\(\int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [135]

Optimal result

Integrand size = 24, antiderivative size = 50 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {x}{a^3}-\frac {i \log (\cos (c+d x))}{a^3 d}+\frac {2 i}{d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

-x/a^3-I*ln(cos(d*x+c))/a^3/d+2*I/d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 50, 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} \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 i}{d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {x}{a^3} \]

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-(x/a^3) - (I*Log[Cos[c + d*x]])/(a^3*d) + (2*I)/(d*(a^3 + I*a^3*Tan[c + d*x]))

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i \left (-\log (i-\tan (c+d x))-\frac {2 a}{a+i a \tan (c+d x)}\right )}{a^3 d} \]

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((-I)*(-Log[I - Tan[c + d*x]] - (2*a)/(a + I*a*Tan[c + d*x])))/(a^3*d)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {2}{a^{3} d \left (\tan \left (d x +c \right )-i\right )}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 a^{3} d}-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a^{3} d}\)	\(56\)
default	\(\frac {2}{a^{3} d \left (\tan \left (d x +c \right )-i\right )}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 a^{3} d}-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a^{3} d}\)	\(56\)
risch	\(\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{3} d}-\frac {2 x}{a^{3}}-\frac {2 c}{a^{3} d}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\)	\(56\)

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

2/a^3/d/(tan(d*x+c)-I)+1/2*I/a^3/d*ln(1+tan(d*x+c)^2)-1/a^3/d*arctan(tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (2 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a^{3} d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{4}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*Integral(sec(c + d*x)**4/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {4 \, {\left (-i \, \tan \left (d x + c\right ) - 1\right )}}{2 i \, a^{3} \tan \left (d x + c\right )^{2} + 4 \, a^{3} \tan \left (d x + c\right ) - 2 i \, a^{3}} - \frac {i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}}}{d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-(4*(-I*tan(d*x + c) - 1)/(2*I*a^3*tan(d*x + c)^2 + 4*a^3*tan(d*x + c) - 2*I*a^3) - I*log(I*tan(d*x + c) + 1)/
a^3)/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. \(2 (44) = 88\).

Time = 0.65 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.00 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{3}} + \frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} + \frac {3 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 i}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{2}}}{d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-(I*log(tan(1/2*d*x + 1/2*c) + 1)/a^3 - 2*I*log(tan(1/2*d*x + 1/2*c) - I)/a^3 + I*log(tan(1/2*d*x + 1/2*c) - 1
)/a^3 + (3*I*tan(1/2*d*x + 1/2*c)^2 + 10*tan(1/2*d*x + 1/2*c) - 3*I)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 3.93 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{a^3\,d}+\frac {2{}\mathrm {i}}{a^3\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

(log(tan(c + d*x) - 1i)*1i)/(a^3*d) + 2i/(a^3*d*(tan(c + d*x)*1i + 1))