3.254 \(\int \frac{\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=75 \[ -\frac{a \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}+\frac{\log (\cos (c+d x)+1)}{2 d (a-b)} \]

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) + Log[1 + Cos[c + d*x]]/(2*(a - b)*d) - (a*Log[b + a*Cos[c + d*x]])/((a^2
- b^2)*d)

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Rubi [A]  time = 0.185584, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {4397, 2668, 706, 31, 633} \[ -\frac{a \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}+\frac{\log (\cos (c+d x)+1)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) + Log[1 + Cos[c + d*x]]/(2*(a - b)*d) - (a*Log[b + a*Cos[c + d*x]])/((a^2
- b^2)*d)

Rule 4397

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 706

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

Rule 31

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

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)]

Rubi steps

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

Mathematica [A]  time = 0.0624528, size = 64, normalized size = 0.85 \[ \frac{(a-b) \log (1-\cos (c+d x))+(a+b) \log (\cos (c+d x)+1)-2 a \log (a \cos (c+d x)+b)}{2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

((a - b)*Log[1 - Cos[c + d*x]] + (a + b)*Log[1 + Cos[c + d*x]] - 2*a*Log[b + a*Cos[c + d*x]])/(2*(a - b)*(a +
b)*d)

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Maple [A]  time = 0.123, size = 75, normalized size = 1. \begin{align*}{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{d \left ( 2\,a-2\,b \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{d \left ( 2\,a+2\,b \right ) }}-{\frac{a\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a-b \right ) \left ( a+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

1/d/(2*a-2*b)*ln(cos(d*x+c)+1)+1/d/(2*a+2*b)*ln(-1+cos(d*x+c))-1/d*a/(a-b)/(a+b)*ln(b+a*cos(d*x+c))

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Maxima [A]  time = 1.06011, size = 99, normalized size = 1.32 \begin{align*} -\frac{\frac{a \log \left (a + b - \frac{{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} - b^{2}} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2 - b^2) - log(sin(d*x + c)/(cos(d*x + c) + 1)
)/(a + b))/d

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Fricas [A]  time = 0.538659, size = 174, normalized size = 2.32 \begin{align*} -\frac{2 \, a \log \left (a \cos \left (d x + c\right ) + b\right ) -{\left (a + b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (a - b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{2} - b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*log(a*cos(d*x + c) + b) - (a + b)*log(1/2*cos(d*x + c) + 1/2) - (a - b)*log(-1/2*cos(d*x + c) + 1/2)
)/((a^2 - b^2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20544, size = 136, normalized size = 1.81 \begin{align*} -\frac{\frac{2 \, a \log \left ({\left | -a - b - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} - b^{2}} - \frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)))
/(a^2 - b^2) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b))/d