3.287 \(\int \cos ^2(e+f x) (a+b \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=57 \[ \frac{(4 a+b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} x (4 a+b)-\frac{b \sin (e+f x) \cos ^3(e+f x)}{4 f} \]

[Out]

((4*a + b)*x)/8 + ((4*a + b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - (b*Cos[e + f*x]^3*Sin[e + f*x])/(4*f)

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Rubi [A]  time = 0.0444805, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3191, 385, 199, 203} \[ \frac{(4 a+b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} x (4 a+b)-\frac{b \sin (e+f x) \cos ^3(e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a + b)*x)/8 + ((4*a + b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - (b*Cos[e + f*x]^3*Sin[e + f*x])/(4*f)

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+(a+b) x^2}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cos ^3(e+f x) \sin (e+f x)}{4 f}+\frac{(4 a+b) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac{(4 a+b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{b \cos ^3(e+f x) \sin (e+f x)}{4 f}+\frac{(4 a+b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{1}{8} (4 a+b) x+\frac{(4 a+b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{b \cos ^3(e+f x) \sin (e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.0799996, size = 46, normalized size = 0.81 \[ \frac{4 (4 a e+4 a f x+b f x)+8 a \sin (2 (e+f x))-b \sin (4 (e+f x))}{32 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(4*a*e + 4*a*f*x + b*f*x) + 8*a*Sin[2*(e + f*x)] - b*Sin[4*(e + f*x)])/(32*f)

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Maple [A]  time = 0.046, size = 70, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( b \left ( -{\frac{\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{8}}+{\frac{fx}{8}}+{\frac{e}{8}} \right ) +a \left ({\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(b*(-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)+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 = 1.49416, size = 93, normalized size = 1.63 \begin{align*} \frac{{\left (f x + e\right )}{\left (4 \, a + b\right )} + \frac{{\left (4 \, a + b\right )} \tan \left (f x + e\right )^{3} +{\left (4 \, a - b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.89235, size = 113, normalized size = 1.98 \begin{align*} \frac{{\left (4 \, a + b\right )} f x -{\left (2 \, b \cos \left (f x + e\right )^{3} -{\left (4 \, a + b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.83485, size = 150, normalized size = 2.63 \begin{align*} \begin{cases} \frac{a x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{a x \cos ^{2}{\left (e + f x \right )}}{2} + \frac{a \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{b x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{b x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{b x \cos ^{4}{\left (e + f x \right )}}{8} + \frac{b \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{b \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin ^{2}{\left (e \right )}\right ) \cos ^{2}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12607, size = 55, normalized size = 0.96 \begin{align*} \frac{1}{8} \,{\left (4 \, a + b\right )} x - \frac{b \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{a \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*a + b)*x - 1/32*b*sin(4*f*x + 4*e)/f + 1/4*a*sin(2*f*x + 2*e)/f