\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx\) [55]

Optimal result

Integrand size = 34, antiderivative size = 57 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {(A-2 B) c x}{a}+\frac {B c \cos (e+f x)}{a f}-\frac {2 (A-B) c \cos (e+f x)}{f (a+a \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3046, 2936, 2718} \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {2 c (A-B) \cos (e+f x)}{f (a \sin (e+f x)+a)}-\frac {c x (A-2 B)}{a}+\frac {B c \cos (e+f x)}{a f} \]

[In]

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

[Out]

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

Rule 2718

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

Rule 2936

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 3046

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]))

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(127\) vs. \(2(57)=114\).

Time = 5.37 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.23 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {\left (-((A-2 B) x)+\frac {B \cos (e) \cos (f x)}{f}-\frac {B \sin (e) \sin (f x)}{f}+\frac {4 (A-B) \sin \left (\frac {f x}{2}\right )}{f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}\right ) (c-c \sin (e+f x))}{a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {2 c \left (-\frac {-2 B +2 A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {B}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}-\left (A -2 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f a}\)	\(67\)
default	\(\frac {2 c \left (-\frac {-2 B +2 A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {B}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}-\left (A -2 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f a}\)	\(67\)
parallelrisch	\(\frac {2 c \left (\frac {B \cos \left (2 f x +2 e \right )}{4}+\left (-\frac {1}{2} f x A +f x B -A +\frac {3}{2} B \right ) \cos \left (f x +e \right )+\left (A -B \right ) \sin \left (f x +e \right )-A +\frac {5 B}{4}\right )}{a f \cos \left (f x +e \right )}\)	\(72\)
risch	\(-\frac {c x A}{a}+\frac {2 c x B}{a}+\frac {B c \,{\mathrm e}^{i \left (f x +e \right )}}{2 a f}+\frac {B c \,{\mathrm e}^{-i \left (f x +e \right )}}{2 a f}-\frac {4 c A}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {4 c B}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}\)	\(104\)
norman	\(\frac {\frac {2 B c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {4 A c -6 B c}{a f}-\frac {\left (A -2 B \right ) c x}{a}+\frac {2 B c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {2 \left (4 A c -5 B c \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {\left (4 A c -4 B c \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {\left (A -2 B \right ) c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}-\frac {2 \left (A -2 B \right ) c x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {2 \left (A -2 B \right ) c x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {\left (A -2 B \right ) c x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {\left (A -2 B \right ) c x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(269\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (57) = 114\).

Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.05 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {{\left (A - 2 \, B\right )} c f x - B c \cos \left (f x + e\right )^{2} + 2 \, {\left (A - B\right )} c + {\left ({\left (A - 2 \, B\right )} c f x + {\left (2 \, A - 3 \, B\right )} c\right )} \cos \left (f x + e\right ) + {\left ({\left (A - 2 \, B\right )} c f x - B c \cos \left (f x + e\right ) - 2 \, {\left (A - B\right )} c\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 828 vs. \(2 (49) = 98\).

Time = 1.06 (sec) , antiderivative size = 828, normalized size of antiderivative = 14.53 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-A*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) - A*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) - A*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) - A*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) - 4*A*c*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) - 4*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) + 2*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) + 2*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) + 2*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) + 2*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) + 4*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*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) + 6*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), Ne(f, 0)), (x*(A + B*sin(e))*(-c*sin(e) + c)/(a*sin(e) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (57) = 114\).

Time = 0.40 (sec) , antiderivative size = 256, normalized size of antiderivative = 4.49 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (B c {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} - A c {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + B c {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {A c}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (57) = 114\).

Time = 0.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.02 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {\frac {{\left (A c - 2 \, B c\right )} {\left (f x + e\right )}}{a} + \frac {2 \, {\left (2 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A c - 3 \, B c\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )} a}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.59 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.93 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {\left (4\,A\,c-4\,B\,c\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-2\,B\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+4\,A\,c-6\,B\,c}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )}-\frac {A\,c\,f\,x-2\,B\,c\,f\,x}{a\,f} \]

[In]

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

[Out]

- (4*A*c - 6*B*c + tan(e/2 + (f*x)/2)^2*(4*A*c - 4*B*c) - 2*B*c*tan(e/2 + (f*x)/2))/(f*(a + a*tan(e/2 + (f*x)/
2) + a*tan(e/2 + (f*x)/2)^2 + a*tan(e/2 + (f*x)/2)^3)) - (A*c*f*x - 2*B*c*f*x)/(a*f)