\(\int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx\) [86]

Optimal result

Integrand size = 23, antiderivative size = 35 \[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A x}{a}-\frac {(A-B) \tan (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

A*x/a-(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4004, 3879} \[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A x}{a}-\frac {(A-B) \tan (c+d x)}{d (a \sec (c+d x)+a)} \]

[In]

Int[(A + B*Sec[c + d*x])/(a + a*Sec[c + d*x]),x]

[Out]

(A*x)/a - ((A - B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]))

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4004

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]

Rubi steps \begin{align*} \text {integral}& = \frac {A x}{a}-(A-B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx \\ & = \frac {A x}{a}-\frac {(A-B) \tan (c+d x)}{d (a+a \sec (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(35)=70\).

Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.06 \[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (A d x \cos \left (\frac {d x}{2}\right )+A d x \cos \left (c+\frac {d x}{2}\right )+2 (-A+B) \sin \left (\frac {d x}{2}\right )\right )}{a d (1+\cos (c+d x))} \]

[In]

Integrate[(A + B*Sec[c + d*x])/(a + a*Sec[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(A*d*x*Cos[(d*x)/2] + A*d*x*Cos[c + (d*x)/2] + 2*(-A + B)*Sin[(d*x)/2]))/(a*d*(1 +
Cos[c + d*x]))

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-A +B \right )+A x d}{a d}\)	\(28\)
norman	\(\frac {A x}{a}-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}\)	\(30\)
derivativedivides	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(45\)
default	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(45\)
risch	\(\frac {A x}{a}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}\)	\(54\)

[In]

int((A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

(tan(1/2*d*x+1/2*c)*(-A+B)+A*x*d)/a/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A d x \cos \left (d x + c\right ) + A d x - {\left (A - B\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

(A*d*x*cos(d*x + c) + A*d*x - (A - B)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

Sympy [F]

\[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)

[Out]

(Integral(A/(sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)/(sec(c + d*x) + 1), x))/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (35) = 70\).

Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {B \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

(A*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))) + B*sin(d*x + c)/(a*(co
s(d*x + c) + 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} A}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \]

[In]

integrate((A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

((d*x + c)*A/a - (A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a)/d

Mupad [B] (verification not implemented)

Time = 13.61 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \sec (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a}-\frac {A\,d\,x}{a}}{d} \]

[In]

int((A + B/cos(c + d*x))/(a + a/cos(c + d*x)),x)

[Out]

-((tan(c/2 + (d*x)/2)*(A - B))/a - (A*d*x)/a)/d