3.797 \(\int \frac {(B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=67 \[ \frac {2 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}+\frac {C x}{b} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3029, 2735, 2659, 205} \[ \frac {2 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}+\frac {C x}{b} \]

Antiderivative was successfully verified.

[In]

Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x]),x]

[Out]

(C*x)/b + (2*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)

Rule 205

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

Rule 2659

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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx &=\int \frac {B+C \cos (c+d x)}{a+b \cos (c+d x)} \, dx\\ &=\frac {C x}{b}-\frac {(-b B+a C) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b}\\ &=\frac {C x}{b}+\frac {(2 (b B-a C)) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}\\ &=\frac {C x}{b}+\frac {2 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 68, normalized size = 1.01 \[ \frac {\frac {2 (a C-b B) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+C (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x]),x]

[Out]

(C*(c + d*x) + (2*(-(b*B) + a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2])/(b*d)

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fricas [A]  time = 0.45, size = 242, normalized size = 3.61 \[ \left [\frac {2 \, {\left (C a^{2} - C b^{2}\right )} d x + {\left (C a - B b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}, \frac {{\left (C a^{2} - C b^{2}\right )} d x - {\left (C a - B b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{{\left (a^{2} b - b^{3}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(C*a^2 - C*b^2)*d*x + (C*a - B*b)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c
)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x
+ c) + a^2)))/((a^2*b - b^3)*d), ((C*a^2 - C*b^2)*d*x - (C*a - B*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) +
b)/(sqrt(a^2 - b^2)*sin(d*x + c))))/((a^2*b - b^3)*d)]

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giac [B]  time = 0.54, size = 296, normalized size = 4.42 \[ -\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} C {\left (2 \, a - b\right )} {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} B b {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} B {\left | a - b \right |} {\left | b \right |} + \sqrt {a^{2} - b^{2}} C {\left | a - b \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} + \frac {{\left (2 \, C a - B b - C b + B {\left | b \right |} - C {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{b^{2} - a {\left | b \right |}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((sqrt(a^2 - b^2)*C*(2*a - b)*abs(a - b) - sqrt(a^2 - b^2)*B*b*abs(a - b) - sqrt(a^2 - b^2)*B*abs(a - b)*abs(
b) + sqrt(a^2 - b^2)*C*abs(a - b)*abs(b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x +
1/2*c)/sqrt((2*a + sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b))))/((a^2 - 2*a*b + b^2)*b^2 + (a^3 - 2*a^2*b + a
*b^2)*abs(b)) + (2*C*a - B*b - C*b + B*abs(b) - C*abs(b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/
2)*tan(1/2*d*x + 1/2*c)/sqrt((2*a - sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b))))/(b^2 - a*abs(b)))/d

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maple [A]  time = 0.18, size = 113, normalized size = 1.69 \[ \frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a C}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

2/d/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B-2/d/b/((a-b)*(a+b))^(1/2)*arcta
n(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*a*C+2/d/b*arctan(tan(1/2*d*x+1/2*c))*C

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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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c)),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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 2.98, size = 344, normalized size = 5.13 \[ \frac {a\,\left (C\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}-C\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}\right )-B\,b\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}+B\,b\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{b\,d\,\left (a^2-b^2\right )}+\frac {2\,C\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*(C*log((b*sin(c/2 + (d*x)/2) - a*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)
/2))*(-(a + b)*(a - b))^(1/2) - C*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 -
a^2)^(1/2))/cos(c/2 + (d*x)/2))*(b^2 - a^2)^(1/2)) - B*b*log((b*sin(c/2 + (d*x)/2) - a*sin(c/2 + (d*x)/2) + c
os(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2))*(-(a + b)*(a - b))^(1/2) + B*b*log((a*sin(c/2 + (d*x)
/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2))*(b^2 - a^2)^(1/2))/(b*d
*(a^2 - b^2)) + (2*C*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)/(a+b*cos(d*x+c)),x)

[Out]

Integral((B + C*cos(c + d*x))*cos(c + d*x)*sec(c + d*x)/(a + b*cos(c + d*x)), x)

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