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

Optimal. Leaf size=100 \[ \frac{2 (a c-b d) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b c-a d) \tan (e+f x)}{f \left (a^2-b^2\right ) (a+b \sec (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.136388, antiderivative size = 100, 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 = {4003, 12, 3831, 2659, 208} \[ \frac{2 (a c-b d) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b c-a d) \tan (e+f x)}{f \left (a^2-b^2\right ) (a+b \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4003

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a*B
)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] &
& LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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))^2} \, dx &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{\int \frac{(-a c+b d) \sec (e+f x)}{a+b \sec (e+f x)} \, dx}{-a^2+b^2}\\ &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{(a c-b d) \int \frac{\sec (e+f x)}{a+b \sec (e+f x)} \, dx}{a^2-b^2}\\ &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{(a c-b d) \int \frac{1}{1+\frac{a \cos (e+f x)}{b}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{(2 (a c-b 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 \left (a^2-b^2\right ) f}\\ &=\frac{2 (a c-b d) \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} f}-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.36182, size = 97, normalized size = 0.97 \[ \frac{\frac{(a d-b c) \sin (e+f x)}{(a-b) (a+b) (a \cos (e+f x)+b)}-\frac{2 (a c-b d) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.068, size = 132, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( -2\,{\frac{ \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 ) }}+2\,{\frac{ac-db}{ \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 ) } \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))^2,x)

[Out]

1/f*(-2*(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*c-b*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)))

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.527234, size = 861, normalized size = 8.61 \begin{align*} \left [\frac{{\left (a b c - b^{2} d +{\left (a^{2} c - a b d\right )} \cos \left (f x + e\right )\right )} \sqrt{a^{2} - b^{2}} \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 ({\left (a^{2} b - b^{3}\right )} c -{\left (a^{3} - a b^{2}\right )} d\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f \cos \left (f x + e\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f\right )}}, \frac{{\left (a b c - b^{2} d +{\left (a^{2} c - a b d\right )} \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}} \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 ) -{\left ({\left (a^{2} b - b^{3}\right )} c -{\left (a^{3} - a b^{2}\right )} d\right )} \sin \left (f x + e\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f \cos \left (f x + e\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\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))^2,x, algorithm="fricas")

[Out]

[1/2*((a*b*c - b^2*d + (a^2*c - a*b*d)*cos(f*x + e))*sqrt(a^2 - b^2)*log((2*a*b*cos(f*x + e) - (a^2 - 2*b^2)*c
os(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)) - 2*((a^2*b - b^3)*c - (a^3 - a*b^2)*d)*sin(f*x + e))/((a^5 - 2*a^3*b^2 + a*b^4)*f*cos(f
*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*f), ((a*b*c - b^2*d + (a^2*c - a*b*d)*cos(f*x + e))*sqrt(-a^2 + b^2)*arcta
n(-sqrt(-a^2 + b^2)*(b*cos(f*x + e) + a)/((a^2 - b^2)*sin(f*x + e))) - ((a^2*b - b^3)*c - (a^3 - a*b^2)*d)*sin
(f*x + e))/((a^5 - 2*a^3*b^2 + a*b^4)*f*cos(f*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*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 )}}{\left (a + b \sec{\left (e + f x \right )}\right )^{2}}\, 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))**2,x)

[Out]

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

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Giac [A]  time = 1.29692, size = 243, normalized size = 2.43 \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 \, 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 c - b d\right )}}{{\left (a^{2} - b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{b c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\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 )}{\left (a^{2} - b^{2}\right )}}\right )}}{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))^2,x, algorithm="giac")

[Out]

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