\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx\) [267]

Optimal result

Integrand size = 33, antiderivative size = 67 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {(B (c-d)+A d) x}{a}-\frac {B d \cos (e+f x)}{a f}-\frac {(A-B) (c-d) \cos (e+f x)}{a f (1+\sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3047, 3102, 2814, 2727} \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {(A-B) (c-d) \cos (e+f x)}{a f (\sin (e+f x)+1)}+\frac {x (A d+B (c-d))}{a}-\frac {B d \cos (e+f x)}{a f} \]

[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 2727

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]

Rule 2814

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 3047

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 3102

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.44 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right ) ((B (c-d)+A d) (e+f x)-B d \cos (e+f x))+(2 A c+B (c-d) (-2+e+f x)+A d (-2+e+f x)-B d \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{a f (1+\sin (e+f x))} \]

[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]))

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A c -d A -B c +d B \right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 d B}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+2 \left (d A +B c -d B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}\)	\(81\)
default	\(\frac {-\frac {2 \left (A c -d A -B c +d B \right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 d B}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+2 \left (d A +B c -d B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}\)	\(81\)
parallelrisch	\(\frac {\left (\left (2 c f x -2 d f x -4 c +3 d \right ) B +4 \left (\left (\frac {f x}{2}-1\right ) d +c \right ) A \right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (\left (2 c f x -2 d f x -3 d \right ) B +2 A d f x \right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-d B \left (\cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )\right )}{2 a f \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}\)	\(130\)
risch	\(\frac {x d A}{a}+\frac {x B c}{a}-\frac {x d B}{a}-\frac {B d \,{\mathrm e}^{i \left (f x +e \right )}}{2 a f}-\frac {B d \,{\mathrm e}^{-i \left (f x +e \right )}}{2 a f}-\frac {2 A c}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {2 d A}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {2 B c}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}-\frac {2 d B}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}\)	\(158\)
norman	\(\frac {\frac {\left (d A +B c -d B \right ) x}{a}+\frac {\left (d A +B c -d B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {\left (d A +B c -d B \right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {\left (d A +B c -d B \right ) x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {2 A c -2 d A -2 B c +4 d B}{a f}-\frac {2 d B \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {2 d B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {2 \left (d A +B c -d B \right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {2 \left (d A +B c -d B \right ) x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {\left (2 A c -2 d A -2 B c +2 d B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {2 \left (2 A c -2 d A -2 B c +3 d B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(319\)

[In]

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

[Out]

2/f/a*(-(A*c-A*d-B*c+B*d)/(tan(1/2*f*x+1/2*e)+1)-d*B/(1+tan(1/2*f*x+1/2*e)^2)+(A*d+B*c-B*d)*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. 154 vs. \(2 (67) = 134\).

Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.30 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {B d \cos \left (f x + e\right )^{2} - {\left (B c + {\left (A - B\right )} d\right )} f x + {\left (A - B\right )} c - {\left (A - B\right )} d - {\left ({\left (B c + {\left (A - B\right )} d\right )} f x - {\left (A - B\right )} c + {\left (A - 2 \, B\right )} d\right )} \cos \left (f x + e\right ) - {\left ({\left (B c + {\left (A - B\right )} d\right )} f x - B d \cos \left (f x + e\right ) + {\left (A - B\right )} c - {\left (A - B\right )} d\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1307 vs. \(2 (49) = 98\).

Time = 1.04 (sec) , antiderivative size = 1307, normalized size of antiderivative = 19.51 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))*(c+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)**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*d*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) - 4*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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (67) = 134\).

Time = 0.30 (sec) , antiderivative size = 256, normalized size of antiderivative = 3.82 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left (B d {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} - B c {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - A d {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac {A c}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (67) = 134\).

Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.25 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {\frac {{\left (B c + A d - B d\right )} {\left (f x + e\right )}}{a} - \frac {2 \, {\left (A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A c - B c - A d + 2 \, B d\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )} a}}{f} \]

[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

Mupad [B] (verification not implemented)

Time = 13.56 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {x\,\left (A\,d+B\,c-B\,d\right )}{a}-\frac {\left (2\,A\,c-2\,A\,d-2\,B\,c+2\,B\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,B\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\,A\,c-2\,A\,d-2\,B\,c+4\,B\,d}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )} \]

[In]

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

[Out]

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