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

Optimal. Leaf size=99 \[ \frac{(b c-a d) \tan (e+f x)}{f \left (c^2-d^2\right ) (c+d \sec (e+f x))}+\frac{2 (a c-b d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f (c-d)^{3/2} (c+d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.143887, antiderivative size = 99, 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} \[ \frac{(b c-a d) \tan (e+f x)}{f \left (c^2-d^2\right ) (c+d \sec (e+f x))}+\frac{2 (a c-b d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f (c-d)^{3/2} (c+d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.360421, size = 97, normalized size = 0.98 \[ \frac{\frac{(b c-a d) \sin (e+f x)}{(c-d) (c+d) (c \cos (e+f x)+d)}-\frac{2 (a c-b d) \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.067, 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 ({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{ac-db}{ \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 ) } \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))/(c+d*sec(f*x+e))^2,x)

[Out]

1/f*(2*(a*d-b*c)/(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)+2*(a*c-b*d)/
(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)))

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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)*(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 [A]  time = 0.529548, size = 861, normalized size = 8.7 \begin{align*} \left [\frac{{\left (a c d - b d^{2} +{\left (a c^{2} - b c d\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^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f \cos \left (f x + e\right ) +{\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f\right )}}, \frac{{\left (a c d - b d^{2} +{\left (a c^{2} - b c d\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^{3} - a c^{2} d - b c d^{2} + a d^{3}\right )} \sin \left (f x + e\right )}{{\left (c^{5} - 2 \, c^{3} d^{2} + c d^{4}\right )} f \cos \left (f x + e\right ) +{\left (c^{4} d - 2 \, c^{2} d^{3} + d^{5}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.23511, size = 242, normalized size = 2.44 \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 )}{\left (a c - b d\right )}}{{\left (c^{2} - d^{2}\right )} \sqrt{-c^{2} + d^{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 (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 )}{\left (c^{2} - d^{2}\right )}}\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))/(c+d*sec(f*x+e))^2,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)))*(a*c - b*d)/((c^2 - d^2)*sqrt(-c^2 + d^2)) + (b*c*tan(1/2*f*x + 1/2*e) - a*d*tan(1/2*f*x
+ 1/2*e))/((c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)*(c^2 - d^2)))/f