3.1529 \(\int \cos ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=143 \[ \frac {(a B+A b) \sin ^6(c+d x)}{6 d}+\frac {(a A-2 b B) \sin ^5(c+d x)}{5 d}-\frac {(a B+A b) \sin ^4(c+d x)}{2 d}-\frac {(2 a A-b B) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin ^2(c+d x)}{2 d}+\frac {a A \sin (c+d x)}{d}+\frac {b B \sin ^7(c+d x)}{7 d} \]

[Out]

a*A*sin(d*x+c)/d+1/2*(A*b+B*a)*sin(d*x+c)^2/d-1/3*(2*A*a-B*b)*sin(d*x+c)^3/d-1/2*(A*b+B*a)*sin(d*x+c)^4/d+1/5*
(A*a-2*B*b)*sin(d*x+c)^5/d+1/6*(A*b+B*a)*sin(d*x+c)^6/d+1/7*b*B*sin(d*x+c)^7/d

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Rubi [A]  time = 0.17, antiderivative size = 143, 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} \[ \frac {(a B+A b) \sin ^6(c+d x)}{6 d}+\frac {(a A-2 b B) \sin ^5(c+d x)}{5 d}-\frac {(a B+A b) \sin ^4(c+d x)}{2 d}-\frac {(2 a A-b B) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin ^2(c+d x)}{2 d}+\frac {a A \sin (c+d x)}{d}+\frac {b B \sin ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(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) - ((2*a*A - b*B)*Sin[c + d*x]^3)/(3*d) - ((A*b + a*B
)*Sin[c + d*x]^4)/(2*d) + ((a*A - 2*b*B)*Sin[c + d*x]^5)/(5*d) + ((A*b + a*B)*Sin[c + d*x]^6)/(6*d) + (b*B*Sin
[c + d*x]^7)/(7*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 ^5(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 )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a A b^4+b^3 (A b+a B) x+b^2 (-2 a A+b B) x^2-2 b (A b+a B) x^3+(a A-2 b B) x^4+\frac {(A b+a B) x^5}{b}+\frac {B x^6}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {a A \sin (c+d x)}{d}+\frac {(A b+a B) \sin ^2(c+d x)}{2 d}-\frac {(2 a A-b B) \sin ^3(c+d x)}{3 d}-\frac {(A b+a B) \sin ^4(c+d x)}{2 d}+\frac {(a A-2 b B) \sin ^5(c+d x)}{5 d}+\frac {(A b+a B) \sin ^6(c+d x)}{6 d}+\frac {b B \sin ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 116, normalized size = 0.81 \[ \frac {\sin (c+d x) \left (35 (a B+A b) \sin ^5(c+d x)+42 (a A-2 b B) \sin ^4(c+d x)-105 (a B+A b) \sin ^3(c+d x)-70 (2 a A-b B) \sin ^2(c+d x)+105 (a B+A b) \sin (c+d x)+210 a A+30 b B \sin ^6(c+d x)\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(Sin[c + d*x]*(210*a*A + 105*(A*b + a*B)*Sin[c + d*x] - 70*(2*a*A - b*B)*Sin[c + d*x]^2 - 105*(A*b + a*B)*Sin[
c + d*x]^3 + 42*(a*A - 2*b*B)*Sin[c + d*x]^4 + 35*(A*b + a*B)*Sin[c + d*x]^5 + 30*b*B*Sin[c + d*x]^6))/(210*d)

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fricas [A]  time = 0.46, size = 88, normalized size = 0.62 \[ -\frac {35 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{6} + 2 \, {\left (15 \, B b \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, A a + B b\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, A a + B b\right )} \cos \left (d x + c\right )^{2} - 56 \, A a - 8 \, B b\right )} \sin \left (d x + c\right )}{210 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/210*(35*(B*a + A*b)*cos(d*x + c)^6 + 2*(15*B*b*cos(d*x + c)^6 - 3*(7*A*a + B*b)*cos(d*x + c)^4 - 4*(7*A*a +
B*b)*cos(d*x + c)^2 - 56*A*a - 8*B*b)*sin(d*x + c))/d

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giac [A]  time = 0.24, size = 145, normalized size = 1.01 \[ -\frac {B b \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {{\left (B a + A b\right )} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {{\left (B a + A b\right )} \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, {\left (B a + A b\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (4 \, A a - 3 \, B b\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, A a - B b\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, A a + B b\right )} \sin \left (d x + c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/448*B*b*sin(7*d*x + 7*c)/d - 1/192*(B*a + A*b)*cos(6*d*x + 6*c)/d - 1/32*(B*a + A*b)*cos(4*d*x + 4*c)/d - 5
/64*(B*a + A*b)*cos(2*d*x + 2*c)/d + 1/320*(4*A*a - 3*B*b)*sin(5*d*x + 5*c)/d + 1/192*(20*A*a - B*b)*sin(3*d*x
+ 3*c)/d + 5/64*(8*A*a + B*b)*sin(d*x + c)/d

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maple [A]  time = 0.46, size = 108, normalized size = 0.76 \[ \frac {B b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {A b \left (\cos ^{6}\left (d x +c \right )\right )}{6}-\frac {a B \left (\cos ^{6}\left (d x +c \right )\right )}{6}+\frac {a A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.53, size = 116, normalized size = 0.81 \[ \frac {30 \, B b \sin \left (d x + c\right )^{7} + 35 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{6} + 42 \, {\left (A a - 2 \, B b\right )} \sin \left (d x + c\right )^{5} - 105 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{4} - 70 \, {\left (2 \, A a - B b\right )} \sin \left (d x + c\right )^{3} + 210 \, A a \sin \left (d x + c\right ) + 105 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{2}}{210 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/210*(30*B*b*sin(d*x + c)^7 + 35*(B*a + A*b)*sin(d*x + c)^6 + 42*(A*a - 2*B*b)*sin(d*x + c)^5 - 105*(B*a + A*
b)*sin(d*x + c)^4 - 70*(2*A*a - B*b)*sin(d*x + c)^3 + 210*A*a*sin(d*x + c) + 105*(B*a + A*b)*sin(d*x + c)^2)/d

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mupad [B]  time = 12.06, size = 118, normalized size = 0.83 \[ \frac {\frac {B\,b\,{\sin \left (c+d\,x\right )}^7}{7}+\left (\frac {A\,b}{6}+\frac {B\,a}{6}\right )\,{\sin \left (c+d\,x\right )}^6+\left (\frac {A\,a}{5}-\frac {2\,B\,b}{5}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {A\,b}{2}-\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {B\,b}{3}-\frac {2\,A\,a}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {A\,b}{2}+\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^2+A\,a\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + B*sin(c + d*x))*(a + b*sin(c + d*x)),x)

[Out]

(sin(c + d*x)^2*((A*b)/2 + (B*a)/2) - sin(c + d*x)^4*((A*b)/2 + (B*a)/2) - sin(c + d*x)^3*((2*A*a)/3 - (B*b)/3
) + sin(c + d*x)^5*((A*a)/5 - (2*B*b)/5) + sin(c + d*x)^6*((A*b)/6 + (B*a)/6) + A*a*sin(c + d*x) + (B*b*sin(c
+ d*x)^7)/7)/d

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sympy [A]  time = 5.84, size = 178, normalized size = 1.24 \[ \begin {cases} \frac {8 A a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A b \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {B a \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {8 B b \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 B b \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {B b \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a + b \sin {\relax (c )}\right ) \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((8*A*a*sin(c + d*x)**5/(15*d) + 4*A*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + A*a*sin(c + d*x)*cos(c
+ d*x)**4/d - A*b*cos(c + d*x)**6/(6*d) - B*a*cos(c + d*x)**6/(6*d) + 8*B*b*sin(c + d*x)**7/(105*d) + 4*B*b*s
in(c + d*x)**5*cos(c + d*x)**2/(15*d) + B*b*sin(c + d*x)**3*cos(c + d*x)**4/(3*d), Ne(d, 0)), (x*(A + B*sin(c)
)*(a + b*sin(c))*cos(c)**5, True))

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