\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx\) [266]

Optimal result

Integrand size = 35, antiderivative size = 143 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx=\frac {\left (2 A (2 c-d) d+B \left (2 c^2-4 c d+3 d^2\right )\right ) x}{2 a}+\frac {2 (A (c-d)-B (2 c-d)) d \cos (e+f x)}{a f}+\frac {(2 A-3 B) d^2 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a+a \sin (e+f x))} \] Output:

1/2*(2*A*(2*c-d)*d+B*(2*c^2-4*c*d+3*d^2))*x/a+2*(A*(c-d)-B*(2*c-d))*d*cos( 
f*x+e)/a/f+1/2*(2*A-3*B)*d^2*cos(f*x+e)*sin(f*x+e)/a/f-(A-B)*cos(f*x+e)*(c 
+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))
 

Mathematica [A] (verified)

Time = 6.86 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.40 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (A-B) (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )+2 \left (2 A (2 c-d) d+B \left (2 c^2-4 c d+3 d^2\right )\right ) (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 d (-A d+B (-2 c+d)) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-B d^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (2 (e+f x))\right )}{4 a f (1+\sin (e+f x))} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x 
]),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(A - B)*(c - d)^2*Sin[(e + f*x)/ 
2] + 2*(2*A*(2*c - d)*d + B*(2*c^2 - 4*c*d + 3*d^2))*(e + f*x)*(Cos[(e + f 
*x)/2] + Sin[(e + f*x)/2]) + 4*d*(-(A*d) + B*(-2*c + d))*Cos[e + f*x]*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2]) - B*d^2*(Cos[(e + f*x)/2] + Sin[(e + f*x 
)/2])*Sin[2*(e + f*x)]))/(4*a*f*(1 + Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3042, 3456, 3042, 3213}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a \sin (e+f x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a \sin (e+f x)+a}dx\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\int (c+d \sin (e+f x)) (a (B (c-2 d)+2 A d)-a (2 A-3 B) d \sin (e+f x))dx}{a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a \sin (e+f x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (c+d \sin (e+f x)) (a (B (c-2 d)+2 A d)-a (2 A-3 B) d \sin (e+f x))dx}{a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a \sin (e+f x)+a)}\)

\(\Big \downarrow \) 3213

\(\displaystyle \frac {\frac {1}{2} a x \left (-\left (d^2 (2 A-3 B)\right )+4 A c d+2 B c (c-2 d)\right )+\frac {2 a d (A c-A d-2 B c+B d) \cos (e+f x)}{f}+\frac {a d^2 (2 A-3 B) \sin (e+f x) \cos (e+f x)}{2 f}}{a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{f (a \sin (e+f x)+a)}\)

Input:

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x]),x]
 

Output:

-(((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]))) 
+ ((a*(2*B*c*(c - 2*d) + 4*A*c*d - (2*A - 3*B)*d^2)*x)/2 + (2*a*d*(A*c - 2 
*B*c - A*d + B*d)*Cos[e + f*x])/f + (a*(2*A - 3*B)*d^2*Cos[e + f*x]*Sin[e 
+ f*x])/(2*f))/a^2
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3213

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co 
s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free 
Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3456

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & 
& NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 0])
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A \,c^{2}-2 A c d +A \,d^{2}-B \,c^{2}+2 B c d -B \,d^{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {2 \left (\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (-A \,d^{2}-2 B c d +B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-A \,d^{2}-2 B c d +B \,d^{2}\right )}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}\)	\(193\)
default	\(\frac {-\frac {2 \left (A \,c^{2}-2 A c d +A \,d^{2}-B \,c^{2}+2 B c d -B \,d^{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {2 \left (\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (-A \,d^{2}-2 B c d +B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-A \,d^{2}-2 B c d +B \,d^{2}\right )}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}\)	\(193\)
parallelrisch	\(\frac {\left (\left (-2 f x A +3 f x B -7 A +7 B \right ) d^{2}+4 \left (\left (-f x -\frac {7}{2}\right ) B +A \left (f x +2\right )\right ) c d -4 \left (\left (-\frac {f x}{2}-1\right ) B +A \right ) c^{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (\left (-2 f x A +3 f x B -A +B \right ) d^{2}+4 c \left (\left (-f x -\frac {1}{2}\right ) B +f x A \right ) d +2 B \,c^{2} f x \right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-d \left (\left (\left (A -\frac {3 B}{4}\right ) d +2 B c \right ) \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (\left (A -\frac {3 B}{4}\right ) d +2 B c \right ) \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )-\frac {B d \left (\cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )-\sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )\right )}{4}\right )}{2 a f \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}\)	\(226\)
risch	\(\frac {2 x A c d}{a}-\frac {x A \,d^{2}}{a}+\frac {x B \,c^{2}}{a}-\frac {2 x B c d}{a}+\frac {3 x B \,d^{2}}{2 a}-\frac {d^{2} {\mathrm e}^{i \left (f x +e \right )} A}{2 a f}-\frac {d \,{\mathrm e}^{i \left (f x +e \right )} B c}{a f}+\frac {d^{2} {\mathrm e}^{i \left (f x +e \right )} B}{2 a f}-\frac {d^{2} {\mathrm e}^{-i \left (f x +e \right )} A}{2 a f}-\frac {d \,{\mathrm e}^{-i \left (f x +e \right )} B c}{a f}+\frac {d^{2} {\mathrm e}^{-i \left (f x +e \right )} B}{2 a f}-\frac {2 A \,c^{2}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {4 A c d}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}-\frac {2 A \,d^{2}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {2 B \,c^{2}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}-\frac {4 B c d}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {2 B \,d^{2}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}-\frac {B \,d^{2} \sin \left (2 f x +2 e \right )}{4 a f}\)	\(348\)
norman	\(\frac {\frac {\left (2 A \,c^{2}-4 A c d -2 B \,c^{2}-2 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {\left (2 A \,c^{2}-4 A c d +2 A \,d^{2}-2 B \,c^{2}+4 B c d -3 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}+\frac {\left (6 A \,c^{2}-12 A c d +2 A \,d^{2}-6 B \,c^{2}+4 B c d -5 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}+\frac {\left (6 A \,c^{2}-12 A c d +4 A \,d^{2}-6 B \,c^{2}+8 B c d -6 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {2 A \,d^{2}+4 B c d -B \,d^{2}}{a f}+\frac {\left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x}{2 a}-\frac {\left (4 A \,d^{2}+8 B c d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a f}-\frac {\left (2 A \,d^{2}+4 B c d +B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a f}+\frac {\left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a}+\frac {3 \left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 a}+\frac {3 \left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 a}+\frac {3 \left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2 a}+\frac {3 \left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2 a}+\frac {\left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{2 a}+\frac {\left (4 A c d -2 A \,d^{2}+2 B \,c^{2}-4 B c d +3 B \,d^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{2 a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(678\)

Input:

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e)),x,method=_RETURNV 
ERBOSE)
 

Output:

2/f/a*(-(A*c^2-2*A*c*d+A*d^2-B*c^2+2*B*c*d-B*d^2)/(tan(1/2*f*x+1/2*e)+1)+( 
1/2*B*d^2*tan(1/2*f*x+1/2*e)^3+(-A*d^2-2*B*c*d+B*d^2)*tan(1/2*f*x+1/2*e)^2 
-1/2*B*d^2*tan(1/2*f*x+1/2*e)-A*d^2-2*B*c*d+B*d^2)/(1+tan(1/2*f*x+1/2*e)^2 
)^2+1/2*(4*A*c*d-2*A*d^2+2*B*c^2-4*B*c*d+3*B*d^2)*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. 303 vs. \(2 (139) = 278\).

Time = 0.09 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.12 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx=\frac {B d^{2} \cos \left (f x + e\right )^{3} - 2 \, {\left (A - B\right )} c^{2} + 4 \, {\left (A - B\right )} c d - 2 \, {\left (A - B\right )} d^{2} + {\left (2 \, B c^{2} + 4 \, {\left (A - B\right )} c d - {\left (2 \, A - 3 \, B\right )} d^{2}\right )} f x - 2 \, {\left (2 \, B c d + {\left (A - B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, {\left (A - B\right )} c^{2} - 4 \, {\left (A - 2 \, B\right )} c d + {\left (4 \, A - 3 \, B\right )} d^{2} - {\left (2 \, B c^{2} + 4 \, {\left (A - B\right )} c d - {\left (2 \, A - 3 \, B\right )} d^{2}\right )} f x\right )} \cos \left (f x + e\right ) - {\left (B d^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A - B\right )} c^{2} + 4 \, {\left (A - B\right )} c d - 2 \, {\left (A - B\right )} d^{2} - {\left (2 \, B c^{2} + 4 \, {\left (A - B\right )} c d - {\left (2 \, A - 3 \, B\right )} d^{2}\right )} f x + {\left (4 \, B c d + {\left (2 \, A - B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e)),x, algorith 
m="fricas")
 

Output:

1/2*(B*d^2*cos(f*x + e)^3 - 2*(A - B)*c^2 + 4*(A - B)*c*d - 2*(A - B)*d^2 
+ (2*B*c^2 + 4*(A - B)*c*d - (2*A - 3*B)*d^2)*f*x - 2*(2*B*c*d + (A - B)*d 
^2)*cos(f*x + e)^2 - (2*(A - B)*c^2 - 4*(A - 2*B)*c*d + (4*A - 3*B)*d^2 - 
(2*B*c^2 + 4*(A - B)*c*d - (2*A - 3*B)*d^2)*f*x)*cos(f*x + e) - (B*d^2*cos 
(f*x + e)^2 - 2*(A - B)*c^2 + 4*(A - B)*c*d - 2*(A - B)*d^2 - (2*B*c^2 + 4 
*(A - B)*c*d - (2*A - 3*B)*d^2)*f*x + (4*B*c*d + (2*A - B)*d^2)*cos(f*x + 
e))*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. 5763 vs. \(2 (117) = 234\).

Time = 2.28 (sec) , antiderivative size = 5763, normalized size of antiderivative = 40.30 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e)),x)
 

Output:

Piecewise((-4*A*c**2*tan(e/2 + f*x/2)**4/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a* 
f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2) 
**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 8*A*c**2*tan(e/2 + f*x/2)**2/(2*a* 
f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2) 
**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 4*A*c* 
*2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 
+ f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) 
+ 4*A*c*d*f*x*tan(e/2 + f*x/2)**5/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e 
/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2 
*a*f*tan(e/2 + f*x/2) + 2*a*f) + 4*A*c*d*f*x*tan(e/2 + f*x/2)**4/(2*a*f*ta 
n(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 
+ 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 8*A*c*d*f* 
x*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)* 
*4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 
 + f*x/2) + 2*a*f) + 8*A*c*d*f*x*tan(e/2 + f*x/2)**2/(2*a*f*tan(e/2 + f*x/ 
2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan( 
e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) + 4*A*c*d*f*x*tan(e/2 + 
f*x/2)/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan( 
e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a 
*f) + 4*A*c*d*f*x/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 606 vs. \(2 (139) = 278\).

Time = 0.13 (sec) , antiderivative size = 606, normalized size of antiderivative = 4.24 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e)),x, algorith 
m="maxima")
 

Output:

(B*d^2*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) 
+ 1)^2 + 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x 
 + e) + 1)^4 + 4)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e 
)^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a*sin 
(f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) 
+ 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) - 4*B*c*d*((sin(f*x + e)/(c 
os(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) - 2*A*d^2*((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) + 2*B*c^2*(arctan(sin(f*x + e)/(cos(f*x + 
 e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) + 1))) + 4*A*c*d*(arctan 
(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) 
+ 1))) - 2*A*c^2/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.50 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx=\frac {\frac {{\left (2 \, B c^{2} + 4 \, A c d - 4 \, B c d - 2 \, A d^{2} + 3 \, B d^{2}\right )} {\left (f x + e\right )}}{a} - \frac {4 \, {\left (A c^{2} - B c^{2} - 2 \, A c d + 2 \, B c d + A d^{2} - B d^{2}\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {2 \, {\left (B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, B c d - 2 \, A d^{2} + 2 \, B d^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} a}}{2 \, f} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e)),x, algorith 
m="giac")
 

Output:

1/2*((2*B*c^2 + 4*A*c*d - 4*B*c*d - 2*A*d^2 + 3*B*d^2)*(f*x + e)/a - 4*(A* 
c^2 - B*c^2 - 2*A*c*d + 2*B*c*d + A*d^2 - B*d^2)/(a*(tan(1/2*f*x + 1/2*e) 
+ 1)) + 2*(B*d^2*tan(1/2*f*x + 1/2*e)^3 - 4*B*c*d*tan(1/2*f*x + 1/2*e)^2 - 
 2*A*d^2*tan(1/2*f*x + 1/2*e)^2 + 2*B*d^2*tan(1/2*f*x + 1/2*e)^2 - B*d^2*t 
an(1/2*f*x + 1/2*e) - 4*B*c*d - 2*A*d^2 + 2*B*d^2)/((tan(1/2*f*x + 1/2*e)^ 
2 + 1)^2*a))/f
 

Mupad [B] (verification not implemented)

Time = 40.11 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.08 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx=\frac {x\,\left (2\,B\,c^2-2\,A\,d^2+3\,B\,d^2+4\,A\,c\,d-4\,B\,c\,d\right )}{2\,a}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,A\,d^2-3\,B\,d^2+4\,B\,c\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,A\,c^2+2\,A\,d^2-2\,B\,c^2-3\,B\,d^2-4\,A\,c\,d+4\,B\,c\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (4\,A\,c^2+6\,A\,d^2-4\,B\,c^2-5\,B\,d^2-8\,A\,c\,d+12\,B\,c\,d\right )+2\,A\,c^2+4\,A\,d^2-2\,B\,c^2-4\,B\,d^2+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A\,d^2-B\,d^2+4\,B\,c\,d\right )-4\,A\,c\,d+8\,B\,c\,d}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )} \] Input:

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

Output:

(x*(2*B*c^2 - 2*A*d^2 + 3*B*d^2 + 4*A*c*d - 4*B*c*d))/(2*a) - (tan(e/2 + ( 
f*x)/2)^3*(2*A*d^2 - 3*B*d^2 + 4*B*c*d) + tan(e/2 + (f*x)/2)^4*(2*A*c^2 + 
2*A*d^2 - 2*B*c^2 - 3*B*d^2 - 4*A*c*d + 4*B*c*d) + tan(e/2 + (f*x)/2)^2*(4 
*A*c^2 + 6*A*d^2 - 4*B*c^2 - 5*B*d^2 - 8*A*c*d + 12*B*c*d) + 2*A*c^2 + 4*A 
*d^2 - 2*B*c^2 - 4*B*d^2 + tan(e/2 + (f*x)/2)*(2*A*d^2 - B*d^2 + 4*B*c*d) 
- 4*A*c*d + 8*B*c*d)/(f*(a + a*tan(e/2 + (f*x)/2) + 2*a*tan(e/2 + (f*x)/2) 
^2 + 2*a*tan(e/2 + (f*x)/2)^3 + a*tan(e/2 + (f*x)/2)^4 + a*tan(e/2 + (f*x) 
/2)^5))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.26 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx=\frac {-2 \sin \left (f x +e \right )^{2} b \,d^{2}+2 \sin \left (f x +e \right ) a \,d^{2}+\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b \,d^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,d^{2}-\cos \left (f x +e \right ) \sin \left (f x +e \right ) b \,d^{2}-8 \cos \left (f x +e \right ) a c d +8 \cos \left (f x +e \right ) b c d +2 a \,d^{2} f x -2 b \,c^{2} f x -3 b \,d^{2} f x +4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b c d -4 a c d f x +4 b c d f x -\sin \left (f x +e \right ) b \,d^{2}+4 \sin \left (f x +e \right ) b c d -4 a \,d^{2}+4 b \,c^{2}+6 b \,d^{2}-4 a \,c^{2}-2 \cos \left (f x +e \right ) a \,d^{2} f x +2 \cos \left (f x +e \right ) b \,c^{2} f x +3 \cos \left (f x +e \right ) b \,d^{2} f x +4 \cos \left (f x +e \right ) a c d f x -4 \cos \left (f x +e \right ) b c d f x -4 \sin \left (f x +e \right ) a c d f x +4 \sin \left (f x +e \right ) b c d f x +\sin \left (f x +e \right )^{3} b \,d^{2}+2 \sin \left (f x +e \right )^{2} a \,d^{2}+4 \sin \left (f x +e \right )^{2} b c d +2 \sin \left (f x +e \right ) a \,d^{2} f x -2 \sin \left (f x +e \right ) b \,c^{2} f x -3 \sin \left (f x +e \right ) b \,d^{2} f x +4 \cos \left (f x +e \right ) a \,c^{2}+4 \cos \left (f x +e \right ) a \,d^{2}-4 \cos \left (f x +e \right ) b \,c^{2}-6 \cos \left (f x +e \right ) b \,d^{2}+8 a c d -8 b c d}{2 a f \left (\cos \left (f x +e \right )-\sin \left (f x +e \right )-1\right )} \] Input:

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

Output:

(cos(e + f*x)*sin(e + f*x)**2*b*d**2 + 2*cos(e + f*x)*sin(e + f*x)*a*d**2 
+ 4*cos(e + f*x)*sin(e + f*x)*b*c*d - cos(e + f*x)*sin(e + f*x)*b*d**2 + 4 
*cos(e + f*x)*a*c**2 + 4*cos(e + f*x)*a*c*d*f*x - 8*cos(e + f*x)*a*c*d - 2 
*cos(e + f*x)*a*d**2*f*x + 4*cos(e + f*x)*a*d**2 + 2*cos(e + f*x)*b*c**2*f 
*x - 4*cos(e + f*x)*b*c**2 - 4*cos(e + f*x)*b*c*d*f*x + 8*cos(e + f*x)*b*c 
*d + 3*cos(e + f*x)*b*d**2*f*x - 6*cos(e + f*x)*b*d**2 + sin(e + f*x)**3*b 
*d**2 + 2*sin(e + f*x)**2*a*d**2 + 4*sin(e + f*x)**2*b*c*d - 2*sin(e + f*x 
)**2*b*d**2 - 4*sin(e + f*x)*a*c*d*f*x + 2*sin(e + f*x)*a*d**2*f*x + 2*sin 
(e + f*x)*a*d**2 - 2*sin(e + f*x)*b*c**2*f*x + 4*sin(e + f*x)*b*c*d*f*x + 
4*sin(e + f*x)*b*c*d - 3*sin(e + f*x)*b*d**2*f*x - sin(e + f*x)*b*d**2 - 4 
*a*c**2 - 4*a*c*d*f*x + 8*a*c*d + 2*a*d**2*f*x - 4*a*d**2 - 2*b*c**2*f*x + 
 4*b*c**2 + 4*b*c*d*f*x - 8*b*c*d - 3*b*d**2*f*x + 6*b*d**2)/(2*a*f*(cos(e 
 + f*x) - sin(e + f*x) - 1))