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*a^2+A*c^2-3*C*a*c)*arctan((c+a*tan(1/2*e*x+1/2*d))/(a^2-c^2)^(1/2))/(a^2-c^2)^(5/2)/e-1/2*B/c/e/(a+c*sin(
e*x+d))^2+1/2*(A*c-C*a)*cos(e*x+d)/(a^2-c^2)/e/(a+c*sin(e*x+d))^2+1/2*(3*A*a*c-C*a^2-2*C*c^2)*cos(e*x+d)/(a^2-
c^2)^2/e/(a+c*sin(e*x+d))

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Rubi [A]  time = 0.25, antiderivative size = 185, normalized size of antiderivative = 1.00, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.91, 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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fricas [B]  time = 1.17, size = 880, normalized size = 4.76 \[ \left [\frac {2 \, B a^{6} - 6 \, B a^{4} c^{2} + 6 \, B a^{2} c^{4} - 2 \, B c^{6} + 2 \, {\left (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}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) + {\left (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} - {\left (2 \, A a^{2} c^{3} - 3 \, C a c^{4} + A c^{5}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (2 \, A a^{3} c^{2} - 3 \, C a^{2} c^{3} + A a c^{4}\right )} \sin \left (e x + d\right )\right )} \sqrt {-a^{2} + c^{2}} \log \left (\frac {{\left (2 \, a^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2} + 2 \, {\left (a \cos \left (e x + d\right ) \sin \left (e x + d\right ) + c \cos \left (e x + d\right )\right )} \sqrt {-a^{2} + c^{2}}}{c^{2} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2}}\right ) + 2 \, {\left (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}\right )} \cos \left (e x + d\right )}{4 \, {\left ({\left (a^{6} c^{3} - 3 \, a^{4} c^{5} + 3 \, a^{2} c^{7} - c^{9}\right )} e \cos \left (e x + d\right )^{2} - 2 \, {\left (a^{7} c^{2} - 3 \, a^{5} c^{4} + 3 \, a^{3} c^{6} - a c^{8}\right )} e \sin \left (e x + d\right ) - {\left (a^{8} c - 2 \, a^{6} c^{3} + 2 \, a^{2} c^{7} - c^{9}\right )} e\right )}}, \frac {B a^{6} - 3 \, B a^{4} c^{2} + 3 \, B a^{2} c^{4} - B c^{6} + {\left (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}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) + {\left (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} - {\left (2 \, A a^{2} c^{3} - 3 \, C a c^{4} + A c^{5}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (2 \, A a^{3} c^{2} - 3 \, C a^{2} c^{3} + A a c^{4}\right )} \sin \left (e x + d\right )\right )} \sqrt {a^{2} - c^{2}} \arctan \left (-\frac {a \sin \left (e x + d\right ) + c}{\sqrt {a^{2} - c^{2}} \cos \left (e x + d\right )}\right ) + {\left (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}\right )} \cos \left (e x + d\right )}{2 \, {\left ({\left (a^{6} c^{3} - 3 \, a^{4} c^{5} + 3 \, a^{2} c^{7} - c^{9}\right )} e \cos \left (e x + d\right )^{2} - 2 \, {\left (a^{7} c^{2} - 3 \, a^{5} c^{4} + 3 \, a^{3} c^{6} - a c^{8}\right )} e \sin \left (e x + d\right ) - {\left (a^{8} c - 2 \, a^{6} c^{3} + 2 \, a^{2} c^{7} - c^{9}\right )} e\right )}}\right ] \]

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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giac [B]  time = 0.27, size = 596, normalized size = 3.22 \[ {\left (\frac {{\left (2 \, A a^{2} - 3 \, C a c + A c^{2}\right )} {\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + c}{\sqrt {a^{2} - c^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} c^{2} + c^{4}\right )} \sqrt {a^{2} - c^{2}}} + \frac {2 \, B a^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} - 3 \, C a^{4} c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 5 \, A a^{3} c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} - 4 \, B a^{3} c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} - 2 \, A a c^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 2 \, B a c^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} - 2 \, C a^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 4 \, A a^{4} c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, B a^{4} c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - 5 \, C a^{3} c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 7 \, A a^{2} c^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - 4 \, B a^{2} c^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - 2 \, C a c^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - 2 \, A c^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, B c^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, B a^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 5 \, C a^{4} c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 11 \, A a^{3} c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 4 \, B a^{3} c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 4 \, C a^{2} c^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 2 \, A a c^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 2 \, B a c^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - 2 \, C a^{5} + 4 \, A a^{4} c - C a^{3} c^{2} - A a^{2} c^{3}}{{\left (a^{6} - 2 \, a^{4} c^{2} + a^{2} c^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a\right )}^{2}}\right )} e^{\left (-1\right )} \]

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)

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maple [B]  time = 0.43, size = 1891, normalized size = 10.22 \[ \text {result too large to display} \]

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^4-2*a^2*c^2+c^4)*a^3*tan(1/2*d+1/2*e*x)^2*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^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+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*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)*C*c^3+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)*arc
tan(1/2*(2*a*tan(1/2*d+1/2*e*x)+2*c)/(a^2-c^2)^(1/2))*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^3*tan(1/2*d+1/2*e*x)^3*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*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*t
an(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^4-5/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-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+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*ta
n(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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*a^2>0)', see `assume?`
for more details)Is 4*c^2-4*a^2 positive or negative?

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mupad [B]  time = 5.27, size = 557, normalized size = 3.01 \[ \frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,A\,a^2-3\,C\,a\,c+A\,c^2\right )\,\left (2\,a^4\,c-4\,a^2\,c^3+2\,c^5\right )}{2\,{\left (a+c\right )}^{5/2}\,{\left (a-c\right )}^{5/2}\,\left (a^4-2\,a^2\,c^2+c^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,a^2-3\,C\,a\,c+A\,c^2\right )}{{\left (a+c\right )}^{5/2}\,{\left (a-c\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,c^2+c^4\right )}{2\,A\,a^2-3\,C\,a\,c+A\,c^2}\right )\,\left (2\,A\,a^2-3\,C\,a\,c+A\,c^2\right )}{e\,{\left (a+c\right )}^{5/2}\,{\left (a-c\right )}^{5/2}}-\frac {\frac {2\,C\,a^3-4\,A\,a^2\,c+C\,a\,c^2+A\,c^3}{a^4-2\,a^2\,c^2+c^4}-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (2\,B\,a^4-2\,A\,c^4+2\,B\,c^4+5\,A\,a^2\,c^2-4\,B\,a^2\,c^2-3\,C\,a^3\,c\right )}{a\,\left (a^4-2\,a^2\,c^2+c^4\right )}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,c^4-2\,B\,a^4-2\,B\,c^4-11\,A\,a^2\,c^2+4\,B\,a^2\,c^2+4\,C\,a\,c^3+5\,C\,a^3\,c\right )}{a\,\left (a^4-2\,a^2\,c^2+c^4\right )}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (2\,A\,c^5+2\,C\,a^5-2\,B\,c^5-7\,A\,a^2\,c^3+4\,B\,a^2\,c^3+5\,C\,a^3\,c^2-4\,A\,a^4\,c-2\,B\,a^4\,c+2\,C\,a\,c^4\right )}{a^2\,\left (a^4-2\,a^2\,c^2+c^4\right )}}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (2\,a^2+4\,c^2\right )+a^2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+a^2+4\,a\,c\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3+4\,a\,c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(d + e*x) + C*sin(d + e*x))/(a + c*sin(d + e*x))^3,x)

[Out]

(atan(((((2*A*a^2 + A*c^2 - 3*C*a*c)*(2*a^4*c + 2*c^5 - 4*a^2*c^3))/(2*(a + c)^(5/2)*(a - c)^(5/2)*(a^4 + c^4
- 2*a^2*c^2)) + (a*tan(d/2 + (e*x)/2)*(2*A*a^2 + A*c^2 - 3*C*a*c))/((a + c)^(5/2)*(a - c)^(5/2)))*(a^4 + c^4 -
2*a^2*c^2))/(2*A*a^2 + A*c^2 - 3*C*a*c))*(2*A*a^2 + A*c^2 - 3*C*a*c))/(e*(a + c)^(5/2)*(a - c)^(5/2)) - ((A*c
^3 + 2*C*a^3 - 4*A*a^2*c + C*a*c^2)/(a^4 + c^4 - 2*a^2*c^2) - (tan(d/2 + (e*x)/2)^3*(2*B*a^4 - 2*A*c^4 + 2*B*c
^4 + 5*A*a^2*c^2 - 4*B*a^2*c^2 - 3*C*a^3*c))/(a*(a^4 + c^4 - 2*a^2*c^2)) + (tan(d/2 + (e*x)/2)*(2*A*c^4 - 2*B*
a^4 - 2*B*c^4 - 11*A*a^2*c^2 + 4*B*a^2*c^2 + 4*C*a*c^3 + 5*C*a^3*c))/(a*(a^4 + c^4 - 2*a^2*c^2)) + (tan(d/2 +
(e*x)/2)^2*(2*A*c^5 + 2*C*a^5 - 2*B*c^5 - 7*A*a^2*c^3 + 4*B*a^2*c^3 + 5*C*a^3*c^2 - 4*A*a^4*c - 2*B*a^4*c + 2*
C*a*c^4))/(a^2*(a^4 + c^4 - 2*a^2*c^2)))/(e*(tan(d/2 + (e*x)/2)^2*(2*a^2 + 4*c^2) + a^2*tan(d/2 + (e*x)/2)^4 +
a^2 + 4*a*c*tan(d/2 + (e*x)/2)^3 + 4*a*c*tan(d/2 + (e*x)/2)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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