\(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\) [250]

Optimal result

Integrand size = 33, antiderivative size = 176 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\frac {a (2 A c+B c-A d-2 B d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \left (c^2-d^2\right )^{3/2} f}+\frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3047, 3100, 2833, 12, 2739, 632, 210} \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\frac {a (2 A c-A d+B c-2 B d) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d) \left (c^2-d^2\right )^{3/2}}-\frac {a \left (A d (c-2 d)+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 d f (c-d) (c+d)^2 (c+d \sin (e+f x))}+\frac {a (B c-A d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

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 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {a A+(a A+a B) \sin (e+f x)+a B \sin ^2(e+f x)}{(c+d \sin (e+f x))^3} \, dx \\ & = \frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {\int \frac {-2 a (A+B) (c-d) d-a (c-d) (A d+B (c+2 d)) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 d \left (c^2-d^2\right )} \\ & = \frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}+\frac {\int \frac {a (c-d) d (2 A c+B c-A d-2 B d)}{c+d \sin (e+f x)} \, dx}{2 d \left (c^2-d^2\right )^2} \\ & = \frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}+\frac {(a (2 A c+B c-A d-2 B d)) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 (c-d) (c+d)^2} \\ & = \frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}+\frac {(a (2 A c+B c-A d-2 B d)) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(c-d) (c+d)^2 f} \\ & = \frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))}-\frac {(2 a (2 A c+B c-A d-2 B d)) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{(c-d) (c+d)^2 f} \\ & = \frac {a (2 A c+B c-A d-2 B d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c-d) (c+d)^2 \sqrt {c^2-d^2} f}+\frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+f x)}{2 (c-d) d (c+d)^2 f (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.99 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.96 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\frac {a (1+\sin (e+f x)) \left (\frac {4 (2 A c+B c-A d-2 B d) \arctan \left (\frac {\sec \left (\frac {f x}{2}\right ) (\cos (e)-i \sin (e)) \left (d \cos \left (e+\frac {f x}{2}\right )+c \sin \left (\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right ) (\cos (e)-i \sin (e))}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}+\frac {\left (2 c^2+d^2\right ) \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cot (e)+d \csc (e) \left (-d \left (A (c-2 d) d+B \left (c^2+2 c d-2 d^2\right )\right ) \cos (e+2 f x)+\left (B c \left (2 c^2+6 c d-5 d^2\right )-A d \left (-4 c^2+6 c d+d^2\right )\right ) \sin (f x)+\left (A d^2 (-2 c+d)+B c \left (2 c^2+2 c d-3 d^2\right )\right ) \sin (2 e+f x)\right )}{d^2 (c+d \sin (e+f x))^2}\right )}{4 (c-d) (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2} \]

[In]

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

[Out]

(a*(1 + Sin[e + f*x])*((4*(2*A*c + B*c - A*d - 2*B*d)*ArcTan[(Sec[(f*x)/2]*(Cos[e] - I*Sin[e])*(d*Cos[e + (f*x
)/2] + c*Sin[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(Cos[e] - I*Sin[e]))/(Sqrt[c^2 - d^2]*S
qrt[(Cos[e] - I*Sin[e])^2]) + ((2*c^2 + d^2)*(A*(c - 2*d)*d + B*(c^2 + 2*c*d - 2*d^2))*Cot[e] + d*Csc[e]*(-(d*
(A*(c - 2*d)*d + B*(c^2 + 2*c*d - 2*d^2))*Cos[e + 2*f*x]) + (B*c*(2*c^2 + 6*c*d - 5*d^2) - A*d*(-4*c^2 + 6*c*d
+ d^2))*Sin[f*x] + (A*d^2*(-2*c + d) + B*c*(2*c^2 + 2*c*d - 3*d^2))*Sin[2*e + f*x]))/(d^2*(c + d*Sin[e + f*x]
)^2)))/(4*(c - d)*(c + d)^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(423\) vs. \(2(167)=334\).

Time = 1.10 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.41

method	result	size
derivativedivides	\(\frac {2 a \left (\frac {-\frac {\left (3 c^{2} d A -2 d^{2} c A -2 A \,d^{3}-B \,c^{3}+2 c^{2} d B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right )}-\frac {\left (2 A \,c^{4}-2 A \,c^{3} d +3 A \,c^{2} d^{2}-4 A c \,d^{3}-2 A \,d^{4}+2 B \,c^{4}-B \,c^{3} d +4 B \,c^{2} d^{2}-2 B c \,d^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right ) c^{2}}-\frac {\left (5 c^{2} d A -6 d^{2} c A -2 A \,d^{3}+B \,c^{3}+6 c^{2} d B -4 d^{2} c B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right )}-\frac {2 A \,c^{2}-2 A c d -A \,d^{2}+2 B \,c^{2}-c d B}{2 \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (2 A c -d A +B c -2 d B \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(424\)
default	\(\frac {2 a \left (\frac {-\frac {\left (3 c^{2} d A -2 d^{2} c A -2 A \,d^{3}-B \,c^{3}+2 c^{2} d B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right )}-\frac {\left (2 A \,c^{4}-2 A \,c^{3} d +3 A \,c^{2} d^{2}-4 A c \,d^{3}-2 A \,d^{4}+2 B \,c^{4}-B \,c^{3} d +4 B \,c^{2} d^{2}-2 B c \,d^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right ) c^{2}}-\frac {\left (5 c^{2} d A -6 d^{2} c A -2 A \,d^{3}+B \,c^{3}+6 c^{2} d B -4 d^{2} c B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right )}-\frac {2 A \,c^{2}-2 A c d -A \,d^{2}+2 B \,c^{2}-c d B}{2 \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (2 A c -d A +B c -2 d B \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{3}+c^{2} d -d^{2} c -d^{3}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(424\)
risch	\(\text {Expression too large to display}\)	\(1073\)

[In]

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

[Out]

2/f*a*((-1/2*(3*A*c^2*d-2*A*c*d^2-2*A*d^3-B*c^3+2*B*c^2*d)/c/(c^3+c^2*d-c*d^2-d^3)*tan(1/2*f*x+1/2*e)^3-1/2*(2
*A*c^4-2*A*c^3*d+3*A*c^2*d^2-4*A*c*d^3-2*A*d^4+2*B*c^4-B*c^3*d+4*B*c^2*d^2-2*B*c*d^3)/(c^3+c^2*d-c*d^2-d^3)/c^
2*tan(1/2*f*x+1/2*e)^2-1/2*(5*A*c^2*d-6*A*c*d^2-2*A*d^3+B*c^3+6*B*c^2*d-4*B*c*d^2)/c/(c^3+c^2*d-c*d^2-d^3)*tan
(1/2*f*x+1/2*e)-1/2*(2*A*c^2-2*A*c*d-A*d^2+2*B*c^2-B*c*d)/(c^3+c^2*d-c*d^2-d^3))/(tan(1/2*f*x+1/2*e)^2*c+2*d*t
an(1/2*f*x+1/2*e)+c)^2+1/2*(2*A*c-A*d+B*c-2*B*d)/(c^3+c^2*d-c*d^2-d^3)/(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)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (167) = 334\).

Time = 0.30 (sec) , antiderivative size = 967, normalized size of antiderivative = 5.49 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 570 vs. \(2 (167) = 334\).

Time = 0.33 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.24 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\frac {\frac {{\left (2 \, A a c + B a c - A a d - 2 \, B a d\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c^{3} + c^{2} d - c d^{2} - d^{3}\right )} \sqrt {c^{2} - d^{2}}} + \frac {B a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, A a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, B a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, A a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, A a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, A a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, B a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, A a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + B a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, A a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, B a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, A a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, B a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, A a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 \, A a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, B a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, A a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, B a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A a c^{4} - 2 \, B a c^{4} + 2 \, A a c^{3} d + B a c^{3} d + A a c^{2} d^{2}}{{\left (c^{5} + c^{4} d - c^{3} d^{2} - c^{2} d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{2}}}{f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.75 (sec) , antiderivative size = 554, normalized size of antiderivative = 3.15 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=-\frac {\frac {A\,a\,d^2-2\,A\,a\,c^2-2\,B\,a\,c^2+2\,A\,a\,c\,d+B\,a\,c\,d}{-c^3-c^2\,d+c\,d^2+d^3}+\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A\,d^3-B\,c^3+6\,A\,c\,d^2-5\,A\,c^2\,d+4\,B\,c\,d^2-6\,B\,c^2\,d\right )}{c\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,A\,d^3+B\,c^3+2\,A\,c\,d^2-3\,A\,c^2\,d-2\,B\,c^2\,d\right )}{c\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^2+2\,d^2\right )\,\left (A\,d^2-2\,A\,c^2-2\,B\,c^2+2\,A\,c\,d+B\,c\,d\right )}{c^2\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2+4\,d^2\right )+c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+c^2+4\,c\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}-\frac {a\,\mathrm {atan}\left (\frac {\left (\frac {a\,\left (2\,A\,c-A\,d+B\,c-2\,B\,d\right )\,\left (-2\,c^3\,d-2\,c^2\,d^2+2\,c\,d^3+2\,d^4\right )}{2\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{3/2}\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}+\frac {a\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A\,c-A\,d+B\,c-2\,B\,d\right )}{{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{3/2}}\right )\,\left (-c^3-c^2\,d+c\,d^2+d^3\right )}{2\,A\,a\,c-A\,a\,d+B\,a\,c-2\,B\,a\,d}\right )\,\left (2\,A\,c-A\,d+B\,c-2\,B\,d\right )}{f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{3/2}} \]

[In]

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

[Out]

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