\(\int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx\) [3]

Optimal result

Integrand size = 22, antiderivative size = 62 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \sec ^6(c+d x)}{6 d}+\frac {a \tan (c+d x)}{d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \]

[Out]

1/6*I*a*sec(d*x+c)^6/d+a*tan(d*x+c)/d+2/3*a*tan(d*x+c)^3/d+1/5*a*tan(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3567, 3852} \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \tan ^5(c+d x)}{5 d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {i a \sec ^6(c+d x)}{6 d} \]

[In]

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

[Out]

((I/6)*a*Sec[c + d*x]^6)/d + (a*Tan[c + d*x])/d + (2*a*Tan[c + d*x]^3)/(3*d) + (a*Tan[c + d*x]^5)/(5*d)

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {i a \sec ^6(c+d x)}{6 d}+a \int \sec ^6(c+d x) \, dx \\ & = \frac {i a \sec ^6(c+d x)}{6 d}-\frac {a \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {i a \sec ^6(c+d x)}{6 d}+\frac {a \tan (c+d x)}{d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \sec ^6(c+d x)}{6 d}+\frac {a \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \]

[In]

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

[Out]

((I/6)*a*Sec[c + d*x]^6)/d + (a*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d

Maple [A] (verified)

Time = 18.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {16 i a \left (20 \,{\mathrm e}^{6 i \left (d x +c \right )}+15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(56\)
derivativedivides	\(\frac {a \left (\tan \left (d x +c \right )+\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(66\)
default	\(\frac {a \left (\tan \left (d x +c \right )+\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(66\)

[In]

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

[Out]

16/15*I*a*(20*exp(6*I*(d*x+c))+15*exp(4*I*(d*x+c))+6*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d*x+c))+1)^6

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (54) = 108\).

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.89 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {16 \, {\left (-20 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 15 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-16/15*(-20*I*a*e^(6*I*d*x + 6*I*c) - 15*I*a*e^(4*I*d*x + 4*I*c) - 6*I*a*e^(2*I*d*x + 2*I*c) - I*a)/(d*e^(12*I
*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I
*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=\begin {cases} \frac {a \left (\frac {\tan ^{5}{\left (c + d x \right )}}{5} + \frac {2 \tan ^{3}{\left (c + d x \right )}}{3} + \tan {\left (c + d x \right )}\right ) + \frac {i a \sec ^{6}{\left (c + d x \right )}}{6}}{d} & \text {for}\: d \neq 0 \\x \left (i a \tan {\left (c \right )} + a\right ) \sec ^{6}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((a*(tan(c + d*x)**5/5 + 2*tan(c + d*x)**3/3 + tan(c + d*x)) + I*a*sec(c + d*x)**6/6)/d, Ne(d, 0)),
(x*(I*a*tan(c) + a)*sec(c)**6, True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {5 i \, a \tan \left (d x + c\right )^{6} + 6 \, a \tan \left (d x + c\right )^{5} + 15 i \, a \tan \left (d x + c\right )^{4} + 20 \, a \tan \left (d x + c\right )^{3} + 15 i \, a \tan \left (d x + c\right )^{2} + 30 \, a \tan \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(5*I*a*tan(d*x + c)^6 + 6*a*tan(d*x + c)^5 + 15*I*a*tan(d*x + c)^4 + 20*a*tan(d*x + c)^3 + 15*I*a*tan(d*x
+ c)^2 + 30*a*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {-5 i \, a \tan \left (d x + c\right )^{6} - 6 \, a \tan \left (d x + c\right )^{5} - 15 i \, a \tan \left (d x + c\right )^{4} - 20 \, a \tan \left (d x + c\right )^{3} - 15 i \, a \tan \left (d x + c\right )^{2} - 30 \, a \tan \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

-1/30*(-5*I*a*tan(d*x + c)^6 - 6*a*tan(d*x + c)^5 - 15*I*a*tan(d*x + c)^4 - 20*a*tan(d*x + c)^3 - 15*I*a*tan(d
*x + c)^2 - 30*a*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 3.71 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.81 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a\,\sin \left (c+d\,x\right )\,\left (30\,{\cos \left (c+d\,x\right )}^5+{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,15{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+6\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,d\,{\cos \left (c+d\,x\right )}^6} \]

[In]

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

[Out]

(a*sin(c + d*x)*(6*cos(c + d*x)*sin(c + d*x)^4 + cos(c + d*x)^4*sin(c + d*x)*15i + 30*cos(c + d*x)^5 + sin(c +
d*x)^5*5i + cos(c + d*x)^2*sin(c + d*x)^3*15i + 20*cos(c + d*x)^3*sin(c + d*x)^2))/(30*d*cos(c + d*x)^6)