\(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx\) [60]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i (a+i a \tan (c+d x))^8}{2 a^3 d}+\frac {4 i (a+i a \tan (c+d x))^9}{9 a^4 d}-\frac {i (a+i a \tan (c+d x))^{10}}{10 a^5 d} \]

[Out]

-1/2*I*(a+I*a*tan(d*x+c))^8/a^3/d+4/9*I*(a+I*a*tan(d*x+c))^9/a^4/d-1/10*I*(a+I*a*tan(d*x+c))^10/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} \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i (a+i a \tan (c+d x))^{10}}{10 a^5 d}+\frac {4 i (a+i a \tan (c+d x))^9}{9 a^4 d}-\frac {i (a+i a \tan (c+d x))^8}{2 a^3 d} \]

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-1/2*I)*(a + I*a*Tan[c + d*x])^8)/(a^3*d) + (((4*I)/9)*(a + I*a*Tan[c + d*x])^9)/(a^4*d) - ((I/10)*(a + I*a*
Tan[c + d*x])^10)/(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)^7 \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a+x)^7-4 a (a+x)^8+(a+x)^9\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i (a+i a \tan (c+d x))^8}{2 a^3 d}+\frac {4 i (a+i a \tan (c+d x))^9}{9 a^4 d}-\frac {i (a+i a \tan (c+d x))^{10}}{10 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 \sec ^{10}(c+d x) (5+23 \cos (2 (c+d x))-22 i \sin (2 (c+d x))) (-i \cos (8 (c+d x))+\sin (8 (c+d x)))}{180 d} \]

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Sec[c + d*x]^10*(5 + 23*Cos[2*(c + d*x)] - (22*I)*Sin[2*(c + d*x)])*((-I)*Cos[8*(c + d*x)] + Sin[8*(c + d
*x)]))/(180*d)

Maple [A] (verified)

Time = 181.82 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.24

method	result	size
risch	\(\frac {128 i a^{5} \left (120 \,{\mathrm e}^{14 i \left (d x +c \right )}+210 \,{\mathrm e}^{12 i \left (d x +c \right )}+252 \,{\mathrm e}^{10 i \left (d x +c \right )}+210 \,{\mathrm e}^{8 i \left (d x +c \right )}+120 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{45 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\)	\(102\)
derivativedivides	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {\sin ^{6}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{6}\left (d x +c \right )}{60 \cos \left (d x +c \right )^{6}}\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {4 \left (\sin ^{5}\left (d x +c \right )\right )}{63 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{5}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{5}}\right )-10 i a^{5} \left (\frac {\sin ^{4}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}\right )-10 a^{5} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {5 i a^{5}}{6 \cos \left (d x +c \right )^{6}}-a^{5} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(295\)
default	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {\sin ^{6}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{6}\left (d x +c \right )}{60 \cos \left (d x +c \right )^{6}}\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {4 \left (\sin ^{5}\left (d x +c \right )\right )}{63 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{5}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{5}}\right )-10 i a^{5} \left (\frac {\sin ^{4}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}\right )-10 a^{5} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {5 i a^{5}}{6 \cos \left (d x +c \right )^{6}}-a^{5} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(295\)

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

128/45*I*a^5*(120*exp(14*I*(d*x+c))+210*exp(12*I*(d*x+c))+252*exp(10*I*(d*x+c))+210*exp(8*I*(d*x+c))+120*exp(6
*I*(d*x+c))+45*exp(4*I*(d*x+c))+10*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d*x+c))+1)^10

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

Time = 0.24 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.79 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {128 \, {\left (-120 i \, a^{5} e^{\left (14 i \, d x + 14 i \, c\right )} - 210 i \, a^{5} e^{\left (12 i \, d x + 12 i \, c\right )} - 252 i \, a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} - 210 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} - 120 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 45 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{5}\right )}}{45 \, {\left (d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

-128/45*(-120*I*a^5*e^(14*I*d*x + 14*I*c) - 210*I*a^5*e^(12*I*d*x + 12*I*c) - 252*I*a^5*e^(10*I*d*x + 10*I*c)
- 210*I*a^5*e^(8*I*d*x + 8*I*c) - 120*I*a^5*e^(6*I*d*x + 6*I*c) - 45*I*a^5*e^(4*I*d*x + 4*I*c) - 10*I*a^5*e^(2
*I*d*x + 2*I*c) - I*a^5)/(d*e^(20*I*d*x + 20*I*c) + 10*d*e^(18*I*d*x + 18*I*c) + 45*d*e^(16*I*d*x + 16*I*c) +
120*d*e^(14*I*d*x + 14*I*c) + 210*d*e^(12*I*d*x + 12*I*c) + 252*d*e^(10*I*d*x + 10*I*c) + 210*d*e^(8*I*d*x + 8
*I*c) + 120*d*e^(6*I*d*x + 6*I*c) + 45*d*e^(4*I*d*x + 4*I*c) + 10*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=i a^{5} \left (\int \left (- i \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int 5 \tan {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 10 \tan ^{3}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int \tan ^{5}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int 10 i \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 5 i \tan ^{4}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**5,x)

[Out]

I*a**5*(Integral(-I*sec(c + d*x)**6, x) + Integral(5*tan(c + d*x)*sec(c + d*x)**6, x) + Integral(-10*tan(c + d
*x)**3*sec(c + d*x)**6, x) + Integral(tan(c + d*x)**5*sec(c + d*x)**6, x) + Integral(10*I*tan(c + d*x)**2*sec(
c + d*x)**6, x) + Integral(-5*I*tan(c + d*x)**4*sec(c + d*x)**6, x))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.32 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {-9 i \, a^{5} \tan \left (d x + c\right )^{10} - 50 \, a^{5} \tan \left (d x + c\right )^{9} + 90 i \, a^{5} \tan \left (d x + c\right )^{8} + 210 i \, a^{5} \tan \left (d x + c\right )^{6} + 252 \, a^{5} \tan \left (d x + c\right )^{5} + 240 \, a^{5} \tan \left (d x + c\right )^{3} - 225 i \, a^{5} \tan \left (d x + c\right )^{2} - 90 \, a^{5} \tan \left (d x + c\right )}{90 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/90*(-9*I*a^5*tan(d*x + c)^10 - 50*a^5*tan(d*x + c)^9 + 90*I*a^5*tan(d*x + c)^8 + 210*I*a^5*tan(d*x + c)^6 +
252*a^5*tan(d*x + c)^5 + 240*a^5*tan(d*x + c)^3 - 225*I*a^5*tan(d*x + c)^2 - 90*a^5*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.81 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.32 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {-9 i \, a^{5} \tan \left (d x + c\right )^{10} - 50 \, a^{5} \tan \left (d x + c\right )^{9} + 90 i \, a^{5} \tan \left (d x + c\right )^{8} + 210 i \, a^{5} \tan \left (d x + c\right )^{6} + 252 \, a^{5} \tan \left (d x + c\right )^{5} + 240 \, a^{5} \tan \left (d x + c\right )^{3} - 225 i \, a^{5} \tan \left (d x + c\right )^{2} - 90 \, a^{5} \tan \left (d x + c\right )}{90 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

-1/90*(-9*I*a^5*tan(d*x + c)^10 - 50*a^5*tan(d*x + c)^9 + 90*I*a^5*tan(d*x + c)^8 + 210*I*a^5*tan(d*x + c)^6 +
252*a^5*tan(d*x + c)^5 + 240*a^5*tan(d*x + c)^3 - 225*I*a^5*tan(d*x + c)^2 - 90*a^5*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 4.05 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5\,\sin \left (c+d\,x\right )\,\left (90\,{\cos \left (c+d\,x\right )}^9+{\cos \left (c+d\,x\right )}^8\,\sin \left (c+d\,x\right )\,225{}\mathrm {i}-240\,{\cos \left (c+d\,x\right )}^7\,{\sin \left (c+d\,x\right )}^2-252\,{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^5\,210{}\mathrm {i}-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^7\,90{}\mathrm {i}+50\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^8+{\sin \left (c+d\,x\right )}^9\,9{}\mathrm {i}\right )}{90\,d\,{\cos \left (c+d\,x\right )}^{10}} \]

[In]

int((a + a*tan(c + d*x)*1i)^5/cos(c + d*x)^6,x)

[Out]

(a^5*sin(c + d*x)*(50*cos(c + d*x)*sin(c + d*x)^8 + cos(c + d*x)^8*sin(c + d*x)*225i + 90*cos(c + d*x)^9 + sin
(c + d*x)^9*9i - cos(c + d*x)^2*sin(c + d*x)^7*90i - cos(c + d*x)^4*sin(c + d*x)^5*210i - 252*cos(c + d*x)^5*s
in(c + d*x)^4 - 240*cos(c + d*x)^7*sin(c + d*x)^2))/(90*d*cos(c + d*x)^10)