3.253 \(\int \frac{1}{a \sin (c+d x)+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=74 \[ \frac{b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}-\frac{\log (\cos (c+d x)+1)}{2 d (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.0798118, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4397, 2721, 801} \[ \frac{b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)}-\frac{\log (\cos (c+d x)+1)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[c + d*x] + b*Tan[c + d*x])^(-1),x]

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) - Log[1 + Cos[c + d*x]]/(2*(a - b)*d) + (b*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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.0845769, size = 63, normalized size = 0.85 \[ \frac{(a-b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-(a+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+b \log (a \cos (c+d x)+b)}{d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[c + d*x] + b*Tan[c + d*x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.099, 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{b\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(1/(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*b/(a-b)/(a+b)*ln(b+a*cos(d*x+c))

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Maxima [A]  time = 1.13642, size = 96, normalized size = 1.3 \begin{align*} \frac{\frac{b \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(1/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

(b*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 = 0.534073, size = 173, normalized size = 2.34 \begin{align*} \frac{2 \, b \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(1/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*b*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{1}{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(1/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

Integral(1/(a*sin(c + d*x) + b*tan(c + d*x)), x)

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Giac [A]  time = 1.21295, size = 135, normalized size = 1.82 \begin{align*} \frac{\frac{2 \, b \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(1/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*b*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