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

Optimal. Leaf size=56 \[ \frac{2 a (A+B) \cos (e+f x)}{f (c-c \sin (e+f x))}-\frac{a x (A+2 B)}{c}+\frac{a B \cos (e+f x)}{c f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.169462, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {2967, 2857, 2638} \[ \frac{2 a (A+B) \cos (e+f x)}{f (c-c \sin (e+f x))}-\frac{a x (A+2 B)}{c}+\frac{a B \cos (e+f x)}{c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 0.862014, size = 125, normalized size = 2.23 \[ \frac{a (\sin (e+f x)+1) \left (\frac{4 (A+B) \sin \left (\frac{f x}{2}\right )}{f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}+x (-(A+2 B))-\frac{B \sin (e) \sin (f x)}{f}+\frac{B \cos (e) \cos (f x)}{f}\right )}{c \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-((A + 2*B)*x) + (B*Cos[e]*Cos[f*x])/f - (B*Sin[e]*Sin[f*x])/f + (4*(A + B)*Sin[(f*x)/2])/(f*(Cos[e/2] - S
in[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])))*(1 + Sin[e + f*x]))/(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^
2)

________________________________________________________________________________________

Maple [A]  time = 0.1, size = 113, normalized size = 2. \begin{align*} -4\,{\frac{aA}{cf \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-4\,{\frac{Ba}{cf \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}+2\,{\frac{Ba}{cf \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 ) A}{cf}}-4\,{\frac{a\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) B}{cf}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/f*a/c/(tan(1/2*f*x+1/2*e)-1)*A-4/f*a/c/(tan(1/2*f*x+1/2*e)-1)*B+2/f*a/c*B/(1+tan(1/2*f*x+1/2*e)^2)-2/f*a/c*
arctan(tan(1/2*f*x+1/2*e))*A-4/f*a/c*arctan(tan(1/2*f*x+1/2*e))*B

________________________________________________________________________________________

Maxima [B]  time = 1.4559, size = 358, normalized size = 6.39 \begin{align*} -\frac{2 \,{\left (B a{\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}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c \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 )}{c}\right )} + A a{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac{1}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + B a{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac{1}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac{A a}{c - \frac{c \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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.40548, size = 289, normalized size = 5.16 \begin{align*} -\frac{{\left (A + 2 \, B\right )} a f x - B a \cos \left (f x + e\right )^{2} - 2 \,{\left (A + B\right )} a +{\left ({\left (A + 2 \, B\right )} a f x -{\left (2 \, A + 3 \, B\right )} a\right )} \cos \left (f x + e\right ) -{\left ({\left (A + 2 \, B\right )} a f x - B a \cos \left (f x + e\right ) + 2 \,{\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 10.4835, size = 830, normalized size = 14.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-A*a*f*x*tan(e/2 + f*x/2)**3/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x
/2) - c*f) + A*a*f*x*tan(e/2 + f*x/2)**2/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*
x/2) - c*f) - A*a*f*x*tan(e/2 + f*x/2)/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/
2) - c*f) + A*a*f*x/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) - 4*A*a*t
an(e/2 + f*x/2)**2/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) - 4*A*a/(c
*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) - 2*B*a*f*x*tan(e/2 + f*x/2)**3
/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) + 2*B*a*f*x*tan(e/2 + f*x/2)
**2/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) - 2*B*a*f*x*tan(e/2 + f*x
/2)/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) + 2*B*a*f*x/(c*f*tan(e/2
+ f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) - 2*B*a*tan(e/2 + f*x/2)**3/(c*f*tan(e/2 +
f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) - 2*B*a*tan(e/2 + f*x/2)**2/(c*f*tan(e/2 +
f*x/2)**3 - c*f*tan(e/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f) - 4*B*a/(c*f*tan(e/2 + f*x/2)**3 - c*f*tan(e
/2 + f*x/2)**2 + c*f*tan(e/2 + f*x/2) - c*f), Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)/(-c*sin(e) + c), Tru
e))

________________________________________________________________________________________

Giac [B]  time = 1.19904, size = 167, normalized size = 2.98 \begin{align*} -\frac{\frac{{\left (A a + 2 \, B a\right )}{\left (f x + e\right )}}{c} + \frac{2 \,{\left (2 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, A a + 3 \, B a\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 )} c}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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