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

Optimal. Leaf size=41 \[ \frac {\log (a \cos (c+d x)+b \sin (c+d x))}{b d}-\frac {\log (\cos (c+d x))}{b d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3102, 3475, 3133} \[ \frac {\log (a \cos (c+d x)+b \sin (c+d x))}{b d}-\frac {\log (\cos (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3102

Int[1/(cos[(c_.) + (d_.)*(x_)]*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])), x_Symbol] :>
Dist[1/b, Int[Tan[c + d*x], x], x] + Dist[1/b, Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rule 3475

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx &=\frac {\int \frac {b \cos (c+d x)-a \sin (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b}+\frac {\int \tan (c+d x) \, dx}{b}\\ &=-\frac {\log (\cos (c+d x))}{b d}+\frac {\log (a \cos (c+d x)+b \sin (c+d x))}{b d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 0.44 \[ \frac {\log (a+b \tan (c+d x))}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a + b*Tan[c + d*x]]/(b*d)

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fricas [A]  time = 0.75, size = 59, normalized size = 1.44 \[ \frac {\log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - \log \left (\cos \left (d x + c\right )^{2}\right )}{2 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.97, size = 19, normalized size = 0.46 \[ \frac {\log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(b*tan(d*x + c) + a))/(b*d)

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maple [A]  time = 0.21, size = 19, normalized size = 0.46 \[ \frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/b*ln(a+b*tan(d*x+c))

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maxima [B]  time = 2.22, size = 103, normalized size = 2.51 \[ \frac {\frac {\log \left (-a - \frac {2 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{b} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.72, size = 62, normalized size = 1.51 \[ -\frac {2\,\mathrm {atanh}\left (\frac {b\,\left (b\,\cos \left (c+d\,x\right )-a\,\sin \left (c+d\,x\right )\right )}{2\,\cos \left (c+d\,x\right )\,a^2+\sin \left (c+d\,x\right )\,a\,b+\cos \left (c+d\,x\right )\,b^2}\right )}{b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a*cos(c + d*x) + b*sin(c + d*x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\left (c + d x \right )}}{a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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