3.1.19 \(\int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^2} \, dx\) [19]

Optimal. Leaf size=120 \[ \frac {2 (a A-b B) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} e}+\frac {C}{b e (a+b \cos (d+e x))}-\frac {(A b-a B) \sin (d+e x)}{\left (a^2-b^2\right ) e (a+b \cos (d+e x))} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4462, 2833, 12, 2738, 211, 2747, 32} \begin {gather*} -\frac {(A b-a B) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}+\frac {2 (a A-b B) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{e (a-b)^{3/2} (a+b)^{3/2}}+\frac {C}{b e (a+b \cos (d+e x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + b*Cos[d + e*x])^2,x]

[Out]

(2*(a*A - b*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*e) + C/(b*e*(a
+ b*Cos[d + e*x])) - ((A*b - a*B)*Sin[d + e*x])/((a^2 - b^2)*e*(a + b*Cos[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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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 2747

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 2833

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 4462

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

\begin {align*} \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^2} \, dx &=C \int \frac {\sin (d+e x)}{(a+b \cos (d+e x))^2} \, dx+\int \frac {A+B \cos (d+e x)}{(a+b \cos (d+e x))^2} \, dx\\ &=-\frac {(A b-a B) \sin (d+e x)}{\left (a^2-b^2\right ) e (a+b \cos (d+e x))}+\frac {\int \frac {-a A+b B}{a+b \cos (d+e x)} \, dx}{-a^2+b^2}-\frac {C \text {Subst}\left (\int \frac {1}{(a+x)^2} \, dx,x,b \cos (d+e x)\right )}{b e}\\ &=\frac {C}{b e (a+b \cos (d+e x))}-\frac {(A b-a B) \sin (d+e x)}{\left (a^2-b^2\right ) e (a+b \cos (d+e x))}+\frac {(a A-b B) \int \frac {1}{a+b \cos (d+e x)} \, dx}{a^2-b^2}\\ &=\frac {C}{b e (a+b \cos (d+e x))}-\frac {(A b-a B) \sin (d+e x)}{\left (a^2-b^2\right ) e (a+b \cos (d+e x))}+\frac {(2 (a A-b B)) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (d+e x)\right )\right )}{\left (a^2-b^2\right ) e}\\ &=\frac {2 (a A-b B) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} e}+\frac {C}{b e (a+b \cos (d+e x))}-\frac {(A b-a B) \sin (d+e x)}{\left (a^2-b^2\right ) e (a+b \cos (d+e x))}\\ \end {align*}

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Mathematica [A]
time = 0.46, size = 115, normalized size = 0.96 \begin {gather*} \frac {\frac {2 (a A-b B) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {\left (a^2-b^2\right ) C-b (A b-a B) \sin (d+e x)}{(a-b) b (a+b) (a+b \cos (d+e x))}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + b*Cos[d + e*x])^2,x]

[Out]

((2*(a*A - b*B)*ArcTanh[((a - b)*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + ((a^2 - b^2)*C - b*
(A*b - a*B)*Sin[d + e*x])/((a - b)*b*(a + b)*(a + b*Cos[d + e*x])))/e

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Maple [A]
time = 0.28, size = 141, normalized size = 1.18

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 \left (A b -a B \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a^{2}-b^{2}}-\frac {2 C}{a -b}}{a \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )+a +b}+\frac {2 \left (a A -B b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{e}\)	\(141\)
default	\(\frac {\frac {-\frac {2 \left (A b -a B \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a^{2}-b^{2}}-\frac {2 C}{a -b}}{a \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )+a +b}+\frac {2 \left (a A -B b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{e}\)	\(141\)
risch	\(\frac {2 i A a b \,{\mathrm e}^{i \left (e x +d \right )}-2 i B \,a^{2} {\mathrm e}^{i \left (e x +d \right )}+2 i A \,b^{2}-2 i B a b -2 C \,a^{2} {\mathrm e}^{i \left (e x +d \right )}+2 C \,b^{2} {\mathrm e}^{i \left (e x +d \right )}}{b e \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (e x +d \right )}+2 a \,{\mathrm e}^{i \left (e x +d \right )}+b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) a A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) e}\)	\(448\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^2,x,method=_RETURNVERBOSE)

[Out]

1/e*(2*(-(A*b-B*a)/(a^2-b^2)*tan(1/2*e*x+1/2*d)-C/(a-b))/(a*tan(1/2*e*x+1/2*d)^2-b*tan(1/2*e*x+1/2*d)^2+a+b)+2
*(A*a-B*b)/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*e*x+1/2*d)/((a-b)*(a+b))^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^2,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*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.43, size = 462, normalized size = 3.85 \begin {gather*} \left [\frac {2 \, C a^{4} - 4 \, C a^{2} b^{2} + 2 \, C b^{4} - {\left (A a^{2} b - B a b^{2} + {\left (A a b^{2} - B b^{3}\right )} \cos \left (x e + d\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (x e + d\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x e + d\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x e + d\right ) + b\right )} \sin \left (x e + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x e + d\right )^{2} + 2 \, a b \cos \left (x e + d\right ) + a^{2}}\right ) + 2 \, {\left (B a^{3} b - A a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sin \left (x e + d\right )}{2 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (x e + d\right ) e + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} e\right )}}, \frac {C a^{4} - 2 \, C a^{2} b^{2} + C b^{4} + {\left (A a^{2} b - B a b^{2} + {\left (A a b^{2} - B b^{3}\right )} \cos \left (x e + d\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (x e + d\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x e + d\right )}\right ) + {\left (B a^{3} b - A a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sin \left (x e + d\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (x e + d\right ) e + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^2,x, algorithm="fricas")

[Out]

[1/2*(2*C*a^4 - 4*C*a^2*b^2 + 2*C*b^4 - (A*a^2*b - B*a*b^2 + (A*a*b^2 - B*b^3)*cos(x*e + d))*sqrt(-a^2 + b^2)*
log((2*a*b*cos(x*e + d) + (2*a^2 - b^2)*cos(x*e + d)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x*e + d) + b)*sin(x*e + d)
- a^2 + 2*b^2)/(b^2*cos(x*e + d)^2 + 2*a*b*cos(x*e + d) + a^2)) + 2*(B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4)*si
n(x*e + d))/((a^4*b^2 - 2*a^2*b^4 + b^6)*cos(x*e + d)*e + (a^5*b - 2*a^3*b^3 + a*b^5)*e), (C*a^4 - 2*C*a^2*b^2
+ C*b^4 + (A*a^2*b - B*a*b^2 + (A*a*b^2 - B*b^3)*cos(x*e + d))*sqrt(a^2 - b^2)*arctan(-(a*cos(x*e + d) + b)/(
sqrt(a^2 - b^2)*sin(x*e + d))) + (B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4)*sin(x*e + d))/((a^4*b^2 - 2*a^2*b^4 +
b^6)*cos(x*e + d)*e + (a^5*b - 2*a^3*b^3 + a*b^5)*e)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.46, size = 173, normalized size = 1.44 \begin {gather*} -2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (A a - B b\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {B a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - A b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - C a - C b}{{\left (a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + a + b\right )} {\left (a^{2} - b^{2}\right )}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^2,x, algorithm="giac")

[Out]

-2*((pi*floor(1/2*(x*e + d)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x*e + 1/2*d) - b*tan(1/2*x*e + 1/2*
d))/sqrt(a^2 - b^2)))*(A*a - B*b)/(a^2 - b^2)^(3/2) - (B*a*tan(1/2*x*e + 1/2*d) - A*b*tan(1/2*x*e + 1/2*d) - C
*a - C*b)/((a*tan(1/2*x*e + 1/2*d)^2 - b*tan(1/2*x*e + 1/2*d)^2 + a + b)*(a^2 - b^2)))*e^(-1)

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Mupad [B]
time = 2.57, size = 126, normalized size = 1.05 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a+b}\,\sqrt {a-b}}\right )\,\left (A\,a-B\,b\right )}{e\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\frac {2\,C}{a-b}+\frac {2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (A\,b-B\,a\right )}{\left (a+b\right )\,\left (a-b\right )}}{e\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+a+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(d + e*x) + C*sin(d + e*x))/(a + b*cos(d + e*x))^2,x)

[Out]

(2*atan((tan(d/2 + (e*x)/2)*(2*a - 2*b))/(2*(a + b)^(1/2)*(a - b)^(1/2)))*(A*a - B*b))/(e*(a + b)^(3/2)*(a - b
)^(3/2)) - ((2*C)/(a - b) + (2*tan(d/2 + (e*x)/2)*(A*b - B*a))/((a + b)*(a - b)))/(e*(a + b + tan(d/2 + (e*x)/
2)^2*(a - b)))

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