\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [93]

Optimal result

Integrand size = 31, antiderivative size = 79 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {(2 A-5 B) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

B*arctanh(sin(d*x+c))/a^2/d+1/3*(2*A-5*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 = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4093, 4083, 3855, 3879} \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {(2 A-5 B) \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {B \text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {(A-B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((2*A - 5*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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{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 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4093

Int[csc[(e_.) + (f_.)*(x_)]^2*(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/(b*f*(2*m + 1))), x] + Dist[1/(b^2*(
2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 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 {\int \frac {\sec (c+d x) (-2 a (A-B)-3 a B \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2} \\ & = -\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 A-5 B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}+\frac {B \int \sec (c+d x) \, dx}{a^2} \\ & = \frac {B \text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {(A-B) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 A-5 B) \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {3 B \text {arctanh}(\sin (c+d x))+\frac {(A-4 B+(2 A-5 B) \sec (c+d x)) \tan (c+d x)}{(1+\sec (c+d x))^2}}{3 a^2 d} \]

[In]

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

[Out]

(3*B*ArcTanh[Sin[c + d*x]] + ((A - 4*B + (2*A - 5*B)*Sec[c + d*x])*Tan[c + d*x])/(1 + Sec[c + d*x])^2)/(3*a^2*
d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\frac {-6 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A -9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a^{2} d}\)	\(74\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
risch	\(-\frac {2 i \left (3 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} A +9 B \,{\mathrm e}^{i \left (d x +c \right )}-A +4 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{2} d}\)	\(110\)
norman	\(\frac {\frac {\left (A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}+\frac {\left (A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 a d}-\frac {\left (5 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\)	\(159\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.63 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {3 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (A - 4 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 5 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

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

\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

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

[Out]

(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**3/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.84 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(B*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(c
os(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/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 = 112, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {6 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.67 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B}{2\,a^2}-\frac {B}{a^2}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}+\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d} \]

[In]

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

[Out]

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