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

Optimal. Leaf size=124 \[ -\frac{2 \left (2 a^2 A b+a^3 (-B)-A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{A x}{a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.207455, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3923, 3919, 3831, 2659, 208} \[ -\frac{2 \left (2 a^2 A b+a^3 (-B)-A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{A x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3923

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 -
b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

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

Mathematica [A]  time = 0.655569, size = 155, normalized size = 1.25 \[ \frac{\frac{A b \left (a^2-b^2\right ) (c+d x)+a A \left (a^2-b^2\right ) (c+d x) \cos (c+d x)-a b (a B-A b) \sin (c+d x)}{a \cos (c+d x)+b}-\frac{2 \left (-2 a^2 A b+a^3 B+A b^3\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{a^2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.091, size = 328, normalized size = 2.7 \begin{align*} 2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-2\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) A}{ad \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+2\,{\frac{b\tan \left ( 1/2\,dx+c/2 \right ) B}{d \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-4\,{\frac{Ab}{d \left ( a+b \right ) \left ( a-b \right ) \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{A{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \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{Ba}{d \left ( a+b \right ) \left ( a-b \right ) \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((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x)

[Out]

2/d*A/a^2*arctan(tan(1/2*d*x+1/2*c))-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)*A+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)*B-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))*A+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))*A*b^3+2/d/(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*a

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 0.585183, size = 1226, normalized size = 9.89 \begin{align*} \left [\frac{2 \,{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \,{\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x -{\left (B a^{3} b - 2 \, A a^{2} b^{2} + A b^{4} +{\left (B a^{4} - 2 \, A a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )\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 (B a^{4} b - A a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac{{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x +{\left (B a^{3} b - 2 \, A a^{2} b^{2} + A b^{4} +{\left (B a^{4} - 2 \, A a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )\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 (B a^{4} b - A a^{3} b^{2} - B a^{2} b^{3} + A a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sec{\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((A+B*sec(d*x+c))/(a+b*sec(d*x+c))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.21517, size = 271, normalized size = 2.19 \begin{align*} \frac{\frac{2 \,{\left (B a^{3} - 2 \, A a^{2} b + 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 )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{{\left (d x + c\right )} A}{a^{2}} + \frac{2 \,{\left (B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(B*a^3 - 2*A*a^2*b + 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)))/((a^4 - a^2*b^2)*sqrt(-a^2 + b^2)) + (d*x + c)*A/a^2 + 2*(B*
a*b*tan(1/2*d*x + 1/2*c) - A*b^2*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*
x + 1/2*c)^2 - a - b)))/d