d*x + c)^3 - 3*cos(d*x + c))*C*b - 12*B*a*cos(d*x + c) - 12*A*b*cos(d*x + c))/d

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Fricas [A]  time = 1.67859, size = 182, normalized size = 2.25 \begin{align*} \frac{2 \, C b \cos \left (d x + c\right )^{3} + 3 \,{\left ({\left (2 \, A + C\right )} a + B b\right )} d x - 3 \,{\left (C a + B b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \,{\left (B a +{\left (A + C\right )} b\right )} \cos \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*C*b*cos(d*x + c)^3 + 3*((2*A + C)*a + B*b)*d*x - 3*(C*a + B*b)*cos(d*x + c)*sin(d*x + c) - 6*(B*a + (A
+ C)*b)*cos(d*x + c))/d

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Sympy [A]  time = 1.7698, size = 189, normalized size = 2.33 \begin{align*} \begin{cases} A a x - \frac{A b \cos{\left (c + d x \right )}}{d} - \frac{B a \cos{\left (c + d x \right )}}{d} + \frac{B b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{B b x \cos ^{2}{\left (c + d x \right )}}{2} - \frac{B b \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{C a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{C a x \cos ^{2}{\left (c + d x \right )}}{2} - \frac{C a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{C b \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{2 C b \cos ^{3}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right ) \left (A + B \sin{\left (c \right )} + C \sin ^{2}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)**2),x)

[Out]

Piecewise((A*a*x - A*b*cos(c + d*x)/d - B*a*cos(c + d*x)/d + B*b*x*sin(c + d*x)**2/2 + B*b*x*cos(c + d*x)**2/2
- B*b*sin(c + d*x)*cos(c + d*x)/(2*d) + C*a*x*sin(c + d*x)**2/2 + C*a*x*cos(c + d*x)**2/2 - C*a*sin(c + d*x)*
cos(c + d*x)/(2*d) - C*b*sin(c + d*x)**2*cos(c + d*x)/d - 2*C*b*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a + b*si
n(c))*(A + B*sin(c) + C*sin(c)**2), True))

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Giac [A]  time = 1.17297, size = 103, normalized size = 1.27 \begin{align*} \frac{1}{2} \,{\left (2 \, A a + C a + B b\right )} x + \frac{C b \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac{{\left (4 \, B a + 4 \, A b + 3 \, C b\right )} \cos \left (d x + c\right )}{4 \, d} - \frac{{\left (C a + B b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))*(A+B*sin(d*x+c)+C*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/2*(2*A*a + C*a + B*b)*x + 1/12*C*b*cos(3*d*x + 3*c)/d - 1/4*(4*B*a + 4*A*b + 3*C*b)*cos(d*x + c)/d - 1/4*(C*
a + B*b)*sin(2*d*x + 2*c)/d