3.442 \(\int \frac{(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=207 \[ \frac{a^2 (3 c-2 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c-d) (c+d)^3 \sqrt{c^2-d^2}}-\frac{a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 d f (c-d) (c+d)^3 (c+d \sin (e+f x))}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d f (c+d)^2 (c+d \sin (e+f x))^2}+\frac{a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.314724, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2762, 2754, 12, 2660, 618, 204} \[ \frac{a^2 (3 c-2 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c-d) (c+d)^3 \sqrt{c^2-d^2}}-\frac{a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 d f (c-d) (c+d)^3 (c+d \sin (e+f x))}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d f (c+d)^2 (c+d \sin (e+f x))^2}+\frac{a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^4,x]

[Out]

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

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && 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+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx &=\frac{a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a \int \frac{-6 a d-a (c+5 d) \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx}{3 d (c+d)}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}+\frac{a \int \frac{10 a (c-d) d+a (c-d) (c+6 d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{6 (c-d) d (c+d)^2}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}-\frac{a \int -\frac{3 a (3 c-2 d) (c-d) d}{c+d \sin (e+f x)} \, dx}{6 (c-d)^2 d (c+d)^3}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}+\frac{\left (a^2 (3 c-2 d)\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 (c-d) (c+d)^3}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}+\frac{\left (a^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 )}{(c-d) (c+d)^3 f}\\ &=\frac{a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}-\frac{\left (2 a^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 )}{(c-d) (c+d)^3 f}\\ &=\frac{a^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 )}{(c-d) (c+d)^3 \sqrt{c^2-d^2} f}+\frac{a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac{a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac{a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.43868, size = 196, normalized size = 0.95 \[ \frac{a^2 \cos (e+f x) \left (-\frac{d (\sin (e+f x)+1)^2}{(c+d \sin (e+f x))^3}-\frac{(3 c-2 d) \left (\frac{6 \tan ^{-1}\left (\frac{\sqrt{d-c} \sqrt{1-\sin (e+f x)}}{\sqrt{-c-d} \sqrt{\sin (e+f x)+1}}\right )}{\sqrt{-c-d} \sqrt{d-c}}-\frac{\sqrt{\cos ^2(e+f x)} ((c+4 d) \sin (e+f x)+4 c+d)}{(c+d \sin (e+f x))^2}\right )}{2 (c+d)^2 \sqrt{\cos ^2(e+f x)}}\right )}{3 f (d-c) (c+d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^4,x]

[Out]

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

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Maple [B]  time = 0.155, size = 2425, normalized size = 11.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x)

[Out]

1/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5
-4/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^
4-4/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*c^3+2/3/f*a^2/(c*tan(1
/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*d^3-18/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*t
an(1/2*f*x+1/2*e)*d+c)^3*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)*d+14/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2
*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)*d^2+2/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+
2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)*d^4-6/f*a^2/(c*tan(1/2*f*x+1/2*e)^2
+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5*d+2/f*a^2/(c*tan(1/2*f*x+1/2*e
)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5*d^4+3/f*a^2/(c*tan(1/2*f*x+1/
2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d-18/f*a^2/(c*tan(1/2*f*
x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d^2+12/f*a^2/(c*tan(1/
2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d^4+4/f*a^2/(c*tan
(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d^5-24/f*a^2/
(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2*d/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+14/f*
a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c*d^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+4
0/3/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c*d^4/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*
e)^3+8/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2*d^5/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x
+1/2*e)^3+8/3/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^3*d^6/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(
1/2*f*x+1/2*e)^3+4/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan
(1/2*f*x+1/2*e)^2*d-8/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*
tan(1/2*f*x+1/2*e)^2+7/3/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*c
^2*d+2/f*a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*c*d^2+4/f*a^2/(c*ta
n(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5*d^3+8/f*a^2/(c*t
an(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d^3-4/f*a^2/(c*
tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*d^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+22/f*a^2/(
c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^2*d^3+8/f*a^2/
(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)*d^3+3/f*a^2/(
c^4+2*c^3*d-2*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))*c-2/f*a^2/(c
^4+2*c^3*d-2*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))*d-1/f*a^2/(c*
tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)-24/f*a^2/(c*
tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^2*d^2+12/f*a^2
/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^2*d^4+4/f*
a^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^2*d^5

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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+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.43455, size = 2874, normalized size = 13.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

[-1/12*(2*(a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6*a^2*c*d^4 + 10*a^2*d^5)*cos(f*x + e)^3 - 6*(a^2*c^5
+ 6*a^2*c^4*d - 8*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x + e)*sin(f*x + e) - 3*(3*a^2*
c^4 - 2*a^2*c^3*d + 9*a^2*c^2*d^2 - 6*a^2*c*d^3 - 3*(3*a^2*c^2*d^2 - 2*a^2*c*d^3)*cos(f*x + e)^2 + (9*a^2*c^3*
d - 6*a^2*c^2*d^2 + 3*a^2*c*d^3 - 2*a^2*d^4 - (3*a^2*c*d^3 - 2*a^2*d^4)*cos(f*x + e)^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)) - 12*(2*a^2*c^5 - a^
2*c^4*d - 2*a^2*c^3*d^2 - a^2*c^2*d^3 + 2*a^2*d^5)*cos(f*x + e))/(3*(c^7*d^2 + 2*c^6*d^3 - c^5*d^4 - 4*c^4*d^5
- c^3*d^6 + 2*c^2*d^7 + c*d^8)*f*cos(f*x + e)^2 - (c^9 + 2*c^8*d + 2*c^7*d^2 + 2*c^6*d^3 - 4*c^5*d^4 - 10*c^4
*d^5 - 2*c^3*d^6 + 6*c^2*d^7 + 3*c*d^8)*f + ((c^6*d^3 + 2*c^5*d^4 - c^4*d^5 - 4*c^3*d^6 - c^2*d^7 + 2*c*d^8 +
d^9)*f*cos(f*x + e)^2 - (3*c^8*d + 6*c^7*d^2 - 2*c^6*d^3 - 10*c^5*d^4 - 4*c^4*d^5 + 2*c^3*d^6 + 2*c^2*d^7 + 2*
c*d^8 + d^9)*f)*sin(f*x + e)), -1/6*((a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6*a^2*c*d^4 + 10*a^2*d^5)*c
os(f*x + e)^3 - 3*(a^2*c^5 + 6*a^2*c^4*d - 8*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x +
e)*sin(f*x + e) - 3*(3*a^2*c^4 - 2*a^2*c^3*d + 9*a^2*c^2*d^2 - 6*a^2*c*d^3 - 3*(3*a^2*c^2*d^2 - 2*a^2*c*d^3)*c
os(f*x + e)^2 + (9*a^2*c^3*d - 6*a^2*c^2*d^2 + 3*a^2*c*d^3 - 2*a^2*d^4 - (3*a^2*c*d^3 - 2*a^2*d^4)*cos(f*x + e
)^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))) - 6*(2*a^2*c^5
- a^2*c^4*d - 2*a^2*c^3*d^2 - a^2*c^2*d^3 + 2*a^2*d^5)*cos(f*x + e))/(3*(c^7*d^2 + 2*c^6*d^3 - c^5*d^4 - 4*c^
4*d^5 - c^3*d^6 + 2*c^2*d^7 + c*d^8)*f*cos(f*x + e)^2 - (c^9 + 2*c^8*d + 2*c^7*d^2 + 2*c^6*d^3 - 4*c^5*d^4 - 1
0*c^4*d^5 - 2*c^3*d^6 + 6*c^2*d^7 + 3*c*d^8)*f + ((c^6*d^3 + 2*c^5*d^4 - c^4*d^5 - 4*c^3*d^6 - c^2*d^7 + 2*c*d
^8 + d^9)*f*cos(f*x + e)^2 - (3*c^8*d + 6*c^7*d^2 - 2*c^6*d^3 - 10*c^5*d^4 - 4*c^4*d^5 + 2*c^3*d^6 + 2*c^2*d^7
+ 2*c*d^8 + d^9)*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+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.76368, size = 1054, normalized size = 5.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x, algorithm="giac")

[Out]

1/3*(3*(3*a^2*c - 2*a^2*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^4 + 2*c^3*d - 2*c*d^3 - d^4)*sqrt(c^2 - d^2)) + (3*a^2*c^6*tan(1/2*f*x + 1/2*e)^5 - 18*a^2*c
^5*d*tan(1/2*f*x + 1/2*e)^5 + 12*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^5 + 6*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^5 - 1
2*a^2*c^6*tan(1/2*f*x + 1/2*e)^4 + 9*a^2*c^5*d*tan(1/2*f*x + 1/2*e)^4 - 54*a^2*c^4*d^2*tan(1/2*f*x + 1/2*e)^4
+ 24*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^4 + 36*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^4 + 12*a^2*c*d^5*tan(1/2*f*x + 1
/2*e)^4 - 72*a^2*c^5*d*tan(1/2*f*x + 1/2*e)^3 + 42*a^2*c^4*d^2*tan(1/2*f*x + 1/2*e)^3 - 12*a^2*c^3*d^3*tan(1/2
*f*x + 1/2*e)^3 + 40*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*c*d^5*tan(1/2*f*x + 1/2*e)^3 + 8*a^2*d^6*tan(
1/2*f*x + 1/2*e)^3 - 24*a^2*c^6*tan(1/2*f*x + 1/2*e)^2 + 12*a^2*c^5*d*tan(1/2*f*x + 1/2*e)^2 - 72*a^2*c^4*d^2*
tan(1/2*f*x + 1/2*e)^2 + 66*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^2 + 36*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^2 + 12*a^
2*c*d^5*tan(1/2*f*x + 1/2*e)^2 - 3*a^2*c^6*tan(1/2*f*x + 1/2*e) - 54*a^2*c^5*d*tan(1/2*f*x + 1/2*e) + 42*a^2*c
^4*d^2*tan(1/2*f*x + 1/2*e) + 24*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e) + 6*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e) - 12*a^
2*c^6 + 7*a^2*c^5*d + 6*a^2*c^4*d^2 + 2*a^2*c^3*d^3)/((c^7 + 2*c^6*d - 2*c^4*d^3 - c^3*d^4)*(c*tan(1/2*f*x + 1
/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^3))/f