\(\int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [117]

Optimal result

Integrand size = 24, antiderivative size = 38 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 x}{a^2}+\frac {2 i \log (\cos (c+d x))}{a^2 d}-\frac {\tan (c+d x)}{a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\tan (c+d x)}{a^2 d}+\frac {2 i \log (\cos (c+d x))}{a^2 d}+\frac {2 x}{a^2} \]

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int \frac {a-x}{a+x} \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (-1+\frac {2 a}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = \frac {2 x}{a^2}+\frac {2 i \log (\cos (c+d x))}{a^2 d}-\frac {\tan (c+d x)}{a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 i \log (i-\tan (c+d x))}{a^2 d}-\frac {\tan (c+d x)}{a^2 d} \]

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {-\tan \left (d x +c \right )-2 i \ln \left (\tan \left (d x +c \right )-i\right )}{d \,a^{2}}\)	\(30\)
default	\(\frac {-\tan \left (d x +c \right )-2 i \ln \left (\tan \left (d x +c \right )-i\right )}{d \,a^{2}}\)	\(30\)
risch	\(\frac {4 x}{a^{2}}+\frac {4 c}{a^{2} d}-\frac {2 i}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(60\)

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.84 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 \, {\left (2 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, d x - {\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i\right )}}{a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

[In]

integrate(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**2,x)

[Out]

-Integral(sec(c + d*x)**4/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {-\frac {2 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {\tan \left (d x + c\right )}{a^{2}}}{d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (36) = 72\).

Time = 0.47 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.63 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2}} - \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{2}} + \frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{2}} + \frac {-i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}}\right )}}{d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.75 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\mathrm {tan}\left (c+d\,x\right )+\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,2{}\mathrm {i}}{a^2\,d} \]

[In]

int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

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