3.78 \(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx\)

Optimal. Leaf size=82 \[ -\frac {i (a+i a \tan (c+d x))^{13}}{13 a^5 d}+\frac {i (a+i a \tan (c+d x))^{12}}{3 a^4 d}-\frac {4 i (a+i a \tan (c+d x))^{11}}{11 a^3 d} \]

[Out]

-4/11*I*(a+I*a*tan(d*x+c))^11/a^3/d+1/3*I*(a+I*a*tan(d*x+c))^12/a^4/d-1/13*I*(a+I*a*tan(d*x+c))^13/a^5/d

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Rubi [A]  time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3487, 43} \[ -\frac {i (a+i a \tan (c+d x))^{13}}{13 a^5 d}+\frac {i (a+i a \tan (c+d x))^{12}}{3 a^4 d}-\frac {4 i (a+i a \tan (c+d x))^{11}}{11 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^8,x]

[Out]

(((-4*I)/11)*(a + I*a*Tan[c + d*x])^11)/(a^3*d) + ((I/3)*(a + I*a*Tan[c + d*x])^12)/(a^4*d) - ((I/13)*(a + I*a
*Tan[c + d*x])^13)/(a^5*d)

Rule 43

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 3487

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*} \int \sec ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x)^2 (a+x)^{10} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (4 a^2 (a+x)^{10}-4 a (a+x)^{11}+(a+x)^{12}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac {4 i (a+i a \tan (c+d x))^{11}}{11 a^3 d}+\frac {i (a+i a \tan (c+d x))^{12}}{3 a^4 d}-\frac {i (a+i a \tan (c+d x))^{13}}{13 a^5 d}\\ \end {align*}

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Mathematica [B]  time = 7.04, size = 234, normalized size = 2.85 \[ \frac {a^8 \sec (c) \sec ^{13}(c+d x) (-1716 \sin (2 c+d x)+1287 \sin (2 c+3 d x)-1287 \sin (4 c+3 d x)+715 \sin (4 c+5 d x)-715 \sin (6 c+5 d x)+286 \sin (6 c+7 d x)-286 \sin (8 c+7 d x)+156 \sin (8 c+9 d x)+26 \sin (10 c+11 d x)+2 \sin (12 c+13 d x)+1716 i \cos (2 c+d x)+1287 i \cos (2 c+3 d x)+1287 i \cos (4 c+3 d x)+715 i \cos (4 c+5 d x)+715 i \cos (6 c+5 d x)+286 i \cos (6 c+7 d x)+286 i \cos (8 c+7 d x)+1716 \sin (d x)+1716 i \cos (d x))}{1716 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^8,x]

[Out]

(a^8*Sec[c]*Sec[c + d*x]^13*((1716*I)*Cos[d*x] + (1716*I)*Cos[2*c + d*x] + (1287*I)*Cos[2*c + 3*d*x] + (1287*I
)*Cos[4*c + 3*d*x] + (715*I)*Cos[4*c + 5*d*x] + (715*I)*Cos[6*c + 5*d*x] + (286*I)*Cos[6*c + 7*d*x] + (286*I)*
Cos[8*c + 7*d*x] + 1716*Sin[d*x] - 1716*Sin[2*c + d*x] + 1287*Sin[2*c + 3*d*x] - 1287*Sin[4*c + 3*d*x] + 715*S
in[4*c + 5*d*x] - 715*Sin[6*c + 5*d*x] + 286*Sin[6*c + 7*d*x] - 286*Sin[8*c + 7*d*x] + 156*Sin[8*c + 9*d*x] +
26*Sin[10*c + 11*d*x] + 2*Sin[12*c + 13*d*x]))/(1716*d)

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fricas [B]  time = 0.47, size = 307, normalized size = 3.74 \[ \frac {1171456 i \, a^{8} e^{\left (20 i \, d x + 20 i \, c\right )} + 2928640 i \, a^{8} e^{\left (18 i \, d x + 18 i \, c\right )} + 5271552 i \, a^{8} e^{\left (16 i \, d x + 16 i \, c\right )} + 7028736 i \, a^{8} e^{\left (14 i \, d x + 14 i \, c\right )} + 7028736 i \, a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 5271552 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 2928640 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 1171456 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 319488 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 53248 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 4096 i \, a^{8}}{429 \, {\left (d e^{\left (26 i \, d x + 26 i \, c\right )} + 13 \, d e^{\left (24 i \, d x + 24 i \, c\right )} + 78 \, d e^{\left (22 i \, d x + 22 i \, c\right )} + 286 \, d e^{\left (20 i \, d x + 20 i \, c\right )} + 715 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 1287 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 1716 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 1716 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 1287 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 715 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 286 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 78 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 13 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/429*(1171456*I*a^8*e^(20*I*d*x + 20*I*c) + 2928640*I*a^8*e^(18*I*d*x + 18*I*c) + 5271552*I*a^8*e^(16*I*d*x +
16*I*c) + 7028736*I*a^8*e^(14*I*d*x + 14*I*c) + 7028736*I*a^8*e^(12*I*d*x + 12*I*c) + 5271552*I*a^8*e^(10*I*d
*x + 10*I*c) + 2928640*I*a^8*e^(8*I*d*x + 8*I*c) + 1171456*I*a^8*e^(6*I*d*x + 6*I*c) + 319488*I*a^8*e^(4*I*d*x
+ 4*I*c) + 53248*I*a^8*e^(2*I*d*x + 2*I*c) + 4096*I*a^8)/(d*e^(26*I*d*x + 26*I*c) + 13*d*e^(24*I*d*x + 24*I*c
) + 78*d*e^(22*I*d*x + 22*I*c) + 286*d*e^(20*I*d*x + 20*I*c) + 715*d*e^(18*I*d*x + 18*I*c) + 1287*d*e^(16*I*d*
x + 16*I*c) + 1716*d*e^(14*I*d*x + 14*I*c) + 1716*d*e^(12*I*d*x + 12*I*c) + 1287*d*e^(10*I*d*x + 10*I*c) + 715
*d*e^(8*I*d*x + 8*I*c) + 286*d*e^(6*I*d*x + 6*I*c) + 78*d*e^(4*I*d*x + 4*I*c) + 13*d*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 4.33, size = 173, normalized size = 2.11 \[ \frac {33 \, a^{8} \tan \left (d x + c\right )^{13} - 286 i \, a^{8} \tan \left (d x + c\right )^{12} - 1014 \, a^{8} \tan \left (d x + c\right )^{11} + 1716 i \, a^{8} \tan \left (d x + c\right )^{10} + 715 \, a^{8} \tan \left (d x + c\right )^{9} + 2574 i \, a^{8} \tan \left (d x + c\right )^{8} + 5148 \, a^{8} \tan \left (d x + c\right )^{7} - 3432 i \, a^{8} \tan \left (d x + c\right )^{6} + 1287 \, a^{8} \tan \left (d x + c\right )^{5} - 4290 i \, a^{8} \tan \left (d x + c\right )^{4} - 3718 \, a^{8} \tan \left (d x + c\right )^{3} + 1716 i \, a^{8} \tan \left (d x + c\right )^{2} + 429 \, a^{8} \tan \left (d x + c\right )}{429 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

1/429*(33*a^8*tan(d*x + c)^13 - 286*I*a^8*tan(d*x + c)^12 - 1014*a^8*tan(d*x + c)^11 + 1716*I*a^8*tan(d*x + c)
^10 + 715*a^8*tan(d*x + c)^9 + 2574*I*a^8*tan(d*x + c)^8 + 5148*a^8*tan(d*x + c)^7 - 3432*I*a^8*tan(d*x + c)^6
+ 1287*a^8*tan(d*x + c)^5 - 4290*I*a^8*tan(d*x + c)^4 - 3718*a^8*tan(d*x + c)^3 + 1716*I*a^8*tan(d*x + c)^2 +
429*a^8*tan(d*x + c))/d

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maple [B]  time = 0.53, size = 475, normalized size = 5.79 \[ \frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{13 \cos \left (d x +c \right )^{13}}+\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{143 \cos \left (d x +c \right )^{11}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{1287 \cos \left (d x +c \right )^{9}}\right )+\frac {4 i a^{8}}{3 \cos \left (d x +c \right )^{6}}-28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{11 \cos \left (d x +c \right )^{11}}+\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{99 \cos \left (d x +c \right )^{9}}+\frac {8 \left (\sin ^{7}\left (d x +c \right )\right )}{693 \cos \left (d x +c \right )^{7}}\right )-56 i a^{8} \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 )+70 a^{8} \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 )+56 i a^{8} \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 )-28 a^{8} \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 )-8 i a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{12}}+\frac {\sin ^{8}\left (d x +c \right )}{30 \cos \left (d x +c \right )^{10}}+\frac {\sin ^{8}\left (d x +c \right )}{120 \cos \left (d x +c \right )^{8}}\right )-a^{8} \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x)

[Out]

1/d*(a^8*(1/13*sin(d*x+c)^9/cos(d*x+c)^13+4/143*sin(d*x+c)^9/cos(d*x+c)^11+8/1287*sin(d*x+c)^9/cos(d*x+c)^9)+4
/3*I*a^8/cos(d*x+c)^6-28*a^8*(1/11*sin(d*x+c)^7/cos(d*x+c)^11+4/99*sin(d*x+c)^7/cos(d*x+c)^9+8/693*sin(d*x+c)^
7/cos(d*x+c)^7)-56*I*a^8*(1/8*sin(d*x+c)^4/cos(d*x+c)^8+1/12*sin(d*x+c)^4/cos(d*x+c)^6+1/24*sin(d*x+c)^4/cos(d
*x+c)^4)+70*a^8*(1/9*sin(d*x+c)^5/cos(d*x+c)^9+4/63*sin(d*x+c)^5/cos(d*x+c)^7+8/315*sin(d*x+c)^5/cos(d*x+c)^5)
+56*I*a^8*(1/10*sin(d*x+c)^6/cos(d*x+c)^10+1/20*sin(d*x+c)^6/cos(d*x+c)^8+1/60*sin(d*x+c)^6/cos(d*x+c)^6)-28*a
^8*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)-8*I*a^8*(1/1
2*sin(d*x+c)^8/cos(d*x+c)^12+1/30*sin(d*x+c)^8/cos(d*x+c)^10+1/120*sin(d*x+c)^8/cos(d*x+c)^8)-a^8*(-8/15-1/5*s
ec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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maxima [B]  time = 0.40, size = 173, normalized size = 2.11 \[ \frac {495 \, a^{8} \tan \left (d x + c\right )^{13} - 4290 i \, a^{8} \tan \left (d x + c\right )^{12} - 15210 \, a^{8} \tan \left (d x + c\right )^{11} + 25740 i \, a^{8} \tan \left (d x + c\right )^{10} + 10725 \, a^{8} \tan \left (d x + c\right )^{9} + 38610 i \, a^{8} \tan \left (d x + c\right )^{8} + 77220 \, a^{8} \tan \left (d x + c\right )^{7} - 51480 i \, a^{8} \tan \left (d x + c\right )^{6} + 19305 \, a^{8} \tan \left (d x + c\right )^{5} - 64350 i \, a^{8} \tan \left (d x + c\right )^{4} - 55770 \, a^{8} \tan \left (d x + c\right )^{3} + 25740 i \, a^{8} \tan \left (d x + c\right )^{2} + 6435 \, a^{8} \tan \left (d x + c\right )}{6435 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/6435*(495*a^8*tan(d*x + c)^13 - 4290*I*a^8*tan(d*x + c)^12 - 15210*a^8*tan(d*x + c)^11 + 25740*I*a^8*tan(d*x
+ c)^10 + 10725*a^8*tan(d*x + c)^9 + 38610*I*a^8*tan(d*x + c)^8 + 77220*a^8*tan(d*x + c)^7 - 51480*I*a^8*tan(
d*x + c)^6 + 19305*a^8*tan(d*x + c)^5 - 64350*I*a^8*tan(d*x + c)^4 - 55770*a^8*tan(d*x + c)^3 + 25740*I*a^8*ta
n(d*x + c)^2 + 6435*a^8*tan(d*x + c))/d

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mupad [B]  time = 5.34, size = 190, normalized size = 2.32 \[ \frac {a^8\,\sin \left (c+d\,x\right )\,\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (-184\,{\sin \left (c+d\,x\right )}^2-184\,{\sin \left (2\,c+2\,d\,x\right )}^2+\frac {\sin \left (2\,c+2\,d\,x\right )\,9867{}\mathrm {i}}{256}-184\,{\sin \left (3\,c+3\,d\,x\right )}^2-184\,{\sin \left (4\,c+4\,d\,x\right )}^2+\frac {\sin \left (4\,c+4\,d\,x\right )\,69069{}\mathrm {i}}{1024}-28\,{\sin \left (5\,c+5\,d\,x\right )}^2-2\,{\sin \left (6\,c+6\,d\,x\right )}^2+\frac {\sin \left (6\,c+6\,d\,x\right )\,42757{}\mathrm {i}}{512}+\frac {\sin \left (8\,c+8\,d\,x\right )\,23023{}\mathrm {i}}{256}+\frac {\sin \left (10\,c+10\,d\,x\right )\,7007{}\mathrm {i}}{512}+\frac {\sin \left (12\,c+12\,d\,x\right )\,1001{}\mathrm {i}}{1024}+429\right )}{429\,d\,{\left ({\sin \left (c+d\,x\right )}^2-1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^8/cos(c + d*x)^6,x)

[Out]

(a^8*sin(c + d*x)*(2*sin(c/2 + (d*x)/2)^2 - 1)*((sin(2*c + 2*d*x)*9867i)/256 + (sin(4*c + 4*d*x)*69069i)/1024
+ (sin(6*c + 6*d*x)*42757i)/512 + (sin(8*c + 8*d*x)*23023i)/256 + (sin(10*c + 10*d*x)*7007i)/512 + (sin(12*c +
12*d*x)*1001i)/1024 - 184*sin(2*c + 2*d*x)^2 - 184*sin(3*c + 3*d*x)^2 - 184*sin(4*c + 4*d*x)^2 - 28*sin(5*c +
5*d*x)^2 - 2*sin(6*c + 6*d*x)^2 - 184*sin(c + d*x)^2 + 429))/(429*d*(sin(c + d*x)^2 - 1)^7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{8} \left (\int \left (- 28 \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int 70 \tan ^{4}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 28 \tan ^{6}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int \tan ^{8}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int 8 i \tan {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 56 i \tan ^{3}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int 56 i \tan ^{5}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 8 i \tan ^{7}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**8,x)

[Out]

a**8*(Integral(-28*tan(c + d*x)**2*sec(c + d*x)**6, x) + Integral(70*tan(c + d*x)**4*sec(c + d*x)**6, x) + Int
egral(-28*tan(c + d*x)**6*sec(c + d*x)**6, x) + Integral(tan(c + d*x)**8*sec(c + d*x)**6, x) + Integral(8*I*ta
n(c + d*x)*sec(c + d*x)**6, x) + Integral(-56*I*tan(c + d*x)**3*sec(c + d*x)**6, x) + Integral(56*I*tan(c + d*
x)**5*sec(c + d*x)**6, x) + Integral(-8*I*tan(c + d*x)**7*sec(c + d*x)**6, x) + Integral(sec(c + d*x)**6, x))

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