\(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx\) [21]

Optimal result

Integrand size = 34, antiderivative size = 56 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {a (A+2 B) x}{c}+\frac {a B \cos (e+f x)}{c f}+\frac {2 a (A+B) \cos (e+f x)}{f (c-c \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 56, 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} \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {2 a (A+B) \cos (e+f x)}{f (c-c \sin (e+f x))}-\frac {a x (A+2 B)}{c}+\frac {a B \cos (e+f x)}{c f} \]

[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 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))}{(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 [A] (verified)

Time = 5.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.91 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {a \cos (e+f x) \left (-2 (A+2 B) \arcsin \left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (e+f x)}+\sqrt {1+\sin (e+f x)} (-2 A-3 B+B \sin (e+f x))\right )}{c f (-1+\sin (e+f x)) \sqrt {1+\sin (e+f x)}} \]

[In]

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

[Out]

(a*Cos[e + f*x]*(-2*(A + 2*B)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]]*Sqrt[1 - Sin[e + f*x]] + Sqrt[1 + Sin[e +
f*x]]*(-2*A - 3*B + B*Sin[e + f*x])))/(c*f*(-1 + Sin[e + f*x])*Sqrt[1 + Sin[e + f*x]])

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.20

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

[In]

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

[Out]

2/f*a/c*(-(2*A+2*B)/(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. 116 vs. \(2 (57) = 114\).

Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.07 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {{\left (A + 2 \, B\right )} a f x - B a \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} a + {\left ({\left (A + 2 \, B\right )} a f x - {\left (2 \, A + 3 \, B\right )} a\right )} \cos \left (f x + e\right ) - {\left ({\left (A + 2 \, B\right )} a f x - B a \cos \left (f x + e\right ) + 2 \, {\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \]

[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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 828 vs. \(2 (48) = 96\).

Time = 1.03 (sec) , antiderivative size = 828, normalized size of antiderivative = 14.79 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\text {Too large to display} \]

[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) - 4*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) + 2*B*a*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) - 6*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), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (57) = 114\).

Time = 0.30 (sec) , antiderivative size = 265, normalized size of antiderivative = 4.73 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (B a {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + A a {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + B a {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {A a}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \]

[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

Giac [B] (verification not implemented)

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

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.09 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {\frac {{\left (A a + 2 \, B a\right )} {\left (f x + e\right )}}{c} + \frac {2 \, {\left (2 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A a + 3 \, B a\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} c}}{f} \]

[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

Mupad [B] (verification not implemented)

Time = 12.74 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.98 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {\left (4\,A\,a+4\,B\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-2\,B\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+4\,A\,a+6\,B\,a}{f\,\left (-c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\right )}-\frac {A\,a\,f\,x+2\,B\,a\,f\,x}{c\,f} \]

[In]

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

[Out]

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