\(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [20]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {4 i (a+i a \tan (c+d x))^5}{5 a^3 d}+\frac {2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac {i (a+i a \tan (c+d x))^7}{7 a^5 d} \]

[Out]

-4/5*I*(a+I*a*tan(d*x+c))^5/a^3/d+2/3*I*(a+I*a*tan(d*x+c))^6/a^4/d-1/7*I*(a+I*a*tan(d*x+c))^7/a^5/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 82, 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} \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {i (a+i a \tan (c+d x))^7}{7 a^5 d}+\frac {2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac {4 i (a+i a \tan (c+d x))^5}{5 a^3 d} \]

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((-4*I)/5)*(a + I*a*Tan[c + d*x])^5)/(a^3*d) + (((2*I)/3)*(a + I*a*Tan[c + d*x])^6)/(a^4*d) - ((I/7)*(a + I*a
*Tan[c + d*x])^7)/(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)^4 \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a+x)^4-4 a (a+x)^5+(a+x)^6\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {4 i (a+i a \tan (c+d x))^5}{5 a^3 d}+\frac {2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac {i (a+i a \tan (c+d x))^7}{7 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \sec ^7(c+d x) (7+22 \cos (2 (c+d x))-20 i \sin (2 (c+d x))) (-i \cos (5 (c+d x))+\sin (5 (c+d x)))}{105 d} \]

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*Sec[c + d*x]^7*(7 + 22*Cos[2*(c + d*x)] - (20*I)*Sin[2*(c + d*x)])*((-I)*Cos[5*(c + d*x)] + Sin[5*(c + d*
x)]))/(105*d)

Maple [A] (verified)

Time = 34.89 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {128 i a^{2} \left (35 \,{\mathrm e}^{8 i \left (d x +c \right )}+35 \,{\mathrm e}^{6 i \left (d x +c \right )}+21 \,{\mathrm e}^{4 i \left (d x +c \right )}+7 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\)	\(69\)
derivativedivides	\(\frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(113\)
default	\(\frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(113\)

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

128/105*I*a^2*(35*exp(8*I*(d*x+c))+35*exp(6*I*(d*x+c))+21*exp(4*I*(d*x+c))+7*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d
*x+c))+1)^7

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. 151 vs. \(2 (64) = 128\).

Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {128 \, {\left (-35 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 35 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 21 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 7 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2}\right )}}{105 \, {\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-128/105*(-35*I*a^2*e^(8*I*d*x + 8*I*c) - 35*I*a^2*e^(6*I*d*x + 6*I*c) - 21*I*a^2*e^(4*I*d*x + 4*I*c) - 7*I*a^
2*e^(2*I*d*x + 2*I*c) - I*a^2)/(d*e^(14*I*d*x + 14*I*c) + 7*d*e^(12*I*d*x + 12*I*c) + 21*d*e^(10*I*d*x + 10*I*
c) + 35*d*e^(8*I*d*x + 8*I*c) + 35*d*e^(6*I*d*x + 6*I*c) + 21*d*e^(4*I*d*x + 4*I*c) + 7*d*e^(2*I*d*x + 2*I*c)
+ d)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**2,x)

[Out]

-a**2*(Integral(tan(c + d*x)**2*sec(c + d*x)**6, x) + Integral(-2*I*tan(c + d*x)*sec(c + d*x)**6, x) + Integra
l(-sec(c + d*x)**6, x))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(15*a^2*tan(d*x + c)^7 - 35*I*a^2*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 - 105*I*a^2*tan(d*x + c)^4 - 3
5*a^2*tan(d*x + c)^3 - 105*I*a^2*tan(d*x + c)^2 - 105*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/105*(15*a^2*tan(d*x + c)^7 - 35*I*a^2*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 - 105*I*a^2*tan(d*x + c)^4 - 3
5*a^2*tan(d*x + c)^3 - 105*I*a^2*tan(d*x + c)^2 - 105*a^2*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 3.70 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.61 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )\,\left (105\,{\cos \left (c+d\,x\right )}^6+{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )\,105{}\mathrm {i}+35\,{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^3\,105{}\mathrm {i}-21\,{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^4+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^5\,35{}\mathrm {i}-15\,{\sin \left (c+d\,x\right )}^6\right )}{105\,d\,{\cos \left (c+d\,x\right )}^7} \]

[In]

int((a + a*tan(c + d*x)*1i)^2/cos(c + d*x)^6,x)

[Out]

(a^2*sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^5*35i + cos(c + d*x)^5*sin(c + d*x)*105i + 105*cos(c + d*x)^6 - 1
5*sin(c + d*x)^6 - 21*cos(c + d*x)^2*sin(c + d*x)^4 + cos(c + d*x)^3*sin(c + d*x)^3*105i + 35*cos(c + d*x)^4*s
in(c + d*x)^2))/(105*d*cos(c + d*x)^7)