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

Optimal. Leaf size=94 \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b d \sqrt{a-b} \sqrt{a+b}}+\frac{A x}{a}+\frac{C \tanh ^{-1}(\sin (c+d x))}{b d} \]

[Out]

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

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Rubi [A]  time = 0.172202, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4050, 3770, 3919, 3831, 2659, 208} \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b d \sqrt{a-b} \sqrt{a+b}}+\frac{A x}{a}+\frac{C \tanh ^{-1}(\sin (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4050

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

Rule 3770

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

Rule 3919

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 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

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 208

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

Rubi steps

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

Mathematica [C]  time = 0.52008, size = 261, normalized size = 2.78 \[ \frac{2 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (2 (\sin (c)+i \cos (c)) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )+\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2} \left (-a C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+a C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+A b d x\right )\right )}{a b d \sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(Sqrt[a^2 - b^2]*(A*b*d*x - a*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]] + a*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Sqrt[(Cos[c] - I*Sin[c])^2] + 2*(A*b^2 + a*(-(b*B) + a*C
))*ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Si
n[c])^2])]*(I*Cos[c] + Sin[c])))/(a*b*Sqrt[a^2 - b^2]*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sqrt
[(Cos[c] - I*Sin[c])^2])

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Maple [B]  time = 0.081, size = 202, normalized size = 2.2 \begin{align*} 2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{ad}}-2\,{\frac{Ab}{ad\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{B}{d\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{aC}{db\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{C}{db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.41009, size = 824, normalized size = 8.77 \begin{align*} \left [\frac{2 \,{\left (A a^{2} b - A b^{3}\right )} d x +{\left (C a^{2} - B a b + A b^{2}\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 (C a^{3} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} b - a b^{3}\right )} d}, \frac{2 \,{\left (A a^{2} b - A b^{3}\right )} d x - 2 \,{\left (C a^{2} - B a b + A b^{2}\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} - C a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (C a^{3} - C a b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} b - a b^{3}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.33198, size = 201, normalized size = 2.14 \begin{align*} \frac{\frac{{\left (d x + c\right )} A}{a} + \frac{C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b} - \frac{C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b} - \frac{2 \,{\left (C a^{2} - B a b + A b^{2}\right )}{\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 )}}{\sqrt{-a^{2} + b^{2}} a b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*A/a + C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b - C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b - 2*(C*a^2 -
B*a*b + A*b^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)))/(sqrt(-a^2 + b^2)*a*b))/d