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

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

[Out]

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

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Rubi [A]  time = 0.080833, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2668, 706, 31, 633} \[ -\frac{b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0609976, size = 64, normalized size = 0.85 \[ \frac{(b-a) \log (1-\sin (c+d x))+(a+b) \log (\sin (c+d x)+1)-2 b \log (a+b \sin (c+d x))}{2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 76, normalized size = 1. \begin{align*} -{\frac{b\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a-b \right ) \left ( a+b \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,a+2\,b \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d \left ( 2\,a-2\,b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.946195, size = 86, normalized size = 1.15 \begin{align*} -\frac{\frac{2 \, b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} + \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.25565, size = 158, normalized size = 2.11 \begin{align*} -\frac{2 \, b \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left (a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a - b\right )} \log \left (-\sin \left (d x + c\right ) + 1\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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*b*log(b*sin(d*x + c) + a) - (a + b)*log(sin(d*x + c) + 1) + (a - b)*log(-sin(d*x + c) + 1))/((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 + b \sin{\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+b*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.19167, size = 96, normalized size = 1.28 \begin{align*} -\frac{\frac{2 \, b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} + \frac{\log \left ({\left | \sin \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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(2*b^2*log(abs(b*sin(d*x + c) + a))/(a^2*b - b^3) - log(abs(sin(d*x + c) + 1))/(a - b) + log(abs(sin(d*x
+ c) - 1))/(a + b))/d