3.70 \(\int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx\)

Optimal. Leaf size=167 \[ \frac{63 i a^5 \sec (c+d x)}{8 d}+\frac{63 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac{21 i a \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d} \]

[Out]

(63*a^5*ArcTanh[Sin[c + d*x]])/(8*d) + (((63*I)/8)*a^5*Sec[c + d*x])/d + (((9*I)/20)*a^2*Sec[c + d*x]*(a + I*a
*Tan[c + d*x])^3)/d + ((I/5)*a*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^4)/d + (((21*I)/20)*a*Sec[c + d*x]*(a^2 + I
*a^2*Tan[c + d*x])^2)/d + (((21*I)/8)*Sec[c + d*x]*(a^5 + I*a^5*Tan[c + d*x]))/d

________________________________________________________________________________________

Rubi [A]  time = 0.126343, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3498, 3486, 3770} \[ \frac{63 i a^5 \sec (c+d x)}{8 d}+\frac{63 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac{21 i a \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(63*a^5*ArcTanh[Sin[c + d*x]])/(8*d) + (((63*I)/8)*a^5*Sec[c + d*x])/d + (((9*I)/20)*a^2*Sec[c + d*x]*(a + I*a
*Tan[c + d*x])^3)/d + ((I/5)*a*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^4)/d + (((21*I)/20)*a*Sec[c + d*x]*(a^2 + I
*a^2*Tan[c + d*x])^2)/d + (((21*I)/8)*Sec[c + d*x]*(a^5 + I*a^5*Tan[c + d*x]))/d

Rule 3498

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3486

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx &=\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac{1}{5} (9 a) \int \sec (c+d x) (a+i a \tan (c+d x))^4 \, dx\\ &=\frac{9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac{1}{20} \left (63 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac{9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac{1}{4} \left (21 a^3\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=\frac{21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac{9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac{21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac{1}{8} \left (63 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac{63 i a^5 \sec (c+d x)}{8 d}+\frac{21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac{9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac{21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac{1}{8} \left (63 a^5\right ) \int \sec (c+d x) \, dx\\ &=\frac{63 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{63 i a^5 \sec (c+d x)}{8 d}+\frac{21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac{9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac{i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac{21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}\\ \end{align*}

Mathematica [A]  time = 1.01146, size = 115, normalized size = 0.69 \[ \frac{a^5 (\cos (5 d x)+i \sin (5 d x)) \left (5040 \tanh ^{-1}\left (\cos (c) \tan \left (\frac{d x}{2}\right )+\sin (c)\right )+i \sec ^5(c+d x) (450 i \sin (2 (c+d x))+325 i \sin (4 (c+d x))+1920 \cos (2 (c+d x))+640 \cos (4 (c+d x))+1344)\right )}{320 d (\cos (d x)+i \sin (d x))^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^5*(Cos[5*d*x] + I*Sin[5*d*x])*(5040*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] + I*Sec[c + d*x]^5*(1344 + 1920*C
os[2*(c + d*x)] + 640*Cos[4*(c + d*x)] + (450*I)*Sin[2*(c + d*x)] + (325*I)*Sin[4*(c + d*x)])))/(320*d*(Cos[d*
x] + I*Sin[d*x])^5)

________________________________________________________________________________________

Maple [B]  time = 0.072, size = 329, normalized size = 2. \begin{align*}{\frac{{\frac{36\,i}{5}}{a}^{5}\cos \left ( dx+c \right ) }{d}}+{\frac{{\frac{10\,i}{3}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}-{\frac{{\frac{10\,i}{3}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{\frac{i}{5}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{\frac{18\,i}{5}}{a}^{5}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{5\,i{a}^{5}}{d\cos \left ( dx+c \right ) }}+{\frac{5\,{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{5\,{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{55\,{a}^{5}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{63\,{a}^{5}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{{\frac{i}{15}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{\frac{i}{5}}{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}+{\frac{{\frac{i}{5}}{a}^{5}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}-5\,{\frac{{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+I*a*tan(d*x+c))^5,x)

[Out]

36/5*I/d*a^5*cos(d*x+c)+10/3*I/d*a^5*sin(d*x+c)^4/cos(d*x+c)-10/3*I/d*a^5*sin(d*x+c)^4/cos(d*x+c)^3+1/5*I/d*a^
5*sin(d*x+c)^6/cos(d*x+c)^5+18/5*I/d*a^5*cos(d*x+c)*sin(d*x+c)^2+5*I/d*a^5/cos(d*x+c)+5/4/d*a^5*sin(d*x+c)^5/c
os(d*x+c)^4-5/8/d*a^5*sin(d*x+c)^5/cos(d*x+c)^2-5/8*a^5*sin(d*x+c)^3/d-55/8*a^5*sin(d*x+c)/d+63/8/d*a^5*ln(sec
(d*x+c)+tan(d*x+c))-1/15*I/d*a^5*sin(d*x+c)^6/cos(d*x+c)^3+1/5*I/d*a^5*sin(d*x+c)^6/cos(d*x+c)+1/5*I/d*a^5*cos
(d*x+c)*sin(d*x+c)^4-5/d*a^5*sin(d*x+c)^3/cos(d*x+c)^2

________________________________________________________________________________________

Maxima [A]  time = 1.11655, size = 290, normalized size = 1.74 \begin{align*} \frac{75 \, a^{5}{\left (\frac{2 \,{\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 600 \, a^{5}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, a^{5} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac{1200 i \, a^{5}}{\cos \left (d x + c\right )} + \frac{800 i \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{5}}{\cos \left (d x + c\right )^{3}} + \frac{16 i \,{\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} a^{5}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(75*a^5*(2*(5*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) + 3*log(sin(d*x +
c) + 1) - 3*log(sin(d*x + c) - 1)) + 600*a^5*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - l
og(sin(d*x + c) - 1)) + 240*a^5*log(sec(d*x + c) + tan(d*x + c)) + 1200*I*a^5/cos(d*x + c) + 800*I*(3*cos(d*x
+ c)^2 - 1)*a^5/cos(d*x + c)^3 + 16*I*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 3)*a^5/cos(d*x + c)^5)/d

________________________________________________________________________________________

Fricas [B]  time = 1.13587, size = 891, normalized size = 5.34 \begin{align*} \frac{1930 i \, a^{5} e^{\left (9 i \, d x + 9 i \, c\right )} + 4740 i \, a^{5} e^{\left (7 i \, d x + 7 i \, c\right )} + 5376 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 2940 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 630 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 315 \,{\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 315 \,{\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{40 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(1930*I*a^5*e^(9*I*d*x + 9*I*c) + 4740*I*a^5*e^(7*I*d*x + 7*I*c) + 5376*I*a^5*e^(5*I*d*x + 5*I*c) + 2940*
I*a^5*e^(3*I*d*x + 3*I*c) + 630*I*a^5*e^(I*d*x + I*c) + 315*(a^5*e^(10*I*d*x + 10*I*c) + 5*a^5*e^(8*I*d*x + 8*
I*c) + 10*a^5*e^(6*I*d*x + 6*I*c) + 10*a^5*e^(4*I*d*x + 4*I*c) + 5*a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x
+ I*c) + I) - 315*(a^5*e^(10*I*d*x + 10*I*c) + 5*a^5*e^(8*I*d*x + 8*I*c) + 10*a^5*e^(6*I*d*x + 6*I*c) + 10*a^
5*e^(4*I*d*x + 4*I*c) + 5*a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) - I))/(d*e^(10*I*d*x + 10*I*c) +
5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{5} \left (\int - 10 \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 5 \tan ^{4}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 5 i \tan{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int - 10 i \tan ^{3}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int i \tan ^{5}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int \sec{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.46856, size = 258, normalized size = 1.54 \begin{align*} \frac{315 \, a^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 315 \, a^{5} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (275 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 200 i \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 750 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1600 i \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 3280 i \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 750 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2240 i \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 275 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 488 i \, a^{5}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{5}}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(315*a^5*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 315*a^5*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(275*a^5*tan
(1/2*d*x + 1/2*c)^9 + 200*I*a^5*tan(1/2*d*x + 1/2*c)^8 - 750*a^5*tan(1/2*d*x + 1/2*c)^7 - 1600*I*a^5*tan(1/2*d
*x + 1/2*c)^6 + 3280*I*a^5*tan(1/2*d*x + 1/2*c)^4 + 750*a^5*tan(1/2*d*x + 1/2*c)^3 - 2240*I*a^5*tan(1/2*d*x +
1/2*c)^2 - 275*a^5*tan(1/2*d*x + 1/2*c) + 488*I*a^5)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d