3.468 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=221 \[ \frac{2 d^2 (3 c+2 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^3 (c+d) \sqrt{c^2-d^2}}-\frac{d \left (c^2-6 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))}-\frac{(c-6 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))} \]

[Out]

(2*d^2*(3*c + 2*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^3*(c + d)*Sqrt[c^2 - d^2]*f)
- (d*(c^2 - 6*c*d - 10*d^2)*Cos[e + f*x])/(3*a^2*(c - d)^3*(c + d)*f*(c + d*Sin[e + f*x])) - ((c - 6*d)*Cos[e
+ f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*(c + d*Sin[e + f*x])) - Cos[e + f*x]/(3*(c - d)*f*(a + a*Sin[e
+ f*x])^2*(c + d*Sin[e + f*x]))

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Rubi [A]  time = 0.417081, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2766, 2978, 2754, 12, 2660, 618, 204} \[ \frac{2 d^2 (3 c+2 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^3 (c+d) \sqrt{c^2-d^2}}-\frac{d \left (c^2-6 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))}-\frac{(c-6 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2),x]

[Out]

(2*d^2*(3*c + 2*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^3*(c + d)*Sqrt[c^2 - d^2]*f)
- (d*(c^2 - 6*c*d - 10*d^2)*Cos[e + f*x])/(3*a^2*(c - d)^3*(c + d)*f*(c + d*Sin[e + f*x])) - ((c - 6*d)*Cos[e
+ f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*(c + d*Sin[e + f*x])) - Cos[e + f*x]/(3*(c - d)*f*(a + a*Sin[e
+ f*x])^2*(c + d*Sin[e + f*x]))

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*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] + Dis
t[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*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

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{1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx &=-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\int \frac{-a (c-4 d)-2 a d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx}{3 a^2 (c-d)}\\ &=-\frac{(c-6 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac{\int \frac{10 a^2 d^2+a^2 (c-6 d) d \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{3 a^4 (c-d)^2}\\ &=-\frac{d \left (c^2-6 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac{(c-6 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\int -\frac{3 a^2 d^2 (3 c+2 d)}{c+d \sin (e+f x)} \, dx}{3 a^4 (c-d)^3 (c+d)}\\ &=-\frac{d \left (c^2-6 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac{(c-6 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac{\left (d^2 (3 c+2 d)\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{a^2 (c-d)^3 (c+d)}\\ &=-\frac{d \left (c^2-6 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac{(c-6 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac{\left (2 d^2 (3 c+2 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^2 (c-d)^3 (c+d) f}\\ &=-\frac{d \left (c^2-6 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac{(c-6 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac{\left (4 d^2 (3 c+2 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^2 (c-d)^3 (c+d) f}\\ &=\frac{2 d^2 (3 c+2 d) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{a^2 (c-d)^3 (c+d) \sqrt{c^2-d^2} f}-\frac{d \left (c^2-6 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac{(c-6 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 1.37593, size = 267, normalized size = 1.21 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\frac{6 d^2 (3 c+2 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}+\frac{3 d^3 \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}{(c+d) (c+d \sin (e+f x))}+2 (c-d) \sin \left (\frac{1}{2} (e+f x)\right )+2 (c-7 d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-(c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 a^2 f (c-d)^3 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2),x]

[Out]

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

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Maple [A]  time = 0.104, size = 361, normalized size = 1.6 \begin{align*} 2\,{\frac{{d}^{4}\tan \left ( 1/2\,fx+e/2 \right ) }{{a}^{2}f \left ( c-d \right ) ^{3} \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}^{3}}{{a}^{2}f \left ( c-d \right ) ^{3} \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 ) }}+6\,{\frac{c{d}^{2}}{{a}^{2}f \left ( c-d \right ) ^{3} \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 ) }+4\,{\frac{{d}^{3}}{{a}^{2}f \left ( c-d \right ) ^{3} \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{c}{{a}^{2}f \left ( c-d \right ) ^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+6\,{\frac{d}{{a}^{2}f \left ( c-d \right ) ^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-{\frac{4}{3\,{a}^{2}f \left ( c-d \right ) ^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{{a}^{2}f \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x)

[Out]

2/f/a^2*d^4/(c-d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)/c*tan(1/2*f*x+1/2*e)+2/f/a^2*d^3/(
c-d)^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)+6/f/a^2*d^2/(c-d)^3/(c+d)/(c^2-d^2)^(1/2)*arcta
n(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*c+4/f/a^2*d^3/(c-d)^3/(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))-2/f/a^2/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*c+6/f/a^2/(c-d)^3/(tan(1/2*f
*x+1/2*e)+1)*d-4/3/f/a^2/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3+2/f/a^2/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2

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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(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.27953, size = 4787, normalized size = 21.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/6*(2*c^5 - 2*c^4*d - 4*c^3*d^2 + 4*c^2*d^3 + 2*c*d^4 - 2*d^5 - 2*(c^4*d - 6*c^3*d^2 - 11*c^2*d^3 + 6*c*d^4
+ 10*d^5)*cos(f*x + e)^3 + 2*(c^5 - 5*c^4*d - 8*c^3*d^2 + c^2*d^3 + 7*c*d^4 + 4*d^5)*cos(f*x + e)^2 - 3*(6*c^2
*d^2 + 10*c*d^3 + 4*d^4 - (3*c*d^3 + 2*d^4)*cos(f*x + e)^3 - (3*c^2*d^2 + 8*c*d^3 + 4*d^4)*cos(f*x + e)^2 + (3
*c^2*d^2 + 5*c*d^3 + 2*d^4)*cos(f*x + e) + (6*c^2*d^2 + 10*c*d^3 + 4*d^4 - (3*c*d^3 + 2*d^4)*cos(f*x + e)^2 +
(3*c^2*d^2 + 5*c*d^3 + 2*d^4)*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*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*c^5 - 5*c^4*d - 16*c^3*d^2 - 8*c^2*d^3 + 14*c*d^4 + 13*d
^5)*cos(f*x + e) - 2*(c^5 - c^4*d - 2*c^3*d^2 + 2*c^2*d^3 + c*d^4 - d^5 - (c^4*d - 6*c^3*d^2 - 11*c^2*d^3 + 6*
c*d^4 + 10*d^5)*cos(f*x + e)^2 - (c^5 - 4*c^4*d - 14*c^3*d^2 - 10*c^2*d^3 + 13*c*d^4 + 14*d^5)*cos(f*x + e))*s
in(f*x + e))/((a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2*d^5 - 2*a^2*c*d^6 + a^2*d^7)*
f*cos(f*x + e)^3 + (a^2*c^7 - 5*a^2*c^5*d^2 + 2*a^2*c^4*d^3 + 7*a^2*c^3*d^4 - 4*a^2*c^2*d^5 - 3*a^2*c*d^6 + 2*
a^2*d^7)*f*cos(f*x + e)^2 - (a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d
^5 - a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e) - 2*(a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*
d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f + ((a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2
*c^2*d^5 - 2*a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e)^2 - (a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*
a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e) - 2*(a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3
*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f)*sin(f*x + e)), 1/3*(c^5 - c^4*d - 2*c^3
*d^2 + 2*c^2*d^3 + c*d^4 - d^5 - (c^4*d - 6*c^3*d^2 - 11*c^2*d^3 + 6*c*d^4 + 10*d^5)*cos(f*x + e)^3 + (c^5 - 5
*c^4*d - 8*c^3*d^2 + c^2*d^3 + 7*c*d^4 + 4*d^5)*cos(f*x + e)^2 + 3*(6*c^2*d^2 + 10*c*d^3 + 4*d^4 - (3*c*d^3 +
2*d^4)*cos(f*x + e)^3 - (3*c^2*d^2 + 8*c*d^3 + 4*d^4)*cos(f*x + e)^2 + (3*c^2*d^2 + 5*c*d^3 + 2*d^4)*cos(f*x +
e) + (6*c^2*d^2 + 10*c*d^3 + 4*d^4 - (3*c*d^3 + 2*d^4)*cos(f*x + e)^2 + (3*c^2*d^2 + 5*c*d^3 + 2*d^4)*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))) + (2*c^5 - 5
*c^4*d - 16*c^3*d^2 - 8*c^2*d^3 + 14*c*d^4 + 13*d^5)*cos(f*x + e) - (c^5 - c^4*d - 2*c^3*d^2 + 2*c^2*d^3 + c*d
^4 - d^5 - (c^4*d - 6*c^3*d^2 - 11*c^2*d^3 + 6*c*d^4 + 10*d^5)*cos(f*x + e)^2 - (c^5 - 4*c^4*d - 14*c^3*d^2 -
10*c^2*d^3 + 13*c*d^4 + 14*d^5)*cos(f*x + e))*sin(f*x + e))/((a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*
c^3*d^4 - a^2*c^2*d^5 - 2*a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e)^3 + (a^2*c^7 - 5*a^2*c^5*d^2 + 2*a^2*c^4*d^3 + 7
*a^2*c^3*d^4 - 4*a^2*c^2*d^5 - 3*a^2*c*d^6 + 2*a^2*d^7)*f*cos(f*x + e)^2 - (a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^
2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e) - 2*(a^2*c^7 - a^2*c^6
*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f + ((a^2*c^6*d - 2*
a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2*d^5 - 2*a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e)^2 - (a^2*c^7 -
a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f*cos(f*x +
e) - 2*(a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*
d^7)*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(1/(a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.36131, size = 432, normalized size = 1.95 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (3 \, c d^{2} + 2 \, d^{3}\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 (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} \sqrt{c^{2} - d^{2}}} + \frac{3 \,{\left (d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c d^{3}\right )}}{{\left (a^{2} c^{5} - 2 \, a^{2} c^{4} d + 2 \, a^{2} c^{2} d^{3} - a^{2} c d^{4}\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 )}} - \frac{3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 15 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, c - 8 \, d}{{\left (a^{2} c^{3} - 3 \, a^{2} c^{2} d + 3 \, a^{2} c d^{2} - a^{2} d^{3}\right )}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

2/3*(3*(3*c*d^2 + 2*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^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*sqrt(c^2 - d^2)) + 3*(d^4*tan(1/2*f*x + 1/2*e) +
c*d^3)/((a^2*c^5 - 2*a^2*c^4*d + 2*a^2*c^2*d^3 - a^2*c*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*
e) + c)) - (3*c*tan(1/2*f*x + 1/2*e)^2 - 9*d*tan(1/2*f*x + 1/2*e)^2 + 3*c*tan(1/2*f*x + 1/2*e) - 15*d*tan(1/2*
f*x + 1/2*e) + 2*c - 8*d)/((a^2*c^3 - 3*a^2*c^2*d + 3*a^2*c*d^2 - a^2*d^3)*(tan(1/2*f*x + 1/2*e) + 1)^3))/f