3.283 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx\)
Optimal. Leaf size=229 \[ \frac{2 d^2 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^3 f (c-d)^3 \sqrt{c^2-d^2}}-\frac{\left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}-\frac{(2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac{(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3} \]
[Out]
(2*d^2*(B*c - A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^3*(c - d)^3*Sqrt[c^2 - d^2]*f) - ((A -
B)*Cos[e + f*x])/(5*(c - d)*f*(a + a*Sin[e + f*x])^3) - ((2*A*c + 3*B*c - 7*A*d + 2*B*d)*Cos[e + f*x])/(15*a*
(c - d)^2*f*(a + a*Sin[e + f*x])^2) - ((B*(3*c^2 - 16*c*d - 2*d^2) + A*(2*c^2 - 9*c*d + 22*d^2))*Cos[e + f*x])
/(15*(c - d)^3*f*(a^3 + a^3*Sin[e + f*x]))
________________________________________________________________________________________
Rubi [A] time = 0.723971, antiderivative size = 229, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{2978, 12, 2660, 618, 204} \[ \frac{2 d^2 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^3 f (c-d)^3 \sqrt{c^2-d^2}}-\frac{\left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}-\frac{(2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac{(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])),x]
[Out]
(2*d^2*(B*c - A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^3*(c - d)^3*Sqrt[c^2 - d^2]*f) - ((A -
B)*Cos[e + f*x])/(5*(c - d)*f*(a + a*Sin[e + f*x])^3) - ((2*A*c + 3*B*c - 7*A*d + 2*B*d)*Cos[e + f*x])/(15*a*
(c - d)^2*f*(a + a*Sin[e + f*x])^2) - ((B*(3*c^2 - 16*c*d - 2*d^2) + A*(2*c^2 - 9*c*d + 22*d^2))*Cos[e + f*x])
/(15*(c - d)^3*f*(a^3 + a^3*Sin[e + f*x]))
Rule 2978
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*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] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 2660
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 618
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 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx &=-\frac{(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac{\int \frac{-a (2 A c+3 B c-5 A d)-2 a (A-B) d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx}{5 a^2 (c-d)}\\ &=-\frac{(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac{(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}+\frac{\int \frac{a^2 \left (B c (3 c-13 d)+A \left (2 c^2-7 c d+15 d^2\right )\right )+a^2 d (2 A c+3 B c-7 A d+2 B d) \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx}{15 a^4 (c-d)^2}\\ &=-\frac{(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac{(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac{\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{\int -\frac{15 a^3 d^2 (B c-A d)}{c+d \sin (e+f x)} \, dx}{15 a^6 (c-d)^3}\\ &=-\frac{(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac{(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac{\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{\left (d^2 (B c-A d)\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^3}\\ &=-\frac{(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac{(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac{\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{\left (2 d^2 (B c-A d)\right ) \operatorname{Subst}\left (\int \frac{1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{a^3 (c-d)^3 f}\\ &=-\frac{(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac{(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac{\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{\left (4 d^2 (B c-A d)\right ) \operatorname{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 )}{a^3 (c-d)^3 f}\\ &=\frac{2 d^2 (B c-A d) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{a^3 (c-d)^3 \sqrt{c^2-d^2} f}-\frac{(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac{(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac{\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end{align*}
Mathematica [B] time = 1.24329, size = 502, normalized size = 2.19 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-\frac{60 d^2 (A d-B c) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}+20 A c^2 \sin \left (\frac{1}{2} (e+f x)\right )-2 A c^2 \sin \left (\frac{5}{2} (e+f x)\right )-10 A c^2 \cos \left (\frac{3}{2} (e+f x)\right )-75 A c d \sin \left (\frac{1}{2} (e+f x)\right )+9 A c d \sin \left (\frac{5}{2} (e+f x)\right )-15 A c d \cos \left (\frac{1}{2} (e+f x)\right )+45 A c d \cos \left (\frac{3}{2} (e+f x)\right )+145 A d^2 \sin \left (\frac{1}{2} (e+f x)\right )+15 A d^2 \sin \left (\frac{3}{2} (e+f x)\right )-22 A d^2 \sin \left (\frac{5}{2} (e+f x)\right )+75 A d^2 \cos \left (\frac{1}{2} (e+f x)\right )-95 A d^2 \cos \left (\frac{3}{2} (e+f x)\right )+15 B c^2 \sin \left (\frac{1}{2} (e+f x)\right )-3 B c^2 \sin \left (\frac{5}{2} (e+f x)\right )+15 B c^2 \cos \left (\frac{1}{2} (e+f x)\right )-15 B c^2 \cos \left (\frac{3}{2} (e+f x)\right )-85 B c d \sin \left (\frac{1}{2} (e+f x)\right )-15 B c d \sin \left (\frac{3}{2} (e+f x)\right )+16 B c d \sin \left (\frac{5}{2} (e+f x)\right )-75 B c d \cos \left (\frac{1}{2} (e+f x)\right )+65 B c d \cos \left (\frac{3}{2} (e+f x)\right )-20 B d^2 \sin \left (\frac{1}{2} (e+f x)\right )+2 B d^2 \sin \left (\frac{5}{2} (e+f x)\right )+10 B d^2 \cos \left (\frac{3}{2} (e+f x)\right )\right )}{30 a^3 f (c-d)^3 (\sin (e+f x)+1)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(15*B*c^2*Cos[(e + f*x)/2] - 15*A*c*d*Cos[(e + f*x)/2] - 75*B*c*d*Cos[(
e + f*x)/2] + 75*A*d^2*Cos[(e + f*x)/2] - 10*A*c^2*Cos[(3*(e + f*x))/2] - 15*B*c^2*Cos[(3*(e + f*x))/2] + 45*A
*c*d*Cos[(3*(e + f*x))/2] + 65*B*c*d*Cos[(3*(e + f*x))/2] - 95*A*d^2*Cos[(3*(e + f*x))/2] + 10*B*d^2*Cos[(3*(e
+ f*x))/2] + 20*A*c^2*Sin[(e + f*x)/2] + 15*B*c^2*Sin[(e + f*x)/2] - 75*A*c*d*Sin[(e + f*x)/2] - 85*B*c*d*Sin
[(e + f*x)/2] + 145*A*d^2*Sin[(e + f*x)/2] - 20*B*d^2*Sin[(e + f*x)/2] - (60*d^2*(-(B*c) + A*d)*ArcTan[(d + c*
Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)/Sqrt[c^2 - d^2] - 15*B*c*d*Sin[(3*
(e + f*x))/2] + 15*A*d^2*Sin[(3*(e + f*x))/2] - 2*A*c^2*Sin[(5*(e + f*x))/2] - 3*B*c^2*Sin[(5*(e + f*x))/2] +
9*A*c*d*Sin[(5*(e + f*x))/2] + 16*B*c*d*Sin[(5*(e + f*x))/2] - 22*A*d^2*Sin[(5*(e + f*x))/2] + 2*B*d^2*Sin[(5*
(e + f*x))/2]))/(30*a^3*(c - d)^3*f*(1 + Sin[e + f*x])^3)
________________________________________________________________________________________
Maple [B] time = 0.138, size = 606, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e)),x)
[Out]
-2/f/a^3*d^3/(c-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))*A+2/f/a^3*d^2/(c
-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))*B*c+4/f/a^3/(c-d)/(tan(1/2*f*x+
1/2*e)+1)^4*A-4/f/a^3/(c-d)/(tan(1/2*f*x+1/2*e)+1)^4*B-8/5/f/a^3/(c-d)/(tan(1/2*f*x+1/2*e)+1)^5*A+8/5/f/a^3/(c
-d)/(tan(1/2*f*x+1/2*e)+1)^5*B+4/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*A*c-6/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e
)+1)^2*A*d-2/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*B*c+4/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*B*d-16/3/f/a^
3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*A*c+20/3/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*A*d+4/f/a^3/(c-d)^2/(tan(1/
2*f*x+1/2*e)+1)^3*B*c-16/3/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*B*d-2/f/a^3/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*A
*c^2+6/f/a^3/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*A*c*d-6/f/a^3/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*A*d^2+2/f/a^3/(c-d)^3
/(tan(1/2*f*x+1/2*e)+1)*B*d^2
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 2.63274, size = 4929, normalized size = 21.52 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e)),x, algorithm="fricas")
[Out]
[1/30*(6*(A - B)*c^4 - 12*(A - B)*c^3*d + 12*(A - B)*c*d^3 - 6*(A - B)*d^4 - 2*((2*A + 3*B)*c^4 - (9*A + 16*B)
*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^3 + 2*(2*(2*A + 3*B)*c^4 -
(18*A + 17*B)*c^3*d + 5*(5*A - 2*B)*c^2*d^2 + (18*A + 17*B)*c*d^3 - (29*A - 4*B)*d^4)*cos(f*x + e)^2 + 15*(4*B
*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^3 - 3*(B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B*c*d^2 - A*d^3)
*cos(f*x + e) + (4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B*c*d^2 - A*d^3)*cos(f*x + e))*si
n(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))
+ 6*((3*A + 2*B)*c^4 - (11*A + 9*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (11*A + 9*B)*c*d^3 - 3*(6*A - B)*d^4)*cos(f*
x + e) - 2*(3*(A - B)*c^4 - 6*(A - B)*c^3*d + 6*(A - B)*c*d^3 - 3*(A - B)*d^4 - ((2*A + 3*B)*c^4 - (9*A + 16*B
)*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^2 - 3*((2*A + 3*B)*c^4 - (
9*A + 11*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (9*A + 11*B)*c*d^3 - (17*A - 2*B)*d^4)*cos(f*x + e))*sin(f*x + e))/(
(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^3 + 3*(a^3*c^5
- 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c
^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^
3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f + ((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3
- 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*
c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3
*d^5)*f)*sin(f*x + e)), 1/15*(3*(A - B)*c^4 - 6*(A - B)*c^3*d + 6*(A - B)*c*d^3 - 3*(A - B)*d^4 - ((2*A + 3*B)
*c^4 - (9*A + 16*B)*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^3 + (2*(
2*A + 3*B)*c^4 - (18*A + 17*B)*c^3*d + 5*(5*A - 2*B)*c^2*d^2 + (18*A + 17*B)*c*d^3 - (29*A - 4*B)*d^4)*cos(f*x
+ e)^2 + 15*(4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^3 - 3*(B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*
(B*c*d^2 - A*d^3)*cos(f*x + e) + (4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B*c*d^2 - A*d^3)
*cos(f*x + e))*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))) + 3*
((3*A + 2*B)*c^4 - (11*A + 9*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (11*A + 9*B)*c*d^3 - 3*(6*A - B)*d^4)*cos(f*x +
e) - (3*(A - B)*c^4 - 6*(A - B)*c^3*d + 6*(A - B)*c*d^3 - 3*(A - B)*d^4 - ((2*A + 3*B)*c^4 - (9*A + 16*B)*c^3*
d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^2 - 3*((2*A + 3*B)*c^4 - (9*A +
11*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (9*A + 11*B)*c*d^3 - (17*A - 2*B)*d^4)*cos(f*x + e))*sin(f*x + e))/((a^3*c
^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^3 + 3*(a^3*c^5 - 3*a^
3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d +
2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*
d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f + ((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a
^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4
+ a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*
f)*sin(f*x + e))]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c+d*sin(f*x+e)),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 1.28868, size = 780, normalized size = 3.41 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e)),x, algorithm="giac")
[Out]
2/15*(15*(B*c*d^2 - A*d^3)*(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)))/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*sqrt(c^2 - d^2)) - (15*A*c^2*tan(1/2*f*x + 1/2*
e)^4 - 45*A*c*d*tan(1/2*f*x + 1/2*e)^4 + 45*A*d^2*tan(1/2*f*x + 1/2*e)^4 - 15*B*d^2*tan(1/2*f*x + 1/2*e)^4 + 3
0*A*c^2*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^3 - 105*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 45*B*c*d
*tan(1/2*f*x + 1/2*e)^3 + 135*A*d^2*tan(1/2*f*x + 1/2*e)^3 - 30*B*d^2*tan(1/2*f*x + 1/2*e)^3 + 40*A*c^2*tan(1/
2*f*x + 1/2*e)^2 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^2 - 135*A*c*d*tan(1/2*f*x + 1/2*e)^2 - 65*B*c*d*tan(1/2*f*x +
1/2*e)^2 + 185*A*d^2*tan(1/2*f*x + 1/2*e)^2 - 40*B*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*A*c^2*tan(1/2*f*x + 1/2*e)
+ 15*B*c^2*tan(1/2*f*x + 1/2*e) - 75*A*c*d*tan(1/2*f*x + 1/2*e) - 55*B*c*d*tan(1/2*f*x + 1/2*e) + 115*A*d^2*t
an(1/2*f*x + 1/2*e) - 20*B*d^2*tan(1/2*f*x + 1/2*e) + 7*A*c^2 + 3*B*c^2 - 24*A*c*d - 11*B*c*d + 32*A*d^2 - 7*B
*d^2)/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*(tan(1/2*f*x + 1/2*e) + 1)^5))/f