3.170 \(\int \frac{\sec ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=55 \[ -\frac{\tan (c+d x)}{a^2 d}+\frac{2 i \log (\sin (c+d x))}{a^2 d}-\frac{2 i \log (\tan (c+d x))}{a^2 d}+\frac{2 x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.0706444, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3088, 848, 77} \[ -\frac{\tan (c+d x)}{a^2 d}+\frac{2 i \log (\sin (c+d x))}{a^2 d}-\frac{2 i \log (\tan (c+d x))}{a^2 d}+\frac{2 x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

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

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.413786, size = 71, normalized size = 1.29 \[ \frac{4 \tan ^{-1}(\tan (d x))+i \sec (c) \sec (c+d x) \left (\cos (d x) \log \left (\cos ^2(c+d x)\right )+\cos (2 c+d x) \log \left (\cos ^2(c+d x)\right )+2 i \sin (d x)\right )}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

(4*ArcTan[Tan[d*x]] + I*Sec[c]*Sec[c + d*x]*(Cos[d*x]*Log[Cos[c + d*x]^2] + Cos[2*c + d*x]*Log[Cos[c + d*x]^2]
+ (2*I)*Sin[d*x]))/(2*a^2*d)

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Maple [A]  time = 0.161, size = 35, normalized size = 0.6 \begin{align*}{\frac{-2\,i\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{2}}}-{\frac{\tan \left ( dx+c \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x)

[Out]

-2*I/d/a^2*ln(tan(d*x+c)-I)-tan(d*x+c)/a^2/d

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Maxima [A]  time = 1.11533, size = 41, normalized size = 0.75 \begin{align*} \frac{-\frac{2 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{\tan \left (d x + c\right )}{a^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.485937, size = 192, normalized size = 3.49 \begin{align*} \frac{4 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, d x +{\left (2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 2 i}{a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

(4*d*x*e^(2*I*d*x + 2*I*c) + 4*d*x + (2*I*e^(2*I*d*x + 2*I*c) + 2*I)*log(e^(2*I*d*x + 2*I*c) + 1) - 2*I)/(a^2*
d*e^(2*I*d*x + 2*I*c) + a^2*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.20462, size = 138, normalized size = 2.51 \begin{align*} \frac{2 \,{\left (-\frac{2 i \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}{a^{2}} + \frac{i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac{i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{-i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{2}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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