3.267 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=67 \[ -\frac{(A-B) (c-d) \cos (e+f x)}{a f (\sin (e+f x)+1)}+\frac{x (A d+B (c-d))}{a}-\frac{B d \cos (e+f x)}{a f} \]

[Out]

((B*(c - d) + A*d)*x)/a - (B*d*Cos[e + f*x])/(a*f) - ((A - B)*(c - d)*Cos[e + f*x])/(a*f*(1 + Sin[e + f*x]))

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Rubi [A]  time = 0.20323, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {2968, 3023, 2735, 2648} \[ -\frac{(A-B) (c-d) \cos (e+f x)}{a f (\sin (e+f x)+1)}+\frac{x (A d+B (c-d))}{a}-\frac{B d \cos (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x]),x]

[Out]

((B*(c - d) + A*d)*x)/a - (B*d*Cos[e + f*x])/(a*f) - ((A - B)*(c - d)*Cos[e + f*x])/(a*f*(1 + Sin[e + f*x]))

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2648

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

Rubi steps

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

Mathematica [A]  time = 0.468425, size = 126, normalized size = 1.88 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right ) ((e+f x) (A d+B (c-d))-B d \cos (e+f x))+\sin \left (\frac{1}{2} (e+f x)\right ) (2 A c+A d (e+f x-2)+B (c-d) (e+f x-2)-B d \cos (e+f x))\right )}{a f (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x]),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(Cos[(e + f*x)/2]*((B*(c - d) + A*d)*(e + f*x) - B*d*Cos[e + f*x]) + (2
*A*c + B*(c - d)*(-2 + e + f*x) + A*d*(-2 + e + f*x) - B*d*Cos[e + f*x])*Sin[(e + f*x)/2]))/(a*f*(1 + Sin[e +
f*x]))

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Maple [B]  time = 0.065, size = 179, normalized size = 2.7 \begin{align*} -2\,{\frac{Bd}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) d}{af}}+2\,{\frac{B\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) c}{af}}-2\,{\frac{B\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) d}{af}}-2\,{\frac{Ac}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{Ad}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{Bc}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-2\,{\frac{Bd}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e)),x)

[Out]

-2/a/f*B*d/(1+tan(1/2*f*x+1/2*e)^2)+2/a/f*A*arctan(tan(1/2*f*x+1/2*e))*d+2/a/f*B*arctan(tan(1/2*f*x+1/2*e))*c-
2/a/f*B*arctan(tan(1/2*f*x+1/2*e))*d-2/a/f/(tan(1/2*f*x+1/2*e)+1)*A*c+2/a/f/(tan(1/2*f*x+1/2*e)+1)*A*d+2/a/f/(
tan(1/2*f*x+1/2*e)+1)*B*c-2/a/f/(tan(1/2*f*x+1/2*e)+1)*B*d

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Maxima [B]  time = 1.47055, size = 346, normalized size = 5.16 \begin{align*} -\frac{2 \,{\left (B d{\left (\frac{\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} - B c{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - A d{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac{A c}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

-2*(B*d*((sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a + a*sin(f*x + e)/(cos(
f*x + e) + 1) + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*
x + e)/(cos(f*x + e) + 1))/a) - B*c*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*
x + e) + 1))) - A*d*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) + 1))) +
A*c/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f

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Fricas [B]  time = 1.9737, size = 354, normalized size = 5.28 \begin{align*} -\frac{B d \cos \left (f x + e\right )^{2} -{\left (B c +{\left (A - B\right )} d\right )} f x +{\left (A - B\right )} c -{\left (A - B\right )} d -{\left ({\left (B c +{\left (A - B\right )} d\right )} f x -{\left (A - B\right )} c +{\left (A - 2 \, B\right )} d\right )} \cos \left (f x + e\right ) -{\left ({\left (B c +{\left (A - B\right )} d\right )} f x - B d \cos \left (f x + e\right ) +{\left (A - B\right )} c -{\left (A - B\right )} d\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

-(B*d*cos(f*x + e)^2 - (B*c + (A - B)*d)*f*x + (A - B)*c - (A - B)*d - ((B*c + (A - B)*d)*f*x - (A - B)*c + (A
- 2*B)*d)*cos(f*x + e) - ((B*c + (A - B)*d)*f*x - B*d*cos(f*x + e) + (A - B)*c - (A - B)*d)*sin(f*x + e))/(a*
f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)

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Sympy [A]  time = 5.03552, size = 1244, normalized size = 18.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e)),x)

[Out]

Piecewise((-2*A*c*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2
) + a*f) - 2*A*c/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + A*d*f*x*ta
n(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + A*d*f*x*t
an(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + A*d*f*x*
tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + A*d*f*x/(a
*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*A*d*tan(e/2 + f*x/2)**2/(a*
f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*A*d/(a*f*tan(e/2 + f*x/2)**3
+ a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + B*c*f*x*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**
3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + B*c*f*x*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)*
*3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + B*c*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2)**3
+ a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + B*c*f*x/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*
x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*B*c*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x
/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*B*c/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2
+ f*x/2) + a*f) - B*d*f*x*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2
+ f*x/2) + a*f) - B*d*f*x*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/
2 + f*x/2) + a*f) - B*d*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2
+ f*x/2) + a*f) - B*d*f*x/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2
*B*d*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - 2*
B*d/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f), Ne(f, 0)), (x*(A + B*sin
(e))*(c + d*sin(e))/(a*sin(e) + a), True))

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Giac [B]  time = 1.23198, size = 216, normalized size = 3.22 \begin{align*} \frac{\frac{{\left (B c + A d - B d\right )}{\left (f x + e\right )}}{a} - \frac{2 \,{\left (A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - A d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + B d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + B d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + A c - B c - A d + 2 \, B d\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

((B*c + A*d - B*d)*(f*x + e)/a - 2*(A*c*tan(1/2*f*x + 1/2*e)^2 - B*c*tan(1/2*f*x + 1/2*e)^2 - A*d*tan(1/2*f*x
+ 1/2*e)^2 + B*d*tan(1/2*f*x + 1/2*e)^2 + B*d*tan(1/2*f*x + 1/2*e) + A*c - B*c - A*d + 2*B*d)/((tan(1/2*f*x +
1/2*e)^3 + tan(1/2*f*x + 1/2*e)^2 + tan(1/2*f*x + 1/2*e) + 1)*a))/f