3.935 \(\int \cos ^2(e+f x) (a+a \sin (e+f x)) (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=79 \[ -\frac {a (c+d) \cos ^3(e+f x)}{3 f}+\frac {a (4 c+d) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {1}{8} a x (4 c+d)-\frac {a d \sin (e+f x) \cos ^3(e+f x)}{4 f} \]

[Out]

1/8*a*(4*c+d)*x-1/3*a*(c+d)*cos(f*x+e)^3/f+1/8*a*(4*c+d)*cos(f*x+e)*sin(f*x+e)/f-1/4*a*d*cos(f*x+e)^3*sin(f*x+
e)/f

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2860, 2669, 2635, 8} \[ -\frac {a (4 c+d) \cos ^3(e+f x)}{12 f}+\frac {a (4 c+d) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {1}{8} a x (4 c+d)-\frac {d \cos ^3(e+f x) (a \sin (e+f x)+a)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*c + d)*x)/8 - (a*(4*c + d)*Cos[e + f*x]^3)/(12*f) + (a*(4*c + d)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - (d*C
os[e + f*x]^3*(a + a*Sin[e + f*x]))/(4*f)

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 2860

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[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \cos ^2(e+f x) (a+a \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=-\frac {d \cos ^3(e+f x) (a+a \sin (e+f x))}{4 f}+\frac {1}{4} (4 c+d) \int \cos ^2(e+f x) (a+a \sin (e+f x)) \, dx\\ &=-\frac {a (4 c+d) \cos ^3(e+f x)}{12 f}-\frac {d \cos ^3(e+f x) (a+a \sin (e+f x))}{4 f}+\frac {1}{4} (a (4 c+d)) \int \cos ^2(e+f x) \, dx\\ &=-\frac {a (4 c+d) \cos ^3(e+f x)}{12 f}+\frac {a (4 c+d) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {d \cos ^3(e+f x) (a+a \sin (e+f x))}{4 f}+\frac {1}{8} (a (4 c+d)) \int 1 \, dx\\ &=\frac {1}{8} a (4 c+d) x-\frac {a (4 c+d) \cos ^3(e+f x)}{12 f}+\frac {a (4 c+d) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {d \cos ^3(e+f x) (a+a \sin (e+f x))}{4 f}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 64, normalized size = 0.81 \[ -\frac {a (24 (c+d) \cos (e+f x)+8 (c+d) \cos (3 (e+f x))-12 f x (4 c+d)-24 c \sin (2 (e+f x))+3 d \sin (4 (e+f x)))}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(a*(-12*(4*c + d)*f*x + 24*(c + d)*Cos[e + f*x] + 8*(c + d)*Cos[3*(e + f*x)] - 24*c*Sin[2*(e + f*x)] + 3
*d*Sin[4*(e + f*x)]))/f

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fricas [A]  time = 0.44, size = 72, normalized size = 0.91 \[ -\frac {8 \, {\left (a c + a d\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (4 \, a c + a d\right )} f x + 3 \, {\left (2 \, a d \cos \left (f x + e\right )^{3} - {\left (4 \, a c + a d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(8*(a*c + a*d)*cos(f*x + e)^3 - 3*(4*a*c + a*d)*f*x + 3*(2*a*d*cos(f*x + e)^3 - (4*a*c + a*d)*cos(f*x +
e))*sin(f*x + e))/f

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giac [A]  time = 0.17, size = 87, normalized size = 1.10 \[ \frac {1}{8} \, {\left (4 \, a c + a d\right )} x - \frac {a d \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {a c \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} - \frac {{\left (a c + a d\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {{\left (a c + a d\right )} \cos \left (f x + e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*a*c + a*d)*x - 1/32*a*d*sin(4*f*x + 4*e)/f + 1/4*a*c*sin(2*f*x + 2*e)/f - 1/12*(a*c + a*d)*cos(3*f*x +
3*e)/f - 1/4*(a*c + a*d)*cos(f*x + e)/f

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maple [A]  time = 0.36, size = 96, normalized size = 1.22 \[ \frac {d a \left (-\frac {\sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )}{4}+\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{8}+\frac {f x}{8}+\frac {e}{8}\right )-\frac {\left (\cos ^{3}\left (f x +e \right )\right ) a c}{3}-\frac {d a \left (\cos ^{3}\left (f x +e \right )\right )}{3}+c a \left (\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(d*a*(-1/4*sin(f*x+e)*cos(f*x+e)^3+1/8*sin(f*x+e)*cos(f*x+e)+1/8*f*x+1/8*e)-1/3*cos(f*x+e)^3*a*c-1/3*d*a*c
os(f*x+e)^3+c*a*(1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e))

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maxima [A]  time = 0.34, size = 74, normalized size = 0.94 \[ -\frac {32 \, a c \cos \left (f x + e\right )^{3} + 32 \, a d \cos \left (f x + e\right )^{3} - 24 \, {\left (2 \, f x + 2 \, e + \sin \left (2 \, f x + 2 \, e\right )\right )} a c - 3 \, {\left (4 \, f x + 4 \, e - \sin \left (4 \, f x + 4 \, e\right )\right )} a d}{96 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*a*c*cos(f*x + e)^3 + 32*a*d*cos(f*x + e)^3 - 24*(2*f*x + 2*e + sin(2*f*x + 2*e))*a*c - 3*(4*f*x + 4*
e - sin(4*f*x + 4*e))*a*d)/f

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mupad [B]  time = 10.24, size = 276, normalized size = 3.49 \[ \frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,c+d\right )}{4\,\left (a\,c+\frac {a\,d}{4}\right )}\right )\,\left (4\,c+d\right )}{4\,f}-\frac {\left (a\,c-\frac {a\,d}{4}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+\left (2\,a\,c+2\,a\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\left (a\,c+\frac {7\,a\,d}{4}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\left (2\,a\,c+2\,a\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\left (-a\,c-\frac {7\,a\,d}{4}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (\frac {2\,a\,c}{3}+\frac {2\,a\,d}{3}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (\frac {a\,d}{4}-a\,c\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {2\,a\,c}{3}+\frac {2\,a\,d}{3}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {a\,\left (4\,c+d\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*atan((a*tan(e/2 + (f*x)/2)*(4*c + d))/(4*(a*c + (a*d)/4)))*(4*c + d))/(4*f) - ((2*a*c)/3 + (2*a*d)/3 + tan(
e/2 + (f*x)/2)^4*(2*a*c + 2*a*d) + tan(e/2 + (f*x)/2)^6*(2*a*c + 2*a*d) + tan(e/2 + (f*x)/2)^2*((2*a*c)/3 + (2
*a*d)/3) + tan(e/2 + (f*x)/2)^7*(a*c - (a*d)/4) - tan(e/2 + (f*x)/2)^3*(a*c + (7*a*d)/4) + tan(e/2 + (f*x)/2)^
5*(a*c + (7*a*d)/4) - tan(e/2 + (f*x)/2)*(a*c - (a*d)/4))/(f*(4*tan(e/2 + (f*x)/2)^2 + 6*tan(e/2 + (f*x)/2)^4
+ 4*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1)) - (a*(4*c + d)*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2))/(4*
f)

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sympy [A]  time = 0.99, size = 199, normalized size = 2.52 \[ \begin {cases} \frac {a c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a c x \cos ^{2}{\left (e + f x \right )}}{2} + \frac {a c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {a c \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {a d x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {a d x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {a d x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {a d \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {a d \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {a d \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (c + d \sin {\relax (e )}\right ) \left (a \sin {\relax (e )} + a\right ) \cos ^{2}{\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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