\(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\) [262]

Optimal result

Integrand size = 35, antiderivative size = 246 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\frac {a^3 \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right ) x}{2 d^4}+\frac {2 a^3 (c-d)^3 (B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^4 \sqrt {c^2-d^2} f}+\frac {a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}-\frac {a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac {(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f} \] Output:

1/2*a^3*(A*d*(2*c^2-6*c*d+7*d^2)-B*(2*c^3-6*c^2*d+7*c*d^2-5*d^3))*x/d^4+2* 
a^3*(c-d)^3*(-A*d+B*c)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/d^ 
4/(c^2-d^2)^(1/2)/f+1/2*a^3*(A*(2*c-5*d)*d-B*(2*c^2-5*c*d+5*d^2))*cos(f*x+ 
e)/d^3/f-1/3*a*B*cos(f*x+e)*(a+a*sin(f*x+e))^2/d/f+1/6*(-3*A*d+3*B*c-5*B*d 
)*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d^2/f
 

Mathematica [A] (verified)

Time = 3.87 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.95 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\frac {a^3 (1+\sin (e+f x))^3 \left (6 \left (A d \left (2 c^2-6 c d+7 d^2\right )+B \left (-2 c^3+6 c^2 d-7 c d^2+5 d^3\right )\right ) (e+f x)+\frac {24 (c-d)^3 (B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}-3 d \left (4 A d (-c+3 d)+B \left (4 c^2-12 c d+15 d^2\right )\right ) \cos (e+f x)+B d^3 \cos (3 (e+f x))-3 d^2 (-B c+A d+3 B d) \sin (2 (e+f x))\right )}{12 d^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6} \] Input:

Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x 
]),x]
 

Output:

(a^3*(1 + Sin[e + f*x])^3*(6*(A*d*(2*c^2 - 6*c*d + 7*d^2) + B*(-2*c^3 + 6* 
c^2*d - 7*c*d^2 + 5*d^3))*(e + f*x) + (24*(c - d)^3*(B*c - A*d)*ArcTan[(d 
+ c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] - 3*d*(4*A*d*(-c + 
 3*d) + B*(4*c^2 - 12*c*d + 15*d^2))*Cos[e + f*x] + B*d^3*Cos[3*(e + f*x)] 
 - 3*d^2*(-(B*c) + A*d + 3*B*d)*Sin[2*(e + f*x)]))/(12*d^4*f*(Cos[(e + f*x 
)/2] + Sin[(e + f*x)/2])^6)
 

Rubi [A] (verified)

Time = 1.71 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.09, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3455, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^2 (a (2 B c+3 A d)-a (3 B c-3 A d-5 B d) \sin (e+f x))}{c+d \sin (e+f x)}dx}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^2 (a (2 B c+3 A d)-a (3 B c-3 A d-5 B d) \sin (e+f x))}{c+d \sin (e+f x)}dx}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {\frac {\int -\frac {3 (\sin (e+f x) a+a) \left ((B c (c-3 d)-A d (c+2 d)) a^2+\left (A (2 c-5 d) d-B \left (2 c^2-5 d c+5 d^2\right )\right ) \sin (e+f x) a^2\right )}{c+d \sin (e+f x)}dx}{2 d}+\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \int \frac {(\sin (e+f x) a+a) \left ((B c (c-3 d)-A d (c+2 d)) a^2+\left (A (2 c-5 d) d-B \left (2 c^2-5 d c+5 d^2\right )\right ) \sin (e+f x) a^2\right )}{c+d \sin (e+f x)}dx}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \int \frac {(\sin (e+f x) a+a) \left ((B c (c-3 d)-A d (c+2 d)) a^2+\left (A (2 c-5 d) d-B \left (2 c^2-5 d c+5 d^2\right )\right ) \sin (e+f x) a^2\right )}{c+d \sin (e+f x)}dx}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \int \frac {\left (A (2 c-5 d) d-B \left (2 c^2-5 d c+5 d^2\right )\right ) \sin ^2(e+f x) a^3+(B c (c-3 d)-A d (c+2 d)) a^3+\left ((B c (c-3 d)-A d (c+2 d)) a^3+\left (A (2 c-5 d) d-B \left (2 c^2-5 d c+5 d^2\right )\right ) a^3\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \int \frac {\left (A (2 c-5 d) d-B \left (2 c^2-5 d c+5 d^2\right )\right ) \sin (e+f x)^2 a^3+(B c (c-3 d)-A d (c+2 d)) a^3+\left ((B c (c-3 d)-A d (c+2 d)) a^3+\left (A (2 c-5 d) d-B \left (2 c^2-5 d c+5 d^2\right )\right ) a^3\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \left (\frac {\int \frac {a^3 d (B c (c-3 d)-A d (c+2 d))-a^3 \left (A d \left (2 c^2-6 d c+7 d^2\right )-B \left (2 c^3-6 d c^2+7 d^2 c-5 d^3\right )\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d}-\frac {a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{d f}\right )}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \left (\frac {\int \frac {a^3 d (B c (c-3 d)-A d (c+2 d))-a^3 \left (A d \left (2 c^2-6 d c+7 d^2\right )-B \left (2 c^3-6 d c^2+7 d^2 c-5 d^3\right )\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d}-\frac {a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{d f}\right )}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \left (\frac {-\frac {2 a^3 (c-d)^3 (B c-A d) \int \frac {1}{c+d \sin (e+f x)}dx}{d}-\frac {a^3 x \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right )}{d}}{d}-\frac {a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{d f}\right )}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \left (\frac {-\frac {2 a^3 (c-d)^3 (B c-A d) \int \frac {1}{c+d \sin (e+f x)}dx}{d}-\frac {a^3 x \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right )}{d}}{d}-\frac {a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{d f}\right )}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \left (\frac {-\frac {4 a^3 (c-d)^3 (B c-A d) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{d f}-\frac {a^3 x \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right )}{d}}{d}-\frac {a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{d f}\right )}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \left (\frac {\frac {8 a^3 (c-d)^3 (B c-A d) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f}-\frac {a^3 x \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right )}{d}}{d}-\frac {a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{d f}\right )}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}-\frac {3 \left (\frac {-\frac {4 a^3 (c-d)^3 (B c-A d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}}-\frac {a^3 x \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right )}{d}}{d}-\frac {a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{d f}\right )}{2 d}}{3 d}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f}\)

Input:

Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x]),x]
 

Output:

-1/3*(a*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(d*f) + ((-3*((-((a^3*(A*d* 
(2*c^2 - 6*c*d + 7*d^2) - B*(2*c^3 - 6*c^2*d + 7*c*d^2 - 5*d^3))*x)/d) - ( 
4*a^3*(c - d)^3*(B*c - A*d)*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^ 
2 - d^2])])/(d*Sqrt[c^2 - d^2]*f))/d - (a^3*(A*(2*c - 5*d)*d - B*(2*c^2 - 
5*c*d + 5*d^2))*Cos[e + f*x])/(d*f)))/(2*d) + ((3*B*c - 3*A*d - 5*B*d)*Cos 
[e + f*x]*(a^3 + a^3*Sin[e + f*x]))/(2*d*f))/(3*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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 3447

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 3455

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[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 
 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
 

Maple [A] (verified)

Time = 2.09 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.54

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\left (-A \,c^{3} d +3 A \,c^{2} d^{2}-3 A c \,d^{3}+A \,d^{4}+B \,c^{4}-3 B \,c^{3} d +3 B \,c^{2} d^{2}-B c \,d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{4} \sqrt {c^{2}-d^{2}}}+\frac {\frac {\left (\frac {1}{2} A \,d^{3}-\frac {1}{2} B c \,d^{2}+\frac {3}{2} B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (A c \,d^{2}-3 A \,d^{3}-B \,c^{2} d +3 B c \,d^{2}-3 B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (2 A c \,d^{2}-6 A \,d^{3}-2 B \,c^{2} d +6 B c \,d^{2}-8 B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-\frac {1}{2} A \,d^{3}+\frac {1}{2} B c \,d^{2}-\frac {3}{2} B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+A c \,d^{2}-3 A \,d^{3}-B \,c^{2} d +3 B c \,d^{2}-\frac {11 B \,d^{3}}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}+\frac {\left (2 A \,c^{2} d -6 A c \,d^{2}+7 A \,d^{3}-2 B \,c^{3}+6 B \,c^{2} d -7 B c \,d^{2}+5 B \,d^{3}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{4}}\right )}{f}\)	\(378\)
default	\(\frac {2 a^{3} \left (\frac {\left (-A \,c^{3} d +3 A \,c^{2} d^{2}-3 A c \,d^{3}+A \,d^{4}+B \,c^{4}-3 B \,c^{3} d +3 B \,c^{2} d^{2}-B c \,d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{4} \sqrt {c^{2}-d^{2}}}+\frac {\frac {\left (\frac {1}{2} A \,d^{3}-\frac {1}{2} B c \,d^{2}+\frac {3}{2} B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (A c \,d^{2}-3 A \,d^{3}-B \,c^{2} d +3 B c \,d^{2}-3 B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (2 A c \,d^{2}-6 A \,d^{3}-2 B \,c^{2} d +6 B c \,d^{2}-8 B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-\frac {1}{2} A \,d^{3}+\frac {1}{2} B c \,d^{2}-\frac {3}{2} B \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+A c \,d^{2}-3 A \,d^{3}-B \,c^{2} d +3 B c \,d^{2}-\frac {11 B \,d^{3}}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}+\frac {\left (2 A \,c^{2} d -6 A c \,d^{2}+7 A \,d^{3}-2 B \,c^{3}+6 B \,c^{2} d -7 B c \,d^{2}+5 B \,d^{3}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{4}}\right )}{f}\)	\(378\)
risch	\(\text {Expression too large to display}\)	\(1143\)

Input:

int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x,method=_RETURNV 
ERBOSE)
 

Output:

2/f*a^3*((-A*c^3*d+3*A*c^2*d^2-3*A*c*d^3+A*d^4+B*c^4-3*B*c^3*d+3*B*c^2*d^2 
-B*c*d^3)/d^4/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2 
-d^2)^(1/2))+1/d^4*(((1/2*A*d^3-1/2*B*c*d^2+3/2*B*d^3)*tan(1/2*f*x+1/2*e)^ 
5+(A*c*d^2-3*A*d^3-B*c^2*d+3*B*c*d^2-3*B*d^3)*tan(1/2*f*x+1/2*e)^4+(2*A*c* 
d^2-6*A*d^3-2*B*c^2*d+6*B*c*d^2-8*B*d^3)*tan(1/2*f*x+1/2*e)^2+(-1/2*A*d^3+ 
1/2*B*c*d^2-3/2*B*d^3)*tan(1/2*f*x+1/2*e)+A*c*d^2-3*A*d^3-B*c^2*d+3*B*c*d^ 
2-11/3*B*d^3)/(1+tan(1/2*f*x+1/2*e)^2)^3+1/2*(2*A*c^2*d-6*A*c*d^2+7*A*d^3- 
2*B*c^3+6*B*c^2*d-7*B*c*d^2+5*B*d^3)*arctan(tan(1/2*f*x+1/2*e))))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 626, normalized size of antiderivative = 2.54 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\left [\frac {2 \, B a^{3} d^{3} \cos \left (f x + e\right )^{3} - 3 \, {\left (2 \, B a^{3} c^{3} - 2 \, {\left (A + 3 \, B\right )} a^{3} c^{2} d + {\left (6 \, A + 7 \, B\right )} a^{3} c d^{2} - {\left (7 \, A + 5 \, B\right )} a^{3} d^{3}\right )} f x + 3 \, {\left (B a^{3} c d^{2} - {\left (A + 3 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, {\left (B a^{3} c^{3} - {\left (A + 2 \, B\right )} a^{3} c^{2} d + {\left (2 \, A + B\right )} a^{3} c d^{2} - A a^{3} d^{3}\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 6 \, {\left (B a^{3} c^{2} d - {\left (A + 3 \, B\right )} a^{3} c d^{2} + {\left (3 \, A + 4 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right )}{6 \, d^{4} f}, \frac {2 \, B a^{3} d^{3} \cos \left (f x + e\right )^{3} - 3 \, {\left (2 \, B a^{3} c^{3} - 2 \, {\left (A + 3 \, B\right )} a^{3} c^{2} d + {\left (6 \, A + 7 \, B\right )} a^{3} c d^{2} - {\left (7 \, A + 5 \, B\right )} a^{3} d^{3}\right )} f x + 3 \, {\left (B a^{3} c d^{2} - {\left (A + 3 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 6 \, {\left (B a^{3} c^{3} - {\left (A + 2 \, B\right )} a^{3} c^{2} d + {\left (2 \, A + B\right )} a^{3} c d^{2} - A a^{3} d^{3}\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) - 6 \, {\left (B a^{3} c^{2} d - {\left (A + 3 \, B\right )} a^{3} c d^{2} + {\left (3 \, A + 4 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right )}{6 \, d^{4} f}\right ] \] Input:

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorith 
m="fricas")
 

Output:

[1/6*(2*B*a^3*d^3*cos(f*x + e)^3 - 3*(2*B*a^3*c^3 - 2*(A + 3*B)*a^3*c^2*d 
+ (6*A + 7*B)*a^3*c*d^2 - (7*A + 5*B)*a^3*d^3)*f*x + 3*(B*a^3*c*d^2 - (A + 
 3*B)*a^3*d^3)*cos(f*x + e)*sin(f*x + e) - 3*(B*a^3*c^3 - (A + 2*B)*a^3*c^ 
2*d + (2*A + B)*a^3*c*d^2 - A*a^3*d^3)*sqrt(-(c - d)/(c + d))*log(((2*c^2 
- d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*((c^2 + c*d)*co 
s(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos(f*x + e))*sqrt(-(c - d)/(c + d)) 
)/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) - 6*(B*a^3*c^2*d 
- (A + 3*B)*a^3*c*d^2 + (3*A + 4*B)*a^3*d^3)*cos(f*x + e))/(d^4*f), 1/6*(2 
*B*a^3*d^3*cos(f*x + e)^3 - 3*(2*B*a^3*c^3 - 2*(A + 3*B)*a^3*c^2*d + (6*A 
+ 7*B)*a^3*c*d^2 - (7*A + 5*B)*a^3*d^3)*f*x + 3*(B*a^3*c*d^2 - (A + 3*B)*a 
^3*d^3)*cos(f*x + e)*sin(f*x + e) - 6*(B*a^3*c^3 - (A + 2*B)*a^3*c^2*d + ( 
2*A + B)*a^3*c*d^2 - A*a^3*d^3)*sqrt((c - d)/(c + d))*arctan(-(c*sin(f*x + 
 e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))) - 6*(B*a^3*c^2*d - 
(A + 3*B)*a^3*c*d^2 + (3*A + 4*B)*a^3*d^3)*cos(f*x + e))/(d^4*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorith 
m="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (233) = 466\).

Time = 0.24 (sec) , antiderivative size = 597, normalized size of antiderivative = 2.43 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorith 
m="giac")
 

Output:

-1/6*(3*(2*B*a^3*c^3 - 2*A*a^3*c^2*d - 6*B*a^3*c^2*d + 6*A*a^3*c*d^2 + 7*B 
*a^3*c*d^2 - 7*A*a^3*d^3 - 5*B*a^3*d^3)*(f*x + e)/d^4 - 12*(B*a^3*c^4 - A* 
a^3*c^3*d - 3*B*a^3*c^3*d + 3*A*a^3*c^2*d^2 + 3*B*a^3*c^2*d^2 - 3*A*a^3*c* 
d^3 - B*a^3*c*d^3 + A*a^3*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + 
arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/(sqrt(c^2 - d^2)*d^4 
) + 2*(3*B*a^3*c*d*tan(1/2*f*x + 1/2*e)^5 - 3*A*a^3*d^2*tan(1/2*f*x + 1/2* 
e)^5 - 9*B*a^3*d^2*tan(1/2*f*x + 1/2*e)^5 + 6*B*a^3*c^2*tan(1/2*f*x + 1/2* 
e)^4 - 6*A*a^3*c*d*tan(1/2*f*x + 1/2*e)^4 - 18*B*a^3*c*d*tan(1/2*f*x + 1/2 
*e)^4 + 18*A*a^3*d^2*tan(1/2*f*x + 1/2*e)^4 + 18*B*a^3*d^2*tan(1/2*f*x + 1 
/2*e)^4 + 12*B*a^3*c^2*tan(1/2*f*x + 1/2*e)^2 - 12*A*a^3*c*d*tan(1/2*f*x + 
 1/2*e)^2 - 36*B*a^3*c*d*tan(1/2*f*x + 1/2*e)^2 + 36*A*a^3*d^2*tan(1/2*f*x 
 + 1/2*e)^2 + 48*B*a^3*d^2*tan(1/2*f*x + 1/2*e)^2 - 3*B*a^3*c*d*tan(1/2*f* 
x + 1/2*e) + 3*A*a^3*d^2*tan(1/2*f*x + 1/2*e) + 9*B*a^3*d^2*tan(1/2*f*x + 
1/2*e) + 6*B*a^3*c^2 - 6*A*a^3*c*d - 18*B*a^3*c*d + 18*A*a^3*d^2 + 22*B*a^ 
3*d^2)/((tan(1/2*f*x + 1/2*e)^2 + 1)^3*d^3))/f
 

Mupad [B] (verification not implemented)

Time = 43.81 (sec) , antiderivative size = 10256, normalized size of antiderivative = 41.69 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\text {Too large to display} \] Input:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c + d*sin(e + f*x)),x)
 

Output:

- ((2*(9*A*a^3*d^2 + 3*B*a^3*c^2 + 11*B*a^3*d^2 - 3*A*a^3*c*d - 9*B*a^3*c* 
d))/(3*d^3) - (tan(e/2 + (f*x)/2)^5*(A*a^3*d - B*a^3*c + 3*B*a^3*d))/d^2 + 
 (4*tan(e/2 + (f*x)/2)^2*(3*A*a^3*d^2 + B*a^3*c^2 + 4*B*a^3*d^2 - A*a^3*c* 
d - 3*B*a^3*c*d))/d^3 + (2*tan(e/2 + (f*x)/2)^4*(3*A*a^3*d^2 + B*a^3*c^2 + 
 3*B*a^3*d^2 - A*a^3*c*d - 3*B*a^3*c*d))/d^3 + (tan(e/2 + (f*x)/2)*(A*a^3* 
d - B*a^3*c + 3*B*a^3*d))/d^2)/(f*(3*tan(e/2 + (f*x)/2)^2 + 3*tan(e/2 + (f 
*x)/2)^4 + tan(e/2 + (f*x)/2)^6 + 1)) - (atan(((((8*(49*A^2*a^6*c^2*d^9 - 
84*A^2*a^6*c^3*d^8 + 64*A^2*a^6*c^4*d^7 - 24*A^2*a^6*c^5*d^6 + 4*A^2*a^6*c 
^6*d^5 + 25*B^2*a^6*c^2*d^9 - 70*B^2*a^6*c^3*d^8 + 109*B^2*a^6*c^4*d^7 - 1 
04*B^2*a^6*c^5*d^6 + 64*B^2*a^6*c^6*d^5 - 24*B^2*a^6*c^7*d^4 + 4*B^2*a^6*c 
^8*d^3 + 70*A*B*a^6*c^2*d^9 - 158*A*B*a^6*c^3*d^8 + 188*A*B*a^6*c^4*d^7 - 
128*A*B*a^6*c^5*d^6 + 48*A*B*a^6*c^6*d^5 - 8*A*B*a^6*c^7*d^4))/d^8 + (8*ta 
n(e/2 + (f*x)/2)*(19*A^2*a^6*c^3*d^9 - 144*A^2*a^6*c^2*d^10 + 116*A^2*a^6* 
c^4*d^8 - 116*A^2*a^6*c^5*d^7 + 48*A^2*a^6*c^6*d^6 - 8*A^2*a^6*c^7*d^5 - 1 
40*B^2*a^6*c^2*d^10 + 189*B^2*a^6*c^3*d^9 - 114*B^2*a^6*c^4*d^8 - 41*B^2*a 
^6*c^5*d^7 + 136*B^2*a^6*c^6*d^6 - 116*B^2*a^6*c^7*d^5 + 48*B^2*a^6*c^8*d^ 
4 - 8*B^2*a^6*c^9*d^3 + 94*A^2*a^6*c*d^11 + 50*B^2*a^6*c*d^11 - 308*A*B*a^ 
6*c^2*d^10 + 258*A*B*a^6*c^3*d^9 + 22*A*B*a^6*c^4*d^8 - 252*A*B*a^6*c^5*d^ 
7 + 232*A*B*a^6*c^6*d^6 - 96*A*B*a^6*c^7*d^5 + 16*A*B*a^6*c^8*d^4 + 140*A* 
B*a^6*c*d^11))/d^9 + ((((32*c^2*d^3 + (8*tan(e/2 + (f*x)/2)*(12*c*d^13 ...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 714, normalized size of antiderivative = 2.90 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)
 

Output:

(a**3*( - 12*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d 
**2))*a*c**2*d + 24*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c 
**2 - d**2))*a*c*d**2 - 12*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d) 
/sqrt(c**2 - d**2))*a*d**3 + 12*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c 
 + d)/sqrt(c**2 - d**2))*b*c**3 - 24*sqrt(c**2 - d**2)*atan((tan((e + f*x) 
/2)*c + d)/sqrt(c**2 - d**2))*b*c**2*d + 12*sqrt(c**2 - d**2)*atan((tan((e 
 + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b*c*d**2 - 2*cos(e + f*x)*sin(e + f*x 
)**2*b*c*d**3 - 2*cos(e + f*x)*sin(e + f*x)**2*b*d**4 - 3*cos(e + f*x)*sin 
(e + f*x)*a*c*d**3 - 3*cos(e + f*x)*sin(e + f*x)*a*d**4 + 3*cos(e + f*x)*s 
in(e + f*x)*b*c**2*d**2 - 6*cos(e + f*x)*sin(e + f*x)*b*c*d**3 - 9*cos(e + 
 f*x)*sin(e + f*x)*b*d**4 + 6*cos(e + f*x)*a*c**2*d**2 - 12*cos(e + f*x)*a 
*c*d**3 - 18*cos(e + f*x)*a*d**4 - 6*cos(e + f*x)*b*c**3*d + 12*cos(e + f* 
x)*b*c**2*d**2 - 4*cos(e + f*x)*b*c*d**3 - 22*cos(e + f*x)*b*d**4 + 6*a*c* 
*3*d*e + 6*a*c**3*d*f*x - 12*a*c**2*d**2*e - 12*a*c**2*d**2*f*x + 2*a*c**2 
*d**2 + 3*a*c*d**3*e + 3*a*c*d**3*f*x - 4*a*c*d**3 + 21*a*d**4*e + 21*a*d* 
*4*f*x - 6*a*d**4 - 6*b*c**4*e - 6*b*c**4*f*x + 12*b*c**3*d*e + 12*b*c**3* 
d*f*x - 2*b*c**3*d - 3*b*c**2*d**2*e - 3*b*c**2*d**2*f*x + 4*b*c**2*d**2 - 
 6*b*c*d**3*e - 6*b*c*d**3*f*x - 4*b*c*d**3 + 15*b*d**4*e + 15*b*d**4*f*x 
- 10*b*d**4))/(6*d**4*f*(c + d))