\(\int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [95]

Optimal result

Integrand size = 23, antiderivative size = 70 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {A x}{a^2}-\frac {(4 A-B) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

A*x/a^2-1/3*(4*A-B)*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^2

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4007, 4004, 3879} \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {(4 A-B) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {A x}{a^2}-\frac {(A-B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

(A*x)/a^2 - ((4*A - B)*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Tan[c + d*x])/(3*d*(a + a*Sec[c +
d*x])^2)

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]

Rule 4007

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-(b
*c - a*d))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[
e + f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f
}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-3 a A+a (A-B) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a^2} \\ & = \frac {A x}{a^2}-\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a} \\ & = \frac {A x}{a^2}-\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-B) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(153\) vs. \(2(70)=140\).

Time = 0.60 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.19 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (9 A d x \cos \left (\frac {d x}{2}\right )+9 A d x \cos \left (c+\frac {d x}{2}\right )+3 A d x \cos \left (c+\frac {3 d x}{2}\right )+3 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 A \sin \left (\frac {d x}{2}\right )+6 B \sin \left (\frac {d x}{2}\right )+12 A \sin \left (c+\frac {d x}{2}\right )-6 B \sin \left (c+\frac {d x}{2}\right )-10 A \sin \left (c+\frac {3 d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )\right )}{24 a^2 d} \]

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*A*d*x*Cos[(d*x)/2] + 9*A*d*x*Cos[c + (d*x)/2] + 3*A*d*x*Cos[c + (3*d*x)/2] + 3
*A*d*x*Cos[2*c + (3*d*x)/2] - 18*A*Sin[(d*x)/2] + 6*B*Sin[(d*x)/2] + 12*A*Sin[c + (d*x)/2] - 6*B*Sin[c + (d*x)
/2] - 10*A*Sin[c + (3*d*x)/2] + 4*B*Sin[c + (3*d*x)/2]))/(24*a^2*d)

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-9 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+6 A x d}{6 a^{2} d}\)	\(49\)
norman	\(\frac {\frac {A x}{a}+\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {\left (3 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}}{a}\)	\(60\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(74\)
risch	\(\frac {A x}{a^{2}}-\frac {2 i \left (6 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )} A -3 B \,{\mathrm e}^{i \left (d x +c \right )}+5 A -2 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(85\)

[In]

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

[Out]

1/6*((A-B)*tan(1/2*d*x+1/2*c)^3+(-9*A+3*B)*tan(1/2*d*x+1/2*c)+6*A*x*d)/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.34 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {3 \, A d x \cos \left (d x + c\right )^{2} + 6 \, A d x \cos \left (d x + c\right ) + 3 \, A d x - {\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 4 \, A - B\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/3*(3*A*d*x*cos(d*x + c)^2 + 6*A*d*x*cos(d*x + c) + 3*A*d*x - ((5*A - 2*B)*cos(d*x + c) + 4*A - 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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.71 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(A*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2) - B*(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 = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.21 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (d x + c\right )} A}{a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*(d*x + c)*A/a^2 + (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x +
1/2*c) + 3*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {3\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,A\,d\,x}{6\,a^2\,d} \]

[In]

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

[Out]

(3*B*tan(c/2 + (d*x)/2) - 9*A*tan(c/2 + (d*x)/2) + A*tan(c/2 + (d*x)/2)^3 - B*tan(c/2 + (d*x)/2)^3 + 6*A*d*x)/
(6*a^2*d)