3.1.79 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx\) [79]

3.1.79.1 Optimal result
3.1.79.2 Mathematica [A] (verified)
3.1.79.3 Rubi [A] (verified)
3.1.79.4 Maple [B] (verified)
3.1.79.5 Fricas [B] (verification not implemented)
3.1.79.6 Sympy [F]
3.1.79.7 Maxima [B] (verification not implemented)
3.1.79.8 Giac [B] (verification not implemented)
3.1.79.9 Mupad [B] (verification not implemented)

3.1.79.1 Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {i (a+i a \tan (c+d x))^{10}}{5 a^2 d}+\frac {i (a+i a \tan (c+d x))^{11}}{11 a^3 d} \]

output
-1/5*I*(a+I*a*tan(d*x+c))^10/a^2/d+1/11*I*(a+I*a*tan(d*x+c))^11/a^3/d
 
3.1.79.2 Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^8 (-i+\tan (c+d x))^{10} (6 i+5 \tan (c+d x))}{55 d} \]

input
Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]
 
output
(a^8*(-I + Tan[c + d*x])^10*(6*I + 5*Tan[c + d*x]))/(55*d)
 
3.1.79.3 Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^4 (a+i a \tan (c+d x))^8dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i \int (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^9d(i a \tan (c+d x))}{a^3 d}\)

\(\Big \downarrow \) 49

\(\displaystyle -\frac {i \int \left (2 a (i \tan (c+d x) a+a)^9-(i \tan (c+d x) a+a)^{10}\right )d(i a \tan (c+d x))}{a^3 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \left (\frac {1}{5} a (a+i a \tan (c+d x))^{10}-\frac {1}{11} (a+i a \tan (c+d x))^{11}\right )}{a^3 d}\)

input
Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]
 
output
((-I)*((a*(a + I*a*Tan[c + d*x])^10)/5 - (a + I*a*Tan[c + d*x])^11/11))/(a 
^3*d)
 

3.1.79.3.1 Defintions of rubi rules used

rule 49
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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[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]
 
3.1.79.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (47 ) = 94\).

Time = 286.48 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.25

method result size
risch \(\frac {1024 i a^{8} \left (55 \,{\mathrm e}^{18 i \left (d x +c \right )}+165 \,{\mathrm e}^{16 i \left (d x +c \right )}+330 \,{\mathrm e}^{14 i \left (d x +c \right )}+462 \,{\mathrm e}^{12 i \left (d x +c \right )}+462 \,{\mathrm e}^{10 i \left (d x +c \right )}+330 \,{\mathrm e}^{8 i \left (d x +c \right )}+165 \,{\mathrm e}^{6 i \left (d x +c \right )}+55 \,{\mathrm e}^{4 i \left (d x +c \right )}+11 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{55 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{11}}\) \(124\)
derivativedivides \(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{11 \cos \left (d x +c \right )^{11}}+\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{99 \cos \left (d x +c \right )^{9}}\right )-56 i a^{8} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )-28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{63 \cos \left (d x +c \right )^{7}}\right )+\frac {2 i a^{8}}{\cos \left (d x +c \right )^{4}}+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}\right )+56 i a^{8} \left (\frac {\sin ^{6}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{6}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{6}}\right )-28 a^{8} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-8 i a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {\sin ^{8}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{8}}\right )-a^{8} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) \(339\)
default \(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{11 \cos \left (d x +c \right )^{11}}+\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{99 \cos \left (d x +c \right )^{9}}\right )-56 i a^{8} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )-28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{63 \cos \left (d x +c \right )^{7}}\right )+\frac {2 i a^{8}}{\cos \left (d x +c \right )^{4}}+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}\right )+56 i a^{8} \left (\frac {\sin ^{6}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{6}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{6}}\right )-28 a^{8} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-8 i a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {\sin ^{8}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{8}}\right )-a^{8} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) \(339\)

input
int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)
 
output
1024/55*I*a^8*(55*exp(18*I*(d*x+c))+165*exp(16*I*(d*x+c))+330*exp(14*I*(d* 
x+c))+462*exp(12*I*(d*x+c))+462*exp(10*I*(d*x+c))+330*exp(8*I*(d*x+c))+165 
*exp(6*I*(d*x+c))+55*exp(4*I*(d*x+c))+11*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d 
*x+c))+1)^11
 
3.1.79.5 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. 269 vs. \(2 (43) = 86\).

Time = 0.24 (sec) , antiderivative size = 269, normalized size of antiderivative = 4.89 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {1024 \, {\left (-55 i \, a^{8} e^{\left (18 i \, d x + 18 i \, c\right )} - 165 i \, a^{8} e^{\left (16 i \, d x + 16 i \, c\right )} - 330 i \, a^{8} e^{\left (14 i \, d x + 14 i \, c\right )} - 462 i \, a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} - 462 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} - 330 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 165 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} - 55 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} - 11 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{8}\right )}}{55 \, {\left (d e^{\left (22 i \, d x + 22 i \, c\right )} + 11 \, d e^{\left (20 i \, d x + 20 i \, c\right )} + 55 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 165 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 330 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 462 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 462 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 330 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 165 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 55 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 11 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

input
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")
 
output
-1024/55*(-55*I*a^8*e^(18*I*d*x + 18*I*c) - 165*I*a^8*e^(16*I*d*x + 16*I*c 
) - 330*I*a^8*e^(14*I*d*x + 14*I*c) - 462*I*a^8*e^(12*I*d*x + 12*I*c) - 46 
2*I*a^8*e^(10*I*d*x + 10*I*c) - 330*I*a^8*e^(8*I*d*x + 8*I*c) - 165*I*a^8* 
e^(6*I*d*x + 6*I*c) - 55*I*a^8*e^(4*I*d*x + 4*I*c) - 11*I*a^8*e^(2*I*d*x + 
 2*I*c) - I*a^8)/(d*e^(22*I*d*x + 22*I*c) + 11*d*e^(20*I*d*x + 20*I*c) + 5 
5*d*e^(18*I*d*x + 18*I*c) + 165*d*e^(16*I*d*x + 16*I*c) + 330*d*e^(14*I*d* 
x + 14*I*c) + 462*d*e^(12*I*d*x + 12*I*c) + 462*d*e^(10*I*d*x + 10*I*c) + 
330*d*e^(8*I*d*x + 8*I*c) + 165*d*e^(6*I*d*x + 6*I*c) + 55*d*e^(4*I*d*x + 
4*I*c) + 11*d*e^(2*I*d*x + 2*I*c) + d)
 
3.1.79.6 Sympy [F]

\[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=a^{8} \left (\int \left (- 28 \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int 70 \tan ^{4}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 28 \tan ^{6}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \tan ^{8}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 8 i \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 56 i \tan ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int 56 i \tan ^{5}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 8 i \tan ^{7}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

input
integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**8,x)
 
output
a**8*(Integral(-28*tan(c + d*x)**2*sec(c + d*x)**4, x) + Integral(70*tan(c 
 + d*x)**4*sec(c + d*x)**4, x) + Integral(-28*tan(c + d*x)**6*sec(c + d*x) 
**4, x) + Integral(tan(c + d*x)**8*sec(c + d*x)**4, x) + Integral(8*I*tan( 
c + d*x)*sec(c + d*x)**4, x) + Integral(-56*I*tan(c + d*x)**3*sec(c + d*x) 
**4, x) + Integral(56*I*tan(c + d*x)**5*sec(c + d*x)**4, x) + Integral(-8* 
I*tan(c + d*x)**7*sec(c + d*x)**4, x) + Integral(sec(c + d*x)**4, x))
 
3.1.79.7 Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (43) = 86\).

Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.44 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {5 \, a^{8} \tan \left (d x + c\right )^{11} - 44 i \, a^{8} \tan \left (d x + c\right )^{10} - 165 \, a^{8} \tan \left (d x + c\right )^{9} + 330 i \, a^{8} \tan \left (d x + c\right )^{8} + 330 \, a^{8} \tan \left (d x + c\right )^{7} + 462 \, a^{8} \tan \left (d x + c\right )^{5} - 660 i \, a^{8} \tan \left (d x + c\right )^{4} - 495 \, a^{8} \tan \left (d x + c\right )^{3} + 220 i \, a^{8} \tan \left (d x + c\right )^{2} + 55 \, a^{8} \tan \left (d x + c\right )}{55 \, d} \]

input
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")
 
output
1/55*(5*a^8*tan(d*x + c)^11 - 44*I*a^8*tan(d*x + c)^10 - 165*a^8*tan(d*x + 
 c)^9 + 330*I*a^8*tan(d*x + c)^8 + 330*a^8*tan(d*x + c)^7 + 462*a^8*tan(d* 
x + c)^5 - 660*I*a^8*tan(d*x + c)^4 - 495*a^8*tan(d*x + c)^3 + 220*I*a^8*t 
an(d*x + c)^2 + 55*a^8*tan(d*x + c))/d
 
3.1.79.8 Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (43) = 86\).

Time = 1.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.44 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {5 \, a^{8} \tan \left (d x + c\right )^{11} - 44 i \, a^{8} \tan \left (d x + c\right )^{10} - 165 \, a^{8} \tan \left (d x + c\right )^{9} + 330 i \, a^{8} \tan \left (d x + c\right )^{8} + 330 \, a^{8} \tan \left (d x + c\right )^{7} + 462 \, a^{8} \tan \left (d x + c\right )^{5} - 660 i \, a^{8} \tan \left (d x + c\right )^{4} - 495 \, a^{8} \tan \left (d x + c\right )^{3} + 220 i \, a^{8} \tan \left (d x + c\right )^{2} + 55 \, a^{8} \tan \left (d x + c\right )}{55 \, d} \]

input
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")
 
output
1/55*(5*a^8*tan(d*x + c)^11 - 44*I*a^8*tan(d*x + c)^10 - 165*a^8*tan(d*x + 
 c)^9 + 330*I*a^8*tan(d*x + c)^8 + 330*a^8*tan(d*x + c)^7 + 462*a^8*tan(d* 
x + c)^5 - 660*I*a^8*tan(d*x + c)^4 - 495*a^8*tan(d*x + c)^3 + 220*I*a^8*t 
an(d*x + c)^2 + 55*a^8*tan(d*x + c))/d
 
3.1.79.9 Mupad [B] (verification not implemented)

Time = 4.73 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.95 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^8\,\left (\frac {\sin \left (9\,c+9\,d\,x\right )}{10}+\frac {\sin \left (11\,c+11\,d\,x\right )}{110}+\frac {\cos \left (c+d\,x\right )\,63{}\mathrm {i}}{1280}+\frac {\cos \left (3\,c+3\,d\,x\right )\,9{}\mathrm {i}}{256}+\frac {\cos \left (5\,c+5\,d\,x\right )\,9{}\mathrm {i}}{512}+\frac {\cos \left (7\,c+7\,d\,x\right )\,3{}\mathrm {i}}{512}-\frac {\cos \left (9\,c+9\,d\,x\right )\,253{}\mathrm {i}}{2560}-\frac {\cos \left (11\,c+11\,d\,x\right )\,23{}\mathrm {i}}{2560}\right )}{d\,{\cos \left (c+d\,x\right )}^{11}} \]

input
int((a + a*tan(c + d*x)*1i)^8/cos(c + d*x)^4,x)
 
output
(a^8*((cos(c + d*x)*63i)/1280 + (cos(3*c + 3*d*x)*9i)/256 + (cos(5*c + 5*d 
*x)*9i)/512 + (cos(7*c + 7*d*x)*3i)/512 - (cos(9*c + 9*d*x)*253i)/2560 - ( 
cos(11*c + 11*d*x)*23i)/2560 + sin(9*c + 9*d*x)/10 + sin(11*c + 11*d*x)/11 
0))/(d*cos(c + d*x)^11)