3.1537 \(\int \cos ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=231 \[ -\frac {\left (-5 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^6}{3 b^6 d}-\frac {\left (a^2-b^2\right ) \left (-5 a^2 B+4 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^6 d}+\frac {\left (a^2-b^2\right )^2 (A b-a B) (a+b \sin (c+d x))^3}{3 b^6 d}+\frac {2 \left (-5 a^3 B+3 a^2 A b+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^6 d}+\frac {(A b-5 a B) (a+b \sin (c+d x))^7}{7 b^6 d}+\frac {B (a+b \sin (c+d x))^8}{8 b^6 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2837, 772} \[ -\frac {\left (-5 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^6}{3 b^6 d}+\frac {2 \left (3 a^2 A b-5 a^3 B+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^6 d}-\frac {\left (a^2-b^2\right ) \left (-5 a^2 B+4 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^6 d}+\frac {\left (a^2-b^2\right )^2 (A b-a B) (a+b \sin (c+d x))^3}{3 b^6 d}+\frac {(A b-5 a B) (a+b \sin (c+d x))^7}{7 b^6 d}+\frac {B (a+b \sin (c+d x))^8}{8 b^6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

((a^2 - b^2)^2*(A*b - a*B)*(a + b*Sin[c + d*x])^3)/(3*b^6*d) - ((a^2 - b^2)*(4*a*A*b - 5*a^2*B + b^2*B)*(a + b
*Sin[c + d*x])^4)/(4*b^6*d) + (2*(3*a^2*A*b - A*b^3 - 5*a^3*B + 3*a*b^2*B)*(a + b*Sin[c + d*x])^5)/(5*b^6*d) -
((2*a*A*b - 5*a^2*B + b^2*B)*(a + b*Sin[c + d*x])^6)/(3*b^6*d) + ((A*b - 5*a*B)*(a + b*Sin[c + d*x])^7)/(7*b^
6*d) + (B*(a + b*Sin[c + d*x])^8)/(8*b^6*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))^2 (A+B \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^2 \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 (\frac {\left (-a^2+b^2\right )^2 (A b-a B) (a+x)^2}{b}+\frac {\left (-a^2+b^2\right ) \left (4 a A b-5 a^2 B+b^2 B\right ) (a+x)^3}{b}-\frac {2 \left (-3 a^2 A b+A b^3+5 a^3 B-3 a b^2 B\right ) (a+x)^4}{b}+\frac {2 \left (-2 a A b+5 a^2 B-b^2 B\right ) (a+x)^5}{b}+\frac {(A b-5 a B) (a+x)^6}{b}+\frac {B (a+x)^7}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\left (a^2-b^2\right )^2 (A b-a B) (a+b \sin (c+d x))^3}{3 b^6 d}-\frac {\left (a^2-b^2\right ) \left (4 a A b-5 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^6 d}+\frac {2 \left (3 a^2 A b-A b^3-5 a^3 B+3 a b^2 B\right ) (a+b \sin (c+d x))^5}{5 b^6 d}-\frac {\left (2 a A b-5 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^6}{3 b^6 d}+\frac {(A b-5 a B) (a+b \sin (c+d x))^7}{7 b^6 d}+\frac {B (a+b \sin (c+d x))^8}{8 b^6 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.50, size = 227, normalized size = 0.98 \[ \frac {840 a^2 A b^6 \sin (c+d x)+140 b^6 \left (a^2 B+2 a A b-2 b^2 B\right ) \sin ^6(c+d x)+168 b^6 \left (a^2 A-4 a b B-2 A b^2\right ) \sin ^5(c+d x)+210 b^6 \left (-2 a^2 B-4 a A b+b^2 B\right ) \sin ^4(c+d x)+280 b^6 \left (-2 a^2 A+2 a b B+A b^2\right ) \sin ^3(c+d x)+a^4 B \left (3 a^4-28 a^2 b^2+210 b^4\right )+120 b^7 (2 a B+A b) \sin ^7(c+d x)+420 a b^6 (a B+2 A b) \sin ^2(c+d x)+105 b^8 B \sin ^8(c+d x)}{840 b^6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^4*(3*a^4 - 28*a^2*b^2 + 210*b^4)*B + 840*a^2*A*b^6*Sin[c + d*x] + 420*a*b^6*(2*A*b + a*B)*Sin[c + d*x]^2 +
280*b^6*(-2*a^2*A + A*b^2 + 2*a*b*B)*Sin[c + d*x]^3 + 210*b^6*(-4*a*A*b - 2*a^2*B + b^2*B)*Sin[c + d*x]^4 + 16
8*b^6*(a^2*A - 2*A*b^2 - 4*a*b*B)*Sin[c + d*x]^5 + 140*b^6*(2*a*A*b + a^2*B - 2*b^2*B)*Sin[c + d*x]^6 + 120*b^
7*(A*b + 2*a*B)*Sin[c + d*x]^7 + 105*b^8*B*Sin[c + d*x]^8)/(840*b^6*d)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 147, normalized size = 0.64 \[ \frac {105 \, B b^{2} \cos \left (d x + c\right )^{8} - 140 \, {\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{6} - 8 \, {\left (15 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 56 \, A a^{2} - 16 \, B a b - 8 \, A b^{2} - 4 \, {\left (7 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/840*(105*B*b^2*cos(d*x + c)^8 - 140*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^6 - 8*(15*(2*B*a*b + A*b^2)*cos(d
*x + c)^6 - 3*(7*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^4 - 56*A*a^2 - 16*B*a*b - 8*A*b^2 - 4*(7*A*a^2 + 2*B*a*
b + A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.38, size = 231, normalized size = 1.00 \[ \frac {B b^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (2 \, B a^{2} + 4 \, A a b - B b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (8 \, B a^{2} + 16 \, A a b + B b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {{\left (10 \, B a^{2} + 20 \, A a b + 3 \, B b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {{\left (2 \, B a b + A b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (4 \, A a^{2} - 6 \, B a b - 3 \, A b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/1024*B*b^2*cos(8*d*x + 8*c)/d - 1/384*(2*B*a^2 + 4*A*a*b - B*b^2)*cos(6*d*x + 6*c)/d - 1/256*(8*B*a^2 + 16*A
*a*b + B*b^2)*cos(4*d*x + 4*c)/d - 1/128*(10*B*a^2 + 20*A*a*b + 3*B*b^2)*cos(2*d*x + 2*c)/d - 1/448*(2*B*a*b +
A*b^2)*sin(7*d*x + 7*c)/d + 1/320*(4*A*a^2 - 6*B*a*b - 3*A*b^2)*sin(5*d*x + 5*c)/d + 1/192*(20*A*a^2 - 2*B*a*
b - A*b^2)*sin(3*d*x + 3*c)/d + 5/64*(8*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c)/d

________________________________________________________________________________________

maple [A]  time = 0.54, size = 199, normalized size = 0.86 \[ \frac {\frac {a^{2} A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {B \,a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{6}-\frac {A a b \left (\cos ^{6}\left (d x +c \right )\right )}{3}+2 B a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )+A \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )+B \,b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{8}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{24}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/5*a^2*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)-1/6*B*a^2*cos(d*x+c)^6-1/3*A*a*b*cos(d*x+c)^6+2*
B*a*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))+A*b^2*(-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))+B*b^2*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24
*cos(d*x+c)^6))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 184, normalized size = 0.80 \[ \frac {105 \, B b^{2} \sin \left (d x + c\right )^{8} + 120 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{7} + 140 \, {\left (B a^{2} + 2 \, A a b - 2 \, B b^{2}\right )} \sin \left (d x + c\right )^{6} + 168 \, {\left (A a^{2} - 4 \, B a b - 2 \, A b^{2}\right )} \sin \left (d x + c\right )^{5} - 210 \, {\left (2 \, B a^{2} + 4 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} + 840 \, A a^{2} \sin \left (d x + c\right ) - 280 \, {\left (2 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 420 \, {\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/840*(105*B*b^2*sin(d*x + c)^8 + 120*(2*B*a*b + A*b^2)*sin(d*x + c)^7 + 140*(B*a^2 + 2*A*a*b - 2*B*b^2)*sin(d
*x + c)^6 + 168*(A*a^2 - 4*B*a*b - 2*A*b^2)*sin(d*x + c)^5 - 210*(2*B*a^2 + 4*A*a*b - B*b^2)*sin(d*x + c)^4 +
840*A*a^2*sin(d*x + c) - 280*(2*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 + 420*(B*a^2 + 2*A*a*b)*sin(d*x + c)^2
)/d

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 180, normalized size = 0.78 \[ \frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )+{\sin \left (c+d\,x\right )}^7\,\left (\frac {A\,b^2}{7}+\frac {2\,B\,a\,b}{7}\right )+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {2\,A\,a^2}{3}+\frac {2\,B\,a\,b}{3}+\frac {A\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^5\,\left (-\frac {A\,a^2}{5}+\frac {4\,B\,a\,b}{5}+\frac {2\,A\,b^2}{5}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {B\,a^2}{2}+A\,a\,b-\frac {B\,b^2}{4}\right )+{\sin \left (c+d\,x\right )}^6\,\left (\frac {B\,a^2}{6}+\frac {A\,a\,b}{3}-\frac {B\,b^2}{3}\right )+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^8}{8}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \]

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))^2,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 10.36, size = 335, normalized size = 1.45 \[ \begin {cases} \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a b \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {8 A b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 A b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {16 B a b \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {8 B a b \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac {B b^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac {B b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a + b \sin {\relax (c )}\right )^{2} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((8*A*a**2*sin(c + d*x)**5/(15*d) + 4*A*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + A*a**2*sin(c + d
*x)*cos(c + d*x)**4/d - A*a*b*cos(c + d*x)**6/(3*d) + 8*A*b**2*sin(c + d*x)**7/(105*d) + 4*A*b**2*sin(c + d*x)
**5*cos(c + d*x)**2/(15*d) + A*b**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - B*a**2*cos(c + d*x)**6/(6*d) + 16*
B*a*b*sin(c + d*x)**7/(105*d) + 8*B*a*b*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + 2*B*a*b*sin(c + d*x)**3*cos(c
+ d*x)**4/(3*d) + B*b**2*sin(c + d*x)**8/(24*d) + B*b**2*sin(c + d*x)**6*cos(c + d*x)**2/(6*d) + B*b**2*sin(c
+ d*x)**4*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*sin(c))**2*cos(c)**5, True))

________________________________________________________________________________________