3.1.32 \(\int (a+b \sin (c+d x)) (A+B \sin (c+d x)+C \sin ^2(c+d x)) \, dx\) [32]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.40, 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 = {3102, 2813} \begin {gather*} -\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} \end {gather*}

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 2813

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]

Rule 3102

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]

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*}

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Mathematica [A]
time = 0.14, size = 92, normalized size = 1.14 \begin {gather*} \frac {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} \end {gather*}

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 = 0.22, size = 104, normalized size = 1.28

method	result	size
risch	\(a x A +\frac {x B b}{2}+\frac {x a C}{2}-\frac {\cos \left (d x +c \right ) A b}{d}-\frac {\cos \left (d x +c \right ) a B}{d}-\frac {3 \cos \left (d x +c \right ) C b}{4 d}+\frac {b C \cos \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (2 d x +2 c \right ) B b}{4 d}-\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\)	\(103\)
derivativedivides	\(\frac {-\frac {C b \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+B b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a C \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A b \cos \left (d x +c \right )-a B \cos \left (d x +c \right )+a A \left (d x +c \right )}{d}\)	\(104\)
default	\(\frac {-\frac {C b \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+B b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a C \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-A b \cos \left (d x +c \right )-a B \cos \left (d x +c \right )+a A \left (d x +c \right )}{d}\)	\(104\)
norman	\(\frac {\left (a A +\frac {1}{2} B b +\frac {1}{2} a C \right ) x +\left (a A +\frac {1}{2} B b +\frac {1}{2} a C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +\frac {3}{2} B b +\frac {3}{2} a C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +\frac {3}{2} B b +\frac {3}{2} a C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (B b +a C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 A b +6 a B +4 C b}{3 d}-\frac {\left (B b +a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 A b +2 a B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (4 A b +4 a B +4 C b \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(224\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 102, normalized size = 1.26 \begin {gather*} \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 {gather*}

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 = 0.39, size = 71, normalized size = 0.88 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (73) = 146\).
time = 0.14, size = 189, normalized size = 2.33 \begin {gather*} \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 {gather*}

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 = 0.45, size = 76, normalized size = 0.94 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 13.76, size = 93, normalized size = 1.15 \begin {gather*} -\frac {6\,A\,b\,\cos \left (c+d\,x\right )+6\,B\,a\,\cos \left (c+d\,x\right )+\frac {9\,C\,b\,\cos \left (c+d\,x\right )}{2}-\frac {C\,b\,\cos \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,B\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,C\,a\,\sin \left (2\,c+2\,d\,x\right )}{2}-6\,A\,a\,d\,x-3\,B\,b\,d\,x-3\,C\,a\,d\,x}{6\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*A*b*cos(c + d*x) + 6*B*a*cos(c + d*x) + (9*C*b*cos(c + d*x))/2 - (C*b*cos(3*c + 3*d*x))/2 + (3*B*b*sin(2*c
+ 2*d*x))/2 + (3*C*a*sin(2*c + 2*d*x))/2 - 6*A*a*d*x - 3*B*b*d*x - 3*C*a*d*x)/(6*d)

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