\(\int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx\) [1]

Optimal result

Integrand size = 22, antiderivative size = 94 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \sec ^{10}(c+d x)}{10 d}+\frac {a \tan (c+d x)}{d}+\frac {4 a \tan ^3(c+d x)}{3 d}+\frac {6 a \tan ^5(c+d x)}{5 d}+\frac {4 a \tan ^7(c+d x)}{7 d}+\frac {a \tan ^9(c+d x)}{9 d} \]

[Out]

1/10*I*a*sec(d*x+c)^10/d+a*tan(d*x+c)/d+4/3*a*tan(d*x+c)^3/d+6/5*a*tan(d*x+c)^5/d+4/7*a*tan(d*x+c)^7/d+1/9*a*t
an(d*x+c)^9/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, 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} \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \tan ^9(c+d x)}{9 d}+\frac {4 a \tan ^7(c+d x)}{7 d}+\frac {6 a \tan ^5(c+d x)}{5 d}+\frac {4 a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {i a \sec ^{10}(c+d x)}{10 d} \]

[In]

Int[Sec[c + d*x]^10*(a + I*a*Tan[c + d*x]),x]

[Out]

((I/10)*a*Sec[c + d*x]^10)/d + (a*Tan[c + d*x])/d + (4*a*Tan[c + d*x]^3)/(3*d) + (6*a*Tan[c + d*x]^5)/(5*d) +
(4*a*Tan[c + d*x]^7)/(7*d) + (a*Tan[c + d*x]^9)/(9*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 ^{10}(c+d x)}{10 d}+a \int \sec ^{10}(c+d x) \, dx \\ & = \frac {i a \sec ^{10}(c+d x)}{10 d}-\frac {a \text {Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {i a \sec ^{10}(c+d x)}{10 d}+\frac {a \tan (c+d x)}{d}+\frac {4 a \tan ^3(c+d x)}{3 d}+\frac {6 a \tan ^5(c+d x)}{5 d}+\frac {4 a \tan ^7(c+d x)}{7 d}+\frac {a \tan ^9(c+d x)}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.84 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \sec ^{10}(c+d x)}{10 d}+\frac {a \left (\tan (c+d x)+\frac {4}{3} \tan ^3(c+d x)+\frac {6}{5} \tan ^5(c+d x)+\frac {4}{7} \tan ^7(c+d x)+\frac {1}{9} \tan ^9(c+d x)\right )}{d} \]

[In]

Integrate[Sec[c + d*x]^10*(a + I*a*Tan[c + d*x]),x]

[Out]

((I/10)*a*Sec[c + d*x]^10)/d + (a*(Tan[c + d*x] + (4*Tan[c + d*x]^3)/3 + (6*Tan[c + d*x]^5)/5 + (4*Tan[c + d*x
]^7)/7 + Tan[c + d*x]^9/9))/d

Maple [A] (verified)

Time = 176.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {256 i a \left (252 \,{\mathrm e}^{10 i \left (d x +c \right )}+210 \,{\mathrm e}^{8 i \left (d x +c \right )}+120 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{315 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\)	\(78\)
derivativedivides	\(\frac {a \left (\tan \left (d x +c \right )+\frac {i \left (\tan ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\tan ^{9}\left (d x +c \right )\right )}{9}+\frac {i \left (\tan ^{8}\left (d x +c \right )\right )}{2}+\frac {4 \left (\tan ^{7}\left (d x +c \right )\right )}{7}+i \left (\tan ^{6}\left (d x +c \right )\right )+\frac {6 \left (\tan ^{5}\left (d x +c \right )\right )}{5}+i \left (\tan ^{4}\left (d x +c \right )\right )+\frac {4 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(108\)
default	\(\frac {a \left (\tan \left (d x +c \right )+\frac {i \left (\tan ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\tan ^{9}\left (d x +c \right )\right )}{9}+\frac {i \left (\tan ^{8}\left (d x +c \right )\right )}{2}+\frac {4 \left (\tan ^{7}\left (d x +c \right )\right )}{7}+i \left (\tan ^{6}\left (d x +c \right )\right )+\frac {6 \left (\tan ^{5}\left (d x +c \right )\right )}{5}+i \left (\tan ^{4}\left (d x +c \right )\right )+\frac {4 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(108\)

[In]

int(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

256/315*I*a*(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. 189 vs. \(2 (82) = 164\).

Time = 0.24 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.01 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {256 \, {\left (-252 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 210 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 120 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 45 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )}}{315 \, {\left (d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-256/315*(-252*I*a*e^(10*I*d*x + 10*I*c) - 210*I*a*e^(8*I*d*x + 8*I*c) - 120*I*a*e^(6*I*d*x + 6*I*c) - 45*I*a*
e^(4*I*d*x + 4*I*c) - 10*I*a*e^(2*I*d*x + 2*I*c) - I*a)/(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 [A] (verification not implemented)

Time = 4.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\begin {cases} \frac {a \left (\frac {\tan ^{9}{\left (c + d x \right )}}{9} + \frac {4 \tan ^{7}{\left (c + d x \right )}}{7} + \frac {6 \tan ^{5}{\left (c + d x \right )}}{5} + \frac {4 \tan ^{3}{\left (c + d x \right )}}{3} + \tan {\left (c + d x \right )}\right ) + \frac {i a \sec ^{10}{\left (c + d x \right )}}{10}}{d} & \text {for}\: d \neq 0 \\x \left (i a \tan {\left (c \right )} + a\right ) \sec ^{10}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sec(d*x+c)**10*(a+I*a*tan(d*x+c)),x)

[Out]

Piecewise(((a*(tan(c + d*x)**9/9 + 4*tan(c + d*x)**7/7 + 6*tan(c + d*x)**5/5 + 4*tan(c + d*x)**3/3 + tan(c + d
*x)) + I*a*sec(c + d*x)**10/10)/d, Ne(d, 0)), (x*(I*a*tan(c) + a)*sec(c)**10, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {63 i \, a \tan \left (d x + c\right )^{10} + 70 \, a \tan \left (d x + c\right )^{9} + 315 i \, a \tan \left (d x + c\right )^{8} + 360 \, a \tan \left (d x + c\right )^{7} + 630 i \, a \tan \left (d x + c\right )^{6} + 756 \, a \tan \left (d x + c\right )^{5} + 630 i \, a \tan \left (d x + c\right )^{4} + 840 \, a \tan \left (d x + c\right )^{3} + 315 i \, a \tan \left (d x + c\right )^{2} + 630 \, a \tan \left (d x + c\right )}{630 \, d} \]

[In]

integrate(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/630*(63*I*a*tan(d*x + c)^10 + 70*a*tan(d*x + c)^9 + 315*I*a*tan(d*x + c)^8 + 360*a*tan(d*x + c)^7 + 630*I*a*
tan(d*x + c)^6 + 756*a*tan(d*x + c)^5 + 630*I*a*tan(d*x + c)^4 + 840*a*tan(d*x + c)^3 + 315*I*a*tan(d*x + c)^2
+ 630*a*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {-63 i \, a \tan \left (d x + c\right )^{10} - 70 \, a \tan \left (d x + c\right )^{9} - 315 i \, a \tan \left (d x + c\right )^{8} - 360 \, a \tan \left (d x + c\right )^{7} - 630 i \, a \tan \left (d x + c\right )^{6} - 756 \, a \tan \left (d x + c\right )^{5} - 630 i \, a \tan \left (d x + c\right )^{4} - 840 \, a \tan \left (d x + c\right )^{3} - 315 i \, a \tan \left (d x + c\right )^{2} - 630 \, a \tan \left (d x + c\right )}{630 \, d} \]

[In]

integrate(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/630*(-63*I*a*tan(d*x + c)^10 - 70*a*tan(d*x + c)^9 - 315*I*a*tan(d*x + c)^8 - 360*a*tan(d*x + c)^7 - 630*I*
a*tan(d*x + c)^6 - 756*a*tan(d*x + c)^5 - 630*I*a*tan(d*x + c)^4 - 840*a*tan(d*x + c)^3 - 315*I*a*tan(d*x + c)
^2 - 630*a*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 4.02 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a\,\left (-{\cos \left (c+d\,x\right )}^{10}\,63{}\mathrm {i}+256\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^9+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+96\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+80\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3+70\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )+63{}\mathrm {i}\right )}{630\,d\,{\cos \left (c+d\,x\right )}^{10}} \]

[In]

int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^10,x)

[Out]

(a*(70*cos(c + d*x)*sin(c + d*x) + 80*cos(c + d*x)^3*sin(c + d*x) + 96*cos(c + d*x)^5*sin(c + d*x) + 128*cos(c
+ d*x)^7*sin(c + d*x) + 256*cos(c + d*x)^9*sin(c + d*x) - cos(c + d*x)^10*63i + 63i))/(630*d*cos(c + d*x)^10)