\(\int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [152]

Optimal result

Integrand size = 24, antiderivative size = 63 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {4 x}{a^4}-\frac {4 i \log (\cos (c+d x))}{a^4 d}+\frac {\tan (c+d x)}{a^4 d}+\frac {4 i}{d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

-4*x/a^4-4*I*ln(cos(d*x+c))/a^4/d+tan(d*x+c)/a^4/d+4*I/d/(a^4+I*a^4*tan(d*x+c))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 63, 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} \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\tan (c+d x)}{a^4 d}+\frac {4 i}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {4 i \log (\cos (c+d x))}{a^4 d}-\frac {4 x}{a^4} \]

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-4*x)/a^4 - ((4*I)*Log[Cos[c + d*x]])/(a^4*d) + Tan[c + d*x]/(a^4*d) + (4*I)/(d*(a^4 + I*a^4*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)^2}{(a+x)^2} \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (1+\frac {4 a^2}{(a+x)^2}-\frac {4 a}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {4 x}{a^4}-\frac {4 i \log (\cos (c+d x))}{a^4 d}+\frac {\tan (c+d x)}{a^4 d}+\frac {4 i}{d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i \left (-4 \log (i-\tan (c+d x))+i \tan (c+d x)+\frac {4 i}{-i+\tan (c+d x)}\right )}{a^4 d} \]

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((-I)*(-4*Log[I - Tan[c + d*x]] + I*Tan[c + d*x] + (4*I)/(-I + Tan[c + d*x])))/(a^4*d)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )}{a^{4} d}-\frac {4 \arctan \left (\tan \left (d x +c \right )\right )}{a^{4} d}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{4} d}+\frac {4}{a^{4} d \left (\tan \left (d x +c \right )-i\right )}\)	\(69\)
default	\(\frac {\tan \left (d x +c \right )}{a^{4} d}-\frac {4 \arctan \left (\tan \left (d x +c \right )\right )}{a^{4} d}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{4} d}+\frac {4}{a^{4} d \left (\tan \left (d x +c \right )-i\right )}\)	\(69\)
risch	\(\frac {2 i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{4} d}-\frac {8 x}{a^{4}}-\frac {8 c}{a^{4} d}+\frac {2 i}{d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {4 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{4} d}\)	\(78\)

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

tan(d*x+c)/a^4/d-4/a^4/d*arctan(tan(d*x+c))+2*I/a^4/d*ln(1+tan(d*x+c)^2)+4/a^4/d/(tan(d*x+c)-I)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 \, {\left (4 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (2 \, d x - i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, {\left (i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i\right )}}{a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{6}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(sec(d*x+c)**6/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**6/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1), x
)/a**4

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {4 \, {\left (\tan \left (d x + c\right )^{2} - 2 i \, \tan \left (d x + c\right ) - 1\right )}}{a^{4} \tan \left (d x + c\right )^{3} - 3 i \, a^{4} \tan \left (d x + c\right )^{2} - 3 \, a^{4} \tan \left (d x + c\right ) + i \, a^{4}} + \frac {4 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {\tan \left (d x + c\right )}{a^{4}}}{d} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

(4*(tan(d*x + c)^2 - 2*I*tan(d*x + c) - 1)/(a^4*tan(d*x + c)^3 - 3*I*a^4*tan(d*x + c)^2 - 3*a^4*tan(d*x + c) +
I*a^4) + 4*I*log(I*tan(d*x + c) + 1)/a^4 + tan(d*x + c)/a^4)/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. 146 vs. \(2 (57) = 114\).

Time = 0.72 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.32 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {2 \, {\left (-\frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{4}} + \frac {4 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{4}} - \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{4}} + \frac {2 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {2 \, {\left (3 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 i\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{2}}\right )}}{d} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

2*(-2*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^4 + 4*I*log(tan(1/2*d*x + 1/2*c) - I)/a^4 - 2*I*log(tan(1/2*d*x + 1/2*
c) - 1)/a^4 + (2*I*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) - 2*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^4) - 2
*(3*I*tan(1/2*d*x + 1/2*c)^2 + 8*tan(1/2*d*x + 1/2*c) - 3*I)/(a^4*(tan(1/2*d*x + 1/2*c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,4{}\mathrm {i}}{a^4\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )}{a^4\,d}+\frac {4{}\mathrm {i}}{a^4\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

int(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

(log(tan(c + d*x) - 1i)*4i)/(a^4*d) + tan(c + d*x)/(a^4*d) + 4i/(a^4*d*(tan(c + d*x)*1i + 1))