\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [94]

Optimal result

Integrand size = 29, antiderivative size = 65 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(A+2 B) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )} \]

[Out]

1/3*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^2+1/3*(A+2*B)*tan(d*x+c)/d/(a^2+a^2*sec(d*x+c))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {4085, 3879} \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {(A+2 B) \tan (c+d x)}{3 d \left (a^2 \sec (c+d x)+a^2\right )}+\frac {(A-B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

((A - B)*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((A + 2*B)*Tan[c + d*x])/(3*d*(a^2 + a^2*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 4085

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/(a*f*(2*m + 1))), x] + Dist[(a*B*m + A*b*
(m + 1))/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x
] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(A+2 B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a} \\ & = \frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(A+2 B) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (3 (A+B) \sin \left (\frac {d x}{2}\right )-3 A \sin \left (c+\frac {d x}{2}\right )+(2 A+B) \sin \left (c+\frac {3 d x}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \]

[In]

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(3*(A + B)*Sin[(d*x)/2] - 3*A*Sin[c + (d*x)/2] + (2*A + B)*Sin[c + (3*d*x)/2]))/(3*
a^2*d*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(-\frac {\left (-3 A -3 B +\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a^{2} d}\)	\(42\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{2 d \,a^{2}}\)	\(60\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{2 d \,a^{2}}\)	\(60\)
risch	\(\frac {2 i \left (3 A \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )} A +3 B \,{\mathrm e}^{i \left (d x +c \right )}+2 A +B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(64\)
norman	\(\frac {-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}-\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (2 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{a \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(89\)

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/6*(-3*A-3*B+(A-B)*tan(1/2*d*x+1/2*c)^2)*tan(1/2*d*x+1/2*c)/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {{\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right ) + A + 2 \, B\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*((2*A + B)*cos(d*x + c) + A + 2*B)*sin(d*x + c)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**2/(sec(c + d*x)
**2 + 2*sec(c + d*x) + 1), x))/a**2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.43 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} + \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(B*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 + A*(3*sin(d*x + c)/(cos(
d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{6 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(A*tan(1/2*d*x + 1/2*c)^3 - B*tan(1/2*d*x + 1/2*c)^3 - 3*A*tan(1/2*d*x + 1/2*c) - 3*B*tan(1/2*d*x + 1/2*c
))/(a^2*d)

Mupad [B] (verification not implemented)

Time = 13.58 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{2\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d} \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)*(a + a/cos(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)*(A + B))/(2*a^2*d) - (tan(c/2 + (d*x)/2)^3*(A - B))/(6*a^2*d)