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

Optimal. Leaf size=202 \[ -\frac{\sin (c+d x) \left (a^2 (-(A-C))-a b B+2 A b^2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{2 \left (3 a^2 A b^2-2 a^3 b B+a^4 C+a b^3 B-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x (2 A b-a B)}{a^3} \]

[Out]

-(((2*A*b - a*B)*x)/a^3) + (2*(3*a^2*A*b^2 - 2*A*b^4 - 2*a^3*b*B + a*b^3*B + a^4*C)*ArcTanh[(Sqrt[a - b]*Tan[(
c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(3/2)*(a + b)^(3/2)*d) - ((2*A*b^2 - a*b*B - a^2*(A - C))*Sin[c + d*x]
)/(a^2*(a^2 - b^2)*d) + ((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.644187, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4100, 4104, 3919, 3831, 2659, 208} \[ -\frac{\sin (c+d x) \left (a^2 (-(A-C))-a b B+2 A b^2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{2 \left (3 a^2 A b^2-2 a^3 b B+a^4 C+a b^3 B-2 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x (2 A b-a B)}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*A*b - a*B)*x)/a^3) + (2*(3*a^2*A*b^2 - 2*A*b^4 - 2*a^3*b*B + a*b^3*B + a^4*C)*ArcTanh[(Sqrt[a - b]*Tan[(
c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(3/2)*(a + b)^(3/2)*d) - ((2*A*b^2 - a*b*B - a^2*(A - C))*Sin[c + d*x]
)/(a^2*(a^2 - b^2)*d) + ((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

Rule 4100

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

Rule 4104

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

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (2 A b^2-a b B-a^2 (A-C)+a (A b-a B+b C) \sec (c+d x)-\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (2 A b^2-a b B-a^2 (A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-\left (a^2-b^2\right ) (2 A b-a B)+a \left (A b^2-a (b B-a C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac{(2 A b-a B) x}{a^3}-\frac{\left (2 A b^2-a b B-a^2 (A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (3 a^2 A b^2-2 A b^4-2 a^3 b B+a b^3 B+a^4 C\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{(2 A b-a B) x}{a^3}-\frac{\left (2 A b^2-a b B-a^2 (A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (3 a^2 A b^2-2 A b^4-2 a^3 b B+a b^3 B+a^4 C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^3 b \left (a^2-b^2\right )}\\ &=-\frac{(2 A b-a B) x}{a^3}-\frac{\left (2 A b^2-a b B-a^2 (A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 \left (3 a^2 A b^2-2 A b^4-2 a^3 b B+a b^3 B+a^4 C\right )\right ) \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^3 b \left (a^2-b^2\right ) d}\\ &=-\frac{(2 A b-a B) x}{a^3}+\frac{2 \left (3 a^2 A b^2-2 A b^4-2 a^3 b B+a b^3 B+a^4 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{3/2} (a+b)^{3/2} d}-\frac{\left (2 A b^2-a b B-a^2 (A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.955774, size = 160, normalized size = 0.79 \[ \frac{-\frac{2 \left (3 a^2 A b^2-2 a^3 b B+a^4 C+a b^3 B-2 A b^4\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{a b \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a \cos (c+d x)+b)}+(c+d x) (a B-2 A b)+a A \sin (c+d x)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*A*b + a*B)*(c + d*x) - (2*(3*a^2*A*b^2 - 2*A*b^4 - 2*a^3*b*B + a*b^3*B + a^4*C)*ArcTanh[((-a + b)*Tan[(c
+ d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + a*A*Sin[c + d*x] - (a*b*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*
x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])))/(a^3*d)

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Maple [B]  time = 0.131, size = 573, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*A/a^2*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-4/d*A/a^3*b*arctan(tan(1/2*d*x+1/2*c))+2/d*B/a^2*arctan(
tan(1/2*d*x+1/2*c))+2/d/a^2*b^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-
b)*A-2/d/a*b^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)*B+2/d*b/(a^2-b
^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)*C+6/d/a/(a+b)/(a-b)/((a+b)*(a-b))^(
1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*b^2-4/d/a^3/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arcta
nh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*b^4-4/d*b/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan
(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+2/d/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c
)/((a+b)*(a-b))^(1/2))*B*b^3+2/d*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-
b))^(1/2))*C

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.726212, size = 1796, normalized size = 8.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(B*a^6 - 2*A*a^5*b - 2*B*a^4*b^2 + 4*A*a^3*b^3 + B*a^2*b^4 - 2*A*a*b^5)*d*x*cos(d*x + c) + 2*(B*a^5*b
- 2*A*a^4*b^2 - 2*B*a^3*b^3 + 4*A*a^2*b^4 + B*a*b^5 - 2*A*b^6)*d*x - (C*a^4*b - 2*B*a^3*b^2 + 3*A*a^2*b^3 + B*
a*b^4 - 2*A*b^5 + (C*a^5 - 2*B*a^4*b + 3*A*a^3*b^2 + B*a^2*b^3 - 2*A*a*b^4)*cos(d*x + c))*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)) + 2*((A - C)*a^5*b + B*a^4*b^2 - (3*A - C)*a^3*b^3
- B*a^2*b^4 + 2*A*a*b^5 + (A*a^6 - 2*A*a^4*b^2 + A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^8 - 2*a^6*b^2 + a^
4*b^4)*d*cos(d*x + c) + (a^7*b - 2*a^5*b^3 + a^3*b^5)*d), ((B*a^6 - 2*A*a^5*b - 2*B*a^4*b^2 + 4*A*a^3*b^3 + B*
a^2*b^4 - 2*A*a*b^5)*d*x*cos(d*x + c) + (B*a^5*b - 2*A*a^4*b^2 - 2*B*a^3*b^3 + 4*A*a^2*b^4 + B*a*b^5 - 2*A*b^6
)*d*x + (C*a^4*b - 2*B*a^3*b^2 + 3*A*a^2*b^3 + B*a*b^4 - 2*A*b^5 + (C*a^5 - 2*B*a^4*b + 3*A*a^3*b^2 + B*a^2*b^
3 - 2*A*a*b^4)*cos(d*x + c))*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 - C)*a^5*b + B*a^4*b^2 - (3*A - C)*a^3*b^3 - B*a^2*b^4 + 2*A*a*b^5 + (A*a^6 - 2*A*a^4*b^2 + A*
a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^8 - 2*a^6*b^2 + a^4*b^4)*d*cos(d*x + c) + (a^7*b - 2*a^5*b^3 + a^3*b^
5)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.29528, size = 551, normalized size = 2.73 \begin{align*} \frac{\frac{2 \,{\left (C a^{4} - 2 \, B a^{3} b + 3 \, A a^{2} b^{2} + B a b^{3} - 2 \, A b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}{\left (a^{4} - a^{2} b^{2}\right )}} + \frac{{\left (B a - 2 \, A b\right )}{\left (d x + c\right )}}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(C*a^4 - 2*B*a^3*b + 3*A*a^2*b^2 + B*a*b^3 - 2*A*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + a
rctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^5 - a^3*b^2)*sqrt(-a^2 + b^2))
+ 2*(A*a^3*tan(1/2*d*x + 1/2*c)^3 - A*a^2*b*tan(1/2*d*x + 1/2*c)^3 + C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - A*a*b^2
*tan(1/2*d*x + 1/2*c)^3 - B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^3*tan(1/2*d*x + 1/2*c)^3 - A*a^3*tan(1/2*d*x
+ 1/2*c) - A*a^2*b*tan(1/2*d*x + 1/2*c) + C*a^2*b*tan(1/2*d*x + 1/2*c) + A*a*b^2*tan(1/2*d*x + 1/2*c) - B*a*b^
2*tan(1/2*d*x + 1/2*c) + 2*A*b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 -
2*b*tan(1/2*d*x + 1/2*c)^2 - a - b)*(a^4 - a^2*b^2)) + (B*a - 2*A*b)*(d*x + c)/a^3)/d