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

Optimal. Leaf size=123 \[ \frac{2 \left (-2 a c^2 d+a d^3+b c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{c^2 f (c-d)^{3/2} (c+d)^{3/2}}-\frac{d (b c-a d) \tan (e+f x)}{c f \left (c^2-d^2\right ) (c+d \sec (e+f x))}+\frac{a x}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.2468, antiderivative size = 123, 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} \[ \frac{2 \left (-2 a c^2 d+a d^3+b c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{c^2 f (c-d)^{3/2} (c+d)^{3/2}}-\frac{d (b c-a d) \tan (e+f x)}{c f \left (c^2-d^2\right ) (c+d \sec (e+f x))}+\frac{a x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/c^2 + (2*(b*c^3 - 2*a*c^2*d + a*d^3)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(c^2*(c - d)^(
3/2)*(c + d)^(3/2)*f) - (d*(b*c - a*d)*Tan[e + f*x])/(c*(c^2 - d^2)*f*(c + d*Sec[e + f*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 (e+f x)}{(c+d \sec (e+f x))^2} \, dx &=-\frac{d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}-\frac{\int \frac{-a \left (c^2-d^2\right )-c (b c-a d) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{c \left (c^2-d^2\right )}\\ &=\frac{a x}{c^2}-\frac{d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac{\left (c^2 (b c-a d)-a d \left (c^2-d^2\right )\right ) \int \frac{\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{c^2 \left (c^2-d^2\right )}\\ &=\frac{a x}{c^2}-\frac{d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac{\left (c^2 (b c-a d)-a d \left (c^2-d^2\right )\right ) \int \frac{1}{1+\frac{c \cos (e+f x)}{d}} \, dx}{c^2 d \left (c^2-d^2\right )}\\ &=\frac{a x}{c^2}-\frac{d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac{\left (2 \left (c^2 (b c-a d)-a d \left (c^2-d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c}{d}+\left (1-\frac{c}{d}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{c^2 d \left (c^2-d^2\right ) f}\\ &=\frac{a x}{c^2}+\frac{2 \left (b c^3-2 a c^2 d+a d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{c^2 (c-d)^{3/2} (c+d)^{3/2} f}-\frac{d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.643414, size = 155, normalized size = 1.26 \[ \frac{\frac{-c d (b c-a d) \sin (e+f x)+a d \left (c^2-d^2\right ) (e+f x)+a c \left (c^2-d^2\right ) (e+f x) \cos (e+f x)}{c \cos (e+f x)+d}-\frac{2 \left (a d \left (d^2-2 c^2\right )+b c^3\right ) \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}}{c^2 f (c-d) (c+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.086, size = 328, normalized size = 2.7 \begin{align*} 2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{2}}}-2\,{\frac{{d}^{2}\tan \left ( 1/2\,fx+e/2 \right ) a}{fc \left ({c}^{2}-{d}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) }}+2\,{\frac{d\tan \left ( 1/2\,fx+e/2 \right ) b}{f \left ({c}^{2}-{d}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) }}-4\,{\frac{ad}{f \left ( c+d \right ) \left ( c-d \right ) \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 ) }+2\,{\frac{a{d}^{3}}{f{c}^{2} \left ( c+d \right ) \left ( c-d \right ) \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 ) }+2\,{\frac{bc}{f \left ( c+d \right ) \left ( c-d \right ) \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.602482, size = 1226, normalized size = 9.97 \begin{align*} \left [\frac{2 \,{\left (a c^{5} - 2 \, a c^{3} d^{2} + a c d^{4}\right )} f x \cos \left (f x + e\right ) + 2 \,{\left (a c^{4} d - 2 \, a c^{2} d^{3} + a d^{5}\right )} f x -{\left (b c^{3} d - 2 \, a c^{2} d^{2} + a d^{4} +{\left (b c^{4} - 2 \, a c^{3} d + a c d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{c^{2} - d^{2}} \log \left (\frac{2 \, c d \cos \left (f x + e\right ) -{\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt{c^{2} - d^{2}}{\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) - 2 \,{\left (b c^{4} d - a c^{3} d^{2} - b c^{2} d^{3} + a c d^{4}\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (c^{7} - 2 \, c^{5} d^{2} + c^{3} d^{4}\right )} f \cos \left (f x + e\right ) +{\left (c^{6} d - 2 \, c^{4} d^{3} + c^{2} d^{5}\right )} f\right )}}, \frac{{\left (a c^{5} - 2 \, a c^{3} d^{2} + a c d^{4}\right )} f x \cos \left (f x + e\right ) +{\left (a c^{4} d - 2 \, a c^{2} d^{3} + a d^{5}\right )} f x +{\left (b c^{3} d - 2 \, a c^{2} d^{2} + a d^{4} +{\left (b c^{4} - 2 \, a c^{3} d + a c d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{-c^{2} + d^{2}} \arctan \left (-\frac{\sqrt{-c^{2} + d^{2}}{\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) -{\left (b c^{4} d - a c^{3} d^{2} - b c^{2} d^{3} + a c d^{4}\right )} \sin \left (f x + e\right )}{{\left (c^{7} - 2 \, c^{5} d^{2} + c^{3} d^{4}\right )} f \cos \left (f x + e\right ) +{\left (c^{6} d - 2 \, c^{4} d^{3} + c^{2} d^{5}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(a*c^5 - 2*a*c^3*d^2 + a*c*d^4)*f*x*cos(f*x + e) + 2*(a*c^4*d - 2*a*c^2*d^3 + a*d^5)*f*x - (b*c^3*d -
2*a*c^2*d^2 + a*d^4 + (b*c^4 - 2*a*c^3*d + a*c*d^3)*cos(f*x + e))*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c
^2 - 2*d^2)*cos(f*x + e)^2 - 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x +
e)^2 + 2*c*d*cos(f*x + e) + d^2)) - 2*(b*c^4*d - a*c^3*d^2 - b*c^2*d^3 + a*c*d^4)*sin(f*x + e))/((c^7 - 2*c^5
*d^2 + c^3*d^4)*f*cos(f*x + e) + (c^6*d - 2*c^4*d^3 + c^2*d^5)*f), ((a*c^5 - 2*a*c^3*d^2 + a*c*d^4)*f*x*cos(f*
x + e) + (a*c^4*d - 2*a*c^2*d^3 + a*d^5)*f*x + (b*c^3*d - 2*a*c^2*d^2 + a*d^4 + (b*c^4 - 2*a*c^3*d + a*c*d^3)*
cos(f*x + e))*sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) - (b*
c^4*d - a*c^3*d^2 - b*c^2*d^3 + a*c*d^4)*sin(f*x + e))/((c^7 - 2*c^5*d^2 + c^3*d^4)*f*cos(f*x + e) + (c^6*d -
2*c^4*d^3 + c^2*d^5)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.46028, size = 282, normalized size = 2.29 \begin{align*} \frac{\frac{2 \,{\left (b c^{3} - 2 \, a c^{2} d + a d^{3}\right )}{\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 )}}{{\left (c^{4} - c^{2} d^{2}\right )} \sqrt{-c^{2} + d^{2}}} + \frac{{\left (f x + e\right )} a}{c^{2}} + \frac{2 \,{\left (b c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (c^{3} - c d^{2}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c - d\right )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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