3.270 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=181 \[ -\frac{2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}+\frac{d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]
[Out]
(-2*(A*d*(2*c + d) - B*(c^2 + c*d + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a*(c - d)*(c^2 -
d^2)^(3/2)*f) + (d*(B*(2*c + d) - A*(c + 2*d))*Cos[e + f*x])/(a*(c - d)^2*(c + d)*f*(c + d*Sin[e + f*x])) - ((
A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x]))
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Rubi [A] time = 0.349443, antiderivative size = 181, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used =
{2978, 2754, 12, 2660, 618, 204} \[ -\frac{2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}+\frac{d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^2),x]
[Out]
(-2*(A*d*(2*c + d) - B*(c^2 + c*d + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a*(c - d)*(c^2 -
d^2)^(3/2)*f) + (d*(B*(2*c + d) - A*(c + 2*d))*Cos[e + f*x])/(a*(c - d)^2*(c + d)*f*(c + d*Sin[e + f*x])) - ((
A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])*(c + d*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 2754
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 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)) (c+d \sin (e+f x))^2} \, dx &=-\frac{(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac{\int \frac{a (2 A d-B (c+d))-a (A-B) d \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{a^2 (c-d)}\\ &=\frac{d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac{\int \frac{a \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right )}{c+d \sin (e+f x)} \, dx}{a^2 (c-d)^2 (c+d)}\\ &=\frac{d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac{\left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{a (c-d)^2 (c+d)}\\ &=\frac{d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}-\frac{\left (2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right )\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 (c-d)^2 (c+d) f}\\ &=\frac{d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}+\frac{\left (4 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right )\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 (c-d)^2 (c+d) f}\\ &=-\frac{2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{a (c-d)^2 (c+d) \sqrt{c^2-d^2} f}+\frac{d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.23584, size = 209, normalized size = 1.15 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\frac{2 \left (B \left (c^2+c d+d^2\right )-A d (2 c+d)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}+\frac{d (B c-A d) \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}+2 (A-B) \sin \left (\frac{1}{2} (e+f x)\right )\right )}{a f (c-d)^2 (\sin (e+f x)+1)} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(A - B)*Sin[(e + f*x)/2] + (2*(-(A*d*(2*c + d)) + B*(c^2 + c*d + d^2
))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/((c + d)*Sqrt[c^2 -
d^2]) + (d*(B*c - A*d)*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x]))))/(
a*(c - d)^2*f*(1 + Sin[e + f*x]))
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Maple [B] time = 0.141, size = 615, normalized size = 3.4 \begin{align*} -2\,{\frac{{d}^{3}\tan \left ( 1/2\,fx+e/2 \right ) A}{af \left ( c-d \right ) ^{2} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) c}}+2\,{\frac{{d}^{2}\tan \left ( 1/2\,fx+e/2 \right ) B}{af \left ( c-d \right ) ^{2} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) }}-2\,{\frac{A{d}^{2}}{af \left ( c-d \right ) ^{2} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) }}+2\,{\frac{Bcd}{af \left ( c-d \right ) ^{2} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) }}-4\,{\frac{Acd}{af \left ( c-d \right ) ^{2} \left ( c+d \right ) \sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{A{d}^{2}}{af \left ( c-d \right ) ^{2} \left ( c+d \right ) \sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }+2\,{\frac{B{c}^{2}}{af \left ( c-d \right ) ^{2} \left ( c+d \right ) \sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }+2\,{\frac{Bcd}{af \left ( c-d \right ) ^{2} \left ( c+d \right ) \sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }+2\,{\frac{B{d}^{2}}{af \left ( c-d \right ) ^{2} \left ( c+d \right ) \sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{A}{af \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{B}{af \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \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))/(c+d*sin(f*x+e))^2,x)
[Out]
-2/a/f/(c-d)^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)*d^3/(c+d)/c*tan(1/2*f*x+1/2*e)*A+2/a/f/(c-d)^
2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)*d^2/(c+d)*tan(1/2*f*x+1/2*e)*B-2/a/f/(c-d)^2/(c*tan(1/2*f*
x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)*d^2/(c+d)*A+2/a/f/(c-d)^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+
c)*d/(c+d)*B*c-4/a/f/(c-d)^2/(c+d)/(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*
c*d-2/a/f/(c-d)^2/(c+d)/(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*d^2+2/a/f/(
c-d)^2/(c+d)/(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^2+2/a/f/(c-d)^2/(c+d
)/(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*d+2/a/f/(c-d)^2/(c+d)/(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*d^2-2/a/f/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)*A+2
/a/f/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)*B
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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))/(c+d*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.45626, size = 3247, normalized size = 17.94 \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))/(c+d*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
[1/2*(2*(A - B)*c^4 - 4*(A - B)*c^2*d^2 + 2*(A - B)*d^4 + 2*((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c
*d^3 - (2*A - B)*d^4)*cos(f*x + e)^2 + (B*c^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 - (B*c^2*d -
(2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e)^2 + (B*c^3 - (2*A - B)*c^2*d - (A - B)*c*d^2)*cos(f*x + e) + (B*c
^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 + (B*c^2*d - (2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e
))*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*co
s(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*((A - B)*c^4 + (A - 2*B)*c^3*d + B*c^2*d^2 - (A - 2*B)*c*d^3 - A*d^4)*cos(f*x + e) - 2*((A - B)*c^4 -
2*(A - B)*c^2*d^2 + (A - B)*d^4 - ((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos
(f*x + e))*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e)^2
- (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e) - (a*c^6 - 3*a*c^4*d^2 +
3*a*c^2*d^4 - a*d^6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e)
+ (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f)*sin(f*x + e)), ((A - B)*c^4 - 2*(A - B)*c^2*d^2 + (A - B)*d^4
+ ((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e)^2 + (B*c^3 - 2*(A - B)
*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 - (B*c^2*d - (2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e)^2 + (B*c^3 -
(2*A - B)*c^2*d - (A - B)*c*d^2)*cos(f*x + e) + (B*c^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 +
(B*c^2*d - (2*A - B)*c*d^2 - (A - B)*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))) + ((A - B)*c^4 + (A - 2*B)*c^3*d + B*c^2*d^2 - (A - 2*B)*c*d^3 - A*d^4)*c
os(f*x + e) - ((A - B)*c^4 - 2*(A - B)*c^2*d^2 + (A - B)*d^4 - ((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B
)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e))*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d
^5 - a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x +
e) - (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*
d^5 - a*d^6)*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f)*sin(f*x + e))]
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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))/(c+d*sin(f*x+e))**2,x)
[Out]
Timed out
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Giac [B] time = 1.30671, size = 983, normalized size = 5.43 \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))/(c+d*sin(f*x+e))^2,x, algorithm="giac")
[Out]
-((B*a*c^5 - 2*A*a*c^4*d + A*a*c^3*d^2 - B*a*c^3*d^2 + 3*A*a*c^2*d^3 - B*a*c^2*d^3 - A*a*c*d^4 - A*a*d^5 + B*a
*d^5)*sqrt(-c^2 + d^2)*log(abs((d + sqrt(-c^2 + d^2))*tan(1/2*f*x + 1/2*e) + c))/(a^2*c^8 - 2*a^2*c^7*d - 2*a^
2*c^6*d^2 + 6*a^2*c^5*d^3 - 6*a^2*c^3*d^5 + 2*a^2*c^2*d^6 + 2*a^2*c*d^7 - a^2*d^8) - (B*a*c^5 - 2*A*a*c^4*d +
A*a*c^3*d^2 - B*a*c^3*d^2 + 3*A*a*c^2*d^3 - B*a*c^2*d^3 - A*a*c*d^4 - A*a*d^5 + B*a*d^5)*sqrt(-c^2 + d^2)*log(
abs(-(d - sqrt(-c^2 + d^2))*tan(1/2*f*x + 1/2*e) - c))/(a^2*c^8 - 2*a^2*c^7*d - 2*a^2*c^6*d^2 + 6*a^2*c^5*d^3
- 6*a^2*c^3*d^5 + 2*a^2*c^2*d^6 + 2*a^2*c*d^7 - a^2*d^8) + 2*(A*c^3*tan(1/2*f*x + 1/2*e)^2 - B*c^3*tan(1/2*f*x
+ 1/2*e)^2 + A*c^2*d*tan(1/2*f*x + 1/2*e)^2 - B*c^2*d*tan(1/2*f*x + 1/2*e)^2 - B*c*d^2*tan(1/2*f*x + 1/2*e)^2
+ A*d^3*tan(1/2*f*x + 1/2*e)^2 + 2*A*c^2*d*tan(1/2*f*x + 1/2*e) - 3*B*c^2*d*tan(1/2*f*x + 1/2*e) + 3*A*c*d^2*
tan(1/2*f*x + 1/2*e) - 3*B*c*d^2*tan(1/2*f*x + 1/2*e) + A*d^3*tan(1/2*f*x + 1/2*e) + A*c^3 - B*c^3 + A*c^2*d -
2*B*c^2*d + A*c*d^2)/((a*c^4 - a*c^3*d - a*c^2*d^2 + a*c*d^3)*(c*tan(1/2*f*x + 1/2*e)^3 + c*tan(1/2*f*x + 1/2
*e)^2 + 2*d*tan(1/2*f*x + 1/2*e)^2 + c*tan(1/2*f*x + 1/2*e) + 2*d*tan(1/2*f*x + 1/2*e) + c)))/f