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

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

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {4072, 3919, 3831, 2659, 208} \[ \frac {B x}{a}-\frac {2 (b B-a C) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

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

Rule 4072

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

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 68, normalized size = 1.01 \[ \frac {\frac {2 (b B-a C) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+B (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(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])/(a*d)

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fricas [A]  time = 0.83, size = 250, normalized size = 3.73 \[ \left [\frac {2 \, {\left (B a^{2} - B b^{2}\right )} d x - {\left (C a - B b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}, \frac {{\left (B a^{2} - B b^{2}\right )} d x + {\left (C a - B b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{{\left (a^{3} - a b^{2}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 3.08, size = 274, normalized size = 4.09 \[ \frac {\frac {{\left (\sqrt {-a^{2} + b^{2}} B {\left (a - 2 \, b\right )} {\left | -a + b \right |} + \sqrt {-a^{2} + b^{2}} C a {\left | -a + b \right |} - \sqrt {-a^{2} + b^{2}} B {\left | a \right |} {\left | -a + b \right |} + \sqrt {-a^{2} + b^{2}} C {\left | a \right |} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b + \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} a^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\left | a \right |}} + \frac {{\left (B a + C a - 2 \, B b + B {\left | a \right |} - C {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b - \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{a^{2} - b {\left | a \right |}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.96, size = 113, normalized size = 1.69 \[ -\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B b}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(cos(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 4.77, size = 573, normalized size = 8.55 \[ \frac {2\,B\,a\,\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 )}{d\,\left (a^2-b^2\right )}-\frac {C\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{d\,\left (a^2-b^2\right )}+\frac {C\,a^2\,\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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}-\frac {C\,b^2\,\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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}-\frac {2\,B\,b^2\,\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 )}{a\,d\,\left (a^2-b^2\right )}+\frac {B\,b^3\,\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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d\,{\left (a^2-b^2\right )}^{3/2}}-\frac {B\,a\,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 {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {B\,b\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{a\,d\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)) - (C*log((a*cos(c/2 + (d*x)/2) + b*cos(c/2
+ (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2))*((a + b)*(a - b))^(1/2))/(d*(a^2 - b^2
)) + (C*a^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 +
(d*x)/2)))/(d*(a^2 - b^2)^(3/2)) - (C*b^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/
2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) - (2*B*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2)))/(a*d*(a^2 - b^2)) + (B*b^3*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a
^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(a*d*(a^2 - b^2)^(3/2)) - (B*a*b*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 +
(d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (B*b*log((a*cos
(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2))*((a + b)*(a
- b))^(1/2))/(a*d*(a^2 - b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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