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

Optimal. Leaf size=145 \[ -\frac{2 b \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2 (A+2 C)-2 a b B+2 A b^2\right )}{2 a^3}-\frac{(A b-a B) \sin (c+d x)}{a^2 d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.452222, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4104, 3919, 3831, 2659, 208} \[ -\frac{2 b \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2 (A+2 C)-2 a b B+2 A b^2\right )}{2 a^3}-\frac{(A b-a B) \sin (c+d x)}{a^2 d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.436987, size = 131, normalized size = 0.9 \[ \frac{2 (c+d x) \left (a^2 (A+2 C)-2 a b B+2 A b^2\right )+\frac{8 b \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a^2 A \sin (2 (c+d x))+4 a (a B-A b) \sin (c+d x)}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*A*b^2 - 2*a*b*B + a^2*(A + 2*C))*(c + d*x) + (8*b*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((-a + b)*Tan[(c +
d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 4*a*(-(A*b) + a*B)*Sin[c + d*x] + a^2*A*Sin[2*(c + d*x)])/(4*a^3*
d)

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Maple [B]  time = 0.123, size = 434, normalized size = 3. \begin{align*} -{\frac{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}Ab}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) b}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{A}{ad}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A{b}^{2}}{d{a}^{3}}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) Bb}{d{a}^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{ad}}-2\,{\frac{A{b}^{3}}{d{a}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{B{b}^{2}}{d{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{Cb}{ad\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^3*A*b+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B+1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+
1/2*c)-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*A*b+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*
x+1/2*c)+1/a/d*A*arctan(tan(1/2*d*x+1/2*c))+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*A*b^2-2/d/a^2*arctan(tan(1/2*d*
x+1/2*c))*B*b+2/a/d*arctan(tan(1/2*d*x+1/2*c))*C-2/d*b^3/a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2
*c)/((a+b)*(a-b))^(1/2))*A+2/d*B/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))
*b^2-2/d*b/a/((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)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.591216, size = 1004, normalized size = 6.92 \begin{align*} \left [\frac{{\left ({\left (A + 2 \, C\right )} a^{4} - 2 \, B a^{3} b +{\left (A - 2 \, C\right )} a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} d x +{\left (C a^{2} b - B a b^{2} + A b^{3}\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 ) +{\left (2 \, B a^{4} - 2 \, A a^{3} b - 2 \, B a^{2} b^{2} + 2 \, A a b^{3} +{\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - a^{3} b^{2}\right )} d}, \frac{{\left ({\left (A + 2 \, C\right )} a^{4} - 2 \, B a^{3} b +{\left (A - 2 \, C\right )} a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} d x - 2 \,{\left (C a^{2} b - B a b^{2} + A b^{3}\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 (2 \, B a^{4} - 2 \, A a^{3} b - 2 \, B a^{2} b^{2} + 2 \, A a b^{3} +{\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - a^{3} b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((A + 2*C)*a^4 - 2*B*a^3*b + (A - 2*C)*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4)*d*x + (C*a^2*b - B*a*b^2 + A*b^3)*
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*B*a^4 - 2*A*a^3*b - 2*B*a^
2*b^2 + 2*A*a*b^3 + (A*a^4 - A*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d), 1/2*(((A + 2*C)*a^4 -
2*B*a^3*b + (A - 2*C)*a^2*b^2 + 2*B*a*b^3 - 2*A*b^4)*d*x - 2*(C*a^2*b - B*a*b^2 + A*b^3)*sqrt(-a^2 + b^2)*arc
tan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (2*B*a^4 - 2*A*a^3*b - 2*B*a^2*b^2 +
2*A*a*b^3 + (A*a^4 - A*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*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 ^{2}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22908, size = 323, normalized size = 2.23 \begin{align*} \frac{\frac{{\left (A a^{2} + 2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2}\right )}{\left (d x + c\right )}}{a^{3}} - \frac{4 \,{\left (C a^{2} b - B a b^{2} + A b^{3}\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 )}}{\sqrt{-a^{2} + b^{2}} a^{3}} - \frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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