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

Optimal. Leaf size=61 \[ \frac {1}{8} x \left (8 a^2+b^2\right )-\frac {a b \cos (2 c+2 d x)}{2 d}-\frac {b^2 \sin (2 c+2 d x) \cos (2 c+2 d x)}{16 d} \]

[Out]

1/8*(8*a^2+b^2)*x-1/2*a*b*cos(2*d*x+2*c)/d-1/16*b^2*cos(2*d*x+2*c)*sin(2*d*x+2*c)/d

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2666, 2644} \[ \frac {1}{8} x \left (8 a^2+b^2\right )-\frac {a b \cos (2 c+2 d x)}{2 d}-\frac {b^2 \sin (2 c+2 d x) \cos (2 c+2 d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2644

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 48, normalized size = 0.79 \[ -\frac {-4 \left (8 a^2+b^2\right ) (c+d x)+16 a b \cos (2 (c+d x))+b^2 \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(-4*(8*a^2 + b^2)*(c + d*x) + 16*a*b*Cos[2*(c + d*x)] + b^2*Sin[4*(c + d*x)])/d

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fricas [A]  time = 2.40, size = 63, normalized size = 1.03 \[ -\frac {8 \, a b \cos \left (d x + c\right )^{2} - {\left (8 \, a^{2} + b^{2}\right )} d x + {\left (2 \, b^{2} \cos \left (d x + c\right )^{3} - b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 46, normalized size = 0.75 \[ \frac {1}{8} \, {\left (8 \, a^{2} + b^{2}\right )} x - \frac {a b \cos \left (2 \, d x + 2 \, c\right )}{2 \, d} - \frac {b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*a^2 + b^2)*x - 1/2*a*b*cos(2*d*x + 2*c)/d - 1/32*b^2*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.24, size = 69, normalized size = 1.13 \[ \frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\left (\cos ^{2}\left (d x +c \right )\right ) a b +a^{2} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)-cos(d*x+c)^2*a*b+a^2*(d*x+c))

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maxima [A]  time = 0.33, size = 48, normalized size = 0.79 \[ a^{2} x - \frac {a b \cos \left (d x + c\right )^{2}}{d} + \frac {{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.03, size = 78, normalized size = 1.28 \[ x\,\left (a^2+\frac {b^2}{8}\right )-\frac {-\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8}+\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{8}+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\,b}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(a^2 + b^2/8) - (a*b + (b^2*tan(c + d*x))/8 - (b^2*tan(c + d*x)^3)/8 + a*b*tan(c + d*x)^2)/(d*(2*tan(c + d*x
)^2 + tan(c + d*x)^4 + 1))

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sympy [A]  time = 0.95, size = 129, normalized size = 2.11 \[ \begin {cases} a^{2} x - \frac {a b \cos ^{2}{\left (c + d x \right )}}{d} + \frac {b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )} \cos {\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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