\(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [19]

Optimal result

Integrand size = 24, antiderivative size = 109 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {4 i (a+i a \tan (c+d x))^6}{3 a^4 d}+\frac {12 i (a+i a \tan (c+d x))^7}{7 a^5 d}-\frac {3 i (a+i a \tan (c+d x))^8}{4 a^6 d}+\frac {i (a+i a \tan (c+d x))^9}{9 a^7 d} \]

[Out]

-4/3*I*(a+I*a*tan(d*x+c))^6/a^4/d+12/7*I*(a+I*a*tan(d*x+c))^7/a^5/d-3/4*I*(a+I*a*tan(d*x+c))^8/a^6/d+1/9*I*(a+
I*a*tan(d*x+c))^9/a^7/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, 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} \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {i (a+i a \tan (c+d x))^9}{9 a^7 d}-\frac {3 i (a+i a \tan (c+d x))^8}{4 a^6 d}+\frac {12 i (a+i a \tan (c+d x))^7}{7 a^5 d}-\frac {4 i (a+i a \tan (c+d x))^6}{3 a^4 d} \]

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((-4*I)/3)*(a + I*a*Tan[c + d*x])^6)/(a^4*d) + (((12*I)/7)*(a + I*a*Tan[c + d*x])^7)/(a^5*d) - (((3*I)/4)*(a
+ I*a*Tan[c + d*x])^8)/(a^6*d) + ((I/9)*(a + I*a*Tan[c + d*x])^9)/(a^7*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)^3 (a+x)^5 \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {i \text {Subst}\left (\int \left (8 a^3 (a+x)^5-12 a^2 (a+x)^6+6 a (a+x)^7-(a+x)^8\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {4 i (a+i a \tan (c+d x))^6}{3 a^4 d}+\frac {12 i (a+i a \tan (c+d x))^7}{7 a^5 d}-\frac {3 i (a+i a \tan (c+d x))^8}{4 a^6 d}+\frac {i (a+i a \tan (c+d x))^9}{9 a^7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \sec ^8(c+d x) (\cos (6 (c+d x))+i \sin (6 (c+d x))) (-40 i+170 i \cos (2 (c+d x))+83 \sec (c+d x) \sin (3 (c+d x))+27 \tan (c+d x))}{504 d} \]

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-1/504*(a^2*Sec[c + d*x]^8*(Cos[6*(c + d*x)] + I*Sin[6*(c + d*x)])*(-40*I + (170*I)*Cos[2*(c + d*x)] + 83*Sec[
c + d*x]*Sin[3*(c + d*x)] + 27*Tan[c + d*x]))/d

Maple [A] (verified)

Time = 114.00 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {64 i a^{2} \left (126 \,{\mathrm e}^{10 i \left (d x +c \right )}+126 \,{\mathrm e}^{8 i \left (d x +c \right )}+84 \,{\mathrm e}^{6 i \left (d x +c \right )}+36 \,{\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{63 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{9}}\)	\(80\)
derivativedivides	\(\frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{4 \cos \left (d x +c \right )^{8}}-a^{2} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )}{d}\)	\(141\)
default	\(\frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{4 \cos \left (d x +c \right )^{8}}-a^{2} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )}{d}\)	\(141\)

[In]

int(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

64/63*I*a^2*(126*exp(10*I*(d*x+c))+126*exp(8*I*(d*x+c))+84*exp(6*I*(d*x+c))+36*exp(4*I*(d*x+c))+9*exp(2*I*(d*x
+c))+1)/d/(exp(2*I*(d*x+c))+1)^9

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. \(2 (85) = 170\).

Time = 0.23 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.73 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {64 \, {\left (-126 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 126 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 84 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 36 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2}\right )}}{63 \, {\left (d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-64/63*(-126*I*a^2*e^(10*I*d*x + 10*I*c) - 126*I*a^2*e^(8*I*d*x + 8*I*c) - 84*I*a^2*e^(6*I*d*x + 6*I*c) - 36*I
*a^2*e^(4*I*d*x + 4*I*c) - 9*I*a^2*e^(2*I*d*x + 2*I*c) - I*a^2)/(d*e^(18*I*d*x + 18*I*c) + 9*d*e^(16*I*d*x + 1
6*I*c) + 36*d*e^(14*I*d*x + 14*I*c) + 84*d*e^(12*I*d*x + 12*I*c) + 126*d*e^(10*I*d*x + 10*I*c) + 126*d*e^(8*I*
d*x + 8*I*c) + 84*d*e^(6*I*d*x + 6*I*c) + 36*d*e^(4*I*d*x + 4*I*c) + 9*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\, dx + \int \left (- 2 i \tan {\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\right )\, dx + \int \left (- \sec ^{8}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate(sec(d*x+c)**8*(a+I*a*tan(d*x+c))**2,x)

[Out]

-a**2*(Integral(tan(c + d*x)**2*sec(c + d*x)**8, x) + Integral(-2*I*tan(c + d*x)*sec(c + d*x)**8, x) + Integra
l(-sec(c + d*x)**8, x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.99 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {28 \, a^{2} \tan \left (d x + c\right )^{9} - 63 i \, a^{2} \tan \left (d x + c\right )^{8} + 72 \, a^{2} \tan \left (d x + c\right )^{7} - 252 i \, a^{2} \tan \left (d x + c\right )^{6} - 378 i \, a^{2} \tan \left (d x + c\right )^{4} - 168 \, a^{2} \tan \left (d x + c\right )^{3} - 252 i \, a^{2} \tan \left (d x + c\right )^{2} - 252 \, a^{2} \tan \left (d x + c\right )}{252 \, d} \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/252*(28*a^2*tan(d*x + c)^9 - 63*I*a^2*tan(d*x + c)^8 + 72*a^2*tan(d*x + c)^7 - 252*I*a^2*tan(d*x + c)^6 - 3
78*I*a^2*tan(d*x + c)^4 - 168*a^2*tan(d*x + c)^3 - 252*I*a^2*tan(d*x + c)^2 - 252*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.99 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {28 \, a^{2} \tan \left (d x + c\right )^{9} - 63 i \, a^{2} \tan \left (d x + c\right )^{8} + 72 \, a^{2} \tan \left (d x + c\right )^{7} - 252 i \, a^{2} \tan \left (d x + c\right )^{6} - 378 i \, a^{2} \tan \left (d x + c\right )^{4} - 168 \, a^{2} \tan \left (d x + c\right )^{3} - 252 i \, a^{2} \tan \left (d x + c\right )^{2} - 252 \, a^{2} \tan \left (d x + c\right )}{252 \, d} \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/252*(28*a^2*tan(d*x + c)^9 - 63*I*a^2*tan(d*x + c)^8 + 72*a^2*tan(d*x + c)^7 - 252*I*a^2*tan(d*x + c)^6 - 3
78*I*a^2*tan(d*x + c)^4 - 168*a^2*tan(d*x + c)^3 - 252*I*a^2*tan(d*x + c)^2 - 252*a^2*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.39 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )\,\left (252\,{\cos \left (c+d\,x\right )}^8+{\cos \left (c+d\,x\right )}^7\,\sin \left (c+d\,x\right )\,252{}\mathrm {i}+168\,{\cos \left (c+d\,x\right )}^6\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^3\,378{}\mathrm {i}+{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^5\,252{}\mathrm {i}-72\,{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^6+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^7\,63{}\mathrm {i}-28\,{\sin \left (c+d\,x\right )}^8\right )}{252\,d\,{\cos \left (c+d\,x\right )}^9} \]

[In]

int((a + a*tan(c + d*x)*1i)^2/cos(c + d*x)^8,x)

[Out]

(a^2*sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^7*63i + cos(c + d*x)^7*sin(c + d*x)*252i + 252*cos(c + d*x)^8 - 2
8*sin(c + d*x)^8 - 72*cos(c + d*x)^2*sin(c + d*x)^6 + cos(c + d*x)^3*sin(c + d*x)^5*252i + cos(c + d*x)^5*sin(
c + d*x)^3*378i + 168*cos(c + d*x)^6*sin(c + d*x)^2))/(252*d*cos(c + d*x)^9)