3.1070 \(\int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=163 \[ -\frac {a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}-\frac {\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}+\frac {b \left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} b x \left (6 a^2+b^2\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d} \]

[Out]

1/16*b*(6*a^2+b^2)*x-1/120*a*(2*a^2+33*b^2)*cos(d*x+c)^3/d+1/16*b*(6*a^2+b^2)*cos(d*x+c)*sin(d*x+c)/d-1/40*(2*
a^2+5*b^2)*cos(d*x+c)^3*(a+b*sin(d*x+c))/d-1/10*a*cos(d*x+c)^3*(a+b*sin(d*x+c))^2/d-1/6*cos(d*x+c)^3*(a+b*sin(
d*x+c))^3/d

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Rubi [A]  time = 0.29, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2862, 2669, 2635, 8} \[ -\frac {a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}-\frac {\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}+\frac {b \left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} b x \left (6 a^2+b^2\right )-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sin[c + d*x]*(a + b*Sin[c + d*x])^3,x]

[Out]

(b*(6*a^2 + b^2)*x)/16 - (a*(2*a^2 + 33*b^2)*Cos[c + d*x]^3)/(120*d) + (b*(6*a^2 + b^2)*Cos[c + d*x]*Sin[c + d
*x])/(16*d) - ((2*a^2 + 5*b^2)*Cos[c + d*x]^3*(a + b*Sin[c + d*x]))/(40*d) - (a*Cos[c + d*x]^3*(a + b*Sin[c +
d*x])^2)/(10*d) - (Cos[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac {1}{6} \int \cos ^2(c+d x) (3 b+3 a \sin (c+d x)) (a+b \sin (c+d x))^2 \, dx\\ &=-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac {1}{30} \int \cos ^2(c+d x) (a+b \sin (c+d x)) \left (21 a b+3 \left (2 a^2+5 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac {1}{120} \int \cos ^2(c+d x) \left (15 b \left (6 a^2+b^2\right )+3 a \left (2 a^2+33 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}-\frac {\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac {1}{8} \left (b \left (6 a^2+b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}+\frac {b \left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}-\frac {\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}+\frac {1}{16} \left (b \left (6 a^2+b^2\right )\right ) \int 1 \, dx\\ &=\frac {1}{16} b \left (6 a^2+b^2\right ) x-\frac {a \left (2 a^2+33 b^2\right ) \cos ^3(c+d x)}{120 d}+\frac {b \left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}-\frac {\left (2 a^2+5 b^2\right ) \cos ^3(c+d x) (a+b \sin (c+d x))}{40 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))^2}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^3}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 138, normalized size = 0.85 \[ \frac {-20 \left (4 a^3+3 a b^2\right ) \cos (3 (c+d x))-120 a \left (2 a^2+3 b^2\right ) \cos (c+d x)+b \left (5 \left (-3 \left (6 a^2+b^2\right ) \sin (4 (c+d x))+72 a^2 c+72 a^2 d x-3 b^2 \sin (2 (c+d x))+b^2 \sin (6 (c+d x))+18 b^2 c+12 b^2 d x\right )+36 a b \cos (5 (c+d x))\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sin[c + d*x]*(a + b*Sin[c + d*x])^3,x]

[Out]

(-120*a*(2*a^2 + 3*b^2)*Cos[c + d*x] - 20*(4*a^3 + 3*a*b^2)*Cos[3*(c + d*x)] + b*(36*a*b*Cos[5*(c + d*x)] + 5*
(72*a^2*c + 18*b^2*c + 72*a^2*d*x + 12*b^2*d*x - 3*b^2*Sin[2*(c + d*x)] - 3*(6*a^2 + b^2)*Sin[4*(c + d*x)] + b
^2*Sin[6*(c + d*x)])))/(960*d)

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fricas [A]  time = 0.70, size = 116, normalized size = 0.71 \[ \frac {144 \, a b^{2} \cos \left (d x + c\right )^{5} - 80 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (6 \, a^{2} b + b^{3}\right )} d x + 5 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} - 2 \, {\left (18 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/240*(144*a*b^2*cos(d*x + c)^5 - 80*(a^3 + 3*a*b^2)*cos(d*x + c)^3 + 15*(6*a^2*b + b^3)*d*x + 5*(8*b^3*cos(d*
x + c)^5 - 2*(18*a^2*b + 7*b^3)*cos(d*x + c)^3 + 3*(6*a^2*b + b^3)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.25, size = 139, normalized size = 0.85 \[ \frac {3 \, a b^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {b^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {1}{16} \, {\left (6 \, a^{2} b + b^{3}\right )} x - \frac {{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )}{8 \, d} - \frac {{\left (6 \, a^{2} b + b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

3/80*a*b^2*cos(5*d*x + 5*c)/d + 1/192*b^3*sin(6*d*x + 6*c)/d - 1/64*b^3*sin(2*d*x + 2*c)/d + 1/16*(6*a^2*b + b
^3)*x - 1/48*(4*a^3 + 3*a*b^2)*cos(3*d*x + 3*c)/d - 1/8*(2*a^3 + 3*a*b^2)*cos(d*x + c)/d - 1/64*(6*a^2*b + b^3
)*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.21, size = 158, normalized size = 0.97 \[ \frac {-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+3 a^{2} b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+3 a \,b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+b^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x)

[Out]

1/d*(-1/3*a^3*cos(d*x+c)^3+3*a^2*b*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1/8*c)+3*a*
b^2*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+b^3*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*si
n(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*c))

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maxima [A]  time = 0.32, size = 108, normalized size = 0.66 \[ -\frac {320 \, a^{3} \cos \left (d x + c\right )^{3} - 90 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} b - 192 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b^{2} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/960*(320*a^3*cos(d*x + c)^3 - 90*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^2*b - 192*(3*cos(d*x + c)^5 - 5*cos(d*x
+ c)^3)*a*b^2 + 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*b^3)/d

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mupad [B]  time = 10.75, size = 425, normalized size = 2.61 \[ \frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2+b^2\right )}{8\,\left (\frac {3\,a^2\,b}{4}+\frac {b^3}{8}\right )}\right )\,\left (6\,a^2+b^2\right )}{8\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^2\,b}{4}+\frac {b^3}{8}\right )+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {4\,a\,b^2}{5}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (6\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^3+\frac {24\,a\,b^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {20\,a^3}{3}+8\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {3\,a^2\,b}{4}+\frac {b^3}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {9\,a^2\,b}{2}+\frac {19\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {9\,a^2\,b}{2}+\frac {19\,b^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {15\,a^2\,b}{4}-\frac {17\,b^3}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {15\,a^2\,b}{4}-\frac {17\,b^3}{24}\right )+\frac {2\,a^3}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {b\,\left (6\,a^2+b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*sin(c + d*x)*(a + b*sin(c + d*x))^3,x)

[Out]

(b*atan((b*tan(c/2 + (d*x)/2)*(6*a^2 + b^2))/(8*((3*a^2*b)/4 + b^3/8)))*(6*a^2 + b^2))/(8*d) - (tan(c/2 + (d*x
)/2)*((3*a^2*b)/4 + b^3/8) + 4*a^3*tan(c/2 + (d*x)/2)^4 + 2*a^3*tan(c/2 + (d*x)/2)^10 + (4*a*b^2)/5 + tan(c/2
+ (d*x)/2)^8*(12*a*b^2 + 6*a^3) + tan(c/2 + (d*x)/2)^2*((24*a*b^2)/5 + 2*a^3) + tan(c/2 + (d*x)/2)^6*(8*a*b^2
+ (20*a^3)/3) - tan(c/2 + (d*x)/2)^11*((3*a^2*b)/4 + b^3/8) - tan(c/2 + (d*x)/2)^5*((9*a^2*b)/2 + (19*b^3)/4)
+ tan(c/2 + (d*x)/2)^7*((9*a^2*b)/2 + (19*b^3)/4) - tan(c/2 + (d*x)/2)^3*((15*a^2*b)/4 - (17*b^3)/24) + tan(c/
2 + (d*x)/2)^9*((15*a^2*b)/4 - (17*b^3)/24) + (2*a^3)/3)/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4
+ 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) -
(b*(6*a^2 + b^2)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(8*d)

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sympy [A]  time = 3.61, size = 340, normalized size = 2.09 \[ \begin {cases} - \frac {a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 a^{2} b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{2} b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 a^{2} b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {a b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {2 a b^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{3} \sin {\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)*(a+b*sin(d*x+c))**3,x)

[Out]

Piecewise((-a**3*cos(c + d*x)**3/(3*d) + 3*a**2*b*x*sin(c + d*x)**4/8 + 3*a**2*b*x*sin(c + d*x)**2*cos(c + d*x
)**2/4 + 3*a**2*b*x*cos(c + d*x)**4/8 + 3*a**2*b*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 3*a**2*b*sin(c + d*x)*co
s(c + d*x)**3/(8*d) - a*b**2*sin(c + d*x)**2*cos(c + d*x)**3/d - 2*a*b**2*cos(c + d*x)**5/(5*d) + b**3*x*sin(c
+ d*x)**6/16 + 3*b**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*b**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + b*
*3*x*cos(c + d*x)**6/16 + b**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) - b**3*sin(c + d*x)**3*cos(c + d*x)**3/(6*d
) - b**3*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sin(c))**3*sin(c)*cos(c)**2, True))

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