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

Optimal. Leaf size=107 \[ -\frac {b \left (16 a^2+b^2\right ) \cos (2 c+2 d x)}{24 d}+\frac {1}{8} a x \left (8 a^2+3 b^2\right )-\frac {5 a b^2 \sin (2 c+2 d x) \cos (2 c+2 d x)}{48 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d} \]

[Out]

1/8*a*(8*a^2+3*b^2)*x-1/24*b*(16*a^2+b^2)*cos(2*d*x+2*c)/d-5/48*a*b^2*cos(2*d*x+2*c)*sin(2*d*x+2*c)/d-1/48*b*c
os(2*d*x+2*c)*(2*a+b*sin(2*d*x+2*c))^2/d

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Rubi [A]  time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2666, 2656, 2734} \[ -\frac {b \left (16 a^2+b^2\right ) \cos (2 c+2 d x)}{24 d}+\frac {1}{8} a x \left (8 a^2+3 b^2\right )-\frac {5 a b^2 \sin (2 c+2 d x) \cos (2 c+2 d x)}{48 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(8*a^2 + 3*b^2)*x)/8 - (b*(16*a^2 + b^2)*Cos[2*c + 2*d*x])/(24*d) - (5*a*b^2*Cos[2*c + 2*d*x]*Sin[2*c + 2*d
*x])/(48*d) - (b*Cos[2*c + 2*d*x]*(2*a + b*Sin[2*c + 2*d*x])^2)/(48*d)

Rule 2656

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*
x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2734

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]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x) \sin (c+d x))^3 \, dx &=\int \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^3 \, dx\\ &=-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d}+\frac {1}{3} \int \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right ) \left (\frac {1}{2} \left (6 a^2+b^2\right )+\frac {5}{2} a b \sin (2 c+2 d x)\right ) \, dx\\ &=\frac {1}{8} a \left (8 a^2+3 b^2\right ) x-\frac {b \left (16 a^2+b^2\right ) \cos (2 c+2 d x)}{24 d}-\frac {5 a b^2 \cos (2 c+2 d x) \sin (2 c+2 d x)}{48 d}-\frac {b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 75, normalized size = 0.70 \[ \frac {-9 \left (16 a^2 b+b^3\right ) \cos (2 (c+d x))+6 a \left (4 \left (8 a^2+3 b^2\right ) (c+d x)-3 b^2 \sin (4 (c+d x))\right )+b^3 \cos (6 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*(16*a^2*b + b^3)*Cos[2*(c + d*x)] + b^3*Cos[6*(c + d*x)] + 6*a*(4*(8*a^2 + 3*b^2)*(c + d*x) - 3*b^2*Sin[4*
(c + d*x)]))/(192*d)

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fricas [A]  time = 1.75, size = 97, normalized size = 0.91 \[ \frac {4 \, b^{3} \cos \left (d x + c\right )^{6} - 6 \, b^{3} \cos \left (d x + c\right )^{4} - 36 \, a^{2} b \cos \left (d x + c\right )^{2} + 3 \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} d x - 9 \, {\left (2 \, a b^{2} \cos \left (d x + c\right )^{3} - a b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(4*b^3*cos(d*x + c)^6 - 6*b^3*cos(d*x + c)^4 - 36*a^2*b*cos(d*x + c)^2 + 3*(8*a^3 + 3*a*b^2)*d*x - 9*(2*a
*b^2*cos(d*x + c)^3 - a*b^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.19, size = 75, normalized size = 0.70 \[ \frac {b^{3} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {3 \, a b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} x - \frac {3 \, {\left (16 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*b^3*cos(6*d*x + 6*c)/d - 3/32*a*b^2*sin(4*d*x + 4*c)/d + 1/8*(8*a^3 + 3*a*b^2)*x - 3/64*(16*a^2*b + b^3)
*cos(2*d*x + 2*c)/d

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maple [A]  time = 0.25, size = 106, normalized size = 0.99 \[ \frac {b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {3 \left (\cos ^{2}\left (d x +c \right )\right ) a^{2} b}{2}+a^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+3*a*b^2*(-1/4*sin(d*x+c)*cos(d*x+c)^3+1/8*sin(d*x+
c)*cos(d*x+c)+1/8*d*x+1/8*c)-3/2*cos(d*x+c)^2*a^2*b+a^3*(d*x+c))

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maxima [A]  time = 0.38, size = 80, normalized size = 0.75 \[ a^{3} x - \frac {3 \, a^{2} b \cos \left (d x + c\right )^{2}}{2 \, d} + \frac {3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2}}{32 \, d} - \frac {{\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} b^{3}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 3/2*a^2*b*cos(d*x + c)^2/d + 3/32*(4*d*x + 4*c - sin(4*d*x + 4*c))*a*b^2/d - 1/12*(2*sin(d*x + c)^6 -
3*sin(d*x + c)^4)*b^3/d

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mupad [B]  time = 3.44, size = 125, normalized size = 1.17 \[ a^3\,x-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (72\,a^2\,b+6\,b^3\right )+36\,a^2\,b+2\,b^3+36\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4-9\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5+9\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left (24\,{\mathrm {tan}\left (c+d\,x\right )}^6+72\,{\mathrm {tan}\left (c+d\,x\right )}^4+72\,{\mathrm {tan}\left (c+d\,x\right )}^2+24\right )}+\frac {3\,a\,b^2\,x}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - (tan(c + d*x)^2*(72*a^2*b + 6*b^3) + 36*a^2*b + 2*b^3 + 36*a^2*b*tan(c + d*x)^4 - 9*a*b^2*tan(c + d*x)
^5 + 9*a*b^2*tan(c + d*x))/(d*(72*tan(c + d*x)^2 + 72*tan(c + d*x)^4 + 24*tan(c + d*x)^6 + 24)) + (3*a*b^2*x)/
8

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sympy [A]  time = 3.36, size = 190, normalized size = 1.78 \[ \begin {cases} a^{3} x - \frac {3 a^{2} b \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {b^{3} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac {b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )} \cos {\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x - 3*a**2*b*cos(c + d*x)**2/(2*d) + 3*a*b**2*x*sin(c + d*x)**4/8 + 3*a*b**2*x*sin(c + d*x)**2
*cos(c + d*x)**2/4 + 3*a*b**2*x*cos(c + d*x)**4/8 + 3*a*b**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 3*a*b**2*sin
(c + d*x)*cos(c + d*x)**3/(8*d) + b**3*sin(c + d*x)**6/(12*d) + b**3*sin(c + d*x)**4*cos(c + d*x)**2/(4*d), Ne
(d, 0)), (x*(a + b*sin(c)*cos(c))**3, True))

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