3.564 \(\int \frac{A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx\)
Optimal. Leaf size=185 \[ \frac{\left (2 a^2 A-3 a c C+A c^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{e \left (a^2-c^2\right )^{5/2}}+\frac{\left (a^2 (-C)+3 a A c-2 c^2 C\right ) \cos (d+e x)}{2 e \left (a^2-c^2\right )^2 (a+c \sin (d+e x))}+\frac{(A c-a C) \cos (d+e x)}{2 e \left (a^2-c^2\right ) (a+c \sin (d+e x))^2}-\frac{B}{2 c e (a+c \sin (d+e x))^2} \]
[Out]
((2*a^2*A + A*c^2 - 3*a*c*C)*ArcTan[(c + a*Tan[(d + e*x)/2])/Sqrt[a^2 - c^2]])/((a^2 - c^2)^(5/2)*e) - B/(2*c*
e*(a + c*Sin[d + e*x])^2) + ((A*c - a*C)*Cos[d + e*x])/(2*(a^2 - c^2)*e*(a + c*Sin[d + e*x])^2) + ((3*a*A*c -
a^2*C - 2*c^2*C)*Cos[d + e*x])/(2*(a^2 - c^2)^2*e*(a + c*Sin[d + e*x]))
________________________________________________________________________________________
Rubi [A] time = 0.246184, antiderivative size = 185, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used =
{4376, 2754, 12, 2660, 618, 204, 2668, 32} \[ \frac{\left (2 a^2 A-3 a c C+A c^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{e \left (a^2-c^2\right )^{5/2}}+\frac{\left (a^2 (-C)+3 a A c-2 c^2 C\right ) \cos (d+e x)}{2 e \left (a^2-c^2\right )^2 (a+c \sin (d+e x))}+\frac{(A c-a C) \cos (d+e x)}{2 e \left (a^2-c^2\right ) (a+c \sin (d+e x))^2}-\frac{B}{2 c e (a+c \sin (d+e x))^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + c*Sin[d + e*x])^3,x]
[Out]
((2*a^2*A + A*c^2 - 3*a*c*C)*ArcTan[(c + a*Tan[(d + e*x)/2])/Sqrt[a^2 - c^2]])/((a^2 - c^2)^(5/2)*e) - B/(2*c*
e*(a + c*Sin[d + e*x])^2) + ((A*c - a*C)*Cos[d + e*x])/(2*(a^2 - c^2)*e*(a + c*Sin[d + e*x])^2) + ((3*a*A*c -
a^2*C - 2*c^2*C)*Cos[d + e*x])/(2*(a^2 - c^2)^2*e*(a + c*Sin[d + e*x]))
Rule 4376
Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Sin[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Cos[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Sin[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] && !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Cos] || EqQ[F, cos])
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])
Rule 2668
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Rule 32
Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]
Rubi steps
\begin{align*} \int \frac{A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx &=B \int \frac{\cos (d+e x)}{(a+c \sin (d+e x))^3} \, dx+\int \frac{A+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx\\ &=\frac{(A c-a C) \cos (d+e x)}{2 \left (a^2-c^2\right ) e (a+c \sin (d+e x))^2}-\frac{\int \frac{-2 (a A-c C)+(A c-a C) \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx}{2 \left (a^2-c^2\right )}+\frac{B \operatorname{Subst}\left (\int \frac{1}{(a+x)^3} \, dx,x,c \sin (d+e x)\right )}{c e}\\ &=-\frac{B}{2 c e (a+c \sin (d+e x))^2}+\frac{(A c-a C) \cos (d+e x)}{2 \left (a^2-c^2\right ) e (a+c \sin (d+e x))^2}+\frac{\left (3 a A c-a^2 C-2 c^2 C\right ) \cos (d+e x)}{2 \left (a^2-c^2\right )^2 e (a+c \sin (d+e x))}+\frac{\int \frac{2 a^2 A+A c^2-3 a c C}{a+c \sin (d+e x)} \, dx}{2 \left (a^2-c^2\right )^2}\\ &=-\frac{B}{2 c e (a+c \sin (d+e x))^2}+\frac{(A c-a C) \cos (d+e x)}{2 \left (a^2-c^2\right ) e (a+c \sin (d+e x))^2}+\frac{\left (3 a A c-a^2 C-2 c^2 C\right ) \cos (d+e x)}{2 \left (a^2-c^2\right )^2 e (a+c \sin (d+e x))}+\frac{\left (2 a^2 A+A c^2-3 a c C\right ) \int \frac{1}{a+c \sin (d+e x)} \, dx}{2 \left (a^2-c^2\right )^2}\\ &=-\frac{B}{2 c e (a+c \sin (d+e x))^2}+\frac{(A c-a C) \cos (d+e x)}{2 \left (a^2-c^2\right ) e (a+c \sin (d+e x))^2}+\frac{\left (3 a A c-a^2 C-2 c^2 C\right ) \cos (d+e x)}{2 \left (a^2-c^2\right )^2 e (a+c \sin (d+e x))}+\frac{\left (2 a^2 A+A c^2-3 a c C\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 c x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{\left (a^2-c^2\right )^2 e}\\ &=-\frac{B}{2 c e (a+c \sin (d+e x))^2}+\frac{(A c-a C) \cos (d+e x)}{2 \left (a^2-c^2\right ) e (a+c \sin (d+e x))^2}+\frac{\left (3 a A c-a^2 C-2 c^2 C\right ) \cos (d+e x)}{2 \left (a^2-c^2\right )^2 e (a+c \sin (d+e x))}-\frac{\left (2 \left (2 a^2 A+A c^2-3 a c C\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-c^2\right )-x^2} \, dx,x,2 c+2 a \tan \left (\frac{1}{2} (d+e x)\right )\right )}{\left (a^2-c^2\right )^2 e}\\ &=\frac{\left (2 a^2 A+A c^2-3 a c C\right ) \tan ^{-1}\left (\frac{c+a \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-c^2}}\right )}{\left (a^2-c^2\right )^{5/2} e}-\frac{B}{2 c e (a+c \sin (d+e x))^2}+\frac{(A c-a C) \cos (d+e x)}{2 \left (a^2-c^2\right ) e (a+c \sin (d+e x))^2}+\frac{\left (3 a A c-a^2 C-2 c^2 C\right ) \cos (d+e x)}{2 \left (a^2-c^2\right )^2 e (a+c \sin (d+e x))}\\ \end{align*}
Mathematica [A] time = 0.956972, size = 174, normalized size = 0.94 \[ \frac{\frac{B \left (c^2-a^2\right )+c (A c-a C) \cos (d+e x)}{c (a-c) (a+c) (a+c \sin (d+e x))^2}+\frac{2 \left (2 a^2 A-3 a c C+A c^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{\left (a^2-c^2\right )^{5/2}}-\frac{\left (a^2 C-3 a A c+2 c^2 C\right ) \cos (d+e x)}{(a-c)^2 (a+c)^2 (a+c \sin (d+e x))}}{2 e} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + c*Sin[d + e*x])^3,x]
[Out]
((2*(2*a^2*A + A*c^2 - 3*a*c*C)*ArcTan[(c + a*Tan[(d + e*x)/2])/Sqrt[a^2 - c^2]])/(a^2 - c^2)^(5/2) + (B*(-a^2
+ c^2) + c*(A*c - a*C)*Cos[d + e*x])/((a - c)*c*(a + c)*(a + c*Sin[d + e*x])^2) - ((-3*a*A*c + a^2*C + 2*c^2*
C)*Cos[d + e*x])/((a - c)^2*(a + c)^2*(a + c*Sin[d + e*x])))/(2*e)
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Maple [B] time = 0.147, size = 1891, normalized size = 10.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x)
[Out]
-2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*C*a^3-1/e/(a*tan(1/2*d+1/2*e*x)^2
+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*A*c^3+2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2
*a^3/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)*B+4/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a
^2*c^2+c^4)*A*a^2*c-1/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*C*a*c^2+2/e/(a
^4-2*a^2*c^2+c^4)/(a^2-c^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d+1/2*e*x)+2*c)/(a^2-c^2)^(1/2))*a^2*A+1/e/(a^4-2*a^
2*c^2+c^4)/(a^2-c^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d+1/2*e*x)+2*c)/(a^2-c^2)^(1/2))*A*c^2+7/e/(a*tan(1/2*d+1/2
*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)^2*A*c^3-4/e/(a*tan(1/2*d+1/2*e*x)^2
+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)^2*B*c^3-4/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*ta
n(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)*C*c^3+2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1
/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*a^3*tan(1/2*d+1/2*e*x)^3*B-2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)
+a)^2/(a^4-2*a^2*c^2+c^4)*a^3*tan(1/2*d+1/2*e*x)^2*C-5/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2*a
^2/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)*C*c+5/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a
^2*c^2+c^4)*a*tan(1/2*d+1/2*e*x)^3*A*c^2-2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^
2+c^4)/a*tan(1/2*d+1/2*e*x)^3*A*c^4-3/e/(a^4-2*a^2*c^2+c^4)/(a^2-c^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d+1/2*e*x)
+2*c)/(a^2-c^2)^(1/2))*a*c*C-4/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*a*tan
(1/2*d+1/2*e*x)^3*B*c^2+2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)/a*tan(1/2*
d+1/2*e*x)^3*B*c^4-3/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*a^2*tan(1/2*d+1
/2*e*x)^3*C*c+4/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*a^2*tan(1/2*d+1/2*e*
x)^2*A*c-2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)/a^2*tan(1/2*d+1/2*e*x)^2*
A*c^5+2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*a^2*tan(1/2*d+1/2*e*x)^2*B*c
+2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)/a^2*tan(1/2*d+1/2*e*x)^2*B*c^5-5/
e/(a*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)*a*tan(1/2*d+1/2*e*x)^2*C*c^2-2/e/(a*
tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/(a^4-2*a^2*c^2+c^4)/a*tan(1/2*d+1/2*e*x)^2*C*c^4+11/e/(a*tan(
1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2*a/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)*A*c^2-2/e/(a*tan(1/2*d+1
/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a)^2/a/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)*A*c^4-4/e/(a*tan(1/2*d+1/2*e*x)
^2+2*c*tan(1/2*d+1/2*e*x)+a)^2*a/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)*B*c^2+2/e/(a*tan(1/2*d+1/2*e*x)^2+2*c*
tan(1/2*d+1/2*e*x)+a)^2/a/(a^4-2*a^2*c^2+c^4)*tan(1/2*d+1/2*e*x)*B*c^4
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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*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.33691, size = 1891, normalized size = 10.22 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x, algorithm="fricas")
[Out]
[1/4*(2*B*a^6 - 6*B*a^4*c^2 + 6*B*a^2*c^4 - 2*B*c^6 + 2*(C*a^4*c^2 - 3*A*a^3*c^3 + C*a^2*c^4 + 3*A*a*c^5 - 2*C
*c^6)*cos(e*x + d)*sin(e*x + d) + (2*A*a^4*c - 3*C*a^3*c^2 + 3*A*a^2*c^3 - 3*C*a*c^4 + A*c^5 - (2*A*a^2*c^3 -
3*C*a*c^4 + A*c^5)*cos(e*x + d)^2 + 2*(2*A*a^3*c^2 - 3*C*a^2*c^3 + A*a*c^4)*sin(e*x + d))*sqrt(-a^2 + c^2)*log
(((2*a^2 - c^2)*cos(e*x + d)^2 - 2*a*c*sin(e*x + d) - a^2 - c^2 + 2*(a*cos(e*x + d)*sin(e*x + d) + c*cos(e*x +
d))*sqrt(-a^2 + c^2))/(c^2*cos(e*x + d)^2 - 2*a*c*sin(e*x + d) - a^2 - c^2)) + 2*(2*C*a^5*c - 4*A*a^4*c^2 - C
*a^3*c^3 + 5*A*a^2*c^4 - C*a*c^5 - A*c^6)*cos(e*x + d))/((a^6*c^3 - 3*a^4*c^5 + 3*a^2*c^7 - c^9)*e*cos(e*x + d
)^2 - 2*(a^7*c^2 - 3*a^5*c^4 + 3*a^3*c^6 - a*c^8)*e*sin(e*x + d) - (a^8*c - 2*a^6*c^3 + 2*a^2*c^7 - c^9)*e), 1
/2*(B*a^6 - 3*B*a^4*c^2 + 3*B*a^2*c^4 - B*c^6 + (C*a^4*c^2 - 3*A*a^3*c^3 + C*a^2*c^4 + 3*A*a*c^5 - 2*C*c^6)*co
s(e*x + d)*sin(e*x + d) + (2*A*a^4*c - 3*C*a^3*c^2 + 3*A*a^2*c^3 - 3*C*a*c^4 + A*c^5 - (2*A*a^2*c^3 - 3*C*a*c^
4 + A*c^5)*cos(e*x + d)^2 + 2*(2*A*a^3*c^2 - 3*C*a^2*c^3 + A*a*c^4)*sin(e*x + d))*sqrt(a^2 - c^2)*arctan(-(a*s
in(e*x + d) + c)/(sqrt(a^2 - c^2)*cos(e*x + d))) + (2*C*a^5*c - 4*A*a^4*c^2 - C*a^3*c^3 + 5*A*a^2*c^4 - C*a*c^
5 - A*c^6)*cos(e*x + d))/((a^6*c^3 - 3*a^4*c^5 + 3*a^2*c^7 - c^9)*e*cos(e*x + d)^2 - 2*(a^7*c^2 - 3*a^5*c^4 +
3*a^3*c^6 - a*c^8)*e*sin(e*x + d) - (a^8*c - 2*a^6*c^3 + 2*a^2*c^7 - c^9)*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*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))**3,x)
[Out]
Timed out
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Giac [B] time = 1.22202, size = 805, normalized size = 4.35 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x, algorithm="giac")
[Out]
((2*A*a^2 - 3*C*a*c + A*c^2)*(pi*floor(1/2*(x*e + d)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x*e + 1/2*d) + c)/sq
rt(a^2 - c^2)))/((a^4 - 2*a^2*c^2 + c^4)*sqrt(a^2 - c^2)) + (2*B*a^5*tan(1/2*x*e + 1/2*d)^3 - 3*C*a^4*c*tan(1/
2*x*e + 1/2*d)^3 + 5*A*a^3*c^2*tan(1/2*x*e + 1/2*d)^3 - 4*B*a^3*c^2*tan(1/2*x*e + 1/2*d)^3 - 2*A*a*c^4*tan(1/2
*x*e + 1/2*d)^3 + 2*B*a*c^4*tan(1/2*x*e + 1/2*d)^3 - 2*C*a^5*tan(1/2*x*e + 1/2*d)^2 + 4*A*a^4*c*tan(1/2*x*e +
1/2*d)^2 + 2*B*a^4*c*tan(1/2*x*e + 1/2*d)^2 - 5*C*a^3*c^2*tan(1/2*x*e + 1/2*d)^2 + 7*A*a^2*c^3*tan(1/2*x*e + 1
/2*d)^2 - 4*B*a^2*c^3*tan(1/2*x*e + 1/2*d)^2 - 2*C*a*c^4*tan(1/2*x*e + 1/2*d)^2 - 2*A*c^5*tan(1/2*x*e + 1/2*d)
^2 + 2*B*c^5*tan(1/2*x*e + 1/2*d)^2 + 2*B*a^5*tan(1/2*x*e + 1/2*d) - 5*C*a^4*c*tan(1/2*x*e + 1/2*d) + 11*A*a^3
*c^2*tan(1/2*x*e + 1/2*d) - 4*B*a^3*c^2*tan(1/2*x*e + 1/2*d) - 4*C*a^2*c^3*tan(1/2*x*e + 1/2*d) - 2*A*a*c^4*ta
n(1/2*x*e + 1/2*d) + 2*B*a*c^4*tan(1/2*x*e + 1/2*d) - 2*C*a^5 + 4*A*a^4*c - C*a^3*c^2 - A*a^2*c^3)/((a^6 - 2*a
^4*c^2 + a^2*c^4)*(a*tan(1/2*x*e + 1/2*d)^2 + 2*c*tan(1/2*x*e + 1/2*d) + a)^2))*e^(-1)