\(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [804]

Optimal result

Integrand size = 32, antiderivative size = 100 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (a B-b C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac {(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4145, 12, 3916, 2738, 214} \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (a B-b C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac {(b B-a C) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))} \]

[In]

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

[Out]

(2*(a*B - b*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*d) - ((b*B -
a*C)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 4145

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)
*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {a (a B-b C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )} \\ & = -\frac {(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {(a B-b C) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2-b^2} \\ & = -\frac {(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {(a B-b C) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{b \left (a^2-b^2\right )} \\ & = -\frac {(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {(2 (a B-b C)) \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 \left (a^2-b^2\right ) d} \\ & = \frac {2 (a B-b C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac {(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.97 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-\frac {2 (a B-b C) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {(-b B+a C) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}}{d} \]

[In]

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

[Out]

((-2*(a*B - b*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + ((-(b*B) + a*C)*Sin
[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])))/d

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {\frac {2 \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}+\frac {2 \left (a B -C b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(132\)
default	\(\frac {\frac {2 \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}+\frac {2 \left (a B -C b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(132\)
risch	\(\frac {2 i \left (-B b +C a \right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C b}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C b}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(396\)

[In]

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

[Out]

1/d*(2*(B*b-C*a)/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)+2*(B*a-C*b)/
(a+b)/(a-b)/((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 = 0.28 (sec) , antiderivative size = 389, normalized size of antiderivative = 3.89 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {{\left (B a b - C b^{2} + {\left (B a^{2} - C a b\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d\right )}}, \frac {{\left (B a b - C b^{2} + {\left (B a^{2} - C a b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d}\right ] \]

[In]

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

[Out]

[1/2*((B*a*b - C*b^2 + (B*a^2 - C*a*b)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*c
os(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)) + 2*(C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*sin(d*x + c))/((a^5 - 2*a^3*b^2 + a*b^4)*d*cos(d
*x + c) + (a^4*b - 2*a^2*b^3 + b^5)*d), ((B*a*b - C*b^2 + (B*a^2 - C*a*b)*cos(d*x + c))*sqrt(-a^2 + b^2)*arcta
n(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*sin
(d*x + c))/((a^5 - 2*a^3*b^2 + a*b^4)*d*cos(d*x + c) + (a^4*b - 2*a^2*b^3 + b^5)*d)]

Sympy [F]

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

[In]

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

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)/(a + b*sec(c + d*x))**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,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 = 0.30 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.74 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (B a - C b\right )}}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{2} - b^{2}\right )}}\right )}}{d} \]

[In]

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

[Out]

2*((pi*floor(1/2*(d*x + c)/pi + 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 - C*b)/((a^2 - b^2)*sqrt(-a^2 + b^2)) - (C*a*tan(1/2*d*x + 1/2*c) - B*b*tan(1/2*d*x
+ 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)*(a^2 - b^2)))/d

Mupad [B] (verification not implemented)

Time = 17.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.06 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a-b}}{\sqrt {a+b}}\right )\,\left (B\,a-C\,b\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,b-C\,a\right )}{d\,\left (a+b\right )\,\left (a-b\right )\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \]

[In]

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

[Out]

(2*atanh((tan(c/2 + (d*x)/2)*(a - b)^(1/2))/(a + b)^(1/2))*(B*a - C*b))/(d*(a + b)^(3/2)*(a - b)^(3/2)) - (2*t
an(c/2 + (d*x)/2)*(B*b - C*a))/(d*(a + b)*(a - b)*(a + b - tan(c/2 + (d*x)/2)^2*(a - b)))