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

Optimal. Leaf size=76 \[ \frac{2 (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b f \sqrt{a-b} \sqrt{a+b}}+\frac{d \tanh ^{-1}(\sin (e+f x))}{b f} \]

[Out]

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

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Rubi [A]  time = 0.126988, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3998, 3770, 3831, 2659, 208} \[ \frac{2 (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b f \sqrt{a-b} \sqrt{a+b}}+\frac{d \tanh ^{-1}(\sin (e+f x))}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A]  time = 0.183553, size = 112, normalized size = 1.47 \[ \frac{\frac{2 (a d-b c) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+d \left (\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 135, normalized size = 1.8 \begin{align*} -2\,{\frac{ad}{fb\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 ) }+2\,{\frac{c}{f\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 ) }+{\frac{d}{fb}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{d}{fb}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53203, size = 694, normalized size = 9.13 \begin{align*} \left [\frac{{\left (a^{2} - b^{2}\right )} d \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (a^{2} - b^{2}\right )} d \log \left (-\sin \left (f x + e\right ) + 1\right ) - \sqrt{a^{2} - b^{2}}{\left (b c - a d\right )} \log \left (\frac{2 \, a b \cos \left (f x + e\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + b^{2}}\right )}{2 \,{\left (a^{2} b - b^{3}\right )} f}, \frac{{\left (a^{2} - b^{2}\right )} d \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (a^{2} - b^{2}\right )} d \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, \sqrt{-a^{2} + b^{2}}{\left (b c - a d\right )} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (f x + e\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \,{\left (a^{2} b - b^{3}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28417, size = 178, normalized size = 2.34 \begin{align*} \frac{\frac{d \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{b} - \frac{d \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{b} - \frac{2 \,{\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 (b c - a d\right )}}{\sqrt{-a^{2} + b^{2}} b}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*log(abs(tan(1/2*f*x + 1/2*e) + 1))/b - d*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b - 2*(pi*floor(1/2*(f*x + e)/p
i + 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)))*(b*c - a
*d)/(sqrt(-a^2 + b^2)*b))/f