3.1530 \(\int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=97 \[ -\frac {(a B+A b) \sin ^4(c+d x)}{4 d}-\frac {(a A-b B) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin ^2(c+d x)}{2 d}+\frac {a A \sin (c+d x)}{d}-\frac {b B \sin ^5(c+d x)}{5 d} \]

[Out]

a*A*sin(d*x+c)/d+1/2*(A*b+B*a)*sin(d*x+c)^2/d-1/3*(A*a-B*b)*sin(d*x+c)^3/d-1/4*(A*b+B*a)*sin(d*x+c)^4/d-1/5*b*
B*sin(d*x+c)^5/d

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Rubi [A]  time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2837, 772} \[ -\frac {(a B+A b) \sin ^4(c+d x)}{4 d}-\frac {(a A-b B) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin ^2(c+d x)}{2 d}+\frac {a A \sin (c+d x)}{d}-\frac {b B \sin ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*Sin[c + d*x])/d + ((A*b + a*B)*Sin[c + d*x]^2)/(2*d) - ((a*A - b*B)*Sin[c + d*x]^3)/(3*d) - ((A*b + a*B)*
Sin[c + d*x]^4)/(4*d) - (b*B*Sin[c + d*x]^5)/(5*d)

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int (a+x) \left (A+\frac {B x}{b}\right ) \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a A b^2+b (A b+a B) x-(a A-b B) x^2-\frac {(A b+a B) x^3}{b}-\frac {B x^4}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {a A \sin (c+d x)}{d}+\frac {(A b+a B) \sin ^2(c+d x)}{2 d}-\frac {(a A-b B) \sin ^3(c+d x)}{3 d}-\frac {(A b+a B) \sin ^4(c+d x)}{4 d}-\frac {b B \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 80, normalized size = 0.82 \[ \frac {\sin (c+d x) \left (-15 (a B+A b) \sin ^3(c+d x)-20 (a A-b B) \sin ^2(c+d x)+30 (a B+A b) \sin (c+d x)+60 a A-12 b B \sin ^4(c+d x)\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]*(60*a*A + 30*(A*b + a*B)*Sin[c + d*x] - 20*(a*A - b*B)*Sin[c + d*x]^2 - 15*(A*b + a*B)*Sin[c + d
*x]^3 - 12*b*B*Sin[c + d*x]^4))/(60*d)

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fricas [A]  time = 0.45, size = 70, normalized size = 0.72 \[ -\frac {15 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B b \cos \left (d x + c\right )^{4} - {\left (5 \, A a + B b\right )} \cos \left (d x + c\right )^{2} - 10 \, A a - 2 \, B b\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(15*(B*a + A*b)*cos(d*x + c)^4 + 4*(3*B*b*cos(d*x + c)^4 - (5*A*a + B*b)*cos(d*x + c)^2 - 10*A*a - 2*B*b
)*sin(d*x + c))/d

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giac [A]  time = 0.20, size = 100, normalized size = 1.03 \[ -\frac {12 \, B b \sin \left (d x + c\right )^{5} + 15 \, B a \sin \left (d x + c\right )^{4} + 15 \, A b \sin \left (d x + c\right )^{4} + 20 \, A a \sin \left (d x + c\right )^{3} - 20 \, B b \sin \left (d x + c\right )^{3} - 30 \, B a \sin \left (d x + c\right )^{2} - 30 \, A b \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(12*B*b*sin(d*x + c)^5 + 15*B*a*sin(d*x + c)^4 + 15*A*b*sin(d*x + c)^4 + 20*A*a*sin(d*x + c)^3 - 20*B*b*
sin(d*x + c)^3 - 30*B*a*sin(d*x + c)^2 - 30*A*b*sin(d*x + c)^2 - 60*A*a*sin(d*x + c))/d

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maple [A]  time = 0.46, size = 88, normalized size = 0.91 \[ \frac {B b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {A b \left (\cos ^{4}\left (d x +c \right )\right )}{4}-\frac {a B \left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

1/d*(B*b*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-1/4*A*b*cos(d*x+c)^4-1/4*a*B*cos(d*x+
c)^4+1/3*a*A*(2+cos(d*x+c)^2)*sin(d*x+c))

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maxima [A]  time = 1.13, size = 80, normalized size = 0.82 \[ -\frac {12 \, B b \sin \left (d x + c\right )^{5} + 15 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{4} + 20 \, {\left (A a - B b\right )} \sin \left (d x + c\right )^{3} - 60 \, A a \sin \left (d x + c\right ) - 30 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(12*B*b*sin(d*x + c)^5 + 15*(B*a + A*b)*sin(d*x + c)^4 + 20*(A*a - B*b)*sin(d*x + c)^3 - 60*A*a*sin(d*x
+ c) - 30*(B*a + A*b)*sin(d*x + c)^2)/d

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mupad [B]  time = 11.99, size = 83, normalized size = 0.86 \[ -\frac {\frac {B\,b\,{\sin \left (c+d\,x\right )}^5}{5}+\left (\frac {A\,b}{4}+\frac {B\,a}{4}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {A\,a}{3}-\frac {B\,b}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (-\frac {A\,b}{2}-\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^2-A\,a\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + B*sin(c + d*x))*(a + b*sin(c + d*x)),x)

[Out]

-(sin(c + d*x)^3*((A*a)/3 - (B*b)/3) - sin(c + d*x)^2*((A*b)/2 + (B*a)/2) + sin(c + d*x)^4*((A*b)/4 + (B*a)/4)
- A*a*sin(c + d*x) + (B*b*sin(c + d*x)^5)/5)/d

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sympy [A]  time = 1.90, size = 128, normalized size = 1.32 \[ \begin {cases} \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A b \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {B a \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {B b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a + b \sin {\relax (c )}\right ) \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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