3.263 \(\int \frac{\sec (e+f x)}{(a+b \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx\)

Optimal. Leaf size=186 \[ -\frac{b^2 \sin (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a \cos (e+f x)+b)}+\frac{2 b \left (-2 a^2 d+a b c+b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f (a-b)^{3/2} (a+b)^{3/2} (b c-a d)^2}+\frac{2 d^2 \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (b c-a d)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.604509, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3988, 3056, 3001, 2659, 208} \[ -\frac{b^2 \sin (e+f x)}{f \left (a^2-b^2\right ) (b c-a d) (a \cos (e+f x)+b)}+\frac{2 b \left (-2 a^2 d+a b c+b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f (a-b)^{3/2} (a+b)^{3/2} (b c-a d)^2}+\frac{2 d^2 \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3988

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[1/g^(m + n), Int[(g*Csc[e + f*x])^(m + n + p)*(b + a*Sin[e + f*x])^m*(d
+ c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && IntegerQ[m] && Inte
gerQ[n]

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^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{\sec (e+f x)}{(a+b \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx &=\int \frac{\cos ^2(e+f x)}{(b+a \cos (e+f x))^2 (d+c \cos (e+f x))} \, dx\\ &=-\frac{b^2 \sin (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (b+a \cos (e+f x))}-\frac{\int \frac{-a b d-\left (a b c-a^2 d+b^2 d\right ) \cos (e+f x)}{(b+a \cos (e+f x)) (d+c \cos (e+f x))} \, dx}{\left (a^2-b^2\right ) (b c-a d)}\\ &=-\frac{b^2 \sin (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (b+a \cos (e+f x))}+\frac{d^2 \int \frac{1}{d+c \cos (e+f x)} \, dx}{(b c-a d)^2}+\frac{\left (b \left (a b c-2 a^2 d+b^2 d\right )\right ) \int \frac{1}{b+a \cos (e+f x)} \, dx}{\left (a^2-b^2\right ) (b c-a d)^2}\\ &=-\frac{b^2 \sin (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (b+a \cos (e+f x))}+\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{c+d+(-c+d) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{(b c-a d)^2 f}+\frac{\left (2 b \left (a b c-2 a^2 d+b^2 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right ) (b c-a d)^2 f}\\ &=\frac{2 b \left (a b c-2 a^2 d+b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} (b c-a d)^2 f}+\frac{2 d^2 \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{\sqrt{c-d} \sqrt{c+d} (b c-a d)^2 f}-\frac{b^2 \sin (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (b+a \cos (e+f x))}\\ \end{align*}

Mathematica [A]  time = 1.01619, size = 176, normalized size = 0.95 \[ \frac{\frac{2 d^2 \left (a^2-b^2\right ) \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}+\frac{2 b \left (-2 a^2 d+a b c+b^2 d\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{b^2 (b c-a d) \sin (e+f x)}{a \cos (e+f x)+b}}{f (b-a) (a+b) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.105, size = 210, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( 2\,{\frac{b}{ \left ( ad-bc \right ) ^{2}} \left ( -{\frac{b \left ( ad-bc \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{ \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}b-a-b \right ) }}-{\frac{2\,{a}^{2}d-abc-{b}^{2}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\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \right ) }+2\,{\frac{{d}^{2}}{ \left ( ad-bc \right ) ^{2}\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+b*sec(f*x+e))^2/(c+d*sec(f*x+e)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+b*sec(f*x+e))**2/(c+d*sec(f*x+e)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.32131, size = 458, normalized size = 2.46 \begin{align*} \frac{2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, c + 2 \, d\right ) + \arctan \left (-\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )} d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-c^{2} + d^{2}}} + \frac{b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (a^{2} b c - b^{3} c - a^{3} d + a b^{2} d\right )}{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a - b\right )}} - \frac{{\left (a b^{2} c - 2 \, a^{2} b d + b^{3} d\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} c^{2} - b^{4} c^{2} - 2 \, a^{3} b c d + 2 \, a b^{3} c d + a^{4} d^{2} - a^{2} b^{2} d^{2}\right )} \sqrt{-a^{2} + b^{2}}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(-2*c + 2*d) + arctan(-(c*tan(1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1/2*e
))/sqrt(-c^2 + d^2)))*d^2/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-c^2 + d^2)) + b^2*tan(1/2*f*x + 1/2*e)/((a^2*
b*c - b^3*c - a^3*d + a*b^2*d)*(a*tan(1/2*f*x + 1/2*e)^2 - b*tan(1/2*f*x + 1/2*e)^2 - a - b)) - (a*b^2*c - 2*a
^2*b*d + b^3*d)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*f*x + 1/2*e) - b*tan(1/2*
f*x + 1/2*e))/sqrt(-a^2 + b^2)))/((a^2*b^2*c^2 - b^4*c^2 - 2*a^3*b*c*d + 2*a*b^3*c*d + a^4*d^2 - a^2*b^2*d^2)*
sqrt(-a^2 + b^2)))/f