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

Optimal. Leaf size=23 \[ \frac {x}{a}+\frac {i \log (\cos (c+d x))}{a d} \]

[Out]

x/a+I*ln(cos(d*x+c))/a/d

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Rubi [A]  time = 0.07, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3092, 3090, 3475} \[ \frac {x}{a}+\frac {i \log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a + (I*Log[Cos[c + d*x]])/(a*d)

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3092

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 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)+i a \sin (c+d x)} \, dx &=-\frac {i \int \sec (c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2}\\ &=-\frac {i \int (i a+a \tan (c+d x)) \, dx}{a^2}\\ &=\frac {x}{a}-\frac {i \int \tan (c+d x) \, dx}{a}\\ &=\frac {x}{a}+\frac {i \log (\cos (c+d x))}{a d}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 23, normalized size = 1.00 \[ \frac {i \log (\cos (c+d x))+c+d x}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x + I*Log[Cos[c + d*x]])/(a*d)

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fricas [A]  time = 0.55, size = 26, normalized size = 1.13 \[ \frac {2 \, d x + i \, \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d*x + I*log(e^(2*I*d*x + 2*I*c) + 1))/(a*d)

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giac [B]  time = 0.20, size = 57, normalized size = 2.48 \[ -\frac {-\frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} + \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a} - \frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-I*log(tan(1/2*d*x + 1/2*c) + 1)/a + 2*I*log(tan(1/2*d*x + 1/2*c) - I)/a - I*log(tan(1/2*d*x + 1/2*c) - 1)/a
)/d

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maple [A]  time = 0.18, size = 22, normalized size = 0.96 \[ -\frac {i \ln \left (i \tan \left (d x +c \right )+1\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 101, normalized size = 4.39 \[ -\frac {-\frac {i \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {i \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {i \, \log \left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.74, size = 41, normalized size = 1.78 \[ -\frac {\left (2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\right )\,1{}\mathrm {i}}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*log(tan(c/2 + (d*x)/2) - 1i) - log(tan(c/2 + (d*x)/2)^2 - 1))*1i)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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