\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [314]

Optimal result

Integrand size = 29, antiderivative size = 76 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{b d}+\frac {2 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} d} \]

[Out]

B*arctanh(sin(d*x+c))/b/d+2*(A*b-B*a)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/b/d/(a-b)^(1/2)/(a+b
)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4083, 3855, 3916, 2738, 214} \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {2 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}+\frac {B \text {arctanh}(\sin (c+d x))}{b d} \]

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(b*d) + (2*(A*b - a*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a
- b]*b*Sqrt[a + b]*d)

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

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 3855

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

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[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]

Rubi steps \begin{align*} \text {integral}& = \frac {B \int \sec (c+d x) \, dx}{b}+\frac {(A b-a B) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b} \\ & = \frac {B \text {arctanh}(\sin (c+d x))}{b d}+\frac {(A b-a B) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{b^2} \\ & = \frac {B \text {arctanh}(\sin (c+d x))}{b d}+\frac {(2 (A b-a B)) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 d} \\ & = \frac {B \text {arctanh}(\sin (c+d x))}{b d}+\frac {2 (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.47 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 (-A b+a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+B \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{b d} \]

[In]

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

[Out]

((2*(-(A*b) + a*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + B*(-Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]))/(b*d)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {2 \left (-A b +B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(92\)
default	\(\frac {\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {2 \left (-A b +B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(92\)
risch	\(\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right ) B a}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) B a}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d b}\)	\(327\)

[In]

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

[Out]

1/d*(B/b*ln(tan(1/2*d*x+1/2*c)+1)-B/b*ln(tan(1/2*d*x+1/2*c)-1)-2/b*(-A*b+B*a)/((a-b)*(a+b))^(1/2)*arctanh((a-b
)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.66 (sec) , antiderivative size = 316, normalized size of antiderivative = 4.16 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\left [-\frac {{\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}, -\frac {2 \, {\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}\right ] \]

[In]

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

[Out]

[-1/2*((B*a - A*b)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*
(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - (B*a^2 - B
*b^2)*log(sin(d*x + c) + 1) + (B*a^2 - B*b^2)*log(-sin(d*x + c) + 1))/((a^2*b - b^3)*d), -1/2*(2*(B*a - A*b)*s
qrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (B*a^2 - B*b^2)*lo
g(sin(d*x + c) + 1) + (B*a^2 - B*b^2)*log(-sin(d*x + c) + 1))/((a^2*b - b^3)*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.67 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b} + \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (B a - A b\right )}}{\sqrt {-a^{2} + b^{2}} b}}{d} \]

[In]

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

[Out]

(B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b - B*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b + 2*(pi*floor(1/2*(d*x + c)/p
i + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))*(B*a - A
*b)/(sqrt(-a^2 + b^2)*b))/d

Mupad [B] (verification not implemented)

Time = 15.38 (sec) , antiderivative size = 573, normalized size of antiderivative = 7.54 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {A\,a^2\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}-\frac {A\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{d\,\left (a^2-b^2\right )}-\frac {2\,B\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,\left (a^2-b^2\right )}-\frac {A\,b^2\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,\left (a^2-b^2\right )}-\frac {B\,a^3\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {B\,a\,b\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {B\,a\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{b\,d\,\left (a^2-b^2\right )} \]

[In]

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

[Out]

(A*a^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x
)/2)))/(d*(a^2 - b^2)^(3/2)) - (A*log((a*cos(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 -
b^2)^(1/2))/cos(c/2 + (d*x)/2))*((a + b)*(a - b))^(1/2))/(d*(a^2 - b^2)) - (2*B*b*atanh(sin(c/2 + (d*x)/2)/co
s(c/2 + (d*x)/2)))/(d*(a^2 - b^2)) - (A*b^2*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)
/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (2*B*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2)))/(b*d*(a^2 - b^2)) - (B*a^3*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*
(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(b*d*(a^2 - b^2)^(3/2)) + (B*a*b*log((a*sin(c/2 + (d*x)/2) - b*sin(c/2
+ (d*x)/2) + cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2)))/(d*(a^2 - b^2)^(3/2)) + (B*a*log((a*c
os(c/2 + (d*x)/2) + b*cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/cos(c/2 + (d*x)/2))*((a + b)*
(a - b))^(1/2))/(b*d*(a^2 - b^2))