\(\int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx\) [101]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {2 i (a-i a \tan (c+d x))^3}{3 a^4 d}-\frac {i (a-i a \tan (c+d x))^4}{4 a^5 d} \]

[Out]

2/3*I*(a-I*a*tan(d*x+c))^3/a^4/d-1/4*I*(a-I*a*tan(d*x+c))^4/a^5/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, 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} \[ \int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {2 i (a-i a \tan (c+d x))^3}{3 a^4 d}-\frac {i (a-i a \tan (c+d x))^4}{4 a^5 d} \]

[In]

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

[Out]

(((2*I)/3)*(a - I*a*Tan[c + d*x])^3)/(a^4*d) - ((I/4)*(a - I*a*Tan[c + d*x])^4)/(a^5*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 (a-x)^2 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a (a-x)^2-(a-x)^3\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = \frac {2 i (a-i a \tan (c+d x))^3}{3 a^4 d}-\frac {i (a-i a \tan (c+d x))^4}{4 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\tan (c+d x) \left (12-6 i \tan (c+d x)+4 \tan ^2(c+d x)-3 i \tan ^3(c+d x)\right )}{12 a d} \]

[In]

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

[Out]

(Tan[c + d*x]*(12 - (6*I)*Tan[c + d*x] + 4*Tan[c + d*x]^2 - (3*I)*Tan[c + d*x]^3))/(12*a*d)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {4 i \left (4 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}\)	\(36\)
derivativedivides	\(-\frac {-\tan \left (d x +c \right )+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}}{a d}\)	\(50\)
default	\(-\frac {-\tan \left (d x +c \right )+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}}{a d}\)	\(50\)

[In]

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

[Out]

4/3*I*(4*exp(2*I*(d*x+c))+1)/d/a/(exp(2*I*(d*x+c))+1)^4

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {4 \, {\left (-4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{3 \, {\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

[In]

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

[Out]

-4/3*(-4*I*e^(2*I*d*x + 2*I*c) - I)/(a*d*e^(8*I*d*x + 8*I*c) + 4*a*d*e^(6*I*d*x + 6*I*c) + 6*a*d*e^(4*I*d*x +
4*I*c) + 4*a*d*e^(2*I*d*x + 2*I*c) + a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {-3 i \, \tan \left (d x + c\right )^{4} + 4 \, \tan \left (d x + c\right )^{3} - 6 i \, \tan \left (d x + c\right )^{2} + 12 \, \tan \left (d x + c\right )}{12 \, a d} \]

[In]

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

[Out]

1/12*(-3*I*tan(d*x + c)^4 + 4*tan(d*x + c)^3 - 6*I*tan(d*x + c)^2 + 12*tan(d*x + c))/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {3 i \, \tan \left (d x + c\right )^{4} - 4 \, \tan \left (d x + c\right )^{3} + 6 i \, \tan \left (d x + c\right )^{2} - 12 \, \tan \left (d x + c\right )}{12 \, a d} \]

[In]

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

[Out]

-1/12*(3*I*tan(d*x + c)^4 - 4*tan(d*x + c)^3 + 6*I*tan(d*x + c)^2 - 12*tan(d*x + c))/(a*d)

Mupad [B] (verification not implemented)

Time = 4.52 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (12\,{\cos \left (c+d\,x\right )}^3-{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,6{}\mathrm {i}+4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^2-{\sin \left (c+d\,x\right )}^3\,3{}\mathrm {i}\right )}{12\,a\,d\,{\cos \left (c+d\,x\right )}^4} \]

[In]

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

[Out]

(sin(c + d*x)*(4*cos(c + d*x)*sin(c + d*x)^2 - cos(c + d*x)^2*sin(c + d*x)*6i + 12*cos(c + d*x)^3 - sin(c + d*
x)^3*3i))/(12*a*d*cos(c + d*x)^4)