3.447 \(\int \frac{\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx\)

Optimal. Leaf size=142 \[ -\frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}+\frac{b \log \left (-(a-c) \tan ^2\left (\frac{x}{2}\right )+a+2 b \tan \left (\frac{x}{2}\right )+c\right )}{b^2-c^2}-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (\tan \left (\frac{x}{2}\right )+1\right )}{b-c} \]

[Out]

(-2*a*c*ArcTanh[(b - (a - c)*Tan[x/2])/Sqrt[a^2 + b^2 - c^2]])/((b^2 - c^2)*Sqrt[a^2 + b^2 - c^2]) - Log[1 - T
an[x/2]]/(b + c) - Log[1 + Tan[x/2]]/(b - c) + (b*Log[a + c + 2*b*Tan[x/2] - (a - c)*Tan[x/2]^2])/(b^2 - c^2)

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Rubi [A]  time = 0.514689, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {4397, 1075, 634, 618, 206, 628, 633, 31} \[ -\frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}+\frac{b \log \left (-(a-c) \tan ^2\left (\frac{x}{2}\right )+a+2 b \tan \left (\frac{x}{2}\right )+c\right )}{b^2-c^2}-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (\tan \left (\frac{x}{2}\right )+1\right )}{b-c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*c*ArcTanh[(b - (a - c)*Tan[x/2])/Sqrt[a^2 + b^2 - c^2]])/((b^2 - c^2)*Sqrt[a^2 + b^2 - c^2]) - Log[1 - T
an[x/2]]/(b + c) - Log[1 + Tan[x/2]]/(b - c) + (b*Log[a + c + 2*b*Tan[x/2] - (a - c)*Tan[x/2]^2])/(b^2 - c^2)

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 1075

Int[((A_.) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (f_.)*(x_)^2)), x_Symbol] :> With[{q =
c^2*d^2 + b^2*d*f - 2*a*c*d*f + a^2*f^2}, Dist[1/q, Int[(A*c^2*d - a*c*C*d + A*b^2*f - a*A*c*f + a^2*C*f + c*(
-(b*C*d) + A*b*f)*x)/(a + b*x + c*x^2), x], x] + Dist[1/q, Int[(c*C*d^2 - A*c*d*f - a*C*d*f + a*A*f^2 - f*(-(b
*C*d) + A*b*f)*x)/(d + f*x^2), x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, f, A, C}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^2(x)}{a+c \sec (x)+b \tan (x)} \, dx &=\int \frac{\sec (x)}{c+a \cos (x)+b \sin (x)} \, dx\\ &=2 \operatorname{Subst}\left (\int \frac{1+x^2}{\left (1-x^2\right ) \left (a+c+2 b x-(a-c) x^2\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=-\frac{\operatorname{Subst}\left (\int \frac{4 c-4 b x}{1-x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{2 \left (b^2-c^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-4 b^2+(-a+c)^2-(a+c)^2-4 b (-a+c) x}{a+c+2 b x+(-a+c) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{2 \left (b^2-c^2\right )}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b-c}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b+c}+\frac{b \operatorname{Subst}\left (\int \frac{2 b+2 (-a+c) x}{a+c+2 b x+(-a+c) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2-c^2}+\frac{(2 a c) \operatorname{Subst}\left (\int \frac{1}{a+c+2 b x+(-a+c) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^2-c^2}\\ &=-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (1+\tan \left (\frac{x}{2}\right )\right )}{b-c}+\frac{b \log \left (a+c+2 b \tan \left (\frac{x}{2}\right )-(a-c) \tan ^2\left (\frac{x}{2}\right )\right )}{b^2-c^2}-\frac{(4 a c) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2-c^2\right )-x^2} \, dx,x,2 b+2 (-a+c) \tan \left (\frac{x}{2}\right )\right )}{b^2-c^2}\\ &=-\frac{2 a c \tanh ^{-1}\left (\frac{b-(a-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}-\frac{\log \left (1-\tan \left (\frac{x}{2}\right )\right )}{b+c}-\frac{\log \left (1+\tan \left (\frac{x}{2}\right )\right )}{b-c}+\frac{b \log \left (a+c+2 b \tan \left (\frac{x}{2}\right )-(a-c) \tan ^2\left (\frac{x}{2}\right )\right )}{b^2-c^2}\\ \end{align*}

Mathematica [A]  time = 0.289371, size = 120, normalized size = 0.85 \[ \frac{\frac{2 a c \tanh ^{-1}\left (\frac{(c-a) \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}}-b \log (a \cos (x)+b \sin (x)+c)+(b-c) \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+(b+c) \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )}{(c-b) (b+c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.059, size = 430, normalized size = 3. \begin{align*}{\frac{ab}{ \left ( b-c \right ) \left ( b+c \right ) \left ( a-c \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}c-2\,b\tan \left ( x/2 \right ) -a-c \right ) }-{\frac{cb}{ \left ( b-c \right ) \left ( b+c \right ) \left ( a-c \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}c-2\,b\tan \left ( x/2 \right ) -a-c \right ) }-2\,{\frac{ac}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }-2\,{\frac{{b}^{2}}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }+2\,{\frac{a{b}^{2}}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( a-c \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }-2\,{\frac{c{b}^{2}}{ \left ( b-c \right ) \left ( b+c \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( a-c \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-c \right ) \tan \left ( x/2 \right ) -2\,b}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }-2\,{\frac{\ln \left ( 1+\tan \left ( x/2 \right ) \right ) }{-2\,c+2\,b}}-2\,{\frac{\ln \left ( \tan \left ( x/2 \right ) -1 \right ) }{2\,b+2\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2/(a+c*sec(x)+b*tan(x)),x)

[Out]

1/(b-c)/(b+c)/(a-c)*ln(a*tan(1/2*x)^2-tan(1/2*x)^2*c-2*b*tan(1/2*x)-a-c)*a*b-1/(b-c)/(b+c)/(a-c)*ln(a*tan(1/2*
x)^2-tan(1/2*x)^2*c-2*b*tan(1/2*x)-a-c)*c*b-2/(b-c)/(b+c)/(-a^2-b^2+c^2)^(1/2)*arctan(1/2*(2*(a-c)*tan(1/2*x)-
2*b)/(-a^2-b^2+c^2)^(1/2))*a*c-2/(b-c)/(b+c)/(-a^2-b^2+c^2)^(1/2)*arctan(1/2*(2*(a-c)*tan(1/2*x)-2*b)/(-a^2-b^
2+c^2)^(1/2))*b^2+2/(b-c)/(b+c)/(-a^2-b^2+c^2)^(1/2)*arctan(1/2*(2*(a-c)*tan(1/2*x)-2*b)/(-a^2-b^2+c^2)^(1/2))
*b^2/(a-c)*a-2/(b-c)/(b+c)/(-a^2-b^2+c^2)^(1/2)*arctan(1/2*(2*(a-c)*tan(1/2*x)-2*b)/(-a^2-b^2+c^2)^(1/2))*b^2/
(a-c)*c-2/(-2*c+2*b)*ln(1+tan(1/2*x))-2/(2*b+2*c)*ln(tan(1/2*x)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+c*sec(x)+b*tan(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 21.9246, size = 1539, normalized size = 10.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(a^2 + b^2 - c^2)*a*c*log((2*a^4 + 3*a^2*b^2 + b^4 - (a^2 - b^2)*c^2 + 2*(a^3 + a*b^2)*c*cos(x) - (
a^4 - b^4 - 2*(a^2 - b^2)*c^2)*cos(x)^2 + 2*((a^2*b + b^3)*c - (a^3*b + a*b^3 - 2*a*b*c^2)*cos(x))*sin(x) + 2*
(2*a*b*c*cos(x)^2 - a*b*c + (a^2*b + b^3)*cos(x) - (a^3 + a*b^2 + (a^2 - b^2)*c*cos(x))*sin(x))*sqrt(a^2 + b^2
- c^2))/(2*a*c*cos(x) + (a^2 - b^2)*cos(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c)*sin(x))) - (a^2*b + b^3 - b*c
^2)*log(2*a*c*cos(x) + (a^2 - b^2)*cos(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c)*sin(x)) + (a^2*b + b^3 - b*c^2
- c^3 + (a^2 + b^2)*c)*log(sin(x) + 1) + (a^2*b + b^3 - b*c^2 + c^3 - (a^2 + b^2)*c)*log(-sin(x) + 1))/(a^2*b^
2 + b^4 + c^4 - (a^2 + 2*b^2)*c^2), 1/2*(2*sqrt(-a^2 - b^2 + c^2)*a*c*arctan((a*c*cos(x) + b*c*sin(x) + a^2 +
b^2)*sqrt(-a^2 - b^2 + c^2)/((a^2*b + b^3 - b*c^2)*cos(x) - (a^3 + a*b^2 - a*c^2)*sin(x))) + (a^2*b + b^3 - b*
c^2)*log(2*a*c*cos(x) + (a^2 - b^2)*cos(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c)*sin(x)) - (a^2*b + b^3 - b*c^2
- c^3 + (a^2 + b^2)*c)*log(sin(x) + 1) - (a^2*b + b^3 - b*c^2 + c^3 - (a^2 + b^2)*c)*log(-sin(x) + 1))/(a^2*b
^2 + b^4 + c^4 - (a^2 + 2*b^2)*c^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (x \right )}}{a + b \tan{\left (x \right )} + c \sec{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2/(a+c*sec(x)+b*tan(x)),x)

[Out]

Integral(sec(x)**2/(a + b*tan(x) + c*sec(x)), x)

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Giac [A]  time = 1.23609, size = 217, normalized size = 1.53 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, c\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - c \tan \left (\frac{1}{2} \, x\right ) - b}{\sqrt{-a^{2} - b^{2} + c^{2}}}\right )\right )} a c}{\sqrt{-a^{2} - b^{2} + c^{2}}{\left (b^{2} - c^{2}\right )}} + \frac{b \log \left (-a \tan \left (\frac{1}{2} \, x\right )^{2} + c \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a + c\right )}{b^{2} - c^{2}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{b - c} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{b + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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